Applied Multivariate Statistical Analysis RICHARD A. JOHNSON University of WisconsinMadison
DEAN W. WICHERN Texas A&M University
>hnson, Richard A. Statistical analysisiRichard A. Johnson.61h ed. Dean W. Winchern p.em. Includes index. ISBN 0131877151 1. Statistical Analysis ~IP
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To the memory of my mother and my father.
R. A. J. To Dorothy, Michael, and An drew. D. W. W.
Introduction 1 Applications of Multivariate Techniques 3 The Organization of Data 5 Arrays,5 Descriptive Statistics, 6 Graphical Techniques, 11
Data Displays and Pictorial Representations 19 Linking Multiple TwoDimensional Scatter Plots, 20 Graphs of Growth Curves, 24 Stars, 26 Chernoff Faces, 27
Introduction 49 Some Basics of Matrix and Vector Algebra 49 Vectors, 49 Matrices, 54
Positive Definite Matrices 60 A SquareRoot Matrix 65 Random Vectors and Matrices 66 Mean Vectors and Covariance Matrices
Partitioning the Covariance Matrix, 73 The Mean Vector and Covariance Matrix for Linear Combinations of Random Variables, 75 Partitioning the Sample Mean Vector and Covariance Matrix, 77
Introduction 149 The Multivariate Normal Density and Its Properties 149 Additional Properties of the Multivariate Normal Distribution, 156
The Sampling Distribution of X and S 173 Properties of the Wishart Distribution, 174
Large Sample Inferences about a Population Mean Vector Multivariate Quality Control Charts 239
Inferences about Mean Vectors when Some Observations Are Missing 251 Difficulties Due to TIme Dependence in Multivariate Observations 256 Supplement 5A: Simultaneous Confidence Intervals and Ellipses as Shadows of the pDimensional Ellipsoids 258 Exercises 261 References 272
Paired Comparisons and a Repeated Measures Design 273 Paired Comparisons, 273 A Repeated Measures Design for Comparing Treatments, 279
Comparing Mean Vectors from Two Populations 284 Assumptions Concerning the Structure of the Data, 284 Further Assumptions When nl and n2 Are Small, 285 Simultaneous Confidence Intervals, 288 The TwoSample Situation When 1:1 oF l;z,291 An Approximation to the Distribution of T2 for Normal Populations When Sample Sizes Are Not Large, 294
Charts for Monitoring a Sample of Individual Multivariate Observations for Stability, 241 Control Regions for Future Individual Observations, 247 Control Ellipse for Future Observations 248 2 ' T Chart for Future Observations, 248 Control Charts Based on Subsample Means, 249 Control Regions for Future SUbsample Observations, 251
Evaluating the Normality of the Univariate Marginal Distributions, 177 Evaluating Bivariate Normality, 182
Simultaneous Confidence Statements, 223 A Comparison of Simultaneous Confidence Intervals with OneataTime Intervals, 229 The Bonferroni Method of Multiple Comparisons, 232
Sampling from a Multivariate Normal Distribution and Maximum Likelihood Estimation 168 The Multivariate Normal Likelihood, 168 Maximum Likelihood Estimation of P and I, 170 Sufficient Statistics, 173
Introduction 210 The Plausibility of Po as a Value for a Normal Population Mean 210 General Likelihood Ratio Method, 219
Introduction 111 The Geometry of the Sample 111 Random Samples and the Expected Values of the Sample Mean and Covariance Matrix 119 Generalized Variance 123
Sample Mean, Covariance, and Correlation As Matrix Operations 137 Sample Values of Linear Combinations of Variables Exercises 144 References 148
Situations in which the Generalized Sample Variance Is Zero, 129 Generalized Variance Determined by I R I and Its Geometrical Interpretation, 134 Another Generalization of Variance, 137
Comparing Several Multivariate Population Means (OneWay Manova) 296 Assumptions about the Structure of the Data for OneWay MANOVA, 296
Contents A Summary of Univariate ANOVA, 297 Multivariate Analysis of Variance (MANOVA), 301
Simultaneous Confidence Intervals for Treatment Effects 308 Testing for Equality of Covariance Matrices 310 1\voWay Multivariate Analysis of Variance 312
Profile Analysis 323 Repeated Measures Designs and Growth Curves 328 Perspectives and a Strategy for Analyzing Multivariate Models 332 Exercises 337 References 358
Inferences from the Estimated Regression Function 378 Model Checking and Other Aspects of Regression 381 Does the Model Fit?, 381 Leverage and Influence, 384 Additional Problems in Linear Regression, 384
Multiple Regression Models with Time Dependent Errors 413 Supplement 7A: The Distribution of the Likelihood Ratio for the Multivariate Multiple Regression Model 418 Exercises  420 References 428
Introduction 481 The Orthogonal Factor Model 482 Methods of Estimation 488 The Pri,!cipal Component (and Principal Factor) Method, 488 A ModifiedApproachthe Principal Factor Solution, 494 The Maximum Likelihood Method, 495 A Large Sample Test for the Number of Common Factors 501
Factor Scores 513 The Weighted Least Squares Method, 514 The Regression Method, 516
Comparing the Two Formulations of the Regression Model 410 Mean Corrected Form of the Regression Model, 410 Relating the Formulations, 412
The Concept of Linear Regression 401 Prediction of Several Variables, 406 Partial Correlation Coefficient, 409
Large Sample Properties of Aj and ej, 456 Testing for the Equal Correlation Structure, 457
The pDimensional Geometrical Interpretation, 468 The nDimensional Geometrical Interpretation, 469
Multivariate Multiple Regression 387 Likelihood Ratio Tests for Regression Parameters, 395 Other Multivariate Test Statistics, 398 Predictions from Multivariate Multiple Regressions, 399
Estimating the Regression Function at Zo, 378 Forecasting a New Observation at Zo, 379
Checking a Given Set of Measurements for Stability, 459 Controlling Future Values, 463
Introduction 360 The Classical Linear Regression Model 360 Least Squares Estimation 364
Inferences Concerning the Regression Parameters, 370 Likelihood Ratio Tests for the Regression Parameters, 374
The Number of Principal Components, 444 Interpretation of the Sample Principal Components, 448 Standardizing the Sample Principal Components, 449
SumoJSquares Decomposition, 366 Geometry of Least Squares, 367 Sampling Properties of Classical Least Squares Estimators, 369
Principal Components Obtained from Standardized Variables 436 Principal Components for Covariance Matrices ' with Special Structures, 439
Univariate TwoWay FixedEffects Model with Interaction, 312 Multivariate Two Way FixedEffects Model with Interaction, 315
Perspectives and a Strategy for Factor Analysis 519 Supplement 9A: Some Computational Details for Maximum Likelihood Estimation 527 Recommended Computational Scheme, 528 Maximum Likelihood Estimators of p = L.L~
+ 1/1. 529
Classification of Normal Populations When l:1 = l:z = :£,584 Scaling, 589 Fisher's Approach to Classification with 1Wo Populations, 590 Is Classification a Good Idea?, 592 Classification of Normal Populations When:£1 =F :£z, 593
Single Linkage, 682 Complete Linkage, 685 Average Linkage, 690 Ward's Hierarchical Clustering Method, 692 Final CommentsHierarchical Procedures, 695
Nonhierarchical Clustering Methods 696 Kmeans Method, 696 Final CommentsNonhierarchical Procedures, 701
.
Algebraic Development of Correspondence Analysis, 718 Inertia,725 Interpretation in Two Dimensions, 726 Final Comments, 726
The Minimum Expected Cost of Misclassification Method, 606 Classification with Normal Populations, 609
Including Qualitative Variables, 644 Classification Trees, 644 Neural Networks, 647 Selection of Variables, 648
.
Evaluating Classification Functions 596 Classification with Several Populations 606
Introduction, 634 The Logit Model, 634 Logistic Regression Analysis, 636 Classification, 638 Logistic Regression with Binomial Responses, 640
Introduction 575 Separation and Classification for Two Populations 576 Classification with 1\vo Multivariate Normal Populations 584
Distances and Similarity Coefficients for Pairs of Items, 673 Similarities and Association Measures for Pairs of Variables, 677 Concluding Comments on Similarity, 678
The Sample Canonical Variates and Sample Canonical Correlations 550 Additional Sample Descriptive Measures 558
Matrices of Errors ofApproximations, 558 Proportions of Explained Sample Variance, 561
Testing for Group Differences, 648 Graphics, 649 Practical Considerations Regarding Multivariate Normality, 649
Introduction 539 Canonical Variates and Canonical Correlations 539 Interpreting the Population Canonical Variables 545 Identifying the {:anonical Variables, 545 Canonical Correlations as Generalizations of Other Correlation Coefficients, 547 The First r Canonical Variables as a Summary of Variability, 548 A Geometrical Interpretation of the Population Canonical Correlation Analysis 549 .
This book originally grew out of our lecture notes for an "Applied Multivariate Analysis" course offered jointly by the Statistics Department and the School of Business at the University of WisconsinMadison. Applied Multivariate StatisticalAnalysis, Sixth Edition, is concerned with statistical methods for describing and analyzing multivariate data. Data analysis, while interesting with one variable, becomes truly fascinating and challenging when several variables are involved. Researchers in the biological, physical, and social sciences frequently collect measurements on several variables. Modem computer packages readily provide the· numerical results to rather complex statistical analyses. We have tried to provide readers with the supporting knowledge necessary for making proper interpretations, selecting appropriate techniques, and understanding their strengths and weaknesses. We hope our discussions wiII meet the needs of experimental scientists, in a wide variety of subject matter areas, as a readable introduction to the statistical analysis of multivariate observations.
Our aim is to present the concepts and methods of muItivariate analysis at a level that is readily understandable by readers who have taken two or more statistics courses. We emphasize the applications of multivariate methods and, consequently, have attempted to make the mathematics as palatable as possible. We avoid the use of calculus. On the other hand, the concepts of a matrix and of matrix manipulations are important. We do not assume the reader is familiar with matrix algebra. Rather, we introduce matrices as they appear naturally in our discussions, and we then show how they simplify the presentation of muItivariate models and techniques. The introductory account of matrix algebra, in Chapter 2, highlights the more important matrix algebra results as they apply to multivariate analysis. The Chapter 2 supplement provides a summary of matrix algebra results for those with little or no previous exposure to the subject. This supplementary material helps make the book selfcontained and is used to complete proofs. The proofs may be ignored on the first reading. In this way we hope to make the book accessible to a wide audience. In our attempt to make the study of muItivariate analysis appealing to a large audience of both practitioners and theoreticians, we have had to sacrifice xv
onsistency of level. Some sections are harder than others. In particular, we
~~ve summarized a volumi?ous amount .of materi~l?n regres~ion ~n Chapter 7.
The resulting presentation IS rather SUCCInct and difficult the fIrst ~Ime throu~h. We hope instructors will be a?le to compensat.e for the une~enness In l~vel by JUdiciously choosing those s~ctIons, and subsectIOns, appropnate for theIr students and by toning them tlown If necessary.
agrams and verbal descriptions to teach the corresponding theoretical developments. If the students have uniformly strong mathematical backgrounds, much of the book can successfully be covered in one term. We have found individual dataanalysis projects useful for integrating material from several of the methods chapters. Here, our rather complete treatments of multivariate analysis of variance (MANOVA), regression analysis, factor analysis, canonical correlation, discriminant analysis, and so forth are helpful, even though they may not be specifically covered in lectures.
The methodological "tools" of multlvariate analysis are contained in Chapters 5 through 12. These chapters represent the heart of the book, but they cannot be assimilated without much of the material in the introd~ctory Chapters 1 thr?~gh 4. Even those readers with a good kno~ledge of matrix algebra or those willing t accept the mathematical results on faIth should, at the very least, peruse Chapo 3 "Sample Geometry," and Chapter 4, "Multivariate Normal Distribution." ter , Our approach in the methodological ~hapters is to ~eep the discussion.dit and uncluttered. Typically, we start with a formulatIOn of the population re~dels delineate the corresponding sample results, and liberally illustrate every:'ing ~ith examples. The exa~ples are of two types: those that are simple and hose calculations can be easily done by hand, and those that rely on realworld ~ata and computer software. These will provide an opportunity to (1) duplicate our analyses, (2) carry out the analyses dictated by exercises, or (3) analyze the data using methods other than the ones we have used or suggest~d. . The division of the methodological chapters (5 through 12) Into three umts llo~s instructors some flexibility in tailoring a course to their needs. Possible a uences for a onesemester (two quarter) course are indicated schematically. seq . . . fr h t Each instructor will undoubtedly omit certam sectIons om some c ap ers to cover a broader collection of topics than is indicated by these two choices. Getting Started Chapters 14
New material. Users of the previous editions will notice several major changes in the sixth edition.
• Twelve new data sets including national track records for men and women, psychological profile scores, car body assembly measurements, cell phone tower breakdowns, pulp and paper properties measurements, Mali family farm data, stock price rates of return, and Concho water snake data. • Thirty seven new exercises and twenty revised exercises with many of these exercises based on the new data sets. • Four new data based examples and fifteen revised examples. • Six new or expanded sections: 1. Section 6.6 Testing for Equality of Covariance Matrices
2. Section 11.7 Logistic Regression and Classification 3. Section 12.5 Clustering Based on Statistical Models 4. Expanded Section 6.3 to include "An Approximation to the, Distribution of T2 for Normal Populations When Sample Sizes are not Large" 5. Expanded Sections 7.6 and 7.7 to include Akaike's Information Criterion 6. Consolidated previous Sections 11.3 and 11.5 on two group discriminant analysis into single Section 11.3
For most students, we would suggest a quick pass through the first four hapters (concentrating primarily on the material in Chapter 1; Sections 2.1, 2.2, ~.3, 2.5, 2.6, and 3.6; and the "assessing normality" material in Chapter ~) followed by a selection of methodological topics. For example, one mIght dISCUSS the comparison of mean vectors, principal components, factor analysis, discriminant analysis and clustering. The di~cussions could feature the many "worke? out" examples included in these sections of the text. Instructors may rely on dI
Web Site. To make the methods of multivariate analysis more prominent in the text, we have removed the long proofs of Results 7.2,7.4,7.10 and 10.1 and placed them on a web site accessible through www.prenhall.comlstatistics. Click on "Multivariate Statistics" and then click on our book. In addition, all full data sets saved as ASCII files that are used in the book are available on the web site. Instructors' Solutions Manual. An Instructors Solutions Manual is available on the author's website accessible through www.prenhall.comlstatistics.For information on additional forsale supplements that may be used with the book or additional titles of interest, please visit the Prentice Hall web site at www.prenhall. corn.
,ACKNOWLEDGMENTS We thank many of our colleagues who helped improve the applied aspect of the book by contributing their own data sets for examples and exercises. A number of individuals helped guide various revisions of this book, and we are grateful for their suggestions: Christopher Bingham, University of Minnesota; Steve Coad, University of Michigan; Richard Kiltie, University of Florida; Sam Kotz, George Mason University; Him Koul, Michigan State University; Bruce McCullough, Drexel University; Shyamal Peddada, University of Virginia; K. Sivakumar University of Illinois at Chicago; Eric Smith, Virginia Tecn; and Stanley Wasserman, University of Illinois at Urbanaciiampaign. We also acknowledge the feedback of the students we have taught these past 35 years in our applied multivariate analysis courses. Their comments and suggestions are largely responsible for the present iteration of this work. We would also like to give special thanks to Wai K wong Cheang, Shanhong Guan, Jialiang Li and Zhiguo Xiao for their help with the calculations for many of the examples. We must thank Dianne Hall for her valuable help with the Solutions Manual, Steve Verrill for computing assistance throughout, and Alison Pollack for implementing a Chernoff faces program. We are indebted to Cliff GiIman for his assistance with the multidimensional scaling examples discussed in Chapter 12. Jacquelyn Forer did most of the typing of the original draft manuscript, and we appreciate her expertise and willingness to endure cajoling of authors faced with publication deadlines. Finally, we would like to thank Petra Recter, Debbie Ryan, Michael Bell, Linda Behrens, Joanne Wendelken and the rest of the Prentice Hall staff for their help with this project. R. A. lohnson [emailprotected] D. W. Wichern [emailprotected]
ASPECTS OF MULTIVARIATE ANALYSIS 1.1 Introduction Scientific inquiry is an iterative learning process. Objectives pertaining to the explanation of a social or physical phenomenon must be specified and then tested by gathering and analyzing data. In turn, an analysis of the data gathered by experimentation or observation will usually suggest a modified explanation of the phenomenon. Throughout this iterative learning process, variables are often added or deleted from the study. Thus, the complexities of most phenomena require an investigator to collect observations on many different variables. This book is concerned with statistical methods designed to elicit information from these kinds of data sets. Because the data include simultaneous measurements on many variables, this body .of methodology is called multivariate analysis. The need to understand the relationships between many variables makes multivariate analysis an inherently difficult subject. Often, the human mind is overwhelmed by the sheer bulk of the data. Additionally, more mathematics is required to derive multivariate statistical techniques for making inferences than in a univariate setting. We have chosen to provide explanations based upon algebraic concepts and to avoid the derivations of statistical results that require the calculus of many variables. Our objective is to introduce several useful multivariate techniques in a clear manner, making heavy use of illustrative examples and a minimum of mathematics. Nonetheless, some mathematical sophistication and a desire to think quantitatively will be required. Most of our emphasis will be on the analysis of measurements obtained without actively controlling or manipulating any of the variables on which the measurements are made. Only in Chapters 6 and 7 shall we treat a few experimental plans (designs) for generating data that prescribe the active manipulation of important variables. Although the experimental design is ordinarily the most important part of a scientific investigation, it is frequently impossible to control the
2 Chapter 1 Aspects of Multivariate Analysis generation of appropriate data in certain disciplines. (This is true, for example, in business, economics, ecology, geology, and sociology.) You should consult [6] and [7] for detailed accounts of design principles that, fortunately, also apply to multivariate situations. It will become increasingly clear that many multivariate methods are based upon an underlying proBability model known as the multivariate normal distribution. Other methods are ad hoc in nature and are justified by logical or commonsense arguments. Regardless of their origin, multivariate techniques must, invariably, be implemented on a computer. Recent advances in computer technology have been accompanied by the development of rather sophisticated statistical software packages, making the implementation step easier. Multivariate analysis is a "mixed bag." It is difficult to establish a classification scheme for multivariate techniques that is both widely accepted and indicates the appropriateness of the techniques. One classification distinguishes techniques designed to study interdependent relationships from those designed to study dependent relationships. Another classifies techniques according to the number of populations and the number of sets of variables being studied. Chapters in this text are divided into sections according to inference about treatment means, inference about covariance structure, and techniques for sorting or grouping. This should not, however, be considered an attempt to place each method into a slot. Rather, the choice of methods and the types of analyses employed are largely determined by the objectives of the investigation. In Section 1.2, we list a smaller number of practical problems designed to illustrate the connection between the choice of a statistical method and the objectives of the study. These problems, plus the examples in the text, should provide you with an appreciation of the applicability of multivariate techniques acroSS different fields. The objectives of scientific investigations to which multivariate methods most naturally lend themselves include the following: L Data reduction or structural simplification. The phenomenon being studied is represented as simply as possible without sacrificing valuable information. It is hoped that this will make interpretation easier. 2. Sorting and grouping. Groups of "similar" objects or variables are created, based upon measured characteristics. Alternatively, rules for classifying objects into welldefined groups may be required. 3. Investigation of the dependence among variables. The nature of the relationships among variables is of interest. Are all the variables mutually independent or are one or more variables dependent on the others? If so, how? 4. Prediction. Relationships between variables must be determined for the purpose of predicting the values of one or more variables on the basis of observations on the other variables. 5. Hypothesis construction and testing. Specific statistical hypotheses, formulated in terms of the parameters of multivariate populations, are tested. This may be done to validate assumptions or to reinforce prior convictions. We conclude this brief overview of multivariate analysis with a quotation from F. H. C Marriott [19], page 89. The statement was made in a discussion of cluster analysis, but we feel it is appropriate for a broader range of methods. You should keep it in mind whenever you attempt or read about a data analysis. It allows one to
maintain a proper perspective and not be overwhelmed by the elegance of some of the theory: If the results disagree with informed opinion, do not admit a simple logical interpreta
tion, and do not show up clearly in a graphical presentation, they are probably wrong. There is no magic about numerical methods, and many ways in which they can break down. They are a valuable aid to the interpretation of data, not sausage machines automatically transforming bodies of numbers into packets of scientific fact.
The published applications of multivariate methods have increased tremendously in recent years. It is now difficult to cover the variety of realworld applications of these methods with brief discussions, as we did in earlier editions of this book. However, in order to give some indication of the usefulness of multivariate techniques, we offer the following short descriptions_of the results of studies from several disciplines. These descriptions are organized according to the categories of objectives given in the previous section. Of course, many of our examples are multifaceted and could be placed in more than one category.
• Using data on several variables related to cancer patient responses to radiotherapy, a simple measure of patient response to radiotherapy was constructed. (See Exercise 1.15.) • ltack records from many nations were used to develop an index of performance for both male and female athletes. (See [8] and [22].) • Multispectral image data collected by a highaltitude scanner were reduced to a form that could be viewed as images (pictures) of a shoreline in two dimensions. (See [23].) • Data on several variables relating to yield and protein content were used to create an index to select parents of subsequent generations of improved bean plants. (See [13].) • A matrix of tactic similarities was developed from aggregate data derived from professional mediators. From this matrix the number of dimensions by which professional mediators judge the tactics they use in resolving disputes was determined. (See [21].)
Sorting and grouping • Data on several variables related to computer use were employed to create clusters of categories of computer jobs that allow a better determination of existing (or planned) computer utilization. (See [2].) • Measurements of several physiological variables were used to develop a screening procedure that discriminates alcoholics from nonalcoholics. (See [26].) • Data related to responses to visual stimuli were used to develop a rule for separating people suffering from a multiplesclerosiscaused visual pathology from those not suffering from the disease. (See Exercise 1.14.)
• The U.S. Internal Revenue Service uses data collected from tax returns to sort taxpayers into two groups: those that will be audited and those that will not. (See [31].)
The preceding descriptions offer glimpses into the use of multivariate methods in widely diverse fields.
• Data on several variables were used to identify factors that were responsible for client success in hiring external consultants. (See [12].) • Measurements of variables related to innovation, on the one hand, and variables related to the business environment and business organization, on the other hand, were used to discover why some firms are product innovators and some firms are not. (See [3].) • Measurements of pulp fiber characteristics and subsequent measurements of . characteristics of the paper made from them are used to examine the relations between pulp fiber properties and the resulting paper properties. The goal is to determine those fibers that lead to higher quality paper. (See [17].) • The associations between measures of risktaking propensity and measures of socioeconomic characteristics for toplevel business executives were used to assess the relation between risktaking behavior and performance. (See [18].) . Prediction
• The associations between test scores, and several high school performance variables, and several college performance variables were used to develop predictors of success in college. (See [10).) • Data on several variables related to the size distribution of sediments were used to develop rules for predicting different depositional environments. (See [7] and [20].) • Measurements on several accounting and financial variables were used to develop a method for identifying potentially insolvent propertyliability insurers. (See [28].) • cDNA microarray experiments (gene expression data) are increasingly used to study the molecular variations among cancer tumors. A reliable classification of tumors is essential for successful diagnosis and treatment of cancer. (See [9].)
1.3 The Organization of Data Throughout this text, we are going to be concerned with analyzing measurements made on several variables or characteristics. These measurements (commonly called data) must frequently be arranged and displayed in various ways. For example, graphs and tabular arrangements are important aids in data analysis. Summary numbers, which quantitatively portray certain features of the data, are also necessary to any description. We now introduce the preliminary concepts underlying these first steps of data organization.
Arrays Multivariate data arise whenever an investigator, seeking to understand a social or physical phenomenon, selects a number p ~ 1 of variables or characters to record . The values of these variables are all recorded for each distinct item, individual, or experimental unit. We will use the notation Xjk to indicate the particular value of the kth variable that is observed on the jth item, or trial. That is, Xjk =
Consequently, n measurements on p variables can be displayed as follows: Variable 1
• Several pollutionrelated variables were measured to determine whether levels for a large metropolitan area were roughly constant throughout the week, or whether there was a noticeable difference between weekdays and weekends. (See Exercise 1.6.) • Experimental data on several variables were used to see whether the nature of the instructions makes any difference in perceived risks, as quantified by test scores. (See [27].) • Data on many variables were used to investigate the differences in structure of American occupations to determine the support for one of two competing sociological theories. (See [16] and [25].) • Data on several variables were used to determine whether different types of firms in newly industrialized countries exhibited different patterns of innovation. (See [15].)
Or we can display these data as a rectangular array, called X, of n rows and p columns: Xll
The array X, then, contains the data consisting of all of .the observations on all of the variables.
Example 1.1 (A data array) A selection of four receipts from a university bookstore was obtained in order to investigate the nature of book sales. Each receipt provided, among other things, the number of books sold and the total amount of each sale. Let the first variable be total dollar sales and the second variable be number of books sold. Then we can re&ard the corresponding numbers on the receipts as four measurements on two variables. Suppose the data, in tabular form, are
If the n measurements represent a subset of the full set of measurements that might have been observed, then Xl is also called the sample mean for the first variable. We adopt this terminology because the bulk of this book is devoted to procedUres designed to analyze samples of measurements from larger collections. The sample mean can be computed from the n measurements on each of the p variables, so that, in general, there will be p sample means:
with four rows and two columns.
k = 1,2, ... ,p
A measure of spread is provided by the sample variance, defined for n measurements on the first variable as
Considering data in the form of arrays facilitates the exposition of the subject matter and allows numerical calculations to be performed in an orderly and efficient manner. The efficiency is twofold, as gains are attained in both (1) describing numerical calculations as operations on arrays and (2) the implementation of the calculations on computers, which now use many languages and statistical packages to perform array operations. We consider the manipulation of arrays of numbers in Chapter 2. At this point, we are concerned only with their value as devices for displaying data.
XiI'S.
k = 1,2, ... ,p
k=I,2, ... ,p
.
1\vo comments are in order. First, many authors define the sample variance with a divisor of n  1 rather than n. Later we shall see that there are theoretical reasons for doing this, and it is particularly appropriate if the number of measurements, n, is small. The two versions of the sample variance will always be differentiated by displaying the appropriate expression. Second, although the S2 notation is traditionally used to indicate the sample variance, we shall eventually consider an array of quantities in which the sample variances lie along the main diagonal. In this situation, it is convenient to use double subscripts on the variances in order to indicate their positions in the array. Therefore, we introduce the notation Skk to denote the same variance computed from measurements on the kth variable, and we have the notational identities
Descriptive Statistics A large data set is bulky, and its very mass poses a serious obstacle to any attempt to visually extract pertinent information. Much of the information contained in the data can be assessed by calculating certain summary numbers, known as descriptive statistics. For example, the arithmetic average, or sample mean, is a descriptive statistic that provides a measure of locationthat is, a "central value" for a set of numbers. And the average of the squares of the distances of all of the numbers from the mean provides a measure of the spread, or variation, in the numbers. We shall rely most heavily on descriptive statistics that measure location, variation, and linear association. The formal definitions of these quantities follow. Let Xll, X2I>"" Xnl be n measurements on the first variable. Then the arithmetic average of these measurements is
The square root of the sample variance, ~, is known as the sample standard deviation. This measure of variation uses the same units as the observations. Consider n pairs of measurements on each of variables 1 and 2:
[X2l], •.. , [Xnl] X22 Xn2
That is, Xjl and Xj2 are observed on the jth experimental item (j = 1,2, ... , n). A measure of linear association between the measurements of variables 1 and 2 is provided by the sample covariance
or the average product of the deviations from their respective means. If large values for one variable are observed in conjunction with large values for the other variable, and the small values also occur together, sl2 will be positive. If large values from one variable occur with small values for the other variable, Sl2 will be negative. If there is no particular association between the values for the two variables, Sl2 will be approximately zero. The sample covariance
= 1,2, ... ,p,
1,2, ... ,p (14)
measures the association between the ·ith and kth variables. We note that the covariance reduces to the sample variance when i = k. Moreover, Sik = Ski for all i and k .. The final descriptive statistic considered here is the sample correlation coefficient (or Pearson's productmoment correlation coefficient, see [14]). This measure of the linear association between two variables does not depend on the units of measurement. The sample correlation coefficient for the ith and kth variables is defined as
The ~u~ntities Sik and rik do not, in general, convey all there is to know about the aSSOCIatIOn between two variables. Nonlinear associations can exist that are not revealed .by these ~es~riptive statistics. Covariance and corr'elation provide measures of lmear aSSOCIatIOn, or association along a line. Their values are less informative ~~r other kinds of association. On the other hand, these quantities can be very sensIttve to "wild" observations ("outIiers") and may indicate association when in fact, little exists. In spite of these shortcomings, covariance and correlation coefficien~s are routi':lel.y calculated and analyzed. They provide cogent numerical summan~s ~f aSSOCIatIOn ~hen the data do not exhibit obvious nonlinear patterns of aSSOCIation and when WIld observations are not present. . Suspect observa.tions must be accounted for by correcting obvious recording mIstakes and by takmg actions consistent with the identified causes. The values of Sik and rik should be quoted both with and without these observations. The sum of squares of the deviations from the mean and the sum of crossproduct deviations are often of interest themselves. These quantities are n
k = 1,2, ... ,p
1,2, ... ,p,
= 1,2, ... , p and k = 1,2, ... , p. Note rik = rki for all i and k. The sample correlation coefficient is a standardized version of the sample covariance, where the product of the square roots of the sample variances provides the standardization. Notice that rik has the same value whether n or n  1 is chosen as the common divisor for Sii, sa, and Sik' The sample correlation coefficient rik can also be viewed as a sample co variance. Suppose the original values 'Xji and Xjk are replaced by standardized values for i
1. The value of r must be between 1 and +1 inclusive. 2. Here r measures the strength of the linear association. If r = 0, this implies a lack of linear association between the components. Otherwise, the sign of r indicates the direction of the association: r < 0 implies a tendency for one value in the pair to be larger than its average when the other is smaller than its average; and r > 0 implies a tendency for one value of the pair to be large when the other value is large and also for both values to be small together. 3. The value of rik remains unchanged if the measurements of the ith variable are changed to Yji = aXji + b, j = 1,2, ... , n, and the values of the kth variable are changed to Yjk = CXjk + d, j == 1,2, ... , n, provided that the constants a and c have the same sign.
k = 1,2, ... ,p
The descriptive statistics computed from n measurements on p variables can also be organized into arrays.
cause both sets are centered at zero and expressed in standard deviation units. The sample correlation coefficient is just the sample covariance of the standardized observations. Although the signs of the sample correlation and the sample covariance are the same, the correlation is ordinarily easier to interpret because its magnitude is bounded. To summarize, the sample correlation r has the following properties:
The sample mean array is denoted by X, the sample variance and covari~nce array by the capital letter Sn, and the sample correlation array by R. The subscrIpt ~ on the array Sn is a mnemonic device used to remind you that n is employed as a divisor for the elements Sik' The size of all of the arrays is determined by the number of variables, p. The arrays Sn and R consist of p rows and p columns. The array x is a single column with p rows. The first subscript on an entry in arrays Sn and R indicates the row; the second subscript indicates the column. Since Sik = Ski and rik = rki for all i and k, the entries in symmetric positions about the main northwestsoutheast diagonals in arrays Sn and R are the same, and the arrays are said to be
symmetric.
Example 1.2 (The arrays ;c, SR' and R for bivariate data) Consider the data intro
duced in Example 1.1. Each. receipt yields a pair of measurements, total dollar sales, and number of books sold. Find the arrays X, Sn' and R. Since there are four receipts, we have a total of four measurements (observations) on each variable. Thesample means are 4
. = .36
data ~re ?lotted as seven points in two dimensions (each axis representIll~ a vanable) III FIgure 1.1. The coordinates of the points are determined by the patr~d measurements: (3,5), (4,5.5), ... , (5,7.5). The resulting twodimensional plot IS known as a scatter diagram or scatter plot.
are im~ortant, but frequently neglected, aids in data analysis. Although it is impossIble to simultaneously plot all the measurements made on several variables and study ~he configurations, plots of individual variables and plots of pairs of variables can stIll be very informative. Sophisticated computer programs and display equipn;tent al.low on~ the luxury of visually examining data in one, two, or three dimenSIOns WIth relatIve ease. On the other hand, many valuable insights can be obtained from !he data by const~uctin~ plots with paper and pencil. Simple, yet elegant and ~ffectIve, met~ods for ~IsplaYIllg data are available in [29]. It is good statistical practIce to plot paIrs of varIables and visually inspect the pattern of association. Consider, then, the following seven pairs of measurements on two variables:
.
I ..
Figure 1.1 A scatter plot and marginal dot diagrams.
Also shown in Figure 1.1 are separate plots of the observed values of variable 1 and the observed values of variable 2, respectively. These plots are called (marginal) dot diagrams. They can be obtained from the original observations or by projecting the points in the scatter diagram onto each coordinate axis. The information contained in the singlevariable dot diagrams can be used to calculate the sample means Xl and X2 and the sample variances SI 1 and S22' (See Exercise 1.1.) The scatter diagram indicates the orientation of the points, and their coordinates can be used to calculate the sample covariance s12' In the scatter diagram of Figure 1.1, large values of Xl occur with large values of X2 and small values of Xl with small values of X2' Hence, S12 will be positive. Dot diagrams and scatter plots contain different kinds of information. The information in the marginal dot diagrams is not sufficient for constructing the scatter plot. As an illustration, suppose the data preceding Figure 1.1 had been paired differently, so that the measurements on the variables Xl and X2 were as follows: Variable 1 Variable 2
for all 16 firms for all firms but Dun & Bradstreet for all firms but Time Warner for all firms but Dun & Bradstreet and Time Warner
Example 1.3 (The effect of unusual observations on sample correlations) Some fi . nancial data representing jobs and productivity for the 16 largest publishing firms appeared in an article in Forbes magazine on April 30, 1990. The data for the pair of variables Xl = employees Gobs) and X2 = profits per employee (productivity) are graphed in Figure 1.3. We have labeled two "unusual" observations. Dun & Bradstreet is the largest firm in terms of number of employees, but is "typical" in terms of profits per employee. TIme Warner has a "typical" number of employees, but comparatively small (negative) profits per employee.
(We have simply rearranged the values of variable 1.) The scatter and dot diagrams for the "new" data are shown in Figure 1.2. Comparing Figures 1.1 and 1.2, we find that the marginal dot diagrams are the same, but that the scatter diagrams are decidedly different. In Figure 1.2, large values of Xl are paired with small values of X2 and small values of Xl with large values of X2' Consequently, the descriptive statistics for the individual variables Xl, X2, SI 1> and S22 remain unchanged, but the sample covariance S12, which measures the association between pairs of variables, will now be negative. The different orientations of the data in Figures 1.1 and 1.2 are not discernible from the marginal dot diagrams alone. At the same time, the fact that the marginal dot diagrams are the same in the two cases is not immediately apparent from the scatter plots. The two types of graphical procedures complement one another; they are nqt competitors. The next two examples further illustrate the information that can be conveyed by a graphic display.
It is clear that atypical observations can have a considerable effect on the sample correlation coefficient.
... XI
Figure 1.2 Scatter plot and dot diagrams for rearranged data.
Example 1.4 (A scatter plot for baseball data) In a July 17,1978, article on money in sports, Sports Illustrated magazine provided data on Xl = player payroll for National League East baseball teams. We have added data on X2 = wonlost percentage "for 1977. The results are given in Table 1.1. The scatter plot in Figure 1.4 supports the claim that a championship team can be bought. Of course, this causeeffect relationship cannot be substantiated, because the experiment did not include a random assignment of payrolls. Thus, statistics cannot answer the question: Could the Mets have won with $4 million to spend on player salaries?
Philadelphia Phillies Pittsburgh Pirates St. Louis Cardinals Chicago Cubs Montreal Expos New York Mets
121.41 127.70 129.20 131.80 135.10 131.50 126.70 115.10 130.80 124.60 118.31 114.20 120.30 115.70 117.51 109.81 109.10 115.10 118.31 112.60 116.20 118.00 131.00 125.70 126.10 125.80 125.50 127.80 130.50 127.90 123.90 124.10 120.80 107.40 120.70 121.91 122.31 110.60 103.51 110.71 113.80
Figure 1.4 Salaries and wonlost percentage from Table 1.1.
To construct the scatter plot in Figure 1.4, we have regarded the six paired observations in Table 1.1 as the coordinates of six points in twodimensional space. The figure allows us to examine visually the grouping of teams with respect to the variables total payroll and wonlost percentage. 
Example I.S (Multiple scatter plots for paper strength measurements) Paper is manufactured in continuous sheets several feet wide. Because of the orientation of fibers within the paper, it has a different strength when measured in the direction produced by the machine than when measured across, or at right angles to, the machine direction. Table 1.2 shows the measured values of
A novel graphic presentation of these data appears in Figure 1.5, page' 16. The scatter plots are arranged as the offdiagonal elements of a covariance array and box plots as the diagonal elements. The latter are on a different scale with this
12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
.841 .816 .840 .842 .820 .802 .828 .819 .826 .802 .810 .802 .832 .796 .759 .770 .759 .772 .806 .803 .845 .822 .971 .816 .836 .815 .822 .822 .843 .824 .788 .782 .795 .805 .836 .788 .772 .776 .758
Source: Data courtesy of SONOCO Products Company.
Cross direction 70.42 72.47 78.20 74.89 71.21 78.39 69.02 73.10 79.28 76.48 70.25 72.88 68.23 68.12 71.62 53.10 50.85 51.68 50.60 53.51 56.53 70.70. 74.35 68.29 72.10 70.64 76.33 76.75 80.33 75.68 78.54 71.91 68.22 54.42 70.41 73.68 74.93 53.52 48.93 53.67 52.42
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on the jth item represent the coordinates of a point in pdimensional space. The coordinate axes are taken to correspond to the variables, so that the jth point is Xjl units along the first axis, Xj2 units along the second, ... , Xjp units along the pth axis. The resulting plot with n points not only will exhibit the overall pattern of variability, but also will show similarities (and differences) among the n items. Groupings of items will manifest themselves in this representation. The next example illustrates a threedimensional scatter plot.
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n Points in p Dimensions (pDimensional Scatter Plot). Consider the natural extension of the scatter plot to p dimensions, where the p measurements
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Example 1.6 (Looking for lowerdimensional structure) A zoologist obtained measurements on n = 25 lizards known scientifically as Cophosaurus texanus. The weight, or mass, is given in grams while the snoutvent length (SVL) and hind limb span (HLS) are given in millimeters. The data are displayed in Table 1.3. Although there are three size measurements, we can ask whether or not most of the variation is primarily restricted to two dimensions or even to one dimension. To help answer questions regarding reduced dimensionality, we construct the threedimensional scatter plot in Figure 1.6. Clearly most of the variation is scatter about a onedimensional straight line. Knowing the position on a line along the major axes of the cloud of poinfs would be almost as good as knowing the three measurements Mass, SVL, and HLS. However, this kind of analysis can be misleading if one variable has a much larger variance than the others. Consequently, we first calculate the standardized values, Zjk = (Xjk  Xk)/~' so the variables contribute equally to the variation
Figure 1.5 Scatter plots and boxplots of paperquality data from Thble 1.2. software so we use only the overall shape to provide information on symme~ry and possible outliers for each individual characteristic. The scatter plots can be mspected for patterns and unusual observations. In Figure 1.5, there is one unusual observation: the density of specimen 25. Some of the scatter plots have patterns suggesting that there are two separate clumps of observations. These scatter plot arrays are further pursued in our discussion of new software graphics in the next section.
In the general multiresponse situation, p variables are simultaneously rec~rded oon items. Scatter plots should be made for pairs of. important variables and, If the task is not too great to warrant the effort, for all pairs. . Limited as we are to a three:dimensional world, we cannot always picture an entire set of data. However, two further geom7tri~ repres~nta~ions of t?e. data provide an important conceptual framework for Vlewmg multIvanable statlstlc~l methods. In cases where it is possible to capture the essence of the data m three dimensions, these representations can actually be graphed.
5.526 10.401 9.213 8.953 7.063 6.610 11.273 2.447 15.493 . 9.004 8.199 6.601 7.622
Source: Data courtesy of Kevin E. Bonine.
Figure 1.8 repeats the scatter plot for the original variables but with males marked by solid circles and females by open circles. Clearly, males are typically larger than females.
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Figure 1.6 3D scatter plot of lizard data from Table 1.3.
er lot. Figure 1.7 gives the threedimensio~al scatter plot for ~he stanin the scatt . Pbl Most of the variation can be explamed by a smgle vanable de. d vana es. dard~ze d b a line through the cloud of points. ternune y
Figure 1.8 3D scatter plot of male and female lizards.
p Points in n Dimensions. The n observations of the p variables can also be regarded as p points in ndimensional space. Each column of X determines one of the points. The ith column,
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• consisting of all n measurements on the ith variable, determines the ith point. In Chapter 3, we show how the closeness of points in n dimensions can be related to measures of association between the corresponding variables .
.
plot of standardized lizard data. 
sional scatter plot can often reveal group structure.
E)(arnpl~ ~.~ ~eresting to see if male and female lizards occupy different parts ~f the ple 1.6, It IS m. I space containing the size data. The gender, by row, for the lizard hree_dimenslona ~ata in Table 1.3 are fmffmfmfmfmfm mmmfmmmffmff
1.4 Data Displays and Pictorial Representations The rapid development of powerful personal computers and workstations has led to a proliferation of sophisticated statistical software for data analysis and graphics. It is often possible, for example, to sit at one's desk and examine the nature of multidimensional data with clever computergenerated pictures. These pictures are valuable aids in understanding data and often prevent many false starts and subsequent inferential problems. As we shall see in Chapters 8 and 12, there are several techniques that seek to represent pdimensional observations in few dimensions such that the original distances (or similarities) between pairs of observations are (nearly) preserved. In general, if multidimensional observations can be represented in two dimensions, then outliers, relationships, and distinguishable groupings can often be discerned by eye. We shall discuss and illustrate several methods for displaying multivariate data in two dimensions. One good source for more discussion of graphical methods is [11].
Linking Multiple TwoDimensional Scatter Plots One of the more exciting new graphical procedures involves electronically connecting many twodimensional scatter plots. Example 1.8 (Linked scatter plots and brushing) To illustrate linked twodimensional scatter plots, we refer to the paperquality data in Thble 1.2. These data represent measurements on the variables Xl = density, X2 = strength in the machine direction, and X3 = strength in the cross direction. Figure 1.9 shows twodimensional scatter plots for pairs of these variables organized as a 3 X 3 array. For example, the picture in the upper lefthand corner of the figure is a scatter plot of the pairs of observations (Xl' X3)' That is, the Xl values are plotted along the horizontal axis, and the X3 values are plotted along the vertical axis. The lower righthand corner of the figure contains a scatter plot of the observations (X3, Xl)' That is, the axes are reversed. Corresponding interpretations hold for the other scatter plots in the figure. Notice that the variables and their threedigit ranges are indicated in the boxes along the SWNE diagonal. The operation of marking (selecting), the obvious outlier in the (Xl, X3) scatter plot of Figure 1.9 creates Figure 1.1O(a), where the outlier is labeled as specimen 25 and the same data point is highlighted in all the scatter plots. Specimen 25 also appears to be an outlierin the (Xl, X2) scatter plot but not in the (Xz, X3) scatter plot. The operation of deleting this specimen leads to the modified scatter plots of Figure 1.10(b). From Figure 1.10, we notice that some points in, for example, the (X2' X3) scatter plot seem to be disconnected from the others. Selecting these points, using the (dashed) rectangle (see page 22), highlights the selected points in all of the other scatter plots and leads to the display in Figure 1.ll(a). Further checking revealed that specimens 1621, specimen 34, and specimens 3841 were actually specimens
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from an older roll of paper that was included in order to have enough plies in the cardboard being manufactured. Deleting the outlier and the cases corresponding to the older paper and adjusting the ranges of the remaining observations leads to the scatter plots in Figure 1.11 (b) . The operation of highlighting points corresponding to a selected range of one of the variables is called brushing. Brushing could begin with a rectangle, as in Figure l.U(a), but then the brush could be moved to provide a sequence of highlighted points. The process can be stopped at any time to provide a snapshot of the current situation. _
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Scatter plots like those in Example 1.8 are extremely useful aids in data analysis. Another important new graphical technique uses software that allows the data analyst to view highdimensional data as slices of various threedimensional perspectives. This can be done dynamically and continuously until informative views are obtained. A comprehensive discussion of dynamic graphical methods is available in [1]. A strategy for online multivariate exploratory graphical analysis, motivated by the need for a routine procedure for searching for structure in multivariate data, is given in [32].
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Example 1.9 (Rotated plots in three dimensions) Four different measurements of lumber stiffness are given in Table 4.3, page 186. In Example 4.14, specimen (board) 16 and possibly specimen (board) 9 are identified as unusual observations. Figures 1.12(a), (b), and (c) contain perspectives of the stiffness data in the XbX2, X3 space. These views were obtained by continually rotating and turning the threedimensional coordinate axes. Spinning the coordinate axes allows one to get a better
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Figure 1.12 Threedimensional perspectives for the lumber stiffness data.
understanding of the threedimensional aspects of the data. Figure 1.12(d) gives one picture of the stiffness data in X2, X3, X4 space. Notice that Figures 1.12(a) and (d) visually confirm specimens 9 and 16 as outliers. Specimen 9 is very large in all three coordinates. A counterclockwiselike rotation of the axes in Figure 1.12(a) produces Figure 1.12(b), and the two unusual observations are masked in this view. A further spinning of the X2, X3 axes gives Figure 1.12(c); one of the outliers (16) is now hidden. Additional insights can sometimes be gleaned from visual inspection of the slowly spinning data. It is this dynamic aspect that statisticians are just beginning to understand and exploit. _
Plots like those in Figure 1.12 allow one to identify readily observations that do not conform to the rest of the data and that may heavily influence inferences based on standard datagenerating models.
When the height of a young child is measured at each birthday, the points can be plotted and then connected by lines to produce a graph. This is an example of a growth curve. In general, repeated measurements of the same characteristic on the same unit or subject can give rise to a growth curve if an increasing, decreasing, or even an increasing followed by a decreasing, pattern is expected. Example 1.10 (Arrays of growth curves) The Alaska Fish and Game Department monitors grizzly bears with the goal of maintaining a healthy population. Bears are shot with a dart to induce sleep and weighed on a scale hanging from a tripod. Measurements of length are taken with a steel tape. Table 1.4 gives the weights (wt) in kilograms and lengths (lngth) in centimeters of seven female bears at 2,3,4, and 5 years of age. . First, for each bear, we plot the weights versus the ages and then connect the weights at successive years by straight lines. This gives an approximation to growth curve for weight. Figure 1.13 shows the growth curves for all seven bears. The noticeable exception to a common pattern is the curve for bear 5. Is this an outlier or just natural variation in the population? In the field, bears are weighed on a scale that
reads pounds. Further inspection revealed that, in this case, an assistant later failed to convert the field readings to kilograms when creating the electronic database. The correct weights are (45, 66, 84, 112) kilograms. B.ecause it can be difficult to inspect visually the individual growth curves in a c.ombmed. plot, the individual curves should be replotted in an array where similaritIes an? dIfferences are easily observed. Figure 1.14 gives the array of seven curves for weIght. Some growth curves look linear and others quadratic. Bear I
Source: Data courtesy of H. Roberts.
Figure 1.13 Combined growth curves for weight for seven female grizzly bears.
Figure 1.14 Individual growth curves for weight for female grizzly bears.
Figure 1.15 gives a growth curve array for length. One bear seemed to get shorter from 2 to 3 years old, but the researcher knows that the steel tape measurement of length can be thrown off by the bear's posture when sedated.
Boston Edison Co. (2)
Central Louisiana Electric Co. (3)
Consolidated Edison Co. (NY) (5) I
Commonwealtb Edison Co. (4)
figure 1.15 Individual growth curves for length for female grizzly bears.
We now turo to two popular pictorial representations of multivariate data in two dimensions: stars and Cherooff faces.
Stars Suppose each data unit consists of .nonnegativ: observations on p. ~ 2.variables. In two dimensions, we can construct crrcles of a fixed (reference) radIUS WIth p equally spaced rays emanating from the center of the circle. The lengths of.the ra~s rep.resent the values of the variables. The ends of the rays can be connected With straight lmes to form a star. Each star represents a multivariate observation, and the stars can be grouped according to their (subjective) siniilarities. It is often helpful, when constructing the stars, to standardize the observations. In this case some of the observations will be negative. The observations can then be reexpressed so. that the center of the circle represents the smallest standardized observation within the entire data set. Example 1.11 (Utility data as stars) Stars representing the first 5 of the ~2 publi.c
utility [rrms in Table 12.4, page 688, are shown in Figure 1.16. There are eight vaflabIes; consequently, the stars are distorted octagons.
figure 1.16 Stars for the first five public utilities.
. The observations on all variables were standardized. Among the first five utilitIes, the smallest standardized observation for any variable was 1.6. TIeating this value ~s ~er~, the variables are plotted on identical scales along eight equiangular rays ongmatmg from the center of the circle. The variables are ordered in a clockwise direction, beginning in the 12 o'clock position. At first glance, none of these utilities appears to be similar to any other. However, beca~se of t~e way the stars are constructed, each variable gets equal weight in the visualImpresslOn. If we concentrate on the variables 6 (sales in kilowatthour [kWh1 use per year) and 8 (total fuel costs in cents per kWh), then Boston Edison and Consolidated Edison are similar (small variable 6, large variable 8), and Arizona Public Service, Central Louisiana Electric, and Commonwealth Edison are similar (moderate • variable 6, moderate variable 8).
Chernoff faces ~eople react to faces. Cherooff [41 suggested representing pdimensional observatIOns as a twodimensional face whose characteristics (face shape, mouth curvature, nose length, eye size, pupil position, and so forth) are determined by the measurements on the p variables.
28 Chapter 1 Aspects of Multivariate Analysis As originally designed, Chernoff faces can handle up to 18 variables. The assignment of variables to facial features is done by the experimenter, and different choices produce different results. Some iteration is usually necessary before satisfactory representations are achieved. Chernoff faces appear to be most useful for verifying (1) an initial grouping suggested by subjectmatter knowledge and intuition or (2) final groupings produced by clustering algorithmS.
Example 1.12 (Utility data as Cher!,!off faces) From the data in Table 12.4, the 22
public utility companies were represented as Chernoff faces. We have the following correspondences: Variable
FIxedcharge coverage Rate of return on capital Cost per kW capacity in place Annual load factor
X5: Peak kWh demand growth from 1974 X6: Sales (kWh use per year) X7: Percent nuclear Xs: Total fuel costs (cents per kWh)
Facial characteristic Halfheight of face Face width Position of center of mouth Slant of eyes Eccentricity
The Chernoff faces are shown in Figure 1.17. We have subjectively grouped "similar" faces into seven clusters. If a smaller number of clusters is desired, we might combine clusters 5,6, and 7 and, perhaps, clusters 2 and 3 to obtain four or five clusters. For our assignment of variables to facial features, the firms group largely according to geographical location. _
Figure 1.17 Cherooff faces for 22 public utilities.
Liquidity ... Profitability
Figure 1.18 Cherooff faces over time.
Example 1.13 (Using Chernoff faces to show changes over time) Figure 1.18 illustrates an additional use of Chernofffaces. (See [24].) In the figure, the faces are used to track the financial wellbeing of a company over time. As indicated, each facial feature represents a single financial indicator, and the longitudinal changes in these indicators are thus evident at a glance. _
Constructing Chernoff faces is a task that must be done with the aid of a computer. The data are ordinarily standardized within the computer program as part of the process for determining the locations, sizes, and orientations of the facial characteristics. With some training, we can use Chernoff faces to communicate similarities or dissimilarities, as the next example indicates.
Cherooff faces have also been used to display differences in m~ltivariate ob~er vations in two dimensions. For example, the twodi~ensional coordInate ~xes ffilght resent latitude and longitude (geographical locatiOn), and the faces mIght repr~ ::~t multivariate measurements on several U.S. cities. Additional examples of thiS
kind are discussed in [30]. .... . There are several ingenious ways to picture multIvanate data m two dimensiOns. We have described some of them. Further advance~ are possible and will almost certainly take advantage of improved computer graphICs.
1.5 Distance Although they may at first appear formida?le, ~ost multiv~ate tec~niques are based
upon the simple concept of distance. StraIght~e, or Euclidean, d~stan~e sh~uld be familiar. If we consider the point P 0= (Xl ,.X2) III th~ plane, the straIghtlIne dIstance, d(O, P), from P to the origin 0 = (0,0) IS, accordmg to the Pythagorean theorem,
The situation is illustrated in Figure 1.19. In general, if the point P has p coo:d.inates so that P = (x), X2, •.. ' x p ), the straightline distance from P to the ongm 0= (O,O, ... ,O)is
Vxr + x~ + ... + x~
(See Chapter 2.) All points (Xl> X2, ... : xp) thatlie a constant squared distance, such as c2, from the origin satisfy the equatIon 2 d2(O, P) = XI + x~ + ... + x~ = c (111) Because this is the equation of a hypersphere (a circle if p = 2), points equidistant from the origin lie on a hypersphere. .. . . The straightline distance between two arbItra~y ~OInts P and Q WIth COordInatesP = (XI,X2, ... ,X p ) andQ 0= (Yl>Y2,···,Yp)lsglVenby d(P,Q) = V(XI  YI)2 + (X2  )'z)2 + ... + (xp  Yp)2
Straightline, or Euclidean, distance is unsatisfactory for most stat~stical purp~s es. This is because each coordinate contributes equally to the calculatlOn of ~uchd ean distance. When the coordinates r~prese~t .measurem~nts that ar~ subject t~ andom fluctuations of differing magmtudes, It IS often deslfable to weIght CO?rdl ~ates subject to a great deal of variability less ~eavily than those that are not highly variable. This suggests a different measure ?f ~lst,~n~e. . Our purpose now is to develop a "staUstlcal distance that ac:counts for dIfferences in variation and, in due course, the presence of correlatlOn. Because our
choice will depend upon the sample variances and covariances, at this point we use the term statistical distance to distinguish it from ordinary Euclidean distance. It is statistical distance that is fundamental to multivariate analysis. To begin, we take as fIXed the set of observations graphed as the pdimensional scatter plOt. From these, we shall construct a measure of distance from the origin to a point P = (Xl, X2, ..• , xp). In our arguments, the coordinates (Xl> X2, ... , xp) of P can vary to produce different locations for the point. The data that determine distance will, however, remain fixed. To illustrate, suppose we have n pairs of measurements on two variables each having mean zero. Call the variables Xl and X2, and assume that the Xl measurements vary independently of the X2 measurements, I In addition, assume that the variability in the X I measurements is larger than the variability in the X2 measurements. A scatter plot of the data would look something like the one pictured in Figure 1.20. X2
Glancing at Figure 1.20, we see that values which are a given deviation from the origin in the Xl direction are not as "surprising" or "unusual" as ~re values equidistant from the origin in the X2 direction. This is because the inherent variability in the Xl direction is greater than the variability in the X2 direction. Consequently, large Xl coordinates (in absolute value) are not as unexpected as large X2 coordinates. It seems reasonable, then, to weight an X2 coordinate more heavily than an Xl coordinate of the same value when computing the "distance" to the origin. . One way to proceed is to divide each coordinate by the sample standard deviatIOn. Therefore, upon division by the standard deviations, we have the "standardized" coordinates x; = xIi";;;; and x; = xz/vS;. The standardized coordinates are now on an equal footing with one another. After taking the differences in variability into account, we determine distance using the standard EucIidean formula. Thus, a statistical distance of the point P = (Xl, X2) from the origin 0 = (0,0) can be computed from its standardized coordinates x~ = xIiVS;; and xi 0= X2/VS; as d(O, P) =V(xD2 = )(
Figure 1.19 Distance given by the Pythagorean theorem.
Figure 1.20 A scatter plot with greater variability in the Xl direction than in the X2 direction.
IAt this point, "independently" means that the Xz measurements cannot be predicted with any accuracy from the Xl measurements, and vice versa.
Comparing (113) with (19), we see that the difference between the two expressions is due to the weights kl = l/s11 and k2 = l/s22 attached to xi and x~ in (1l3). Note that if the sample variances are the same, kl = k 2 , then xI and x~ will receive the same weight. In cases where the weights are the same, it is convenient to ignore the common divisor and use the usual Euc1idean distance formula. In other words, if the variability in thexl direction is the same as the variability in the X2 direction, and the Xl values vary independently of the X2 values, Euc1idean distance is appropriate. Using (113), we see that all points which have coordinates (Xl> X2) and are a constant squared distance c2 from the origin must satisfy
(114) . Equation (114) is the equation of an ellipse centered at the origin whose major and minor axes coincide with the coordinate axes. That is, the statistical distance in (113) has an ellipse as the locus of all points a constant distance from the origin. This general case is shown in Figure 1.21.
. XI + x~ DIstance'. 4 1
. A pl?t ?f the equation xt/4 + xVI = 1 is an ellipse centered at (0,0) whose major. aXIS he~ along the Xl coordinate axis and whose minor axis lies along the X2 coordmate aXIS. The halflengths of these major and minor axes are v'4 = 2 and VI = 1, :espectively. The ellipse of unit distance is plotted in Figure 1.22. All points on the ellIpse are regarded as being the same statistical distance from the originin this case, a distance of 1. • x,
.
statistical distance d 2(O,P) = xI!sll + X~/S22 = c 2.
Example 1.14 (Calculating a statistical distance) A set of paired measurements (Xl, X2) on two variables yields Xl = X2 = 0, Sll = 4, and S22 = 1. Suppose the Xl
measurements are unrelated to the x2 measurements; that is, measurements within a pair vary independently of one another. Since the sample variances are unequal, we measure the square of the distance of an arbitrary point P = (Xl, X2) to the origin 0= (0,0) by
All points (Xl, X2) that are a constant distance 1 from the origin satisfy the equation x2
The coordinates of some points a unit distance from the origin are presented in the following table:
1.
The expression in (113) can be generalized to accommodate the calculation of statistical distance from an arbitrary point P = (Xl, X2) to any fIXed point Q = (YI, )'z). ~f we assume that .the coordinate variables vary independently of one another, the dIstance from P to Q is given by d(P, Q) =
.The extension of this statistical distance to more than two dimensions is straIghtforward. Let the points P and Q have p coordinates such that P = (x~, X2,···, xp) and Q = (Yl,)'z, ... , Yp). Suppose Q is a fixed point [it may be the ongm 0 = (0,0, ... , O)J and the coordinate variables vary independently of one another. Let Su, s22,"" spp be sample variances constructed from n measurements on Xl, X2,"" xp, respectively. Then the statistical distance from P to Q is d(P,Q) =
Yl? + (X2  )'z)2 + ... + (xp  Yp)2 s22 spp
All points P that are a constan t squared distance from Q rle on ad'hyperellipsoid t d at Q whose major and minor axes are parallel to the coor ma e ax es. We centere . note th~ followmg: 1. The distance of P to the origin 0 is obtained by setting Yl = )'2 = ... = YP =
in (116). 
_ .,. =
• The distance in (116) still does not include most of the i~porta~cases ~erSphl~! f the assumption of indepen dent coordmates. e sca e enc~unteri ~;c::;~~ a twodimensional situation in which the xl ~easur~m~nts ~o io FIgure. . f h X measurements. In fact, the coordmates 0 t e p~Irs o.ot vary mdepen dently 0 t e 2 mall together and the sample correlatIOn ) h'b't a tendenc y to b e 1arge or s ' h (.~lf~~ie~ i~ ;ositive . Moreov er, the variability in the X2 direction is larger than t e e co . d' f variability.m the Xl . Ifgfec ::~asure of distance when the variability in the Xl direcWhat IS a meamn u . bles X and X . h variability in the X2 directio n an d t h e vana 1 2 tion is dl~~r~~t :~~a:lyewe can use what we have already intro~uced, provided t~at are corre a e . . . '. wa From Fi ure 1.23, we see that If we rotate the ong;,e ihe angle: while keeping the scatter fixed and lOa) cO ~ d the scatter in terms of the new axes looks very ~uc . the r?tat~d axe; ou 2~ay wish to turn the book to place the Xl and X2 a.xes m tha~ 10 FIgure . ~sitions.) This suggests that we calculate the .sample vananc~ s theIr cust~mar~ coordin ates and measure distance as in EquatIOn (113). That.Is, using the Xl an 2 h ~ d X axes we define the distance from the pomt 'th reference to t e Xl an 2 ' ; =' (Xl, X2) to the origin 0 = (0,0) as
;0 c;. f
The relation between the original coordin ates (Xl' Xz) and the rotated coordinates (Xl, X2) is provide d by
Given the relation s in (118), we can formally substitu te for Xl and X2 in (117) and express the distance in terms of the original coordinates. After some straight forward algebraic manipul ations, the distance from P = (Xl, X2) to the origin 0 = (0,0) can be written in terms of the original coordinates Xl and X2 of Pas d(O,P) = Val1x1 + 2al2xlx2 + a22x~ (119) where the a's are number s such that the distance is nonnega tive for all possible values of Xl and X2. Here all, a12, and a22 are dete,rmined by the angle 8, and Sll, s12, and S22 calculat ed from the original data. 2 The particul ar forms for all, a12, and a22 are not importa nt at this point. What is importa nt is the appeara nce of the crossproduct term 2a12xlxZ necessit ated by the nonzero correlat ion r12' Equatio n (119) can be compar ed with (113). The expressi on in (113) can be regarde d as a special case of (119) with all = 1/s , a22 = 1/s , and a12 = O. ll 22 In general, the statistic al distance ofthe point P = (x], X2) from the fvced point Q = (Yl,)'2) for situatio ns in which the variable s are correlat ed has the general form d(P,Q) = Val1(X I 
denote the sample variances comput ed with the Xl arid X2 where Sl1 and sn measurements. X2
.
. ,.. I. I
(121) By definition, this is the equatio n of an ellipse centere d at Q. The graph of such an equatio n is displayed in Figure 1.24. The major (long) and minor (short) axes are indicated. They are parallel to the Xl and 1'2 axes. For the choice of all, a12, and a22 in footnote 2, the Xl and X2 axes are at an angle () with respect to the Xl and X2 axes. The general ization of the distance formula s of (119) and (120) to p dimensions is straight forward . Let P = (Xl,X2 ,""X ) be a point whose coordin ates p represe nt variable s that are correlat ed and subject to inheren t variability. Let
)'2) + azz(x2 )'2? (120) and can always be comput ed once all, a12, and a22 are known. In addition , the coordinates of all points P = (Xl, X2) that are a constan t squared distance 2 c from Q satisfy
. ). . t the Euclidean distance formula m (112 IS appropna e.
Figure 1.23 A scatter plot for positively correlated measurements and a rotated coordinate system.
sin2(6) all = coS1(O)SIl + 2sin(6)co s(/I)SI2 + sin2(O)s12 + cos2(8)S22  2sin(8)oo s(8)sl2 + sin2(8}slI 2 sin2(/I} oos (8) a22 = cos2(8}SII + 2 sin(lI}cOS(8}SI2 + sin2(6)S22 + cos2(9)sn  2sin(8)oos (/I}SI2 + sin2(8)sll
cos(lI) sin(/I} sin(6} oos(/I} al2 = cos2(II)SIl + 2 sin(8) cos(8)sl2 + sin2(8)~2  cog2(/J)S22  2 sin(/J} ooS(6)812 + sin2(/I}sll
• • •••. . .. . . ... ..••.... •. .. •••••• ... ...
.•....• :•••®:.. .
Figure 1.24 Ellipse of points a constant distance from the point Q.
________________~________~______~__~~~ allx1 + a22x~ + ... + appx~ + 2a12xlx2 + 2a13Xlx3 + ... + 2a p_l,px p_IX p (122)
Y2)2 + .. , + app(xp Yp)2 + 2an(xI YI)(X 2__ Y2) 2a13(XI  YI)(X3  Y:l) + ... + 2apl,p(xp1  YpI)(X p Yp)] (123) .
where the a's are numbers such that the distances are always nonnegatIve. . We note that the distances in (122) and (123) are completely dete~~llned by .) .  1, 2 , ... , p, k.  1,'2 , ... , P. These coeffIcIents can . the coeffiCIents (weIghts aik> I be set out in the rectangular array
We have attempted to motivate the study of multivariate analysis and to provide you with some rudimentary, but important, methods for organizing, summarizing, and displaying data. In addition, a general concept of distance has been introduced that will be used repeatedly in later chapters.
lJbe 81 ebraic expressions for the squares of the distances in ,<1.22) .and (1.~) are known as qu~. . ~ gand in particular positive definite quadratic forms. It IS possible to display these quadrahc dratlCJorms" . S . 23 fCh t 2 forms in a simpler manner using matrix algebra; we shall do so iD echon . 0 ap er .
the a· 's with i k are displayed twice, since they are multiplied by 2 in the h were ,k . . h' 'fy the distance func distance formulas. Consequently, the entnes m t IS array specI . The a. 's cannot be arbitrary numbers; they must be such that the computed t1OnS. ,k . f . (S E . 110 ) distance is nonnegative for every paIr 0 pomts. ee xerclse . . Contours of constant distances computed from (122) a~d. \123) .are ereIlipsoids. A hypereIIipsoid resembles a football when p = 3; It IS Impossible hY P . . . to visualize in more than three dlmens~ons.
Figure 1.25 A cluster of points relative to a point P and the origin.
The need to consider statistical rather than Euclidean distance is illustrated heuristically in Figure 1.25. Figure 1.25 depicts a cluster of points whose center of gravity (sample mean) is indicated by the point Q. Consider the Euclidean distances from the point Q to the point P and the origin O. The Euclidean distance from Q to P is larger than the Euclidean distance from Q to O. However, P appears to be more like the points in the cluster than does the origin. If we take into account the variability of the points in the cluster and measure distance by the statistical distance in (120), then Q will be closer to P than to O. This result seems reasonable, given the nature of the scatter. Other measures of distance can be advanced. (See Exercise 1.12.) At times, it is useful to consider distances that are not related to circles or ellipses. Any distance measure d(P, Q) between two points P and Q is valid provided that it satisfies the following properties, where R is any other intermediate point:
0) denote the origin, and let Q = (YI, Y2, ... , Yp) be a speC!"fd le fix;d p~i~i.·Then the distances from P to 0 and from Pto Q have the general
Exercises 1.1.
Calculate the sample means Xl and x2' the sample variances S]l and covariance Sl2 .
3S Chapter 1 Aspects of Multivariate Analysis .1.2. A morning newspaper lists the following usedcar prices for a foreign compact with age XI measured in years and selling price X2 measured in thousands of dollars:
(b) Infer the sign of the sampkcovariance sl2 from the scatter plot. (c) Compute the sample means XI and X2 and the sample variartces SI I and S22' Compute the sample covariance SI2 and the sample correlation coefficient '12' Interpret these quantities. (d) Display the sample mean array i, the sample variancecovariance array Sn, and the sample correlation array R using (I8). 1.3. The following are five measurements on the variables
Find the arrays i, Sn, and R. 1.4. The world's 10 largest companies yield the following data: The World's 10 Largest Companiesl (billions)
Company Citigroup General Electric American Int! Group Bank of America HSBCGroup ExxonMobil Royal Dutch/Shell BP INGGroup Toyota Motor
IFrom www.Forbes.compartiallybasedonForbesTheForbesGlobaI2000, April 18,2005.
(a) Plot the scatter diagram and marginal dot diagrams for variables Xl and ment on the appearance of the diagrams. (b) Compute Xl>
1.5. Use the data in Exercise 1.4. (a) Plot the scatter diagrams and dot diagrams for (X2, thepattems. (b) Compute the i, Sn, and R arrays for (XI' X2, X3).
1.6. The data in Table 1.5 are 42 measurements on airpollution variables recorded at 12:00 noon in the Los Angeles area on different days. (See also the airpollution data on the web at www.prenhall.com/statistics. ) (a) Plot the marginal dot diagrams for all the variables. (b) Construct the i, Sn, and R arrays, and interpret the entries in R.
(a) Construct a scatter plot of the data and marginal dot diagrams.
8 7 7 10 6 8 9 5 7 8 6 6 7 10 10 9 8 8 9 9 10 9 8 5 6 8 6 8 6 10 8 7 5 6 10 8 5 5 7 7 6 8
98 107 103 88 91 90 84 72 82 64 71 91 72 70 72 77 76 71 67 69 62 88 80 30 83 84 78 79 62 37 71 52 48 75 35 85 86 86 79 79 68 40
Source: Data courtesy of Professor O. C. Tiao.
2 3 3 2 2 2 4 4 1 2 4 2 4 2 1 1 1 3 2 3 3 2 2 3 1 2 2 1 3 1 1 1 5 1 1 1 1 2 4 2 2 3
1.7.
You are given the following n = 3 observations on p = 2 variables: Variable 1: Xll
(a) Plot the pairs of observations in the twodimensional "variable space." That is, construct a twodimensional scatter plot of the data. (b) Plot the data as two points in the threedimensional "item space." 1.8.
1.9.
Evaluate the distance of the point P = (1, 1) to the point Q = (I,?) usin~ the Euclidean distance formula in (112) with p = 2 and using the statistic~1 dIstance m (120) 'th  1/3 a 2 = 4/27 and aI2'= 1/9. Sketch the focus of pomts that are a conWI all , 2 .' . stant squared statistical distance 1 from the pomt Q.
Assume the distance along a street between two intersections in either the NS or EW direction is 1 unit. Define the distance between any two intersections (points) on the grid to be the "city block" distance. [For example, the distance between intersections (D, 1) and (C,2), which we might call deeD, 1), (C, 2», is given by deeD, 1), (C, 2» = deeD, 1), (D, 2» + deeD, 2), (C, 2» = 1 + 1 = 2. Also, deeD, 1), (C, 2» = deeD, 1), (C, 1» + d«C, 1), (C, 2» = 1 + 1 = 2.] 3
(a) Plot the data as a scatter diagram, and compute SII, S22, and S12: ~ ~ (b) Using (118), calculate the corr~sponding measurements on vanables XI and ~' as: uming that the original coordmate axes are rotated through an angle of ()  26 0 [given cos (26 0 ) = .899 and sin (26 ) = .438]. . (c) Using the Xl and X2 measurements from (b), compute the sample vanances Sll and S22' (d) Consider the new pair of measurements (Xl>X2) = (4, 2) Transform these to easurements on xI and X2 using (118), and calculate the dIstance d(O, P) of the :ewpointP = (xl,~)from,.!heoriginO = (0,0) using (117). Note: You will need SIl and S22 from (c). (e) Calculate the distance from P = (4,.2) to the origin 0 = (0,0) using (119) and the expressions for all' a22, and al2 m footnote 2. Note: You will need SIl, Sn, and SI2 from (a). . ~ ~ Compare the distance calculated here with the distance calculated USIng the XI and X2 values in (d). (Within rounding error, the numbers should be the same.) 1.10. Are the following distance functions valid for distance from the origin? Explain.
(a) xi + 4x~ + XIX2 = (distance)2 (b) xi  2x~ = (distance)2 Verify that distance defined by (120) with a 1.1 = 4'~22.= l,an~a.12 = 1 s~tisfiesthe 1.11. first three conditions in (125). (The triangle mequahty IS more dIfficult to venfy.) 1.12. DefinethedistancefromthepointP =
d(O, P) = max(lxd, I X21) (a) Compute the distance from P = (3,4) to the origin. (b) Plot the locus of points whose squared distance from the origin is (c) Generalize the foregoing distance expression to points in p dimenSIOns.
I 13 A I ge city has major roads laid out in a grid pattern, as indicated in the following dia• • ar Streets 1 through 5 run northsouth (NS), and streets A through E run eastwest f~;j. Suppose there are retail stores located at intersections (A, 2), (E, 3), and (C, 5).
Locate a supply facility (warehouse) at an intersection such that the sum of the distances from the warehouse to the three retail stores is minimized. The following exercises contain fairly extensive data sets. A computer may be necessary for the required calculations. 1.14. Table 1.6 contains some of the raw data discussed in Section 1.2. (See also the multiplesclerosis data on the web at www.prenhall.com/statistics.) Two different visual stimuli (SI and S2) produced responses in both the left eye (L) and the right eye (R) of subjects in the study groups. The values recorded in the table include Xl (subject's age); X2 (total response of both eyes to stimulus SI, that is, SIL + SIR); X3 (difference between responses of eyes to stimulus SI, ISIL  SIR I); and so forth. (a) Plot the twodimensional scatter diagram for the variables X2 and X4 for the multiplesclerosis group. Comment on the appearance of the diagram. (b) Compute the X, Sn, and R arrays for the nonmultipleSclerosis and multiplesclerosis groups separately. 1.15. Some of the 98 measurements described in Section 1.2 are listed in Table 1.7 (See also the radiotherapy data on the web at www.prenhall.com/statistics.)The data consist of average ratings over the course of treatment for patients undergoing radiotherapy. Variables measured include XI (number of symptoms, such as sore throat or nausea); X2 (amount of activity, on a 15 scale); X3 (amount of sleep, on a 15 scale); X4 (amount of food consumed, on a 13 scale); Xs (appetite, on a 15 scale); and X6 (skin reaction, on a 03 scale). (a) Construct the twodimensional scatter plot for variables X2 and X3 and the marginal dot diagrams (or histograms). Do there appear to be any errors in the X3 data? (b) Compute the X, Sn, and R arrays. Interpret the pairwise correlations. 1.16. At the start of a study to determine whether exercise or dietary supplements would slow bone loss in older women, an investigator measured the mineral content of bones by photon absorptiometry. Measurements were recorded for three bones on the dominant and nondominant sides and are shown in Table 1.8. (See also the mineralcontent data on the web at www.prenhall.comlstatistics.) Compute the i, Sn, and R arrays. Interpret the pairwise correlations.
1.103 .842 .925 .857 .795 .787 .933 .799 .945 .921 .792 .815 .755 .880 .900 .764 .733 .932 .856 .890 .688 .940 .493 .835 .915
1.052 .859 .873 .744 .809 .779 .880 .851 .876 .906 .825 .751 .724 .866 .838 .757 .748 .898 .786 .950 .532 .850 .616 .752 .936
Dominant humerus 2.139 1.873 1.887 1.739 1.734 1.509 1.695 1.740 1.811 1.954 1.624 2.204 1.508 1.786 1.902 1.743 1.863 2.028 1.390 2.187 1.650 2.334 1.037 1.509 1.971
Humerus 2.238 1.741 1.809 1.547 1.715 1.474 1.656 1.777 1.759 2.009 1.657 1.846 1.458 1.811 1.606 1.794 1.869 2.032 1.324 2.087 1.378 2.225 1.268 1.422 1.869
Dominant ulna .873 .590 .767 .706 .549 .782 .737 .618 .853 .823 .686 .678 .662 .810 .723 .586 .672 .836 .578 .758 .533 .757 .546 .618 .869
Ulna .872 .744 .713 .674 .654 .571 .803 .682 .777 .765 .668 .546 .595 .819 .677 .541 .752 .805 .610 .718 .482 .731 .615 .664 .868
Source: Data courtesy of Everett Smith .
Source: Data courtesy of Dr. G. G. Celesia. Table 1.7 Radiotherapy Data Xl
Source: Data courtesy of Mrs. Annette Tealey, R.N. Values of X2 and x3less than 1.0 are du~ to errors in the datacollection process. Rows containing values of X2 and X3 less than 1.0 may be omItted.
1.17. Some of the data described in Section 1.2 are listed in Table 1.9. (See also the nationaltrackrecords data on the web at www.prenhall.comJstatistics.) The national track records for women in 54 countries can be examined for the relationships among the running eventl> Compute the X, Sn, and R arrays. Notice the magnitudes of the correlation coefficients as you go from the shorter (lOOmeter) to the longer (marathon) ruHning distances. Interpret ihese pairwise correlations. 1.18. Convert the national track records for women in Table 1.9 to speeds measured in meters per second. For example, the record speed for the lOOm dash for Argentinian women is 100 m/1l.57 sec = 8.643 m/sec. Notice that the records for the 800m, 1500m, 3000m and marathon runs are measured in minutes. The marathon is 26.2 miles, or 42,195 meters, long. Compute the X, Sn, and R arrays. Notice the magnitudes of the correlation coefficients as you go from the shorter (100 m) to the longer (marathon) running distances. Interpret these pairwise correlations. Compare your results with the results you obtained in Exercise 1.17. 1.19. Create the scatter plot and boxplot displays of Figure l.5 for (a) the mineralcontent data in Table 1.8 and (b) the nationaltrackrecords data in Table 1.9.
Argentina Australia Austria Belgium Bermuda Brazil Canada Chile China Columbia Cook Islands Costa Rica Czech Republic Denmark Dominican Republic Finland France Germany Great Britain Greece Guatemala Hungary India Indonesia Ireland Israel Italy Japan Kenya Korea, South Korea, North Luxembourg Malaysia Mauritius Mexico Myanmar(Burma) Netherlands New Zealand Norway Papua New Guinea Philippines Poland Portugal Romania Russia Samoa
11.57 11.12 11.15 11.14 11.46 11.17 10.98 11.65 10.79 11.31 12.52 11.72 11.09 11.42 11.63 11.13 10.73 10.81 11.10 10.83 11.92 11.41 11.56 11.38 11.43 11.45 11.14 11.36 11.62 11.49 11.80 11.76 11.50 11.72 11.09 11.66 11.08 11.32 11.41 11.96 11.28 10.93 11.30 11.30 10.77 12.38
22.94 22.23 22.70 22.48 23.05 22.60 22.62 23.84 22.01 22.92 25.91 23.92 21.97 23.36 23.91 22.39 21.99 21.71 22.10 22.67 24.50 23.06 23.86 22.82 23.02 23.15 22.60 23.33 23.37 23.80 25.10 23.96 23.37 23.83 23.13 23.69 22.81 23.13 23.31 24.68 23.35 22.13 22.88 22.35 21.87 25.45
52.50 48.63 50.62 51.45 53.30 50.62 49.9153.68 49.81 49.64 61.65 52.57 47.99 52.92 53.02 50.14 48.25 47.60 49.43 50.56 55.64 51.50 55.08 51.05 51.07 52.06 51.31 51.93 51.56 53.67 56.23 56:07 52.56 54.62 48.89 52.96 51.35 51.60 52.45 55.18 54.75 49.28 51.92 49.88 49.11 56.32
2.05 1.98 1.94 1.97 2.07 1.97 1.97 2.00 1.93 2.04 2.28 2.10 1.89 2.02 2.09 2.01 1.94 1.92 1.94 2.00 2.15 1.99 2.10 2.00 2.01 2.07 1.96 2.01 1.97 2.09 1.97 2.07 2.12 2.06 2.02 2.03 1.93 1.97 2.03 2.24 2.12 1.95 1.98 1.92 1.91 2.29
4.25 4.02 4.05 4.08 4.29 4.17 4.00 4.22 3.84 4.34 4.82 4.52 4.03 4.12 4.54 4.10 4.03 3.96 3.97 4.09 4.48 4.02 4.36 4.10 3.98 4.24 3.98 4.16 3.96 4.24 4.25 4.35 4.39 4.33 4.19 4.20 4.06 4.10 4.01 4.62 4.41 3.99 3.96 3.90 3.87 5.42
9.19 8.63 8.78 8.82 9.81 9.04 8.54 9.26 8.10 9.37 11.10 9.84 8.87 8.71 9.89 8.69 8.64 8.51 8.37 8.96 9.71 8.55 9.50 9.11 8.36 9.33 8.59 8.74 8.39 9.01 8.96 9.21 9.31 9.24 8.89 9.08 8.57 8.76 8.53 10.21 9.81 8.53 8.50 8.36 8.38 13.12
150.32 143.51 154.35 143.05 174.18 147.41 148.36 152.23 139.39 155.19 212.33 164.33 145.19 149.34 166.46 148.00 148.27 141.45 135.25 153.40 171.33 148.50 154.29 158.10 142.23 156.36 143.47 139.41 138.47 146.12 145.31 149.23 169.28 167.09 144.06 158.42 143.43 146.46 141.06 221.14 165.48 144.18 143.29 142.50 141.31 191.58 (continues)
Country Singapore Spain Sweden Switzerland Taiwan . Thailand Thrkey U.S.A.
Source: IAAFIATFS T,ack and Field Ha])dbook fo, Helsinki 2005 (courtesy of Ottavio Castellini).
1.20. Refer to the bankruptcy data in Table 11.4, page 657, and on the following website www.prenhall.com/statistics.Using appropriate computer software, (a) View the entire data set in Xl, X2, X3 space. Rotate the coordinate axes in various directions. Check for unusual observations. (b) Highlight the set of points corresponding to the bankrupt firms. Examine various threedimensional perspectives. Are there some orientations of threedimensional space for which the bankrupt firms can be distinguished from the nonbankrupt firins? Are there observations in each of the two groups that are likely to have a significant impact on any rule developed to classify firms based on the sample mearis, variances, and covariances calculated from these data? (See Exercise 11.24.) 1.21. Refer to the milk transportationcost data in Thble 6.10, page 345, and on the web at www.prenhall.com/statistics.Using appropriate computer software,
(a) View the entire data set in three dimensions. Rotate the coordinate axes in various directions. Check for unusual observations. (b) Highlight the set of points corresponding to gasoline trucks. Do any of the gasolinetruck points appear to be multivariate outliers? (See Exercise 6.17.) Are there some orientations of Xl, X2, X3 space for which the set of points representing gasoline trucks can be readily distinguished from the set of points representing diesel trucks? 1.22. Refer to the oxygenconsumption data in Table 6.12, page 348, and on the web at www.prehhall.com/statistics.Using appropriate computer software, (a) View the entire data set in three dimensions employing various combinations of . three variables to represent the coordinate axes. Begin with the Xl, X2, X3 space. (b) Check this data set for outliers. 1.23. Using the data in Table 11.9, page 666, and on the web at www.prenhall.coml statistics, represent the cereals in each of the following ways. (a) Stars. (b) Chemoff faces. (Experiment with the assignment of variables to facial characteristics.) 1.24. Using the utility data in Table 12.4, page 688, and on the web at www.prenhalI. cornlstatistics, represent the public utility companies as Chemoff faces with assignments of variables to facial characteristics different from those considered in Example 1.12. Compare your faces with the faces in Figure 1.17. Are different groupings indicated?
1.25. Using the data in Table 12.4 and on the web at www.prenhall.com/statistics.represent the 22 public utility companies as stars. Visually group the companies into four or five clusters. 1.26. The data in Thble 1.10 (see the bull data on the web at www.prenhaIl.com!statistics) are the measured characteristics of 76 young (less than two years old) bulls sold at auction. Also included in the taBle are the selling prices (SalePr) of these bulls. The column headings (variables) are defined as follows: I Angus Breed = 5 Hereford { 8 Simental
SalePr 2200 2250 . 1625 4600 2150
(c) Would the correlation in Part b change if you measure size in square miles instead of acres? Explain. Table 1.11 Attendance and Size of National Parks N ationaI Park
(a) Compute the X, Sn, and R arrays. Interpret the pairwise correlations. Do some of these variables appear to distinguish one breed from another? (b) View the data in three dimensions using the variables Breed, Frame, and BkFat. Rotate the coordinate axes in various directions. Check for outliers. Are the breeds well separated in this coordinate system? (c) Repeat part b using Breed, FtFrBody, and SaleHt. WhichthreedimensionaI display appears to result in the best separation of the three breeds of bulls?
(b) Identify the park that is unusual. Drop this point andrecaIculate the correlation coefficient. Comment on the effect of this one point on correlation.
Source: Data courtesy of Mark EIIersieck. 1.27. Table 1.11 presents the 2005 attendance (millions) at the fIfteen most visited national parks and their size (acres).
(a) Create a scatter plot and calculate the correlliltion coefficient.
Arcadia Bruce Canyon Cuyahoga Valley Everglades Grand Canyon Grand Teton Great Smoky Hot Springs Olympic Mount Rainier Rocky Mountain Shenandoah . Yellowstone Yosemite Zion
47.4 35.8 32.9 1508.5 1217.4 310.0 521.8 5.6 922.7 235.6 265.8 199.0 2219.8 761.3 146.6
References 1. Becker, R. A., W. S. Cleveland, and A. R. Wilks. "Dynamic Graphics for Data Analysis." Statistical Science, 2, no. 4 (1987),355395.
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Chapter 1 Aspects of Multivariate Analysis 11. Everitt, B. Graphical Techniques for Multivariate Data. New York: NorthHolland, 1978. 12. Gable, G. G. "A Multidimensional Model of Client Success when Engaging External Consultants." Management Science, 42, no. 8 (1996) 11751198. 13. Halinar, 1. C. "Principal Component Analysis in Plant Breeding." Unpublished report based on data collected by Dr. F. A. Bliss, University of Wisconsin, 1979. 14. Johnson, R. A., and 6. K. Bhattacharyya. Statistics: Principles and Methods (5th ed.). New York: John Wiley, 2005. 15. Kim, L., and Y. Kim. "Innovation in a Newly Industrializing Country: A Multiple Discriminant Analysis." Management Science, 31, no. 3 (1985) 312322. 16. Klatzky, S. R., and R. W. Hodge. "A Canonical Correlation Analysis of Occupational Mobility." Journal of the American Statistical Association, 66, no. 333 (1971),1622. 17. Lee, 1., "Relationships Between Properties of PulpFibre and Paper." Unpublished doctoral thesis, University of Toronto. Faculty of Forestry (1992). 18. MacCrimmon, K., and D. Wehrung. "Characteristics of Risk Taking Executives." Management Science, 36, no. 4 (1990),422435. 19. Marriott, F. H. C. The Interpretation of Multiple Observations. London: Academic Press, 1974. 20. Mather, P. M. "Study of Factors Influencing Variation in Size Characteristics in FIuvioglacial Sediments." Mathematical Geology, 4, no. 3 (1972),219234. 21. McLaughlin, M., et al. "Professional Mediators' Judgments of Mediation Tactics: Multidimensional Scaling and Cluster Analysis." Journal of Applied Psychology, 76, no. 3 (1991),465473. 22. Naik, D. N., and R. Khattree. "Revisiting Olympic Track Records: Some Practical Considerations in the Principal Component Analysis." The American Statistician, 50, no. 2 (1996),140144. 23. Nason, G. "Threedimensional Projection Pursuit." Applied Statistics, 44, no. 4 (1995), 411430. 24. Smith, M., and R. Taffler. "Improving the Communication Function of Published Accounting Statements." Accounting and Business Research, 14, no. 54 (1984), 139...:146. 25. Spenner, K.1. "From Generation to Generation: The nansmission of Occupation." Ph.D. dissertation, University of Wisconsin, 1977. 26. Tabakoff, B., et al. "Differences in Platelet Enzyme Activity between Alcoholics and Nonalcoholics." New England Journal of Medicine, 318, no. 3 (1988),134139. 27. Timm, N. H. Multivariate Analysis with Applications in Education and Psychology. Monterey, CA: Brooks/Cole, 1975. 28. Trieschmann, J. S., and G. E. Pinches. "A Multivariate Model for Predicting Financially Distressed PL Insurers." Journal of Risk and Insurance, 40, no. 3 (1973),327338. 29. Thkey, 1. W. Exploratory Data Analysis. Reading, MA: AddisonWesley, 1977. 30. Wainer, H., and D. Thissen. "Graphical Data Analysis." Annual Review of Psychology, 32, (1981), 191241. 31. Wartzman, R. "Don't Wave a Red Flag at the IRS." The Wall Street Journal (February 24, 1993), Cl, C15. 32. Weihs, C., and H. Schmidli. "OMEGA (On Line Multivariate Exploratory Graphical Analysis): Routine Searching for Structure." Statistical Science, 5, no. 2 (1990), 175226.
MATRIX ALGEBRA AND RANDOM VECTORS 2.1 Introduction We saw in Chapter 1 that multivariate data can be conveniently displayed as an array of numbers. In general, a rectangular array of numbers with, for instance, n rows and p columns is called a matrix of dimension n X p. The study of multivariate methods is greatly facilitated by the use of matrix algebra. The matrix algebra results presented in this chapter will enable us to concisely state statistical models. Moreover, the formal relations expressed in matrix terms are easily programmed on computers to allow the routine calculation of important statistical quantities. We begin by introducing some very basic concepts that are essential to both our geometrical interpretations and algebraic explanations of subsequent statistical techniques. If you have not been previously exposed to the rudiments of matrix algebra, you may prefer to follow the brief refresher in the next section by the more detailed review provided in Supplement 2A.
2.2 Some Basics of Matrix and Vector Algebra Vectors An array x of n real numbers
Xl, X2, • •. , Xn
(Xl> X2, ... ,
where the prime denotes the operation of transposing a column to a row. 49
1\vo vectors may be added. Addition of x and y is defined as 2 _________________ ~,,/ ;__
[.
Figure 2.1 The vector x' = [1,3,2].
A vector x can be represented geometrically as a directed line in n dimensions with component along the first axis, X2 along the second axis, .,. , and Xn along the nth axis. This is illustrated in Figure 2.1 for n = 3. A vector can be expanded or contracted by mUltiplying it by a constant c. In particular, we define the vector c x as
. xn
so that x + y is the vector with ith element Xi + Yi' The sum of two vectors emanating from the origin is the diagonal of the parallelogram formed with the two original vectors as adjacent sides. This geometrical interpretation is illustrated in Figure 2.2(b). A vector has both direction and length. In n = 2 dimensions, we consider the vector x =
The length of x, written L., is defined to be L. =
Geometrically, the length of a vector in two dimensions can be viewed as the hypotenuse of a right triangle. This is demonstrated schematicaIly in Figure 2.3. The length of a vector x' = X2,"" xn], with n components, is defined by
.
+ x~ + ... + x~
Multiplication of a vector x by a scalar c changes the length. From Equation (21),
Le. = v'c2xt + c2X~ + .. , + c2x~ That is, cx is the vector obtained by multiplying each element of x by c. [See Figure 2.2(a).]
= Ic Iv'XI + x~ + ... + x~ = Ic ILx Multiplication by c does not change the direction of the vector x if c > O. However, a negative value of c creates a vector with a direction opposite that of x. From
/elL.
it is clear that x is expanded if I cl> 1 and contracted if 0 < Ic I < 1. [Recall Figure 2.2(a).] Choosing c = L;I, we obtain the unit vector which has length 1 and lies in the direction of x.
Figure 2.2 Scalar multiplication and vector addition.
Length of x = v'xi + x~.
Using the inner product, we have the natural extension of length and angle to vectors of n components:
Since, again, cos (8) = 0 only if x/y = 0, we say that x and y are perpendicular whenx/y = O.
A second geometrical conc~pt is angle. Consider. two vectors in a plane and the le 8 between them, as in Figure 2.4. From the figure, 8 can be represented. as ang difference between the angles 81 and 82 formed by the two vectors and the fITSt the inate axis. Since, . b d f· .. y e ImtJon, coord YI COS(02) = L
Example 2.1 (Calculating lengths of vectors and the angle between them) Given the vectors x' = [1,3,2) and y' = [2,1, IJ, find 3x and x + y. Next, determine the length of x, the length of y, and the angle between x and y. Also, check that the length of 3x is three times the length of x. First,
We find it convenient to introduce the inner product of two vectors. For n dimensions, the inner product of x and y is x'y = XIYl
Next, x'x = l z + 32 + 22 = 14, y'y 1(2) + 3(1) + 2(1) = 1. Therefore,
.
CIX Since cos(900) = cos (270°) = 0 and cos(O) = 0 only if x'y = 0, x and y are e endicular when x'y = O. . P rpFor an arbitrary number of dimensions n, we define the Inner product of x andyas
1be inner product is denoted by either x'y or y'x.
A pair of vectors x and y of the same dimension is said to be linearly dependent if there exist constants Cl and C2, both not zero, such that
x'y x'y cos(O) = L L =. ~. ~ x.y vx'x vy'y
x/y = XIYI + XzY2 + ... + xnYn
showing L 3x = 3L x.
so 0 = 96.3°. Finally,
A set of vectors Xl, Xz, ... , Xk is said to be linearly dependent if there exist constants Cl, Cz, ... , Cb not all zero, such that (27) Linear dependence implies that at least one vector in the set can be written as a linear combination of the other vectors. Vectors of the same dimension that are not linearly dependent are said to be linearly independent.
Example 2.2 (Identifying linearly independent vectors) Consider the set of vectors
Many of the vector concepts just introduced have direct generalizations to matrices. The transpose operation A' of a matrix changes the columns into rows, so that the first column of A becomes the first row of A', the second column becomes the second row, and so forth. Example 2.3 (The transpose of a matrix) If
with the unique solution Cl = C2 = C3 = O. As we cannot find three constants Cl, C2, and C3, not all zero, such that Cl Xl + C2 X2 + C3 x3 = 0, the vectors Xl, x2, and X3 are linearly independent. •
A matrix may also be multiplied by a constant c. The product cA is the matrix that results from multiplying each element of A by c. Thus
where the vector L~ly has unit length. The length of the projection is
..
...
: : '. can 1 ca n 2 ...
1\vo matrices A and B of the same dimensions can be added. The sum A (i,j)th entry aij + bij .
where B is the angle between x and y. (See Figure 2.5.) Example 2.4 (The sum of two matrices and multiplication of a matrix by a constant) If
Figure 2.5 The projection of x on y.
A matrix is any rectangular array of real numbers. We denote an arbitrary array of n rows and p columns by
all a21 . :
a12 a22 . :
It is also possible to define the multiplication of two matrices if the dimensions of the matrices conform in the following manner: When A is (n X k) and B is (k X p), so that the number of elements in a row of A is the same as the number of elements in a column of B, we can form the matrix product AB. An element of the new matrix AB is formed by taking the inner product of each row of A with each column ofB.
When a matrix B consists of a single column, it is customary to use the lowercase b vector notation.
the (n X p) matrix whose entry in the ith row and jth column is the inner product of the ith row of A and the jth column of B
+ ... + aikbkj =
When k = 4, we have four products to add for each· entry in the matrix AB. Thus, a12
.
b11 ... ...
Then Ab,bc',b'c, and d'Ab are typical products.
... ...
The product A b is a vector with dimension equal to the number of rows of A.
~ Row { . (a" ~I + a,,1>,1 + a,,1>,1 + a"b,J.. ]
1 vector or a single number, here 13.
The product b c' is a matrix whose row dimension equals the dimension of band whose column dimension equals that of c. This product is unlike b' c, which is a single number.
1 vector or a single number, here 26.
Square matrices will be of special importance in our development of statistical methods. A square matrix is said to be symmetric if A = A' or aij = aji for all i andj.
[ is AI. We note that
is not symmetric.
When two square matrices A and B are of the same dimension, both products AB and BA are defined, although they need not be equal. (See Supplement 2A.) If we let I denote the square matrix with ones on the diagonal and zeros elsewhere, it follows from the definition of matrix multiplication that the (i, j)th entry of AI is ail X 0 + ... + ai.jI X 0 + aij X 1 + ai.j+1 X 0 + .. , + aik X 0 = aij, so AI = A. Similarly, lA = A, so I
The matrix I acts like 1 in ordinary multiplication (1· a = a '1= a), so it is called the identity matrix. The fundamental scalar relation about the existence of an inverse number aI such that ala = aaI = 1 if a =f. 0 has the following matrix algebra extension: If there exists a matrix B such that
implies that Cl = C2 = 0, so the columns of A are linearly independent. This • confirms the condition stated in (212). A method for computing an inverse, when one exists, is given in Supplement 2A. The routine, but lengthy, calculations are usually relegated to a computer, especially when the dimension is greater than three. Even so, you must be forewarned that if the column sum in (212) is nearly 0 for some constants Cl, .•. , Ck, then the computer may produce incorrect inverses due to extreme errors in rounding. It is always good to check the products AAI and AI A for equality with I when AI is produced by a computer package. (See Exercise 2.10.) Diagonal matrices have inverses that are easy to compute. For example,
then B is called the inverse of A and is denoted by AI. The technical condition that an inverse exists is that the k columns aI, a2, ... , ak of A are linearly indeperident. That is, the existence of AI is equivalent to
if all the aH =f. O. Another special class of square matrices with which we shall become familiar are the orthogonal matrices, characterized by
The name derives from the property that if Q has ith row qi, then QQ' = I implies that qiqi ;: 1 and qiqj = 0 for i =f. j, so the rows have unit length and are mutually perpendicular (orthogonal).According to the condition Q'Q = I, the columns have the same property. We conclude our brief introduction to the elements of matrix algebra by introducing a concept fundamental to multivariate statistical analysis. A square matrix A is said to have an eigenvalue A, with corresponding eigenvector x =f. 0, if
Ordinarily, we normalize x so that it has length unity; that is, 1 = x'x. It is convenient to denote normalized eigenvectors bye, and we do so in what follows. Sparing you the details of the derivation (see [1 D, we state the following basic result: Let A be a k X k square symmetric matrix. Then A has k pairs of eigenvalues and eigenvectorsnamely,
multivariate analysis. In this section, we consider quadratic forms that are always nonnegative and the associated positive definite matrices. Results involving quadratic forms and symmetric matrices are, in many cases, a direct consequence of an expansion for symmetric matrices known as the spectral decomposition. The spectral decomposition of a k X k symmetric matrix A is given by1
(215) The eigenvectors can be chosen to satisfy 1 = e; el = ... = e"ek and be mutually perpendicular. The eigenvectors· are unique unless two or more eigenvalues are equal.
.
+ ..1.2 e2 ez + ... + Ak ek eA: (kX1)(lXk)
where AI, A2, ... , Ak are the eigenvalues of A and el, e2, ... , ek are the associated normalized eigenvectors. (See also Result 2A.14 in Supplement 2A). Thus, eiei = 1 for i = 1,2, ... , k, and e:ej = 0 for i j.
Example 2.1 0 (The spectral decomposition of a matrix) Consider the symmetric matrix
The eigenvalues obtained from the characteristic equation I A  AI I = 0 are Al = 9, A2 = 9, and ..1.3 = 18 (Definition 2A.30). The corresponding eigenvectors el, e2, and e3 are the (normalized) solutions of the equations Aei = Aiei for i = 1,2,3. Thus, Ael = Ae1 gives
or is its corresponding normalized eigenvector. You may wish to show that a second eigenvalueeigenvector pair is ..1.2 = 4, = [1/v'2,I/\I2]. •
A method for calculating the A's and e's is described in Supplement 2A. It is instructive to do a few sample calculations to understand the technique. We usually rely on a computer when the dimension of the square matrix is greater than two or three.
2.3 Positive Definite Matrices The study of the variation and interrelationships in multivariate data is often based upon distances and the assumption that the data are multivariate normally distributed. Squared distances (see Chapter 1) and the multivariate normal density can be expressed in terms of matrix products called quadratic forms (see Chapter 4). Consequently, it should not be surprising that quadratic forms play a central role in
Moving the terms on the right of the equals sign to the left yields three homogeneous equations in three unknowns, but two of the equations are redundant. Selecting one of the equations and arbitrarily setting el1 = 1 and e21 = 1, we find that e31 = O. Consequently, the normalized eigenvector is e; = [1/VI2 + 12 + 02, I/VI2 + 12 + 02, 0/V12 + 12 + 02] = [1/\12, 1/\12,0], since the sum of the squares of its elements is unity. You may verify that ez = [1/v18, 1/v'I8, 4/v'I8] is also an eigenvector for 9 = A2 , and e3 = [2/3, 2/3, 1/3] is the normalized eigenvector corresponding to the eigenvalue A3 = 18. Moreover, e:ej = 0 for i j.
lA proof of Equation (216) is beyond the scope ofthis book. The interested reader will find a proof in [6), Chapter 8.
Example 2.11 (A positive definite matrix and quadratic form) Show that the matrix
O.
where el and e2 are the normalized and orthogonal eigenvectors associated with the eigenvalues Al = 4 and Az = 1, respectively. Because 4 and 1 are scalars, premuItiplication and postmultiplication of A by x/ and x, respectively, where x/ = (XI' xz] is any non zero vector, give
We now show that YI and Yz are not both zero and, consequently, that x/ Ax = 4YI + y~ > 0, or A is positive definite. From the definitions of Y1 and Yz, we have
for all x/ = (XI' Xz, ... , xd, both the matrix A and the quadratic form are said to be nonnegative definite. If equality holds in (217) only for the vector x/ = (0,0, ... ,0], then A or the quadratic form is said to be positive definite. In other words, A is positive definite if (218) 0< x/Ax ~
By Definition 2A.30, the eigenvalues of A are the solutions of the equation  AI I = 0, or (3  A)(2  A)  2 = O. The solutions are Al = 4 and Az = l. Using the spectral decomposition in (216), we can write
The spectral decomposition is an important analytical tool. With it, we are very easily able to demonstrate certain statistical results. The first of these is a matrix explanation of distance, which we now develop. Because x/ Ax has only squared terms xt and product terms XiXb it is caIled a quadratic form. When a k X k symmetric matrix A is such that (217) Os x/A x
as you may readily verify.
To illustrate the general approach, we first write the quadratic form in matrix notation as
Now E is an orthogonal matrix and hence has inverse E/. Thus, x = E/y. But x is a nonzero vector, and 0 ~ x = E/y implies that y ~ O. • Using the spectral decomposition, we can easily show that a k X k symmetric matrix A is a positive definite matrix if and only if every eigenvalue of A is positive. (See Exercise 2.17.) A is a nonnegative definite matrix if and only if all of its eigenvalues are greater than or equal to zero. Assume for the moment that the p elements XI, Xz, ... , Xp of a vector x are realizations of p random variables XI, Xz, ... , Xp. As we pointed out in Chapter 1,
64 we can regard these elements as the coordinates of a point in pdimensional space, and the "distance" of the point [XI> X2,···, xpJ' to the origin can, and in this case should, be interpreted in terms of standard deviation units. In this way, we can account for the inherent uncertainty (variability) in the observations. Points with the same associated "uncertainty" are regarded as being at the same distance from the origin. If we use the distance formula introduced in Chapter 1 [see Equation (122»), the distance from the origin satisfies the general formula (distance)2 = allxI + a22x~
+ ... + appx~ + 2(a12xlx2 + a13 x l x 3 + ... + ap1.p x plXp)
provided that (distance)2 > 0 for all [Xl, X2,···, Xp) ~ [0,0, ... ,0). Setting a·· = ti·· . . . ' I) Jl' I ~ J, I = 1,2, ... ,p, ] = 1,2, ... ,p, we have
a2p [Xl] X2 .. alP] . . .. .. . ... a pp Xp or 0< (distancef
From (219), we see that the p X P symmetric matrix A is positive definite. In sum, distance is determined from a positive definite quadratic form x' Ax. Conversely, a positive definite quadratic form can be interpreted as a squared distance. Com~~nt.
L~t the squ~re of the dista~ce from the point x' = [Xl, X2, ... , Xp) to the ongm be gIven by x A x, where A IS a p X P symmetric positive definite
matrix. Then the square of the distance from x to an arbitrary fixed point po I = [p.1> P.2, ... , p.p) is given by the general expression (x  po)' A( x  po). Expressing distance as the square root of a positive definite quadratic form allows us to give a geometrical interpretation based on the eigenvalues and eigenvectors of the matrix A. For example, suppose p = 2. Then the points x' = [XI, X2) of constant distance c from the origin satisfy x' A x = a1lx1
Ifp > 2, the points x' = [XI,X2,.·.,X p ) a constant distancec = v'x'Axfrom the origin lie on hyperellipsoids c2 = AI (x'el)2 + ... + A (x'e )2 whose axes are . b . PP' gIven y the elgenvectors of A. The halflength in the direction e· is equal to cl Vi . 1,2, ... , p, where AI, A , ... , Ap are the eigenvalues of A. . " I = 2
2.4 A SquareRoot Matrix The spect.ral ~ecomposition allows us to express the inverse of a square matrix in term~ of Its elgenvalues and eigenvectors, and this leads to a useful squareroot ~~
.
2: Aieie;. Let the normalized eigenvectors be the columns of another matrix
P = [el, e2,.'·' ed. Then
Now, c2 = AIYI + A2Y~ is an ellipse in YI = x'el and Y2 = x'e2 because AI> A2 > 0 when A is positive definite. (See Exercise 2.17.) We easily verify that x = cA I l/2el . f·Ies x 'A x = "l ' (Clll ' 1/2' satIs elel )2 = 2 . S·ImiI arIy, x = cA1/2· 2 e2 gIves the appropriate distance in the e2 direction. Thus, the points at distance c lie on an ellipse whose axes are given by the eigenvectors of A with lengths proportional to the reciprocals of the square roots of the eigenvalues. The constant of proportionality is c. The situation is illustrated in Figure 2.6.
= PAP'(PAIp') = PP' = I. Next, let A 1/2 denote the diagonal matrix with VX; as the ith diagonal element. k . The matrix L VX; eje; = P A l/2p; is called the square root of A and is denoted by
AI/2.
if Xij is a continuous random variable with probability density functionfu(xij) if Xij is a discrete random variable with probability function Pij( Xij)
Example 2.12 (Computing expected values for discrete random variables) Suppose P = 2 and,! = 1, and consider the random vector X' = [XI ,X2 ]. Let the discrete random vanable XI have the following probability function:
(1)(.3) + (0)(.3) + (1)(.4) ==.1.
1. (N/ 2 )' = AI/2 (that is, AI/2 is symmetric).
2. AI/2 AI/2 = A. 3. (AI/2) I =
eiej = P A1/2p', where A1j2 is a diagonal matrix with vA j 1/ VX; as the ith diagorial element. j=1
4. A I/2A I/2
(AI/2rl.
+ (1) (.2) == .2.
2.5 Random Vectors and Matrices A random vector is a vector whose elements are random variables. Similarly, a random matrix is a matrix whose elements are random variables. The expected value of a random matrix (or vector) is the matrix (vector) consisting of the expected values of each of its elements. Specifically, let X = {Xij} be an n X P random matrix. Then the expected value of X, denoted by E(X), is the n X P matrix of numbers (if they exist)
'!Wo results involving the expectation of sums and products of matrices follow directly from the definition of the expected value of a random matrix and the univariate properties of expectation, E(XI + Yj) == E(XI) + E(Yj) and E(cXd = cE(XI)' Let X and Y be random matrices of the same dimension, and let A and B be conformable matrices of constants. Then (see Exercise 2.40) E(X + Y) == E(X) + E(Y)
(223) 2If you are unfamiliar with calculus, you should concentrate on the interpretation of the expected value and, ~ventu~lIy, variance. Our development is based primarily on the properties of expectation rather than Its partIcular evaluation for continuous or discrete random variables.
for all pairs of values xi, Xk, then X; and X k are said to be statistically independent. When X; and X k are continuous random variables with joint density fik(Xi, xd and marginal densities fi(Xi) and fk(Xk), the independence condition becomes
2.6 Mean Vectors and Covariance Matrices SupposeX' = [Xl, x 2, .. ·, Xp] isap x 1 random vector.TheneachelementofXisa random variable with its own marginal probability distripution; (See Example 2.12.) The marginal means JLi and variances (Tf are defined as JLi = E (X;) and (Tt = E (Xi  JLi)2, i = 1, 2, ... , p, respectively. Specifically,
fik(Xi, Xk) = fi(Xi)fk(Xk) for all pairs (Xi, Xk)' The P continuous random variables Xl, X 2, ... , Xp are mutually statistically independent if their joint density can be factored as
x. [.( x) dx. if Xi is a continuous random variable with probability '" 'density function fi( x;)
.
for all ptuples (Xl> X2,.'" xp). Statistical independence has an important implication for covariance. The factorization in (228) implies that Cov (X;, X k ) = O. Thus,
(x.  JLlt..(x) dx. if Xi is a continuous random vari.able '" 'with probability density function fi(Xi)
The converse of (229) is not true in general; there are situations where Cov(Xi , X k ) = 0, but X; and X k are not independent. (See [5].) The means and covariances of the P X 1 random vector X can be set out as matrices. The expected value of each element is contained in the vector of means /L = E(X), and the P variances (T;i and the pep  1)/2 distinct covariances (Tik(i < k) are contained in the symmetric variancecovariance matrix .I = E(X  /L)(X  /L)'. Specifically,
It will be convenient in later sections to denote the marginal variances by (T;; rather and consequently, we shall adopt this notation .. than the more traditional The behavior of any pair of random variables, such as X; and Xb is described by their joint probability function, and a measure of the linear association between them is provided by the covariance
if X;, X k are continuous random variables with the joint density functionfik(x;, Xk) all
if X;, X k are discrete random variables with joint probability function Pike Xi, Xk) (226)
and JL; and JLk, i, k = 1,2, ... , P, are the marginal means. When i = k, the covariance becomes the marginal variance. More generally, the collective behavior of the P random variables Xl, X 2, ... , Xp or, equivalently, the random vector X' = [Xl, X 2, ... , Xp], is described by a joint probability density function f(XI' X2,.'" xp) = f(x). As we have already noted in this book,f(x) will often be the multivariate normal density function. (See Chapter 4.) If the joint probability P[ Xi :5 X; and X k :5 Xk] can be written as the product of the corresponding marginal probabilities, so that (227)
.. , (Xl  JLI)(Xp  JLP)] .... (X2  JL2);(Xp ~ JLp) (Xp  JLp) E(XI  JLl)(Xp  JLP)] E(X2  ILz)(Xp  JLp) E(Xp  JLp)2
l: = E(X  JL)(X  JL)' Example 2.13 (Computing the covariance matrix) Find the covariance matrix for
the two random variables XI and X 2 introduced ill Example 2.12 when their joint probability function pdxJ, X2) "is represented by the entries in the body of the following table:
We note that the computation of means, variances, and covariances for discrete random variables involves summation (as in Examples 2.12 and 2.13), while analogous computations for continuous random variables involve integration. Because lTik = E(Xi  JLi) (Xk  JLk) = ITki, it is convenient to write the matrix appearing in (231) as
= E(XI) = .1 and iL2 = E(X2) = .2. (See Exam
... .,.
+ .. , + (1  .1)(1  .2)(.00) 1T21
We shall refer to JL and l: as the population mean (vector) and population variancecovariance (matrix), respectively. The multivariate normal distribution is completely specified once the mean vector JL and variancecovariance matrix l: are given (see Chapter 4), so it is not surprising that these quantities play an important role in many multivariate procedures. It is frequently informative to separate the information contained in variances lTii from that contained in measures of association and, in particular, the measure of association known as the population correlation coefficient Pik' The correlation coefficient Pik is defined in terms of the covariance lTik and variances ITii and IT kk as
The correlation coefficient measures the amount of linear association between the random variables Xi and X k . (See,for example, [5].)
Mean Vectors and Covariance Matrices. 73
~~ Yu;YU;; Consequently, from (237), the correlation matrix p is given by (234)
(236) and (237) obtained from · "can be obtained from Vl/2 and p, whereas p can be Th a t IS,..... . .' II l:. Moreover, the expression of these relationships in terms of matrIX operatIOns a ows the calculations to be conveniently implemented on a computer.
3 Obtain Vl/2 and p.
• Often, the characteristics measured on individual trials will fall naturally into two or more groups. As examples, consider measurements of variables representing consumption and income or variables representing personality traits and physical characteristics. One approach to handling these situations is to let the characteristics defining the distinct groups be subsets of the total collection of characteristics. If the total collection is represented by a (p X 1)dimensional random vector X, the subsets can be regarded as components of X and can be sorted by partitioning X. In general, we can partition the p characteristics contained in the p X 1 random vector X into, for instance, two groups of size q and p  q, respectively. For example, we can write
Note that 1: 1z = 1: 21 , The covariance matrix of X(I) is 1: , that of X(2) is 1:22 , and 11 that of element s from X(!) and X(Z) is 1:12 (or 1: ), 21 It is sometimes conveni ent to use the COy (X(I), X(Z» notation where COy
Upon taking the expectation of the matrix (X(I)  JL(I»)(X(2)  ,.,.(2»', we get UI,q+1 E(X(l)  JL(I»)(X(Z)  JL(Z»'
lTI,q+2 ... lTZt Z :..
The Mean Vector and Covariance Matrix for linear Combinations of Random Variables Recal1 that if a single random variable, such as XI, is multiplied by a E(cXd
If X 2 is a second random variable and a and b are constants, then, using addition al Cov(aXI ,bX2)
1".....
which gives al1 the covariances,lTi;, i = 1,2, ... , q, j = q + 1, q + 2, ... , p, between a compon ent of X(!) and a component of X(2). Note that the matrix 1:12 is not necessarily symmetric or even square. Making use of the partitioning in Equation (238), we can easily demons trate that
is a matrix containi ng all of the covariances between a compon ent of X(!) and a compon ent of X(Z).
[~~J = e',.,.
....
be the variancecovariance matrix o~X, Equation (241) becomes Var(aXl since c'l:c = [a
The preceding results can be extended to a linear combination of p random variables:
in terms of Px and l:x. Here
The linear combination c'X·= CIXI + '" + c~Xp has mean = E( c'X) = c' Pvariance = Var(c'X) = c'l:c
where p == E(X) and l: == Cov (X). In general, consider the q 1·mear combinations of the p random variables
Xj, ... ,Xp:
+ C12X2 + .,. + CjpXp + CnX2 + .:. + C2pXp Note that if all = a22 that is, if Xl and X 2 have equal variancestheoffdiagona} terms in :tz vanish. This demonstrates the wellknown result that the sum and difference of two random variables with identical variances are uncorrelated. , •
Many of the matrix results in this section have been expressed in terms of population means and variances (covariances). The results in (236), (237), (238), and (240) also hold if the population quantities are replaced by their appropriately defined sample counterparts.
Let x' = [XI, X2,"" xp] be the vector of sample averages constructed from n observations on p variables XI, X 2 , •.. , X p , and let .
the mean vector and variancecovar~ance matrix o~ Xc~sr:;,c)v:here Px and l:x. are228 for the computation of the offdiagonal terms m x.
tIvel~~s~ea~;:;;I~:a~ilY on the result in (245) in our discussions of principal com
ponents and factor analysis in Chapters 8 and 9. E l 2 IS (Means and covariances of linear combinations) Let X'. = [Xl> X~} e· . vector with mean vector Px , _ [/LI, p,z } and variancecovanance matrIX bexamp a random
.•. n L
.. .
... _ )2
be the corresponding sample variancecovariance matrix.
The sample mean vector and the covariance matrix can be partitioned in order to distinguish quantities corresponding to groups of variables. Thus,
Proof. The inequality is obvious if either b = 0 or d = O. Excluding this possibility, consider the vector b  X d, where x is an arbitrary scalar. Since the length of b  xd is positive for b  xd * 0, in this case
The last expression is quadratic in x. If we complete the square by adding and subtracting the scalar (b'd)2/ d 'd, we get (b'd)2 (b'd)2 0< b'b   +    2 (b'd) + 2(d'd) d'd d'd x x
Sqp .
(b'd)2 O
where x(1) and x(Z) are the sample mean vectors constructed from observations x(1) = [Xi>"" x q]' and x(Z) = [Xq+b"" .xp]', re~pective!y; SII is the sample c~vari ance matrix computed from observatIOns x( ); SZ2 IS the sample covanance matrix computed from observations X(2); and S12 = S:n is the sample covariance matrix for elements of x(I) and elements of x(Z).
A simple, ~ut important, extension of the CauchySchwarz inequality follows directly. Extended CauchySchwarz Inequality. Let band let B be a positive definite matrix. Then (pXl)
with equality if and only if b = c B1d (or d = cB b) for some constant c. Proof. The inequality is obvious when b = 0 or d = O. For cases other than these, consider the squareroot matrix Bl/2 defined in terms of its eigenvalues A; and
2.1 Matrix Inequalities and Maximization Maximization principles play an important role in several multivariate techniques. Linear discriminant analysis, for example, is concerned with allocating observations to predetermined groups. The allocation rule is often a linear function of measurements that maximizes the separation between groups relative to their withingroup variability. As another example, principal components are linear combinations of measurements with maximum variability. The matrix inequalities presented in this section will easily allow us to derive certain maximization results, which will be referenced in later chapters. CauchySchwarz Inequality. Let band d be any two p (b'd)2 with equality if and only if b
= cd (or d = cb) for some constant c.
2: VX; e;ej. If we set [see also (222)] ;=1
and the proof is completed by applying the CauchySchwarz inequality to the vectors (Bl/2b) and (B1/2d). •
1 vectors. Then
The extended CauchySchwarz inequality gives rise to the following maximization result.
..... Matrix Inequalities and Maximization 81
Maximization Lemma . Let
be a given vector.
Then, for an arbitrar y nonzero vector x , (pXl) ( 'd)2 max 2.....x>,o x'Bx with the maximum attained when x (pXI)
* O.
proof. By the extende d CauchySchwarz inequality, (x'd)2
$: (x'Bx) (d'BId ). Because x 0 and B is positive definite, x'Bx > O. Dividing both sides of the inequality by the positive scalar x'Bx yields the upper bound
'd)2 ::; ( __ _x d'B1d x'Bx Taking the maximum over x gives Equatio n (250) because the bound is attained for x = CBId.
eigenvalues.
Maximization of Quadratic Forms for Points on the Unit Sphere. Let B be a (pXp) positive definite matrix with eigenvalues Al ~ A2 ~ ... ~ Ap ~ 0 and associated normalized eigenvectors el, e2,' .. , e po Then x'Bx max , == Al x>'O x.x x'Bx min =A x>'o x'x p
== ye'e 1 i 1 + Y2e;e2 ' + ... + ypeje p == Yi,
k Therefo re, for x perpend icular to the first k . inequality in (253) become s elgenvectors e;, the lefthan d side of the $:
k = 1,2, ... , P  1)
be the orthogonal matrix whose columns are the eigenvectoIS el, e2,"" e and A be the diagonal matrix with eigenvalues AI, A2 ,···, Ap along the p main diagonal . Let Bl/2 = PA 1/2P' [see (222)] and (plO) v = (pxp)(px P' x. l) Consequently, x#,O implies Y O. Thus, x'Bx x'B1(2B1/2x x'PA 1/2P'PA 1(2P'x =y'Ay y'y y'y x'pP'x x'x (pxp)
A similar ar~ument produce s the second part of (251). Now, x  Py == Ylel + Y2e2 + ... + ype , so x .1 eh'" ek . p Implies
where the symbol .1 is read "is perpendicular to." Proof. Let
Taking Yk+I=I Yk , +2  .. ,  Yp == O· gIVes the asserted maximum.
a fixed x 0 x' B / I x' ==For xo/Vx& xo is ~f u~it l~n x~ x~o has the same .value as x'Bx, where largest eigenvalue A I'S the gt: onsequently, EquatIOn (251) says that the . ' 1, maXImu of th ' pomts x whose distance from the ori inmi value . y . . e quad rahc form x'Bx for all the quadratic form for all pOI'nts g s. ufmt . SImIlarly, Ap is the smallest value of . x one umt rom the ori' Th I elgenvalues thus represen t extreme values f I gm.. e argest and smallest The "interm ediate" eigenvalues of the X 0 x ~ x for ~~mts on the unit sphere. interpre tation as extreme values hP. pOSItIve d~flmte matrix B also have an the earlier choices. w en x IS urther restncte d to be perpend icular to
Definition 2A.3 (Vector addition). The sum of two vectors x and y, each having the same number of entries, is that vector z = x
Taking the zero vector, 0, to be the mtuple (0,0, ... ,0) and the vector x to be the mtuple (Xl,  X2, ... ,  xm), the two operations of scalar multiplication and vector addition can be combined in a useful manner.
Definition 2A.4. The space of all real mtuples, with scalar multiplication and vector addition as just defined, is called a vector space.
Vectors Many concepts, such as a person's health, intellectual abilities, or p~rsonality, cannot be adequately quantified as a single number. Rather, several different measurements Xl' Xz,· .. , Xm are required. Definition 2A.1. An mtuple of real numbers (Xl> Xz,·.·, Xi,"" Xm) arranged in a column is called a vector and is denoted by a boldfaced, lowercase letter. Examples of vectors are
Definition 2A.S. The vector y = alxl + azxz + ... + akXk is a linear combination of the vectors Xl, Xz, ... , Xk' The set of all linear combinations of Xl, Xz, ... ,Xk, is called their linear span. Definition 2A.6. A set of vectors xl, Xz, ... , Xk is said to be linearly dependent if there exist k numbers (ai, az, ... , ak), not all zero, such that alxl
a2x Z + ...
Otherwise the set of vectors is said to be linearly independent. If one of the vectors, for example, Xi, is 0, the set is linearly dependent. (Let ai be the only nonzero coefficient in Definition 2A.6.) The familiar vectors with a one as an entry and zeros elsewhere are lirIearly independent. For m = 4,
Vectors are said to be equal if their corresponding entries are the same.
. Definition 2A.2 (Scalar multiplication). Let c be an arbitrary scalar. Then the product cx is a vector with i~h ~ntr.y CXi' To illustrate scalar multiplIcatiOn, take Cl = Sand Cz = 1.2. Then
= a2 = a3 = a4 = O.
.....
~84 Chapter 2 Matrix Algebra and Random Vectors As another example, let k
Definition 2A.9. Th e angI e () between two vectors x and y both h . .. defined from . , avmg m entfles, IS
+ X2)'2 + ... +
where Lx = length of x and L = len th of and YI, )'2, ... , Ym are the elem:nts Of:' y, xl, X2, ... , Xm are the elements of x,
Thus, x I, x2, x3 are a linearly dependent set of vectors, since anyone can be written as a linear combination of the others (for example, x2 = 2xI + 3X3)·
Definition 2A.T. Any set of m linearly independent vectors is called a basis for the vector space of all mtuples of real numbers. Result 2A.I. Every vector can be expressed as a unique linear combination of a
fixed basis.
Then the length of x, the len th of d . vectors are g y, an the cosme of the angle between the two length ofx =
and These four vectors were shown to be linearly independent. Any vector x can be uniquely expressed as
A vector consisting of m elements may be regarded geometrically as a point in mdimensional space. For example, with m = 2, the vector x may be regarded as representing the point in the plane with coordinates XI and X2· Vectors have the geometrical properties of length and direction. 2 •
pefinition 2A.IO. The inner (or dot) d number of entries is defined as the pro uct of two vectors x and y with the same sum 0 f component products: XIYI
+ ... +
We use the notation x'y or y'x to denoteth·IS mner . pro d uct.
th With the x'y notation we ma the angle between two vedtors as y express e length ?f a vector and the cosine of
Definition 2A.S. The length of a vector of m elements emanating from the origin is given by the Pythagorean formula:
Consequently, () = 135°.
VXI + x~ + ... + x~
= length of x = V xI + x~ + ... + x~ = ~ cos«() =
Definition 2A.II. When the angle between two vectors x, y is 8 = 9(}" or 270°, we say that x and y are perpendicular. Since cos (8) = 0 only if 8 = 90° or 270°, the condition becomes x and Yare perpendicular if x' Y = 0 We write x .1 y. ~
Uj/~. In this
of Xk on the linear span of Xl , X2, ... , Xkl'
we take are mutually perpendicular. Also, each has length unity. The same construction holds for any number of entries m. Result 2A.2. (a) z is perpendicular to every vector if and only if z = O. (b) If z is perpendicular to each vector XI, X2,"" Xb then Z is perpendicular to
their linear span. (c) Mutually perpendicular vectors are linearly independent.
Definition 2A.12. The projection (or shadow) of a vector x on a vector y is projection ofx on y =
If Yhas unit length so that Ly = 1, , projection ofx on Y = (x'y)y If YJ, Y2, ... , Yr are mutually perpendicular, the projection (or shadow) of a vector x on the linear span ofYI> Y2, ... , Yr is
+ ,Y2 + .,. ,Yr Y2Y2
Result 2A.l (GramSchmidt Process). Given linearly independent vectors Xl, X2, ... , Xk, there exist mutually perpendicular vectors UI, U2, ... , Uk with the same linear span. These may be constructed sequentially by setting
Definition 2A.ll. An m X k matrix, generally denoted by a boldface uppercase letter such as A, R, l;, and so forth, is a rectangular array of elements having m rows and k columns.
1/~J.
In our work, the matrix elements will be real numbers or functions taking on values in the real numbers. Definition 2A.14. The dimension (abbreviated dim) of an rn x k matrix is the ordered pair (rn, k); "m is the row dimension and k is the column dimension. The dimension of a matrix is frequentIyindicated in parentheses below the letter representing the matrix. Thus, the rn X k matrix A is denoted by A . (mXk)
Definition 2A.17 (Scalar multiplication). Let c be an arbitrary scalar and A .= {aij}. Then
... alkl
Definition 2A.18 (Matrix subtraction). Let A
written A = B,ifaij = bij,i = 1,2, ... ,rn,j = 1,2, ... ,k.Thatis,two matrices are equal if (a) Their dimensionality is the same. (b) Every corresponding element is the same. Definition 2A.16 (Matrix addition). Let the matrices A and B both be of dimension rn X k with arbitrary elements aij and b ij , i = 1,2, ... , rn, j = 1,2, ... , k, respectively. The sum of the matrices A and B is an m X k matrix C, written C = A + B, such that the arbitrary element of C is given by i = 1,2, ... , m, j
matrices of equal dimension. Then the difference between A and B, written A  B, is an m x k matrix C = {c;j} given by
index j refers to the column. An rn X 1 matrix is referred to as a column vector. A 1 X k matrix is referred to as a row vector. Since matrices can be considered as vectors side by side, it is natural to define multiplication by a scalar and the addition of two matrices with the same dimensions. (mXk)
i = 1,2, ... , m,
Multiplication of a matrix by a scalar produces a new matrix whose elements are the elements of the original matrix, each multiplied by the scalar. For example, if c = 2,
j = 1,2, ... , k.
In the preceding examples, the dimension of the matrix I is 3 X 3, and this information can be conveyed by wr:iting I .
.•.
= 1,2, ... , k
Note that the addition of matrices is defined only for matrices of the same dimension.
... ,m,j
... ,k.
Definition 2A.19. Consider the rn x k matrix A with arbitrary elements aij, i = 1, 2, ... , rn, j = 1, 2, ... , k. The transpose of the matrix A, denoted by A', is the k X m matrix with elements aji, j = 1,2, ... , k, i = 1,2, ... , rn. That is, the transpose of the matrix A is obtained from A by interchanging the rows and columns. As an example, if
Result 2A.4. For all matrices A, B, and C (of equal dimension) and scalars c and d, the following hold: (a) (A
Definition 2A.20. If an arbitrary matrix A has the same number of rows and columns, then A is called a square matrix. The matrices l;, I, and E given after Definition 2A.13 are square matrices.
Definition 2A.21. Let A be a k X k (square) matrix. Then A is said to be symmetric if A = A'. That is:A is symmetric if aij = aji, i = 1,2, ... , k, j = 1,2, ... , k.
Definition 2A.23 (Matrix multiplication). The product AB of an m X n matrix A = {aij} and an n X k matrix B = {biJ is the m X k matrix C whose elements are n
with ones on the main (NWSE) diagonal and zeros elsewhere. The 3 matrix is shown before this definition.
As an additional example, consider the product of two vectors. Let
Definition 2A.22. The k
i ='l,2" .. ,m j = 1,2, ... ,k
Note that for the product AB to be defined, the column dimension of A must equal the row dimension of B. If that is so, then the row dimension of AB equals the row dimension of A, and the column dimension of AB equals the column dimension of B.
Note that the product xy is undefined, since x is a 4 X 1 matrix and y is a 4 X 1 matrix, so the column dim of x, 1, is unequal to the row dim of y, 4. If x and y are vectors of the same dimension, such as n X 1, both of the products x'y and xy' are defined. In particular, y'x = x'y = XIYl + X2Y2 + '" + XnY,,, and xy' is an n X n matrix with i,jth element XiYj' Result 2A.S. For all matrices A, B, and C (of dimensions such that the indicated products are defined) and a scalar c, (a) c(AB) = (c A)B
There are several important differences between the algebra of matrices and the algebra of real numbers. TWo of these differences are as follows:
Definition 2A.24. The determinant of the square k by 1A I, is the scalar
1. Matrix multiplication is, in general, not commutative. That is, in g.eneral, AB #0 BA. Several examples will illustrate the failure of the commutatIve law (for matriceJ).
row.
2. Let 0 denote the zero matrix, that is, the matrix with zero for every element. In the algebra of real numbers, if the product of two numbers, ab, is zero, the~ a = 0 or b = O. In matrix algebra, however, the product of two nonzero matn~ ces may be the zero matrix. Hence,
If I is the k X k identity matrix, 1I 1 = 1.
= 0 or B = O. For example,
 a11 /azz aZ3 !(_1)2 + a12la21 aZ31(_1)3 + al3la21 a ZZ I(_1)4 a32 a33 a31 a33 an a32
where Ali is the (k  1) X (k  1) matrix obtained by deleting the first row and but
is not defined.
It is true, however, that if either A B = 0 . (mXn)(nxk)
The determinant of any 3 X 3 matrix can be computed by summing the products of elements along the solid lines and subtracting the products along the dashed
lines in the following diagram. This procedure is not valid for matrices of higher dimension, but in general, Definition 2A.24 can be employed to evaluate these determinants.
Definition 2A.26. A square matrix A
0 . If a matrix fails to be nonsingular, it is called singUlar. Equivalently,
a square matrix is nonsingular if its rank is equal to the number of rows (or columns) it has. Note iliat Ax = X13I + X232 + ... + Xk3b where 3i is the ith column of A, so that the condition of nonsingularity is just the statement that the columns of A are linearly independent. Result 2A.T. Let A be a nonsingular square matrix of dimension k X k. Then there is a unique k X k matrix B such that AB = BA = I
We next want to state a result that describes some properties of the determinant. However, we must first introduce some notions related to matrix inverses.
k identity matrix.
Definition 2A.2T. The B such that AB = BA = I is called the inverse of A and is denoted by AI. In fact, if BA = I or AB = I, then B = AI, and both products must equal I.
For example, Definition 2A.2S. The row rank of a matrix is the maximum number of linearly independent rows, considered as vectors .( that is, row vectors). The column rank of a matrix
is the rank of its set of columns, consIdered as vectors.
Result 2A.S.
The rows of A, written as vectors, were shown to be linearly dependent after Definition 2A.6. Note that the column rank of A is also 2, since
. A=[:~: is given by
but columns 1 and 2 are linearly independent. This is no coincidence, as the following result indicates. Result 2A.6. The row rank and the column rank of a matrix are equal.
Thus, the rank of a matrix is either the row rank or the column rank.
Result 2A.12. Let A and B be k X k matrices and c be a scalar.
In both (a) and (b), it is clear that IA I "# 0 if the inverse is to exist. j (c) In general, KI has j, ith entry [lA;NIAIJ(lr , where A;j is the matrix obtained from A by deleting the ith row and jth column. _
Definition 2A.29. A square matrix A is said to be orthogonal if its rows, considered as vectors, are mutually perpendicular and have unit lengths; that is, AA' = I.
Result 2A.9. For a square matrix A of dimension k X k, the following are equivalent:
Result 2A.13. A matrix A is orthogonal if and only if AI = A'. For an orthogonal matrix, AA' = A' A = I, so the columns are also mutually perpendicular and have unit lengths. _
0 (A is nonsingular).
o.
.
Result 2A.1 o. Let A and B be square matrices of the same dimension, and let the indicated inverses exist. Then the following hold: (a) (AI), = (AT
(b) (ABt l = B1AI The determinant has the following properties.
(a) IAI = lA' I (b)· If each element of a row (column) of A is zero, then I A I = 0 (d) If A is nonsingular, then IA I = 1/1 AI I; that is, IA II AI I = 1. (e) IABI = IAIIBI = ck I A I, where c is a scalar.
You are referred to [6} for proofs of parts of Results 2A.9 and 2A.ll. Some of these proofs are rather complex and beyond the scope of this book. _
X k square matrix. The trace of the matrix A, k
so A' = AI, and A must be an orthogonal matrix. Square matrices are best understood in terms of quantities called eigenvalues and eigenvectors.
Definition 2A.2B. Let A
= A',soAA' = A'A = AA. We verify that AA = I = AA' = A'A,or 2
Result 2A.II. Let A and B be k X k square matrices.
Definition 2A.30. Let A be a k X k square matrix and I be the k X k identity matrix. Then the scalars AI, Az, ... , Ak satisfying the polynomial equation I A  All = 0 are called the eigenvalues (or characteristic roots) of a matrix A. The equation IA  AI I = 0 (as a function of A) is called the characteristic equation. For example, let
......
implies that there are two roots, Al = 1 and A2 and 1. Let
~ 3. The eigenva lues of A are 3
or Xl =  2X2 There are many solution s for Xl and X2' Setting X2 = 1 (arbitrar ily) gives Xl = 2, and hence,
is an eigenve ctor correspo nding to the eigenva lue 1. From the second
has three roots: Al = 9, A2 = 9, and A3 = 18; that is, 9, 9, and 18 are the eigenva lues ofA. Definition 2A.31. Let A be a square matrix of dimension k X k and let A be an eigenvalue of A. If x is a nonzero vector ( x 0) such that (kXI) (kXI) (kXl) Ax = Ax then x is said to be an eigenvector (characteristic vector) of the matrix A associat ed with the eigenvalue A.
An equivalent condition for A to be a solution of the eigenval ueeige nvector equation is IA  AI I = O. This follows because the stateme nt that A x = Ax for some A and x 0 implies that
colj(A  AI) + ... +
That is, the columns of A  AI are linearly depende nt so, by Result 2A.9(b) ,  AI I = 0, as asserted. Following Definiti on 2A.30, we have shown that the eigenvalues of
is an. eigenve ctor correspo nding to the eigenva lue 3. It is usual practice to determi ne an elge~vector so that It has length unity. That is, ifAx = Ax, we take e = x/YX'X as the elgenve ctor correspo nding to A. For example , the eigenve ctor for A = 1 is
[2/v'S , 1/v'S].
are Al = 1 and A2 = 3. The eige~vectors ~ssociated with these eigenva lues can be determin ed by solving the followmg equatIOns:
Definition2A.32. A quadraticform Q(x) in thekvar iables Xl,x2," " where x' = [Xl, X2, ••. , Xk] and A is a k X k symmetr ic matrix.
.
A~y symmet ric square matrix can be reconst ructured from its eigenva lues and elg~nvector~. The particul ar express ion reveals the relative importa nce of e~ch paIr accordm g to the relative size of the eigenva lue and the directio n of the elgenve ctor. '
Result 2A.14. The Spectral Decomposition. Let A be a k x k symmetric matrix. Then A can be expressed in terms of its k eigenvalueeigenvector pairs (Ai, e;) as
with At, A~, ... , A~ > 0 = A~+l>A~+2"'" A~, (for m> k).Then Vi = A~IA'ui.Alter natively, the Vi are the eigenvectors of A' A with the same nonzero eigenvalues At. The matrix expansion for the singularvalue decomposition written in terms of the full dimensional matrices U, V, A is
The ideas that lead to the spectral decomposition can be extended to provide a decomposition for a rectangular, rather than a square, matrix. If A is a rectangular matrix, Uten the vectors in the expansion of A are the eigenvectors of the square matrices AA' and A' A. Result 2A.1 S. SingularValue Decomposition. Let A be an m X k matrix of real numbers. Then there exist an m X m orthogonal matrix U and a k X k orthogonal matrix V such that A = UAV' where Ute m X k matrix A has (i, i) entry Ai ~ 0 for i = 1, 2, ... , mine m, k) and the other entries are zero. The positive constants Ai are called the singular values of A. • The singularvalue decomposition can also be expressed as a matrix expansion that depends on the rank r of A. Specifically, there exist r positive constants AI, A2, ... , An r orthogonal m X 1 unit vectors U1, U2, ... , Un and r orthogonal k X Lunit vectors VI, Vz, ... , V" such that
Vi V2' respectively. J
so IA' A  ')'1 I = _,),3  22')'2  120')' = ')'( ')'  12)(')'  10), and the eigenvalues are ')'1 = AI = 12, ')'2 = A~ = 10, and ')'3 = A~ = O. The nonzero eigenvalues are the same as those of AA'. A computer calculation gives the eigenvectors VII
where U r = [UI> U2, ... , Ur], Vr = [VI' V2,"" Vr ], and Ar is an r X r diagonal matrix with diagonal entries Ai'
You may verify Utat the eigenvalues ')' = A2 of AA' satisfy the equation ')'2  22,), + 120 = (y 12)(')'  10), and consequently, the eigenvalues are
where U has m orthogonal eigenvectors of AA' as its columns, V has k orthogonal eigenvectors of A' A as its columns, and A is specified in Result 2A.15. For example, let
so A has eigenvalues Al = 3 and A2 = 2. The corresponding eigenvectors are et = [1/VS, 2/VS] and ez = [2/VS, l/VS], respectively. Consequently,
2.1.
1].
(b) F~nd (i) ~e length of x, (ii) the angle between x and y, and (iii) the projection of y on x. (c) Smce x = 3 and y = 1, graph [5  3,1  3,3  3] = [2 2 DJ and [11,31,11J=[2,2,OJ. ' ,
Letx' = [5, 1, 3] andy' = [1, . (a) Graph the two vectors.
2.2. Given the matrices
v'2 The equality may be checked by carrying out the operations on the righthand side. The singularvalue decomposition is closely connected to a result concerning the approximation of a rectangular matrix by a lowerdimensional matrix, due to Eckart and Young ([2]). If a m X k matrix A is approximated by B, having the same dimension but lower rank, the sum of squared differences m
Result 2A.16. Let A be an m X k matrix of real numbers with m ~ k and singular value decomposition VAV'. Lets < k = rank (A). Then
perform the indicated multiplications. (a) 5A (b) BA (c) A'B' (d) C'B (e) Is AB defined?
2.3. Verify the following properties of the transpose when A
is the ranks least squares approximation to A. It minimizes tr[(A  B)(A  B)') over all m X k matrices B having rank no greater than s. The minimum value, or k
AT.
(A')' = A (C,)l = (C I )' (AB)' = B' A' For general A and B , (AB)' = B'A' (mXk)
2,4. When AI and B exist, prove each of the following. . (a) (A,)l = (AI), (b) (AB)I = BIA I
Hint: Part a can be proved br noting that AAI = I, I'; 1', and (AAi)' = (AI),A'. Part b follows from (B 1A )AB = BI(AIA)B = BIB = I.
is an orthogonal matrix.
.
2.5. Check that
= V'BV. Clearly, the minimum occurs when Cij
the s largest singular values. The other Cu = O. That is, UBV' = As or B =
2.6. Let
(a) Is A symmetric? (b) Show that A is positive definite.
Chapter 2 Matrix Algebra and Random Vectors 2.7.
Let A be as given in Exercise 2.6. (a) Determine the eigenvalues and eigenvectors of A. (b) Write the spectral decomposition of A. (c) Find AI.
2.17. Prove that every eigenvalue of a k x k positive definite matrix A is positive. Hint: Consider the definition of an eigenvalue, where Ae = Ae. Multiply on the left by e' so that e' Ae = Ae' e. 2.18. Consider the sets of points (XI, x2) whose "distances" from the origin are given by
(d) Find the eigenvaiues and eigenvectors of AI.
2.8. Given the matrix A =
for c = 1 and for c = 4. Determine the major and minor axes of the ellipses of constant distances and their associated lengths. Sketch the ellipses of constant distances and comment on their pOSitions. What will happen as c2 increases?
find the eigenvalues Al and A2 and the associated nonnalized eigenvectors el and e2. Determine the spectral decomposition (216) of A. 2.9. Let A be as in Exercise 2.8. (a) Find AI.
(b) Compute the eigenvalues and eigenvectors of AI. (c) Write the spectral decomposition of AI, and compare it with that of A from Exercise 2.8.
2.19. Let AI/2
I. (The A.'s and the e.'s are '
the eigenvalues and associated normalized eigenvectors of the matrix A.) Show Properties (1)(4) of the squareroot matrix in (222). 2.20. Determine the squareroot matrix AI/2, using the matrix A in Exercise 2.3. Also, deter. mine AI/2, and show that A I/2A I/2 = A 1f2A1/ 2 = I. 2.21. (See Result 2AIS) Using the matrix
2.10. Consider the matrices
These matrices are identical except for a small difference in the (2,2) position. Moreover, the columns of A (and B) are nearly linearly dependent. Show that AI ='= (3)B I. Consequently, small changesperhaps caused by roundingcan give substantially different inverses.
(a) Calculate A' A and obtain its eigenvalues and eigenvectors. (b) Calculate AA' and obtain its eigenvalues and eigenvectors. Check that the nonzero eigenvalues are the same as those in part a. (c) Obtain the singularvalue decomposition of A.
2.11. Show that the determinant of the p X P diagonal matrix A = {aij} with aij = 0, i * j, is given by the product of the diagonal elements; thus, 1A 1 = a" a22 ... a p p. Hint: By Definition 2A24, I A I = a" A" + 0 + ... + O. Repeat for the submatrix All obtained by deleting the first row and first column of A.
2.22. (See Result 2A1S) Using the matrix
2.12. Show that the determinant of a square symmetric p x p matrix A can be expressed as the product of its eigenvalues AI, A2, ... , Ap; that is, IA I = Ai. Hint: From (216) and (220), A = PAP' with P'P = I. From Result 2A.1I(e), lA I = IPAP' I = IP IIAP' I = IP 11 A liP' I = I A 1111, since III = IP'PI = IP'IIPI. Apply Exercise 2.11.
(a) Calculate AA' and obtain its eigenvalues and eigenvectors. (b) Calculate A' A and obtain its eigenvalues and eigenvectors. Check that the nonzero eigenvalues are the same as those in part a. (c) Obtain the singularval~e decomposition of A. 2.23. Verify the relationships V I/ 2pV I!2 = I and p = (Vlf2rII(VI/2rl, where I is the p X .P popul~tion cov~riance matrix [E~uation (232)], p is the p X P population correlatIOn matnx [EquatIOn (234)], and V /2 is the population standard deviation matrix [Equation (235)].
2.13. Show that I Q I = + 1 or 1 if Q is a p X P orthogonal matrix. Hint: I QQ' I = II I. Also, from Result 2A.11, IQ" Q' I = IQ 12. Thus, IQ 12 use Exercise 2.11. 2.14. Show that Q'
= II I. Now
Q and A have the same eigenvalues if Q is orthogonal.
2.24. Let X have covariance matrix
Hint: Let A be an eigenvalue of A. Then 0 = 1A  AI I. By Exercise 2.13 and Result 2A.11(e), we can write 0 = IQ' 11 A  AlII Q I = IQ' AQ  All, since Q'Q = I. 2.1 S. A quadratic form x' A x is said to be positive definite if the matrix A is positive definite. . Is the quadratic form 3xt + 3x~  2XIX2 positive definite? 2.16. Consider an arbitrary n X p matrix A. Then A' A is a symmetric p that A' A is necessarily nonnegative definite. Hint: Set y = A x so that y'y = x' A' A x.
matrix. Show
Find (a) II (b) The eigenvalues and eigenvectors of I. (c) The eigenvalues and eigenvectors of II.
2.25. Let X have covariance matrix
2.29. Consider the arbitrary random vector X' ,.,: = [ILl> IL2. IL3, IL4, Jl.sJ· Partition X into
(a) Determine p a~d V 1/2. (b) Multiply your matrices to check the relation VI/2pVI/2 =
(a) Findpl3' (b) Find the correlation between XI and ~X2 + ~X3' 2.27. Derive expressions for the mean and variances of the following linear combinations in terms of the means and covariances of the random variables XI, X 2, and X 3. (a) XI  2X2 (b) XI + 3X2 (c) XI + X 2 + X3 (e) XI + 2X2  X3 (f) 3XI  4X2 if XI and X 2 are independent random variables. 2.28. Show that
where Cl = [CJl, cl2, ... , Cl PJ and ci = [C2l> C22,' .. , C2 pJ. This verifies the offdiagonal elements CIxC' in (245) or diagonal elements if Cl = C2' Hint: By (243),ZI  E(ZI) = Cl1(XI  ILl) + '" + Clp(Xp  ILp) and Z2  E(Z2) = C21(XI  ILl) + ... + C2p(Xp  ILp).SOCov(ZI,Zz) = E[(ZI  E(Zd)(Z2  E(Z2»J = E[(cll(XI  ILl) + '" + CIP(Xp  ILp»(C21(XI  ILd + C22(X2  IL2) + ... + C2p(Xp  ILp»J. The product (Cu(XI  ILl) + CdX2  IL2) + .. ,
+ Clp(Xp  IL p»(C21(XI  ILl) + C22(X2  IL2) + ... + C2p(Xp  ILp»
Let I be the covariance matrix of X with general element (Tik' Partition I into the covariance matrices of X(l) and X(2) and the covariance matrix of an element of X(1) and an element of X (2). 2.30. You are given the random vector X' = [XI' X 2, X 3, X 4 J with mean vector Jl.x = [4,3,2, 1J and variancecovariance matrix
I.
2.26. Use I as given in Exercise 2.25.
and consider the linear combinations AX(!) and BX(2). Find (a) E(X(J) (b) E(AX(l) (c) Cov(X(l) (d) COY (AX(!) (e) E(X(2) (f) E(BX(2) (g) COY (X(2) (h) Cov (BX(2) (i) COY (X(l), X (2) (j) COY (AX(J), BX(2) 2 .31. Repeat Exercise 2.30, but with A and B replaced by
Verify the last step by the definition of matrix multiplication. The same steps hold for all elements.
Chapter 2 Matrix Algebra and Random Vectors 2.32. You are given the random vector X' = [XI, X 2 , ... , Xs] with mean vector IJ.'x = [2,4, 1,3,0] and variancecovariance matrix 4
1 1.
2.3S. Using the vecto~s b' = [4,3] and d' = [1,1]' verify the extended CauchySchwarz inequality (b'd) s (b'Bb)(d'B1d) if
2.36. Fmd the maximum and minimum values of the quadratic form 4x~ + 4x~ + all points x' = [x I , X2] such that x' x = 1.
2.37. With A as given in Exercise 2.6, fmd the maximum value of x' A x for x' x = 1. 2.38. Find the maximum and minimum values of the ratio x' Ax/x'x for any nonzero vectors x' = [Xl> X2, X3] if
2.39. Show that s
and consider the linear combinations AX(I) and BX(2). Find (a) E(X(l)
2.40. Verify (224): E(X + Y) = E(X) + E(Y) and E(AXB) = AE(X)B. Hint: X. + ~ has Xij + Yij as its (i,j~th element. Now,E(Xij + Yij ) = E(X ij ) + E(Yi) by a umvanate property of expectation, and this last quantity is the (i, j)th element of
+ E(Y). Next (see Exercise 2.39),AXB has (i,j)th entry ~ ~ aieXCkbkj, and by the additive property of expectation, C k
E(~e ~ aiCXCkbkj) = ~ ~ aj{E(XCk)bkj k e k which is the (i, j)th element of AE(X)B. 2.41. You are given the random vector X' = [Xl, X 2, X 3 , X 4 ] with mean vector IJ.x = [3,2, 2,0] and variancecovariance matrix
2.33. Repeat Exercise 2.32, but with X partitioned as
2.34. Consider the vectorsb' = [2, 1,4,0] and d' = [1,3, 2, 1]. Verify the CauchySchwan inequality (b'd)2 s (b'b)(d'd).
1 (a) Find E (AX), the mean of AX. (b) Find Cov (AX), the variances and covariances ofAX. (c) Which pairs of linear combinations have zero covariances?
References 1. BeIlman, R. Introduction to Mat~ix Analysis (2nd ed.) Philadelphia: Soc for Industrial &
Applied Math (SIAM), 1997. . 2. Eckart, C, and G. young. "The Approximation of One Matrix by Another of Lower Rank." Psychometrika, 1 (1936),211218. 3. Graybill, F. A. Introduction to Matrices with Applications in Statistics. Belmont, CA: Wadsworth,1969. 4. Halmos, P. R. FiniteDimensional Vector Spaces. New York: SpringerVeriag, 1993. 5. Johnson, R. A., and G. K. Bhattacharyya. Statistics: Principles and Methods (5th ed.) New York: John Wiley, 2005. 6. Noble, B., and 1. W. Daniel. Applied Linear Algebra (3rd ed.). Englewood Cliffs, NJ: Prentice Hall, 1988.
SAMPLE GEOMETRY AND RANDOM SAMPLING 3.1 Introduction With the vector concepts introduced in the previous chapter, we can now delve deeper into the geometrical interpretations of the descriptive statistics K, Sn, and R; we do so in Section 3.2. Many of our explanations use the representation of the columns of X as p vectors in n dimensions. In Section 3.3 we introduce the assumption that the observations constitute a random sample. Simply stated, random sampling implies that (1) measurements taken on different items (or trials) are unrelated to one another and (2) the joint distribution of all p variables remains the same for all items. Ultimately, it is this structure of the random sample that justifies a particular choice of distance and dictates the geometry for the ndimensional representation of the data. Furthermore, when data can be treated as a random sample, statistical inferences are based on a solid foundation. Returning to geometric interpretations in Section 3.4, we introduce a single number, called generalized variance, to describe variability. This generalization of variance is an integral part of the comparison of multivariate means. In later sections we use matrix algebra to provide concise expressions for the matrix products and sums that allow us to calculate x and Sn directly from the data matrix X. The connection between K, Sn, and the means and covariances for linear combinations of variables is also clearly delineated, using the notion of matrix products.
3.2 The Geometry of the Sample A single multivariate observation is the collection of measurements on p different variables taken on the same item or trial. As in Chapter 1, if n observations have been obtained, the entire data set can be placed in an n X p array (matrix):
"' .~.
Each row of X represents a multivariate observation. Since the entire set of measurements is often one particular realization of what might have been observed, we say that the data are a sample of size n from a "population." The sample then consists of n measurements, each of which has p components. As we have seen, the data can be ploUed in two different ways. For the. pdimensional scatter plot, the rows of X represent n points in pdimensional space. We can write
.. .
Figure 3.1 A plot of the data matrix X as n = 3 points in p = 2 space.
The row vector xj, representing the jth observation, contains the coordinates of point. . . . . . The scatter plot of n points in pdlmensIOnal space provIdes mformatlOn on the . locations and variability of the points. If the points are regarded as solid spheres, the sample mean vector X, given by (18), is the center of balance. Variability occurs in more than one direction, and it is quantified by the sample variancecovariance matrix Sn. A single numerical measure of variability is provided by the determinant of the sample variancecovariance matrix. When p is greate: tha~ 3, this scaUer plot representation cannot actually be graphed. Yet the conslde~atlOn ?f the data as n points in p dimensions provides insights that are not readIly avallable from algebraic expressions. Moreover, the concepts illustrated for p = 2 or p = 3 remain valid for the other cases.
.
The alternative geometrical representation is constructed by considering the data as p vectors in ndimensional space. Here we take the elements of the columns of the data matrix to be the coordinates of the vectors. Let
". '"
data matrix.
Plot the n = 3 data points in p = 2 space, and locate xon the resulting diagram. The first point, Xl> has coordinates xi = [4,1). Similarly, the remaining two points are xi = [1,3] andx3 = [3,5). Finally,
Then the coordinates of the first point yi = [Xll, XZI, ... , xnd are the n measurements on the first variable. In general, the ith point yi = [Xli, X2i,"" xnd is determined by the ntuple of all measurements on the ith variable. In this geometrical representation, we depict Yb"" YP as vectors rather than points, as in the pdimensional scatter plot. We shall be manipulating these quantities shortly using the algebra of vectors discussed in Chapter 2.
Example 3.2 (Data as p vectors in n dimensions) Plot the following data as p = 2 vectors in n = 3 space:
where XiI is perpendicular to Yi  XiI. The deviation, or mean corrected, vector is
Figure 3.2 A plot of the data matrix X as p = 2 vectors in n = 3space.
[1,3,5]. These vectors are shown in Figure 3.2. _
The elements of d i are the deviations of the measurements on the ith variable from their sample mean. Decomposition of the Yi vectors into mean components and deviation from the mean components is shown in Figure 3.3 for p = 3 and n = 3. 3
Many of the algebraic expressions we shall encounter in multivariate analysis can be related to the geometrical notions of length, angle, and volume. This is important because geometrical representations ordinarily facilitate understanding and lead to further insights. Unfortunately, we are limited to visualizing objects in three dimensions, and consequently, the ndimensional representation of the data matrix X may not seem like a particularly useful device for n > 3. It turns out, however, that geometrical relationships and the associated statistical concepts depicted for any three vectors remain valid regardless of their dimension. This follows because three vectors, even if n dimensional, can span no more than a threedimensional space, just as two vectors with any number of components must lie in a plane. By selecting an appropriate threedimensional perspectivethat is, a portion of the ndimensional space containing the three vectors of interesta view is obtained that preserves both lengths and angles. Thus, it is possible, with the right choice of axes, to illustrate certain algebraic statistical concepts in terms of only two or three vectors of any dimension n. Since the specific choice of axes is not relevant to the geometry, we shall always . label the coordinate axes 1,2, and 3. It is possible to give a geometrical interpretation of the process of finding a sample mean. We start by defining the n X 1 vector 1;, = (1,1, ... ,1]. (To simplify the notation, the subscript n will be dropped when the dimension of the vector 1" is clear from the context.) The vector 1 forms equal angles with each of the n coordinate axes, so the vector (l/Vii)I has unit length in the equalangle direction. Consider the vector Y; = [Xli, x2i,"" xn;]. The projection of Yi on the unit vector (1/ vn)I is, by (28),
Figure 3.3 The decomposition of Yi into a mean component XiI and a deviation component d i = Yi  XiI, i = 1,2,3.
Example 3.3 (Decomposing a vector into its mean and deviation components) Let us carry out the decomposition of Yi into xjI and d i = Yi  XiI, i = 1,2, for the data given in Example 3.2:
That is, the sample mean Xi = (Xli + x2i + .. , + xn;}/n = yjI/n corresponds to the multiple of 1 required to give the projection of Yi onto the line determined by 1.
We have translated the deviation vectors to the origin without changing their lengths or orientations. Now consider the squared lengths of the deviation vectors. Using (25) and (34), we obtain
From (13), we see that the squared length is proportional to the variance of the measurements on the ith variable. Equivalently, the length is proportional to the standard deviation. Longer vectors represent more variability than shorter vectors. For any two deviation vectors d i and db n
x21. The decomposition is
Let fJ ik denote the angle formed by the vectors d i and d k . From (26), we get
pm~ml:] so that [see (15)] For the time being, we are interested in the deviation (or residual) vectors d; = Yi  xiI. A plot of the deviation vectors of Figur,e 3.3 is given in Figure 3.4.
The cosine of the angle is the sample correlation coefficient. Thus, if the two deviation vectors have nearly the same orientation, the sample correlation will be close to 1. If the two vectors are nearly perpendicular, the sample correlation will be approximately zero. If the two vectors are oriented in nearly opposite directions, the sample correlation will be close to 1.
Example 3.4 (Calculating Sn and R from deviation vectors) Given the deviation vectors in Example 3.3, let us compute the sample variancecovariance matrix Sn and sample correlation matrix R using the geometrical concepts just introduced. From Example 3.3,
Figure 3.4 The deviation vectors d i from Figure 3.3.
The concepts of length, angle, and projection have provided us with a geometrical interpretation of the sample. We summarize as follows:
Geometrical Interpretation of the Sample X onto the equal angular vector 1 is the vector XiI. The vector XiI has length Vii 1Xi I. Therefore, the ith sample mean, Xi, is related to the length of the projection of Yi on 1. 2. The information comprising Sn is obtained from the deviation vectors d i = Yi  XiI = [Xli  Xi,X2i  x;"",Xni  Xi)" The square of the length ofdi is nSii, and the (inner) product between d i and d k is nSik.1 3. The sample correlation rik is the cosine of the angle between d i and d k • 1. The projection of a column Yi of the data matrix
These vectors, translated to the origin, are shown in Figure 3.5. Now,
3.3 Random Samples and the Expected Values of the Sample Mean and Covariance Matrix In order to study the sampling variability of statistics such as xand Sn with the ultimate aim of making inferences, we need to make assumptions about the variables whose oDserved values constitute the data set X. Suppose, then, that the data have not yet been observed, but we intend to collect n sets of measurements on p variables. Before the measurements are made, their values cannot, in general, be predicted exactly. Consequently, we treat them as random variables. In this context, let the (j, k )th entry in the data matrix be the random variable X jk • Each set of measurements Xj on p variables is a random vector, and we have the random matrix
¥. Also,
.
rX~J ~2 ..
or S22 = ~. Finally,
or S12 = ~. Consequently,
XIPJ x.2P = .
A random sample can now be defined. If the row vectors Xl, Xl, ... , X~ in (38) represent independent observations from a common joint distribution with density function f(x) = f(xl> X2,"" xp), then Xl, X 2 , ... , Xn are said to form a random sample from f(x). Mathematically, Xl> X 2, ••. , Xn form a random sample if their joint density function is given by the product f(Xl)!(X2)'" f(xn), where f(xj) = !(Xj!, Xj2"'" Xjp) is the density function for the jth row vector. Two points connected with the definition of random sample merit special attention: 1. The measurements of the p variables in a single trial, such as Xi = [Xjl , X j2 , ... , Xjp], will usually be correlated. Indeed, we expect this to be the case. The measurements from different trials must, however, be independent. 1 The square of the length and the inner product are (n  l)s;; and (n  I)s;k, respectively, when the divisor n  1 is used in the definitions of the sample variance and covariance.
Chapter 3 Sample Geometry and Random Sampling 2. The independence of measurements from trial to trial may not hold when the variables are likely to drift over time, as with sets of p stock prices or p economic indicators. Violations of the tentative assumption of independence can have a serious impact on the quality of statistical inferences. The following eJglmples illustrate these remarks. Example 3.5 (Selecting a random sample) As a preliminary step in designing a permit system for utilizing a wilderness canoe area without overcrowding, a naturalresource manager took a survey of users. The total wilQerness area was divided into subregions, and respondents were asked to give information on the regions visited, lengths of stay, and other variables. The method followed was to select persons randomly (perhaps using a random· number table) from all those who entered the wilderness area during a particular week. All persons were e~ually likely to be in the sample, so the more popular entrances were represented by larger proportions of canoeists. Here one would expect the sample observations to conform closely to the criterion for a random sample from the population of users or potential users. On the other hand, if one of the samplers had waited at a campsite far in the interior of the area and interviewed only canoeists who reached that spot, successive measurements would not be independent. For instance, lengths of stay in the wilderness area for dif• ferent canoeists from this group would all tend to be large. Example 3.6 (A nonrandom sample) Because of concerns with future solidwaste disposal, an ongoing study concerns the gross weight of municipal solid waste generated per year in the United States (Environmental Protection Agency). Estimated amounts attributed to Xl = paper and paperboard waste and X2 = plastic waste, in millions of tons, are given for selected years in Table 3.1. Should these measurements on X t = [Xl> X 2 ] be treated as a random sample of size n = 7? No! In fact, except for a slight but fortunate downturn in paper and paperboard waste in 2003, both variables are increasing over time.
If the n components are not independent or the marginal distributions are not identical, the influence of individual measurements (coordinates) on location is asymmetrical. We would then be led to consider a distance function in which the coordinates were weighted unequally, as in the "statistical" distances or quadratic forms introduced in Chapters 1 and 2. Certain conclusions can be reached concerning the sampling distributions of X and Sn without making further assumptions regarding the form of the underlying joint distribution of the variables. In particular, we can see how X and Sn fare as point estimators of the corresponding population mean vector p. and covariance matrix l:. Result 3.1. Let Xl' X 2 , .•• , Xn be a random sample from a joint distribution that has mean vector p. and covariance matrix l:. Then X is an unbiased estimator of p., and its covariance matrix is
E(X) = p.
so [n/(n  1) ]Sn is an unbiased estimator of l:, while Sn is a biased estimator with (bias) = E(Sn)  l: = (l/n)l:.
Proof. Now, X = (Xl + X 2 + ... + Xn)/n. The repeated use of the properties of expectation in (224) for two vectors gives
• As we have argued heuristically in Chapter 1, the notion of statistical independence has important implications for measuring distance. Euclidean distance appears appropriate if the components of a vector are independent and have the same vari= [Xlk' X 2k>'.·' X nk ] ances. Suppose we consider the location ofthe kthcolumn of X, regarded as a point in n dimensions. The location of this point is determined by the joint probability distribution !(Yk) = !(Xlk,X2k> ... ,Xnk)' When the measurements X lk , X 2k , ... , X nk are a random sample, !(Yk) = !(Xlk, X2k,"" Xnk) = !k(Xlk)!k(X2k)'" !k(Xnk) and, consequently, each coordinate Xjk contributes equally to the location through the identical marginal distributions !k( Xj k)'
E(X) = E ;;Xl + ;;X2 + .,. + ;;Xn
E(~Xl) + E(~X2) + .. , + E(~Xn) 1
= ;;E(Xd + ;;E(X2 ) + ... + ;;:E(Xn) =;;p. +;;p. + ... + ;;p. =p. Next, n (X  p.)(X  p.)' = ( 1 ~ (Xj  p.) ) n j~l
For j "# e, each entry in E(Xj  IL )(Xe  IL)' is zero because the entry is the covariance between a component of Xi and a component of Xe, and these are independent. [See Exercise 3.17 and (229).] Therefore,
[nl (n  1) ]Sn is an unbiased estimator of (Fi k' However, the individual sample standard deviations VS;, calculated with either n or n  1 as a divisor, are not unbiased estimators of the corresponding population quantities VU;;. Moreover, the correlation coefficients rik are not unbiased estimators of the population quantities Pik' However, the bias E (~)  VU;;, or E(rik)  Pik> can usually be ignored if the sample size n is moderately large. Consideration of bias motivates a slightly modified definition of the sample variancecovariance matrix. Result 3.1 provides us with an unbiased estimator S of :I:
(:I + :I + .,. + :I) , n terms
To obtain the expected value of Sn' we first note that (Xii  XJ (Xik  X k ) is the (i, k)th element of (Xi  X) (Xj  X)'. The matrix representing sums of squares and cross products can then be written as
X k ).
This definition of sample covariance is commonly used in many multivariate test statistics. Therefore, it will replace Sn as the sample covariance matrix in most of the material throughout the rest of this book.
With a single variable, the sample variance is often used to describe the amount of variation in the measurements on that variable. When p variables are observed on each unit, the variation is described by the sample variancecovariance matrix
2: X;. Therefore, its expected value is i=1
For any random vector V with E(V) = ILv and Cov (V) = :Iv, we have E(VV') :Iv + ILvlLv· (See Exercise 3.16.) Consequently, E(XjXj) = :I
The sample covariance matrix contains p variances and !p(p  1) potentially different covariances. Sometimes it is desirable to assign a single numerical value for the variation expressed by S. One choice for a value is the determinant of S, which reduces to the usual sample variance of a single characteristic when p = 1. This determinant 2 is called the generalized sample variance:
2 Definition 2A.24 defines "determinant" and indicates one method for calculating the value of a determinant.
employee (X2) for the 16 largest publishing firms in the United States are shown in Figure 1.3. The sample covariance matrix, obtained from the data in the April 30, 1990, magazine article, is
Evaluate the generalized variance. In this case, we compute /S/
The generalized sample variance provides one way of writing the information on all variances and covariances as a single number. Of course, when p > 1, some information about the sample is lost in the process. A geometrical interpretation of / S / will help us appreciate its strengths and weaknesses as a descriptive summary. Consider the area generated within the plane by two deviation vectors d l = YI  XII and d 2 = Yz  x21. Let Ldl be the length of d l and Ldz the length of d z . By elementary geometry, we have the diagram
Figure 3.6 (a) "Large" generalized sample variance for p = 3.
= 3.
and the area of the trapezoid is / Ld J sin ((1) /L d2 . Since cosz( (1) express this area as
Assuming now that / S / = (n  l)(pl) (volume )2 holds for the volume generated in n space by the p  1 deviation vectors d l , d z, ... , d p  l , we can establish the following general result for p deviation vectors by induction (see [1],p. 266): GeneraIized sample variance = /S/ = (n 1)P(volume)Z
Equation (315) says that the generalized sample variance, for a fixed set of data, is 3 proportional to the square of the volume generated by the p deviation vectors d l = YI  XII, d 2 = Yz  x21, ... ,dp = Yp  xpl. Figures 3.6(a) and (b) show trapezoidal regions, generated by p = 3 residual vectors, corresponding to "large" and "small" generalized variances. . For a fixed sample size, it is clear from the geometry that volume, or / S /, will increase when the length of any d i = Yi  XiI (or ~) is increased. In addition, volume will increase if the residual vectors of fixed length are moved until they are at right angles to one another, as in Figure 3.6(a). On the other hand, the volume, or / S /, will be small if just one of the Sii is small or one of the deviation vectors lies nearly in the (hyper) plane formed by the others, or both. In the second case, the trapezoid has very little height above the plane. This is the situation in Figure 3.6(b), where d 3 1ies nearly in me plane formed by d 1 and d 2 . 3 If generalized variance is defmed in terms of the samplecovariance matrix S. = [en  l)/njS, then, using Result 2A.11,ISnl = I[(n  1)/n]IpSI = I[(n l)/njIpIlSI = [en  l)/nJPISI. Consequently, using (315), we can also write the following: Generalized sample variance = IS.I = n volume? .
Generalized Variance Generalized variance also has interpretations in the pspace scatter plot representa_ tion of the data. The most intuitive interpretation concerns the spread of the scatter about the sample mean point x' = [XI, X2,"" xpJ. Consider the measure of distance_ given in the comment below (219), with xplaying the role of the fixed point p. and SI playing the role of A. With these choices, the coordinates x/ = [Xl> X2"'" xp) of the points a constant distance c from x satisfy (x  x)'SI(X  i) =
.
..
[When p = 1, (x  x)/SI(x.  x) = (XI  XI,2jSll is the squared distance from XI to XI in standard deviation units.] Equation (316) defines a hyperellipsoid (an ellipse if p = 2) centered at X. It can be shown using integral calculus that the volume of this hyperellipsoid is related to 1S I. In particular, Volume of {x: (x  x)'SI(x  i)
where the constant kp is rather formidable. A large volume corresponds to a large generalized variance. Although the generalized variance has some intuitively pleasing geometrical interpretations, it suffers from a basic weakness as a descriptive summary of the sample covariance matrix S, as the following example shows.
Example 3.8 (Interpreting the generalized variance) Figure 3.7 gives three scatter
plots with very different patterns of correlation. All three data sets have x' = [2,1 J, and the covariance matrices are
. ••.... • ••• • .. ..'. • •
Figure 3.7 Scatter plots with three different orientations.
Each covariance matrix S contains the information on the variability of the component variables and also the information required to calculate the correlation coefficient. In this sense, S captures the orientation and size of the pattern of scatter. The eigenvalues and eigenvectors extracted from S further describe the pattern in the scatter plot. For S=
at z.
For those who are curious, kp = 2u1'/2/ p r(p/2). where f(z) denotes the gamma function evaluated
:n~we d~term[in.~ !,he eigenva]lueeigenvector pairs Al = 9 ei = [1/\1'2 1/\/2] and "2  1,e2 = 1/ v2, 1/\/2 . " The meancentered ellipse, with center x' = [2 , 1] £or a I1 three cases, IS . (x  x),SI(X  x) ::s c2 To describe this ellipse as in S ti 2 3 ' I eigenvalueeigenvecto; air fo~c on . ,,:,::th ~ = S~ , we notice that if (A, e) is an SI That' if S _ A P S, .the? (A ,e) IS an elgenvalueeigenvector pair for I' _ ,!? The  e, the? mu1tlplymg on the left by SI givesSISe = ASle or S e " e erefore usmg t h · I ' extends cvX; in the dir;ction of eiefr~:~~a ues from S, we know that the e11ipse
In p = 2 dimensions, the choice C Z = 5.99 will produce an ellipse that contains approximately 95% of the observations. The vectors 3v'5.99 el and V5.99 ez are drawn in Figure 3.8(a). Notice how the directions are the natural axes for the ellipse, and observe that the lengths of these scaled eigenvectors are comparable to the size of the pattern in each direction. Next,for
s=[~ ~J.
and we arbitrarily choose the eigerivectors so that Al = 3, ei = [I, 0] and A2 = 3, ei ,: [0, 1]. The vectors v'3 v'5]9 el and v'3 v'5:99 ez are drawn in Figure 3.8(b).
.
and we determine theeigenvalueeigenvectorpairs Al = 9, el = [1/V2, 1/V2J and A2 = 1, ei = [1/V2, 1/V2J. The scaled eigenvectors 3V5.99 el and V5.99 e2 are drawn in Figure 3.8(c). In two dimensions, we can often sketch the axes of the meancentered ellipse by eye. However, the eigenvector approach also works for high dimensions where the data cannot be examined visually. Note: Here the generalized variance 1SI gives the same value, 1S I = 9, for all three patterns. But generalized variance does not contain any information on the orientation of the patterns. Generalized variance is easier to interpret when the two or more samples (patterns) being compared have nearly the same orientations. Notice that our three patterns of scatter appear to cover approximately the same area. The ellipses that summarize the variability (x  i)'SI(X  i) :5 c2 do have exactly the same area [see (317)], since all have IS I = 9.
As Example 3.8 demonstrates, different correlation structures are not detected by IS I. The situation for p > 2 can be even more obscure. . Consequently, it is often desirable to provide more than the single number 1S I _as a summary of S. From Exercise 2.12, IS I can be expressed as the product AIAz'" Ap of the eigenvalues of S. Moreover, the meancentered ellipsoid based on SI [see (316)] has axes. whose lengths are proportional to the square roots of the A;'s (see Section 2.3). These eigenvalues then provide information on the variability in all directions in the pspace representation of the data. It is useful, therefore, to report their individual values, as well as their product. We shall pursue this topic later when we discuss principal components.
The generalized sample variance will be zero in certain situations. A generalized variance of zero is indicative of extreme degeneracy, in the sense that at least one column of the matrix of deviations,
.. .
..
Figure 3.8 Axes of the meancentered 95% ellipses for the scatter plots in Figure 3.7.
..
can be expressed as a linear combination of the other columns. As we have shown geometrically, this is a case where one of the deviation vectorsfor instance, di = [Xli  Xi'"'' Xni  xdlies in the (hyper) plane generated by d 1 ,· .. , dil> di+l>"" d p .
Result 3.2. The generalized variance is zero when, and only when, at least one deviation vector lies in the (hyper) plane formed by all linear combinations of the othersthat is, when the columns of the matrix of deviations in (318) are linearly dependent. Proof. If the ct>lumns of the deviation matrix (X  li') are linearly dependent, there is a linear combination of the columns such that 0= al coll(X  li') + ... + apcolp(X  li')
figure 3.9 A case where the threedimensional volume is zero (/SI = 0).
so the same a corresponds to a linear dependency, al coll(S) + ... + ap colp(S) = Sa = 0, in the columns of S. So, by Result 2A.9, 1S 1 = O. In the other direction, if 1S 1 = 0, then there is some linear combination Sa of the columns of S such that Sa = O. That is, 0 = (n  1)Sa = (X  Ix')' (X  li') a. Premultiplying by a' yields
and, for the length to equal zero, we must have (X  li')a = O. Thus, the columns of (X  li') are linearly dependent. Example 3.9 (A case where the generalized variance is zero) Show that 1 S 1 = 0 for
and determine the degeneracy. Here x' = [3,1, 5J, so 1 3
The deviation (column) vectors are di = [2,1, 1J, d z = [1,0, 1], and = d l + 2d2 , there is column degeneracy. (Note that there 3 is row degeneracy also.) This means that one of the deviation vectorsfor example, d lies in the plane generated by the other two residual vectors. Consequently, the threedimensional volume is zero. This case is illustrated in Figure 3.9 and may be verified algebraically by showing that IS I = O. We have d = [0,1, IJ. Since d3
When large data sets are sent and received electronically, investigators are sometimes unpleasantly surprised to find a case of zero generalized variance, so that S does not have an inverse. We have encountered several such cases, with their associated difficulties, before the situation was unmasked. A singular covariance matrix occurs when, for instance, the data are test scores and the investigator has included variables that are sums of the others. For example, an algebra score and a geometry score could be combined to give a total math score, or class midterm and final exam scores summed to give total points. Once, the total weight of a number of chemicals was included along with that of each component. This common practice of creating new variables that are sums of the original variables and then including them in the data set has caused enough lost time that we emphasize the necessity of being alert to avoid these consequences. Example 3.10 (Creating new variables that lead to a zero generalized variance) Consider the data matrix
where the third column is the sum of first two columns. These data could be the number of successful phone solicitations per day by a parttime and a fulltime employee, respectively, so the third column is the total number of successful solicitations per day. Show that the generalized variance 1S 1 = 0, and determine the nature of the dependency in the data.
. [2.5 0 2.5]' 2.5 2.5 S= 0 2.5 2.5 5.0
In general, if the three columns of the data matrix X satisfy a linear constraint al xjl + a2Xj2 + a3xj3 = c, a constant for all j, then alxl + a2 x2+ a3 x3 = c, so that
for all j. That is,
and the columns of the mean corrected data matrix are linearly dependent. Thus, the inclusion of the third variable, which is linearly related to the first two, has led to the case of a zero generalized variance. Whenever the columns of the mean corrected data matrix are linearly dependent,
(n  I)Sa = (X  li/)/(X li/)a = (X  li/)O = 0 and Sa = 0 establishes the linear dependency of the columns of S. Hence, IS I = o. Since Sa = 0 = 0 a, we see that a is a scaled eigenvector of S associated with an eigenvalue of zero. This gives rise to an important diagnostic: If we are. unaware of any extra variables that are linear combinations of the others, we. can fID? them by calculating the eigenvectors of S and identifying the one assocIated WIth a zero eigenvalue. That is, if we were unaware of the dependency in this example, a computer calculation would find an eigenvalue proportional to a/ = [1,1, 1), since 2.5
In addition, the sum of the first two variables minus the third is a constant c for all n units. Here the third variable is actually the sum of the first two variables, so the columns of the original data matrix satisfy a linear constraint with c = O. Because we have the special case c = 0, the constraint establishes the fact that the columns of the data matrix are linearly dependent. 
allj (c = a/x) , (3) a/xj = c for ...,...
The linear combination of the mean corrected data, using a, is zero.
The linear combination of the original data, using a, is a constant.
We showed that if condition (3) is satisfiedthat is, if the values for one variable can be expressed in terms of the othersthen the generalized variance is zero because S has a zero eigenvalue. In the other direction, if condition (1) holds, then the eigenvector a gives coefficients for the linear dependency of the mean corrected data. In any statistical analysis, IS I = 0 means that the measurements on some variables should be removed from the study as far as the mathematical computations are concerned. The corresponding reduced data matrix will then lead to a covariance matrix of full rank and a nonzero generalized variance. The question of which measurements to remove in degenerate cases is not easy to answer. When there is a choice, one should retain measurements on a (presumed) causal variable instead of those on a secondary characteristic. We shall return to this subject in our discussion of principal components. At this point, we settle for delineating some simple conditions for S to be of full rank or of reduced rank.
Result 3.3. If n :s; p, that is, (sample size) :s; (number of variables), then IS I = 0 for all samples. Proof. We must show that the rank of S is less than or equal to p and then apply Result 2A.9. For any fixed sample, the n row vectors in (318) sum to the zero vector. The existence of this linear combination means that the rank of X  li' is less than or equal to n  1, which, in turn, is less than or equal to p  1 because n :s; p. Since
the kth column of S, colk(S), can be written as a linear combination of the columns of (X  li/)'. In particular,
Let us summarize the important equivalent conditions for a generalized variance to be zero that we discussed in the preceding example. Whenever a nonzero vector a satisfies one of the following three conditions, it satisfies all of them:
ais a scaled eigenvector of S with eigenvalue O.
+ ... + (Xnk  Xk) coln(X  li/)'
Since the column vectors of (X  li')' sum to the zero vector, we can write, for example, COlI (X  li')' as the negative of the sum of the remaining column vectors. After substituting for rowl(X  li')' in the preceding equation, we can express colk(S) as a linear combination of the at most n  1 linearly independent row vectorscol2(X li')', ... ,coln(X li/)'.TherankofSisthereforelessthanorequal to n  1, whichas noted at the beginning of the proofis less than or equal to p  1, and S is singular. This implies, from Result 2A.9, that IS I = O. •
134 Chapter 3 Sample Geometry and Random Sampling Result 3.4. Let the p X 1 vectors Xl> X2,' •. , Xn , where xj is the jth row of the data matrix X, be realizations of the independent random vectors X I, X 2, ... , X n • Then
1. If the linear combination a/Xj has positive variance for each constant vector a
then, provided that p < n, S has full rank with probability 1 and 1SI> o. 2: If, with probability 1, a/Xj is a constant (for example, c) for all j, then 1S 1 = O. Proof. (Part 2). If a/Xj
when two or more of these vectors are in almost the same direction. Employing the argument leading to (37), we readily find that the cosine of the angle ()ik between (Yi  xi1 )/Vi;; and (Yk  xkl)/vSkk is the sample correlation coefficient rik' Therefore, we can make the statement that 1R 1 is large when all the rik are nearly zero and it is small when one or more of the rik are nearly + 1 or 1. In sum, we have the following result: Let Xli 
= alXjl + a2 X j2 + .,. + apXjp = c with probability 1,
= c for all j, imd the sample mean of this linear combination is c = + a2 x j2 + .,. + apxjp)/n = alxl + a2x2 + ... + apxp = a/x. Then J
a/x.
i = 1,2, ... , p
be the deviation vectors of the standardized variables. The ith deviation vectors lie in the direction of d;, but all have a squared length of n  1. The volume generated in pspace by the deviation vectors can be related to the generalized sample variance. The saine steps that lead to (315) produce
indicating linear dependence; the conclusion follows fr.om Result 3.2. The proof of Part (1) is difficult and can be found m [2].
Generalized Variance Determined by IRI and Its Geometrical Interpretation The generalized sample variance is unduly affected by the ~ari.ability of measu~e ments on a single variable. For example, suppose some Sii IS either large or qUIte small. Then, geometrically, the corresponding deviation vector di = (Yi  XiI) will be very long or very short and will therefore clearly be an important factor in determining volume. Consequently, it is sometimes useful to scale all the deviation vectors so that they have the same length. Scaling the residual vectors is equivalent to replacing each original observation x. by its standardized value (Xjk  Xk)/VS;;;· The sample covariance matrix of the si:ndardized variables is then R, the sample correlation matrix of the original variables. (See Exercise 3.13.) We define Generalized sample variance) = R ( of the standardized variables 1 1
Generalized sample variance) 1R 1 (2 = n  1) P( volume) ( ofthe standardized variables =
all have length ~, the generalized sample variance of the standardized variables will be large when these vectors are nearly perpendicular and will be small
The volume generated by deviation vectors of the standardized variables is illustrated in Figure 3.10 for the two sets of deviation vectors graphed in Figure 3.6. A comparison of Figures 3.10 and 3.6 reveals that the influence of the d 2 vector (large variability in X2) on the squared volume 1S 1 is much greater than its influence on the squared volume 1R I. 3
\,..
...... .> \
the standardized variables.
We concludethis discussion by mentioning another generalization of variance. Specifically, we define the total sample variance as the sum of the diagonal elements of the sample varianceco)(ariance matrix S. Thus,
(322) [The proof of (321) is left to the reader as Exercise 3.12.] Interpreting (322) in terms of volumes, we see from (315) and (320) that the squared volume (n  1)pISI is proportional to th<; squared volume (n  I)PIRI. The constant of proportionality is the product of the variances, which, in turn, is proportional to the product of the squares of the lengths (n  l)sii of the d i . Equation (321) shows, algebraically, how a change in the· measurement scale of Xl> for example, will alter the relationship between the generalized variances. Since IR I is based on standardized measurements, it is unaffected by the change in scale. However, the relative value of IS I will be changed whenever the multiplicative factor SI I changes. Example 3.11 (Illustrating the relation between IS I and I R I) Let us illustrate the
relationship in (321) for the generalized variances IS I and IR I when p Suppose S
3.
Geometrically, the total sample variance is the sum of the squared lengths of the = (YI  xII), ... , d p = (Yp  xpI), divided by n  1. The total sample variance criterion pays no attention to the orientation (correlation structure) of the residual vectors. For instance, it assigns the same values to both sets ofresidual vectors (a) and (b) in Figure 3.6. p deviation vectors d I
= 1. Moreover,
Example 3.12· (Calculating the total sample variance) Calculate the total sample variance for the variancecovariance matrices S in Examples 3.7 and 3.9. From Example 3.7.
+ ... + spp
3.5 Sample Mean, Covariance, and Correlation as Matrix Operations We have developed geometrical representations of the data matrix X and the derived descriptive statistics i and S. In addition, it is possible to link algebraically the calculation of i and S directly to X using matrix operations. The resulting expressions, which depict the relation between i, S, and the full data set X concisely, are easily programmed on electronic computers.
1 + X2i'l + ... + Xni '1)ln
= yj1/n. Therefore,
111')'(1  111') =111 I , 11 1 , +1 11" 11 =1111' (I  n n. n n n2 n
That is, x is calculated from the transposed data matrix by postmultiplying by the vector 1 and then multiplying the result by the constant l/n. Next, we create an n X p matrix of means by transposing both sides of (324) and premultiplying by 1; that is,
...
...
The result for Sn is similar, except that I/n replaces l/(n  1) as the first factor. The relations in (327) show clearly how matrix operations on the data matrix X lead to x and S. Once S is computed, it can be related to the sample correlation matrix R. The resulting expression can also be "inverted" to relate R to S. We fIrst defIne the p X P sample standard deviation matrix Dl/2 and compute its inverse, (D J/ 2 l = D I/2. Let
o Now, the matrix (n  I)S representing sums of squares and cross products is just the transpose of the matrix (326) times the matrix itself, or
.
Postmultiplying and premultiplying both sides of (329) by nl/2 and noting that n l/2nI/2 = n l/2n l/2 = I gives S
That is, R can be optained from the information in S, whereas S can be obtained from nl/2 and R. Equations (329) and (330) are sample analogs of (236) and (237).
It follows from (332) and (333) that the sample mean and variance of these derived observations are Sample mean of b'X = b'i Sample variance of b'X = b'Sb Moreover, the sample covariance computed from pairs of observations on b'X and c'X is Sample covariance = (b'xI  b'i)(e'x!  e'i)
We have introduced linear combinations of p variables in Section 2.6. In many multivariate procedures, we are led naturally to consider a linear combination of the foim c'X
+ (b'X2  b'i)(e'x2  e'i) + ... + (b'xn  b'i)(e'xn  e'i)
+ c2X2 + .,. + cpXp
+ b'(X2  i)(X2  i)'e + ... + b'(xn  i)(x n  i)'e n1
+ (X2  i)(X2  i)' + ... + (XII  i)(xlI  i),Je n1
whose observed value on the jth trial is j = 1,2, ... , n
(C'XI + e'x2 + ... + e'x n) n
+ X2 + ... + xn) l
Equations (332) and (333) are sample analogs of (243). They correspond to substituting the sample quantities i and S for the "population" quantities /L and 1;, respectively, in (243). Now consider a second linear combination
+ c2X2 + ... + cpXp
Sample mean of b'X Sample mean of e'X Samplevarianceofb'X Sample variance of e'X Samplecovarianceofb'Xande'X
Example 3.13 (Means and covariances for linear combinations) We shall consider two linear combinations and their derived values for the n = 3 observations given in Example 3.9 as
b'X = blXI + hzX2 + ... + bpXp
j = 1,2, ... , n
blXI + hzX2 + ... + bpXp CIXI
e'(xI i)(xI  i)'e + C'(X2  i)(X2  i)'e + ... + e'(xn  i)(x n  i)'e
Result 3.5. The linear combinations
. (e'xI  e'i)2 + (e'x2  e'i)2 + ... + (e'xn  e'i/ Sample vanance = n  1
In sum, we have the following result.
Consequently, using (336), we find that the two sample means for the derived observations are
The means, variances, and covariance will first be evaluate.d directly and then be evaluated by (336). Observations on these linear combinations are obtained by replacing Xl, X 2 , and X3 with their observed values. For example, the n = 3 observations on b'X are b'XI = b'X2 = b'X3 =
. Sample vanance .
C'XI = 1Xll  .1X12 + 3x13 = 1(1)  1(2) + 3(5) = 14 C'X2 = 1(4)  1(1) + 3(6) = 21 C'X3 = 1(4)  1(0) + 3(4) = 16
Moreover, the sample covariance, computed from the pairs of observations (b'XI, c'xd, (b'X2, C'X2), and (b'X3, C'X3), is Sample covariance (1  3)(14 17) + (4  3)(21  17) 3 1
Alternatively, we use the sample mean vector i and sample covariance matrix S derived from the original data matrix X to calculate the sample means, variances, and covariances for the linear combinations. Thus, if only the descriptive statistics are of interest, we do not even need to calculate the observations b'xj and C'Xj. From Example 3.9,
As indi~ated, these last results check with the corresponding sample quantities _ computed directly from the observations on the linear combinations. . The sampl~ m~an and ~variance relations in Result 3.5 pertain to any number of lInear combmatlOns. ConSider the q linear combinations
i = 1,2, ... , q
Chapter 3 Sample Geometry and Random Sampling These can be expressed in matrix notation as
+ ... + + .,. +
aq 2X 2 + .,. +
(c) Graph (to scale) the triangle formed by Yl> xII, and YI  xII. Identify the length of each component in your graph. (d) Repeat Parts ac for the variable X 2 in Table 1.1. (e) Graph (to scale) the two deviation vectors YI  xII and Y2  x21. Calculate the value of the angle between them.
'" k' th 'th roW of A a' to be b' and the kth row of A, ale, to be c', we see that ~a lng e l ' " 1 '  d th e It. h and . (336) imply that the ith row ofAX has samp e mean ajX an EquatIOns . , N h 's . h (. k)th eIekth rows ofAX have sample covariance ajS ak' ote t at aj ak IS t e I,
3.S. Calculate the generalized sample variance 1SI for (a) the data matrix X in Exercise 3.1 and (b) the data matrix
in Exercise 3.2.
3.6. Consider the data matrix
ment of ASA'.
(a) Calculate the matrix of deviations (residuals), X  lX'. Is this matrix of full rank? Explain. (b) Determine S and calculate the generalized sample variance 1S I. Interpret the latter geometrically. (c) Using the results in (b), calculate the total sample variance. [See (323).]
Th q linear combinations AX in (338) have sample mean vector Ai Resu It 3 .6. e ., • and sample covariance matnx ASA .
Exercises 3.1.
3.7.
Sketch the solid ellipsoids (x  X)'SI(x  x) s 1 [see (316)] for the three matrices
X'[Hl lot in p = 2 dimensions. Locate the sample mean on your diagram. (a) Graph the sca tter p . . . 3 dimensional representatIon of the data, and plot the deVIatIOn (b) Sketch the n_ _ vectors YI  xII and Y2  x21. . ti'on vectors in (b) emanating from the origin. Calculate the lengths .. (c) Sketch th e d eVIa e cosine of the angle between them. Relate these quantIties to of these vect ors and th Sn and R. 3.2. Given the data matrix
(Note that these matrices have the same generalized variance 1 SI.) 3.S. Given
(a) Calculate the total sample variance for each S. Compare the results. (b) Calculate the gene'ralized sample variance for each S, and compare the results. Comment on the discrepancies, if any, found between Parts a and b. 3.9. The following data matrix contains data on test scores, with XI = score on first test, X2 = score on second test, and X3 = total score on the two tests:
3.3.
(a) Graph the scatter plot in p = 2 dimensions, and locate the sample mean on.y~ur diagram. space representation of the data, and plot the deVIatIOn vectors  3_ (b) Sk etch ten h YI  XII and Y2  x21. . . . . viation vectors in (b) emanatmg from the ongm. Calculate their lengths () c Sketch the de I h .. t S d R . of the angle between them. Re ate t ese quantIties 0 n an . and t hecosme . Perform the decomposition of YI into XII and YI  XII using the first column of the data matrix in Example 3.9. . bse rvat'lons on the. variable XI ' in units of millions, from Table 1.1. Useth esIXO (a) Find the projection on I' = [1,1,1,1,1,1]. (b) Calculate the deviation vector YI  XII. Relate its length to the sample standard deviation.
(a) Obtain the mean corrected data matrix, and verify that the columns are linearly dependent. Specify an a' = [ai, a2, a3] vector that establishes the linear dependence. (b) Obtain the sample covariance matrix S,and verify that the generalized variance is zero. Also, show that Sa = 0, so a can be rescaled to be an eigenvector corresponding to eigenvalue zero. (c) Verify that the third column of the data matrix is the sum of the first two columns. That is, show that there is linear dependence, with al = 1, a2 = 1, and Q3 = 1.
the generalized variance is zero, it is the columns of the mean corrected data 3I.0. Wh en Xc = X  lx' that are linearly depend ' ly t h ose 0 f t h e data ent, not necessan matrix matrix itself. Given the data
(a) Obtain the mea~ corre~ted d~ta matrix, and verify that. the columns are linearly dependent. Specify an a = [ai, a2, a3] vector that estabhshes the dependence.. (b) Obtain the sample covariance matrix S, and verify that the generalized variance is zero. (c) Show that the columns of the data matrix are linearly independent in this case. 3.
11 U the sample covariance obtained in Example 3.7 to verify (329) and (330), which _ D1/2SD1/2 and D l/2RD 1/2 = S. . se state that R 
3.12. ShowthatlSI = (SIIS22"· S pp)IRI· . . 1S 1 = · t" From Equation (330), S = D 1/2 RD 1/2...., . la k'mg d etermmants gIves H m. ~I 2 I IDl/211 R 11 D / 1· (See Result 2A.l1.) Now examine 1D . 3.13. Given a data matrix X and the resulting sample correlation matrix R, the standardized observations (Xjk  Xk)/~' k = 1,2, ... , p, I'der cons .. h ' j = 1, 2, ... , n. Show that these standar d'Ize d quantities ave sampi e covanance matrix R. 'der the data matrix X in Exercise 3.1. We have n = 3 observations on p = 2 vari3. 14 • ConSl . b" abies Xl and X 2 • FOTID the hnear com matIons c'X=[1
( ) E aluate the sample means, variances, and covariance of b'X and c'X from first a pr~nciples. That is, calculate the observed values of b'X and c'X, and then use the sample mean, variance, and covariance fOTlDulas. (b) Calculate the sample means, variances, and covariance of b'X and c'X using (336). Compare the results in (a) and (b). 3.1 S. Repeat Exercise 3.14 using the data matrix
3.16. Let V be a vector random variable with mean vector E(V) = /Lv and covariance matrix E(V  /Lv)(V  /Lv)'= Iv· ShowthatE(VV') = Iv + /Lv/Lv,
3.17. Show that, if X and Z are independent then each component of X is (pXl)
independent of each component of Z. Hint:P[Xl:S Xl,X2 :s X2""'Xp :S x p andZ1 :s ZI,""Zq:s Zq] = P[Xl:s Xl,X 2 :s X2""'X p :S xp]·P[ZI:S Zj, ... ,Zq:s Zq]
by independence. Let
Zl' So Xl and ZI are independent: Repeat for other pairs.
3.IS. Energy consumption in 2001, by state, from the major sources Xl
is recorded in quadrillions (1015) of BTUs (Source: Statistical Abstract of the United States 2006), The resulting mean and covariance matrix are
766 0.508 J O. 0.438 0.161
O. 856 S =
(a) Using the summary statistics, determine the sample mean and variance of a state's total energy consumption for these major sources. (b) Determine the sample mean and variance of the excess of petroleum consumption over natural gas consumption. Also find the sample covariance of this variable with the total variable in part a. 3.19. Using the summary statistics for the first three variables in Exercise 3.18, verify the relation
th climates roads must be cleared of snow quickly following a storm. One torm se~erity is Xl = its duration in hours, while the effectiveness of snow 3.20. In nor em f measure 0 s d h' uantified by X2 = the number of hours crews, men, an mac me, spend removal can be q .. . W' . to clear snoW. Here are the results for 25 mCldents m Isconsm.
(a) Find the sam~l~ mean and variance of the difference X2  Xl by first obtaining the summary statIstIcs. (b) Obtain the mean and variance by first obtaining the .individual values Xf2  Xjh 25 and then calculating the mean and vanance. Compare these values . 1 2 for]  , , ... , with those obtained in part a.
References T W. An Introduction to Multivariate Statistical Analysis (3rd ed.). New York: 1. An derson,. John Wiley, 2003. . 2 Eaton, M., adnM· PerIman ."The NonSingularity of Generalized Sample Covanance . Matrices." Annals of Statistics, 1 (1973),710717.
A generalization of the familiar bellshaped normal density to several dimensions plays a fundamental role in multivariate analysis. In fact, most of the techniques encountered in this book are based on the assumption that the data were generated from a multivariate normal distribution. While real data are never exactly multivariate normal, the normal density is often a useful approximation to the "true" population distribution. One advantage of the multivariate normal distribution stems from the fact that it is mathematically tractable and "nice" results can be obtained. This is frequently not the case for other datagenerating distributions. Of course, mathematical attractiveness per se is of little use to the practitioner. It turns out, however, that normal distributions are useful in practice for two reasons: First, the normal distribution serves as a bona fide population model in some instances; second, the sampling distributions of many multivariate statistics are approximately normal, regardless of the form of the parent population, because of a central limit effect. To summarize, many realworld problems fall naturally within the framework of normal theory. The importance of the normal distribution rests on its dual role as both population model for certain natural phenomena and approximate sampling distribution for many statistics.
4.2 The Multivariate Normal Density and Its Properties The multivariate normal density is a generalization of the univariate normal density to p ~ 2 dimensions. Recall that the univariate normal distribution, with mean ft and variance u 2 , has the probability density functio~ 00
Fi~re 4.1 A normal density with mean /L and variance (T2 and selected areas under the curve.
A plot of this function yields the familiar bellshaped curve shown in Figure 4.1. Also shown in the figure are app~oximate areas under the curve within ± 1 standard deviatio ns and ±2 standard deviations of the mean. These areas represen t probabi lities, and thus, for the normal random variable X,
Example 4.1 (Bivariatenormal density) L density in terms of the ·nd· ·d al et us evaluate the p = 2variat e normal I IVI paramet ers /L  E(X ) z (T11 = Var(X ), (TZ2 = Var(X ) andU _ 1 I, /L2 == E(X ), I Using Result l , Xz)· 2A.8, we findzthat thP1.Z = Corr(X e mverse of the covarian ce matrix
Intr~ducing the correlat ion ent Pl2 b writin obtam (T11(T22  (T12 = (T (T coeffici (1 _ 2) Y squared g ya:;, we 11 Z2 Pl2 , and the dIstance become s
It is conveni ent to denote the normal density function with mean /L and variance (Tz by N(/L, (TZ). Therefore, N(lO, 4) refers to the function in (41) with /L = 10 and (T = 2. This notation will be extended to the multivariate case later.
(42) in the exponen t of the univariate normal density function measure s the square of the distance from x to /L in standard deviatio n units. This can be generali zed for a p X 1 vector x of observations on several variables as (43) The p x 1 vector /L represents the expected value of the random vector X, and the p X P matrix I is the variancecovariance matrix ofX. [See (230) and (231).] We shall assume that the symmetric matrix I is positive definite, so the expressi on in (43) is the square of th.e generalized distance from x to /L. The multivariate normal density is obtained by replacing the univaria te distance in (42) by the multivariate generalized distance of (43) in the density function of (41). When this replacement is made, the univariate normali zing constan t (27T rl/2( (Tzrl/2 must be changed to a more general constant that makes the volume under the surface of the multivariate density function unity for any p. This is necessary because, in the multivariate case, probabilities are represen ted by volumes under the surface over regions defined by intervals of the Xi values. It can be shown (see [1]) that this constant is (27TF/zl Irl/2, and consequently, a pdimen sional normal density for the random vector X' = [XI' X z,···, Xp] has the form (44)
where CXJ < Xi < CXJ, i = 1,2, ... , p. We shall denote this pdimen sional normal density by Np(/L, I), which is analogous to the normal density in the univaria te case.
. terms of the standard ized wn.ttenm values (Xl  I1d/VC;:;; and
 P12), 2 and Next, III i since . (44)II I = (Tll (T22  (T2 = (T 11 (T we can substItu te for II n to get the expressIOn fo th b· . ( involvin g the individu al parame ter r e Ivanate p = 2) normal density s 111> 112, (T11> (T22, and PI2:
12.
.
. .
The expresSIOn m (46) is somewh at . Id (44) is more informa tive in man wa unWIe y, and the compac t general form in useful for discussi ng certain pro/ertiZs~7~ the other ~an?, th.e express ion in (46) is random variable s X and X t e normal dIstnbut ion. For example if the . I 2 are uncorre lated so that . . .' e wntten as the product of two un.. ' ~~2  0 , t hofe the Jomt denSity can b Ivanate normal denSItIes each form of (41).
That is, !(X1, X2) = !(X1)!(X2) and Xl and X 2 are independent. [See (228).] This result is true in general. (See Result 4.5.) Two bivariate distributions with CT11 = CT22 are shown in FIgure 4.2. In FIgure 4.2(a), Xl and X 2 are independent (P12 = 0). In Figure 4.2(b), P12 = .75. Notice how the presence of correlation causes the probability to concentrate along a line. •
From the expression in (44) for the density of a pdimensional normal variable, it should be clear that the paths of x values yielding a constant height for the density are ellipsoids. That is, the multivariate normal density is constant on surfaces where the square of the distance (x  J.l)' l:1 (x  J.l) is constant. These paths are called contours: Constant probability density contour
The axes of each ellipsoid of constant density are in the direction of the eigenvectors of l:1, and their lengths are proportional to the reciprocals of the square roots of the eigenvalues of l:1. Fortunately, we can avoid the calculation of l:1 when determining the axes, since these ellipsoids are also determined by the eigenvalues and eigenvectors of l:. We state the correspondence formally for later reference.
Result 4.1. If l: is positive definite, so that l:1 exists, then
so (A, e) is an eigenvalueeigenvector pair for l: corresponding to the pair (1/ A, e) for l:1. Also, l:1 is positive definite. Proof. For l: positive definite and e oF 0 an eigenvector, we have 0 < e'l:e = e' (l:e) = e'(Ae) = Ae'e = A. Moreover, e = r1(l:e) = l:l(Ae), or e = U;le, and divi
sion by A> 0 gives l:le = (l/A)e. Thus, (l/A, e) is an eigenvalueeigenvector pair for l:1. Also, for any p X 1 x, by (221) (a)
since each term Ai1(x'e;)2 is nonnegative. In addition, x'ej = 0 for all i only if p
O. So x
positive definite.
The following summarizes these concepts: Contours of constant density for the pdimensional normal distribution are ellipsoids defined by x such the that (47) These ellipsoids are centered at J.l and have axes ±cv'X;ej, where l:ej for i = 1, 2, ... , p.
Figure 4.2 '!Wo bivariate normal distributions. (a) CT1! (b)CTll = CT22andp12 = .75.
and P12 = O.
A contour of constant density for a bivariate normal distribution with CTU = CT22 is obtained in the following example.
Example 4.2 (Contours of the bivariate normal d.ensi.ty) We shall ~bt~in ~e axes of constant probability density contours for a blvan?te normal dlst~lbutlOn when O"u = 0"22' From (47), these axes are given by the elgenvalues and elgenvectors of :£. Here 1:£  All = 0 becomes
When the covariance (correlation) is negative, A2 = 0"11  0"12 will be the largest eigenvalue, and the major axes of the constantdensity ellipses will lie along a line at right angles to the 45° line through /L. (These results are true only for 0"11
To summarize, the axes of the ellipses of constant density for a bivariate normal distribution with 0"11 = 0"22 are determined by
• We show in Result 4.7 that the choice c 2 = x~(a), where x~(a) is the upper (looa)th percentile of a chisquare distribution with p degrees of freedom,leads to contours that contain (1  a) X 100% of the probability. Specifically, the following is true for a pdimensional normal distribution:
eigenvector pair is has probability 1  a.
The constantdensity contours containing 50% and 90% of the probability under the bivariate normal surfaces in Figure 4.2 are pictured in Figure 4.4. Similarly, A2 = 0"11  (112 yields the eigen:ector ei. = [1("!2, 1/\12). . When the covariance (112 (or correlatIOn pn) IS pOSItive, AI = 0"11 + ~12 IS the largest eigenvalue, and its associated eigenvect.or. e; = [1/\12, 1/~) hes along the 45° line through the point p: = [ILl' 1Lz)· 11llS IS true for any p~sltIve. value of the covariance (correlation). Since the axes of the constantdensity elhpses are iven by ±cVA, e and ±cVX; e2 [see (47)], and the eigenvectors each have fength unity, th~ ~ajor axis will be associated with the largest .eigen~alue. For positively correlated normal random variable~, then, the major a~ls of the constantdensity ellipses wiil be along the 45° lme through /L. (See Figure 4.3.)
Figure 4.4 The 50% and 90% contours for the bivariate normal distributions in Figure 4.2.
Figure 4.3 A constantdensity contour for a bivariate normal distribution with Cri I = (122 and (112) 0 (or P12 > 0).
The pvariate normal density in (44) has a maximum value when the squared distance in (43) is zerothat is, when x = /L. Thus, /L is the point of maximum density, or mode, as well as the expected value of X, or mean. The fact that /L is the mean of the multivariate normal distribution follows from the symmetry exhibited by the constantdensity contours: These contours are centered, or balanced, at /L.
Additional Properties of the Multivariate Normal Distribution Certain properties of the normal distribution will be needed repeatedly in OUr explanations of statistical models and methods. These properties make it possible to manipulate normal distributions easily and, as we suggested in Section 4.1, are partly responsible for the popularity of the normal distribution. The key properties, which we shall soon discuss in some mathematical detail, can be stated rather simply. . The following are true for a.random vector X having a multivariate normal distribution:
a ~a  [1,0, ... ,0]
3. Zero covariance implies that the corresponding components are independently .distributed. 4. The conditional distributions of the components are (multivariate) normal.
These statements are reproduced mathematically in the results that follow. Many of these results are illustrated with examples. The proofs that are included should help improve your understanding of matrix manipulations and also lead you to an appreciation for the manner in which the results successively build on themselves. Result 4.2 can be taken as a working definition of the normal distribution. With this in hand, the subsequ~nt properties are almost immediate. Our partial proof of Result 4.2 indicates how the linear combination definition of a normal density relates to the multivariate density in (44). Result 4.2. If X is distributed as Np(/L, ~), then any linear combination of variables a'X = alXl + a2X2 + .. , + apXp is distributed as N(a' /L, a'~a). Also, if a'X is distributed as N(a' /L, a'~a) for every a, then X must be Np(/L, ~). Proof. The expected value and variance of a'X follow from (243). Proving that a'Xis normally distributed if X is multivariate normal is more difficult. You can find • a proof in [1 J. The second part of result 4.2 is also demonstrated in [1].
Example 4.3 (The distribution of a linear combination of the components of a normal random vector) Consider the linear combination a'X of a m.ultivariate normal random vector determined by the choice a' = [1,0, .. ,,0]. Since
'" (JIP1 [11 0_ '" .
1. Linear combinations of the components of X are normally distributed.
and it fol!ows ~ro~ R~sult 4.2 that Xl is distributed as N (/JI, 0"11)' More generally, • the margmal dlstnbutlOn of any component Xi of X is N(/Ji, O"ii)' The next result considers several linear combinations of a multivariate normal vectorX. Result 4.3. If X is distributed as Nip" ~), the q linear combinations
are distributed as Nq(Ap" A~A'). Also, constants, is distributed as Np(/L
+ d, I).
Proof. The expected value E(AX) and the covariance matrix ofAX follow from (245). Any linear combination b'(AX) is a linear combination of X of the form a'X with a = A'b. Thus, the conclusion concerning AX follows direc~ly from Result 4.2. The second part of the result can be obtained by considering a'(X + d) = a'~ +.(a'd), where a'~ is distributed as N(a'p"a'Ia). It is known from the umvanate case that addmg a constant a'd to the random variable a'X leaves the varianc~ unchanged and translates the mean to a' /L + a'd = a'(p, + d). Since a • was arbItrary, X + d is distributed as Np(/L + d, ~). Example 4.4 (The distribution of two linear combinations of the components of a normal random vector) For X distributed as N3 (/L, ~), find the distribution of
By Result 4.3, the distribution ofAX is multivariate normal with mean Example 4.5 (The distribution of a subset of a normal random vector) 0J [::] 1
J.
and note that with this assignment, X, /L, and :t can respectively be rearranged and as
or Alternatively, the mean vector AIL and covariance matrix A:tA' may be verified by direct calculation of the means and covariances of the two random variables • YI = XI  X 2 and Yi = X 2  X 3 · We have mentioned that all subsets of a multivariate normal random vector X are themselves normally distributed. We state this property formally as Result 4.4.
Result 4.4. All subsets of X are normally distributed. If we respectively partition X, its mean vector /L, and its covariance matrix :t as we have the distribution
We are now in a position to state that zero correlation between normal random variables or sets of normal random variables is equivalent to statistical independence.
Result 4.5. (ql XI)
It is clear from this example that the normal distribution for any subset can be expressed by simply selecting the appropriate means and covariances from the original /L and :to The formal process of relabeling and partitioning is unnecessary_ _
Proof. Set
in Result 4.3, and the conclusion follows.
To apply Result 4.4 to an arbitrary subset of the components of X, we simply relabel the subset of interest as Xl and select the corresponding component means and covariances as ILl and :t ll , respectively. 
zeros. ( b) If [ XI] IS . Nq1 + q2 ([ILl] i :t12]) , then XI and X 2 are independent ".If , [:t11 .jX2 IL2 :t21: :t22 and only if:t12 = o.
Chapter 4 The Multivariate Normal Distribution (c) If Xl and X 2 are independent and are distributed as Nq1(PI, Ill) and . N (P2, I q2
has the multivariate normal distribution.
Note that the covariance does not depend on the value X2 of the conditioning variable.
Proof. We shall give an indirect proof. (See Exercise 4.13, which uses the densities directly.) Take Proof. (See Exercise 4.14 for partial proofs based upon factoring the density function when I12 = 0.)
Example 4.6. (The equivalence of zero covariance and independence for normal variables) Let X be N3 (p, I) with
Are XI and X 2 independent? What about (X I ,X2) and X3? Since Xl and X 2 have covariance Ul2 = 1, they are not mdependent. However, partitioning X and I as
m~trix. I12 =[?J. Therefore,
and X are independent by Result 4.5. This unphes X3 IS mdependent of ( X I, X) 2 3 • Xl and also of X 2· We pointed out in our discussion of the bivariate nor~~l distri?ution t~at P12 = 0 (zero correlation) implied independence because ~he Jo(mt de~)sl~y fu.n~tJo~ [see (46)] could then be written as the product of the ~arg~al n~rm.a ensItJes.o Xl and X . This fact, which we encouraged you to verIfy dIrectly, IS SImply a speCial 2 case of Result 4.5 with ql = q2 = l. Result 4.6. Let X I =
Example 4.7 (The conditional density of a bivariate normal distribution) The conditional density of Xl' given that X 2 = X2 for any bivariate distribution, is defined by f( Xl IX2 ) =
Since Xl  PI  I12Iz1 (X2  P2) and X 2  P2 have zero covariance, they are independent. Moreover, the quantity Xl  PI  I12Iz1 (X2  P2) has distribution Nq(O, III  I12I21I21)' Given that X 2 = X2, Pl + I12Iz1 (X2  P2) is a constant. Because XI  ILl  I12I21 (X2  IL2) and X 2  IL2 are independent, the conditional distribution of Xl  ILl  I12Izi (X2  IL2) is the same as the unconditional distribution of Xl  ILl  I12I21 (X2  P2)' Since Xl  ILl  I12Iz1 (X2  P2) is Nq(O, III  I 12I 2iI21 ), so is the random vector XI  PI  I12Iz1 (X2  P2) when X 2 has the particular value x2' Equivalently, given that X 2 = X2, Xl is distributed as Nq(ILI + I12Izi (X2  P2), III  I12Izi I2d· •
~...;.:.~:.:..
> O. Then the conditional distribution of Xl> given
·· Id . f . enslty 0 Xl gIven that X 2 = {cond ItIona
where f(X2) is the marginal distribution of X 2. If f(x!> X2) is the bivariate normal density, show that f(xII X2) is N ( PI
162 Chapter 4 The MuJtivariate Normal Distribution Here Ull  Urz/U22 = ull(1  PI.2)' The two te?Ds involving Xl : ILl in the expothe bivariate normal density [see Equation (46)] become, apart from the nen t of 2 multiplicative constant 1/2( 1  PI2), (Xl  ILl?
1. All conditional distributions are (multivariate) normal. 2. The conditional mean is of the form
• r .
= UI2/~ ya;, or Pl2vU;Jvu:;;. = Ulz/ U22, the complete expo
.
...
.
. ..
(b) The Np(p" I) distribution assigns probability 1  a to the solid ellipsoid {x: (x  p,)'II(x  p,) :5 x~(a)}, where ~(a) denotes the upper (l00a)th percentile of the ~ distribution. Proof. We know that ~ is defined as the distribution of the sum Zt + Z~ + ... + Z~, where Zl, Z2,"" Zp are independent N(O,l) random variables. Next, by the spectral decomposition [see Equations (216) and (221) with A = I, and see
(a) (X  p,)':II(X  p,) is distributed as X~, where ~ denotes the chisquare distribution with p degrees of freedom.
... ...
Result 4.7. Let X be distributed as Np(IL, I) with II 1 > O. Then
We conclude this section by presenting two final properties of multivariate normal random vectors. One has to do with the probability content of the ellipsoids of constant density. The other discusses the distribution of another form of linear combinations. The chisquare distribution determines the variability of the sample variance S2 = SJ1 for samples from a univariate normal population. It also plays a basic role in the multivariate case.
3. The conditional covariance, I11  II2I2"~I2 1> does not depend upon the value(s) of the conditioning variable(s).
.... 12.... 22 
Thus, with our customary notation, the conditional distribution of Xl given that X = x is N(ILl + (U12/Un) (X2  IL2)' uu(l PI2»' Now, III  I 12I21I 21 = U:l  !rz/U22 = uu(1  PI2) and I12I2"! = Ulz/U22, agreeing with Result 4.6, which we obtained by an indirect method. 
Zr, for instance. Now, we can write Z = A(X 
+ ... +
The squared statistical distance is calculated as if, first, the random vector X were transformed to p independent standard normal random variables and then the usual squared distance, the sum of the squares of the variables, were applied. Next, consider the linear combination of vector random variables and X  /L is distributed as Np(O, I). Therefore, by Result 4.3, Z = A(X  /L) is distributed as Np(O, AIA'), where
ClX l + C2X2 + .,. + cnXn = [Xl
i X 2 i ... i (pXn)
This linear combination differs from the linear combinations considered earlier in that it defines a p. x 1 vector random variable that is a linear combination of vectors. Previously, we discussed a single random variable that could be written as a linear combination of other univariate random variables.
Result 4.8. Let Xl, X 2, ... , Xn be mutually independent with Xj distributed as Np(/Lj, I). (Note that each Xj has the same covariance matrix I.) Then
Remark: (Interpretation of statistical distance) Result 4.7 provides an interpretation of a squared statistical distance. When X is distributed as Np(/L, I),
CY)I). Moreover, Vl and V2 = blX 1 + b 2 X 2
+ .. , + bnXn are jointly multivariate normal with covariance matrix
By Result 4.5, Zl, Z2, ... , Zp are independent standard normal variables, and we conclude that (X  /L )'Il(X  /L) has a x;,distribution. For Part b, we note that P[ (X  /L ),Il(X  /L) :5 c ] is the probability assigned to the ellipsoid (X  /L)'Il(X  /L):5 c2 by the density Np(/L,I). But from Part a, P[(X  /L),Il(X  /L) :5 x~(a)] = 1  a, and Part b holds. •
= ClX l + C2X2 + ... + cnXn
. (b'c)I ]
O.
Proof. By Result 4.5(c), the np component vector
(X  /L)'Il(X  /L) is the squared statistical distance from X to the population mean vector /L. If one component has a much larger variance than another, it will contribute less to the squared distance. Moreover, two highly correlated random variables will contribute less than two variables that are nearly uncorrelated. Essentially, the use of the inverse of the covariance matrix, (1) standardizes all of the variables and (2) eliminates the effects of correlation. From the proof of Result 4.7, eX  /L),Il(X  /L) = Z1
+ Z~ + .. ' + Z~
is multivariate normal. In particular,
... I
which is itself a random vector. Here each term in the sum is a constant times a random vector. Now consider two linear combinations of random vectors X
and AX is normal N2p (AIL, Al:,A') by Result 4.3. Straightforward block multiplication shows that Al:.A' has the first block diagonal term
Find the mean vector and covariance matrix for each linear combination of vectors and also the covariance between them. By Result 4.8 with Cl = C2 = C3 = C4 = 1/2, the first linear combination has mean vector
The offdiagonal term is [CIl:, c2l:, ... , cnIJ [b l I, b2I, ... , bnIJ' =
2:. cjbj =
0 ,VI and V2 are independent by Result 4.5(b). •
For the second linear combination of random vectors, we apply Result 4.8 with bl = bz = b3 = 1 and b4 = 3 to get mean vector
. For sums of the type in (410), the property of zero correlation is equivalent to requiring the coefficient vectors band c to be perpendicular. Example 4.8 (Linear combinations of random vectors) Let XI. X 2 , X 3 , and X 4 be independent and identically distributed 3 X 1 random vectors with
We first consider a linear combination a'XI of the three components of Xl. This is a random variable with mean
Finally, the covariance matrix for the two linear combinations of random vectors is
a'l: a = 3af + a~ + 2aj  2ala2 + 2ala3 That is, a linear combination a'X I of the components of a random vector is a single random variable consisting of a sum of terms that are each a constant times a variable. This is very different from a linear combination of random vectors, say, CIX I
Every Component of the first linear combination of random vectors has zero covariance with every component of the second linear combination of random vectors. If, in addition, each X has a trivariate normal distribution, then the two linear combinations have a joint sixvariate normal distribution, and the two linear combinations of vectors are independent. _
Sampling from a Muitivariate Normal Distribution and Maximum Likelihood Estimation
4.3 Sampling from a Multivariate Normal Distribution and Maximum likelihood Estimation We discussed sampling and selecting random samples briefly in Chapter 3. In this section, we shallbe concerned with samples from ~multivariate normal populationin particular, with the sampling distribution of X and S.
The Multivariate Normal likelihood Let us assume that the p X 1 vectors Xl, X 2, .. ·, Xn represent a random sample from a multivariate normal population with mean vector p. and covariance matrix l:. Since Xl, X 2 , ..• , Xn are mutually independent and each has distribution Np(p., l:), the joint density function of all the observations is the product of the marginal normal densities: Joint density } = { ofX 1,X 2"",X n
.
since the trace of a sum of matrices is equal to the sum of the traces of the matrices, according to Result 2A.12(b). We can add and subtract i = {l/n) term
p. )(Xj  p.)' to give
= tr(BC).
Result 4.9. Let A be a k x k symmetric matrix and x be a k X 1 vector. Then
When the numerical values of the observations become available, they may be substituted for the x . in Equation (411). The resulting expression, now considered as a function of p. and l: Jfor the fixed set of observations Xl, X2, ... , Xn, is called the likelihood. Many good statistical procedures employ values for the popUlation parameters that "best" explain the observed data. One meaning of best is to select the parameter values that maximize the joint density evaluated at the observations. This technique is called maximum likelihood estimation, and the maximizing parameter values are called maximum likelihood estimates. At this point, we shall consider maximum likelihood estimation of the parameters p. and l: for a muItivariate normal population. To do so, we take the observations Xl'X2'''',Xn as fixed and consider the joint density of Equation (411) evaluated at these values. The result is the likelihood function. In order to simplify matters we rewrite the likelihood function in another form. We shaH need some additionai properties for the trace of a square matrix. (The trace .of a mat~ix is t~e .s~m of its diagonal elements, and the properties of the trace are discussed m DefmlUon 2A.28 and Result 2A.12.)
p. )(Xj
are both matrices of zeros. (See Exercise 4.15.) Consequently, using Equations (413) and (414), we can write the joint density of a random sample from a multivariate normal population as
Now the exponent in the joint density in (411) can be simplified. By Result 4.9(a),
Let x' be the matrix B with rn = 1, and let Ax play the role of the matrix C. Then tr(x'(Ax» = tr«Ax)x'),and the result follows. Part b is proved by using the spectral decomposition of (220) to write A = P' AP, where pp' = I and A is a diagonal matrix with entries AI, A , ••• , A • 2 k • Therefore, tr(A) = tr(P'AP) = tr(APP') = tr(A) = Al + A2 + ... + A •
2.: Ai, where the Ai are the eigenvalues of A. i=1
Proof. For Part a, we note thatx'Ax is a scalar,sox'Ax = tr(x'Ax). We pointed out in Result 2A.12 that tr(BC) = tr(CB) for any two matrices Band C of
dimensions. m X k and k X rn, respectively. This follows because BC has
Sampling from a Multivariate Normal Distribution and Maximum Likelihood Estimation
Substituting the observed values Xl, X2, ... , Xit into the joint density yields the likelihood function. We shall denote this function by L(iL, l:), to stress the fact that it is a function of the (unknown) population parameters iL and l:. Thus, when the vectors Xj contain the specific numbers actually observed, we have
It will be convenient in later sections of this book to express the exponent in the likelihood function (416) in different ways. In particular, we shall make use of the identity
But the function 17berJ/2 has a maximum, with respect to 17, of (2b )beb, occurrjng at 17 = 2b. The choice 17; = 2b, for each i, therefore gives
The upper bound is uniquely attained when l: = (1/2b )B, since, for this choice, B1/2l:1B 1/2 = Bl/2(2b )B1B 1/2 = (2b) I
The next result will eventually allow us to obtain the maximum likelihood estimators of p. and l:.
Proof. Let Bl/2 be the symmetric square root of B [see Equation (222)], Bl/2Bl/2 = I,
and Bl/2Bl /2 = B1. Then tr(l:IB) = tr [(l:1 Bl/2)Bl/2] = tr [Bl/2(l:IBl/2)]. Let 17 be an eigenvalue of B l/2l:1Bl/2. This matrix is positive definite because y'Bl/2l:1BI/2y = (B1/ 2y)'l:I(B l /2y) > 0 if BI/2y 0 or, equivalently, y O. Thus, the eigenvaiues 17; of Bl/2l: I B 1/ 2 are positive
Result 4.1 I. Let X I, X 2, ... , Xn be a random sample from a normal population with mean p. and covariance l:. Then
values, x and (l/n) 2: (Xj  x) (Xj  x)', are called the maximum likelihood esti
p • IT 17; by Exercise 2.12. From the properties of determinants ;=1
The maximum likelihood estimates of p. and l: are those valuesdenoted by ji, and ithat maximize the function l:) in (416). The estimates ji, and i will depend on the observed values XI, X2, ... , Xn through the summary statistics i and S.
by Exercise 2.17. Result 4.9(b) then gives
Straightforward substitution for tr[l:IB 1and 1/1l: Ib yields the bound asserted.
for all positive definitel: , with equality holding only for l: = (1/2b )B.
Result 4.10. Given a p X P symmetric positive definite matrix B and a scalar b > 0, it follows that
mates of p. and l:.
Proof. The exponent in the likelihood function [see Equation (416)], apart from the multiplicative factor is [see (417)]
By Result 4.1, :t l is positive definite, so the distance (x  /L )':tl(x  /L} > 0 unless /L = X. Thus, the likelihood is maximized with respect to /L at jl = X. It remains to maximize
Sufficient Statistics From expression (415), the joint density depends on the whole set of observations XI, x2, ..., xn only through the sample mean x and the sumofsquaresandcrossn
over :to By Result 4.10 with b = nl2 and B = L(Xj : x)(Xj  x)', the maximum j=l
x)(Xj  x)' = (n  l)S. We express this fact by saying j=l that x and (n  l)S (or S) are sufficient statistics: products matrix
x)(Xj  x)', as stated. j=l The maximum likelihood estimators are random quantities. They are optained by replacing the observations Xl, X2, ... , Xn in the expressions for jl and :t with the corresponding random vectors, Xl> X 2,···, X n • • We note that the maximum likelihood estimator X is a random vector and the maximum likelihood estimator i is a random matrix. The maximum likelihood estimates are their particular values for the given data set. In addition, the maximum of the likelihood is L( ~ /L,
Let Xl, X 2, ... , Xn be a random sample from a multivariate normal population with mean JL and covariance:t. Then
The importance of sufficient statistics for normal populations is that all of the information about /L and :t in the data matrix X is contained in x and S, regardless of the sample size n. This generally is not true for nonnormal populations. Since many multivariate techniques begin with sample means and covariances, it is prudent to check on the adequacy of the multivariate normal assumption. (See Section 4.6.) If the data cannot be regarded as multivariate normal, techniques that depend solely on x and S may be ignoring other useful sample information.
The generalized variance determines the "peakedness" of the likelihood function and, consequently, is a natural measure of variability when the parent population is multivariate normal. ~ Maximum likelihood estimators possess an invariance property. Let 8 be the maximum likelihood estimator of 8, and consider estimating the parameter h(8), which is a function of 8. Then the maximum likelihood estimate of h(8)
4.4 The Sampling Distribution of X and S The tentative assumption that Xl> X 2, ... , Xn constitute a random sample from a normal population with mean /L and covariance :t completely determines the sampling distributions of X and S. Here we present the results on the sampling distributions of X and S by drawing a parallel with the familiar univariate conclusions. In the univariate case (p = 1), we know that X is normal with mean /L = (population mean) and variance 1 n
1. The maximum likelihood estimator of /L':tl/L isjl'iljl, where jl = X and
i = l)ln)S are the maximum likelihood estimators of /L and :t, respectively. 2. The maximum likelihood estimator of ~ is ~, where ~ 1 ~  2 l7ii = n .£J (Xij  Xi) j=l
The result for the multivariate case (p ~ 2) is analogous in that X has a normal distribution with mean /L and covariance matrix (lln ):t. For the sample variance, recall that (n  1 )s2 =
~ times a chisquare variable having n  1 degreesJ~f freedom (dJ.). In turn, this chisquare is the distribution of a sum of squares of independent standard normal random variables. That is, (n  1)s2 is distributed as 172( Z1 + ... + Z~l) = (17 Zl)2 + ... + (I7Zn lf The individual terms 17Zi are independently distributed as N(O, ~). It is this latter form that is suitably generalized to the basic sampling distribution for the sample covariance matrix.
variance matr ix is calle d the Wish an . 'b' samp1e cO The sam plin g dlstn utiOn 0f .the . f' d s the sum of inde pend ent prod ucts . of distribution, afte r ItS d'ISCoverer, Itt IS de me a s Specifically, mul tiva riate norm al rand om vec or . . . 'f (422) · hart distributIOn with m d .. W (. \ '1) == WIS . m In
Sup pose the quan tity X is dete rmin ed by a larg e num ber of inde pend ent causes VI, V2 ,.· . , Vn , whe re the rand om vari able s V; repr esen ting the caus es have appr oximate ly the sam e variability. If X is the sum
X= ltJ. +V2 +" ·+v "
. dentl whe re the Z j are each mde P n d' y distributed as Np( 0, '1). We sum mar ize the samp ng IS tribution results as follows:
le of size n from a pva riate norm X al samp . Let X I, 2, ... , X n be adrandom rianc e matrJ X t. The n distr ibut ion with mea n po an cova 1. X is distr ibut ed as Np (p.,{l/ n ).'l). random matrix with n  1 d.f. 2. (n  l)S is distributed as a WIsh art
X and S are independent.
. 'b' . dire ctly to mak e the dlstn utlOn of X cannot be used Bec ause '1 IS unk now n,· . 'd' dependent informatiOn d abou t ~ S ~, an th e provI es III infe renc es abo ut iJ. However, . . f Tb' allow s us to cons truc t a stati stic or on p.. IS distr ibut ion of S d oes no t depend e .' mak ing infe renc es abou t p., as w shall see in Chapter 5. e further results from multlvanabl~ . . ' dlstn~utiOn For the pres ent, we record. som the Wishart distribution are derI ved direc tly theo ry. The following propertieS ?fde endent products, ZjZ j. Proo fs can be foun d from its defi nitio n as a sum of the III P in [1].
Pro erties of the Wishart Distribution . . . p .' t independently of A 2, which IS dlstnbu~ If Al is distr Ibut ed as W",,(AI I .). ed as d W (A + A2 \ '1). Tha t IS, the 1. \ A + W"'2(A 2 '1), then
",,+1>12 I 1 (424 ) degr ees of free dom add. \ ) h CAC ' is distr ibute d as Wm (CA C' \ C'lC ') . . d' 'b t d sW (A t ,t en 2. If A IS IStn u e a m arlicular need for the probabilit~ density Alth oug h we do not have ~ny ~ be of som e inte rest to see ItS rath er unct ion of the Wis hart distributIOn, It f tmax~lst unless the sample size n is grea ter does no e com plic ated form . The . densl.ty Whe . fi . n it does exist, its value at the posi tive than the num ber of van abies p. de mte mat rix A is
then the cent ral limi t theo rem appl ies, and we conc lude that X has a distr ibut ion that is near ly non nal. This is true for virtually any pare nt distr ibut ion of the V;'s, provided that n is larg e enou gh. The univ aria te cent ral limi t theo rem also tells us that the sam plin g distr ibut ion of the sam ple mea n, X for a larg e sam ple size is near ly non nal, wha teve r the form of the unde rlyin g popu latio n distr ibut ion. A simi lar resu lt hold s for man y othe r imp orta nt univ aria te statistics . It turn s out that cert ain muI tivar iate statistics, like X and S, have larg esa mpl e prop ertie s anal ogou s to thei r univ aria te coun terp arts. As the sam ple size is increa sed with out boun d, cert ain regu larit ies gove rn the sam plin g vari atio n in X and S, irres pect ive of the form of the pare nt popu latio n. The refo re, the conc lusio ns present ed in this sect ion do not requ ire mul tivar iate norm al popu latio ns. The only requ irem ents are that the pare nt popu latio n, wha teve r its form , have a mea n p. and a finite cova rian ce :to
Res ult 4.12 (Law of larg e num bers ). Let YI , 12, ... ,1';, be inde pend ent obse rvation s from a popU latio n with mea n E(Y;) = /L. The n }j z +" ·+ 1';, Y = ~+Y =". n conv erge s in prob abil ity to /L as n incr ease s with out boun d. Tha t is, for any pres crib ed accu racy e > 0, P[ e < Y  /L < e) appr oach es unit y as n + 00.
Proof. See [9). As a dire ct cons eque nce of the law of larg e num bers , which says that each conv erge s in prob abil ity to JLi, i = 1,2, ... , p,
X conv erge s in prob abil ity to po Also, each sam ple covariance Sik conv erges in probability to (Fib i, k
S (or i = Sn) conv erge s in prob abil ity to:t Stat eme nt (427) follows from writ ing j=1
~.
is the gamma function. (See [11 and [11].)
= 1,2, ... , p, and
3.
Letting Yj = (Xii  J.Li)(Xik  J.Lk), with E(Yj) = (Fib we see that the first term in Sik converges to (Fik and the second term converges to zero, by applying the law of large numbers. The practical interpretation of statements (426) and (427) is that, with high probability, X will be close to I' an~ S will be close to I whene.ver the sampl~ si~e is large. The statemellt concerning X is made even more precIse by a multtvanate version of the central limit theorem. Result 4.13 (The central limit theorem). Let X I, X 2 , ... , Xn be independent observations from any population with mean I' and finite covariance I. Then
Vii eX  1') has an approximate NP(O,I) distribution for large sample sizes. Here n should also be large relative to p.
Proof. See [1].
The approximation provided by the central limit theorem applies to discrete, as well as continuous, multivariate populations. Mathematically, the limit is exact, and the approach to normality is often fairly rapid. Moreover, from the results in Section 4.4, we know that X is exactly normally distributed when the underlying population is normal. Thus, we would expect the central limit theorem approximation to be quite good for moderate n when the parent population is nearly normal. As we have seen, when n is large, S is close to I with high probability. Consequently, replacing I by S in the approximating normal distribution for X will have a 2 . • . negligible effect on subsequent probabili~ caIcul~tions.:... Result 4.7 can be used to show that n(X  1') r l (X  1') has a Xp dlstnbutlOn when X is distributed as
Np(O, I) distribution. The X~ distribution is .approximately the sampling distribution of n(X  1')' II (X  1') when X is approximately normally distributed. Replacing II by SI does not seriously affect this approximation for n large and much greater than p. We summarize the major conclusions of this section as follows: Let XI, X 2 , ... , Xn be independent observations from a population with mean JL and finite (nonsingular) covariance I. Then
for n  p large. In the next three sections, we consider ways of verifying the assumption of normality and methods for transforming nonnormal observations into observations that are approximately normal.
4.6 Assessing the Assumption of Normality As we have pointed out, most of the statistical techniques discussed in subsequent chapters assume that each vector observation Xi comes from a multivariate normal distribution. On the other hand, in situations where the sample size is large and the techniques depend solely on the behavior of X, or distances involving X of the form n(X  I' )'SI(X  1'), the assumption of normality for the individual observations is less crucial. But to some degree, the quality of inferences made by these methods depends on how closely the true parent population resembles the multivariate normal form. It is imperative, then, that procedures exist for detecting cases where the data exhibit moderate to extreme departures from what is expected under muItivariate normality. We want to answer this question: Do the observations Xi appear to violate the assumption that they came from a normal population? Based on the properties of normal distributions, we know that all linear combinations of normal variables are normal and the contours of the multivariate normal density are ellipsoids. Therefore, we address these questions:
1. Do the marginal distributions of the elements of X appear to be normal? What about a few linear combinations of the components Xi? 2. Do the scatter plots of pairs of observations on different characteristics give the elliptical appearance expected from normal populations? 3. Are there any "wild" observations that should be checked for accuracy? It will become clear that our investigations of normality will concentrate on the behavior of the observations in one or two dimensions (for example, marginal distributions and scatter plots). As might be expected, it has proved difficult to construct a "good" overall test of joint normality in more than two dimensions because of the large number of things that can go wrong. To some extent, we must pay a price for concentrating on univariate and bivariate examinations of normality: We can never be sure that we have not missed some feature that is revealed only in higher dimensions. (It is possible, for example, to construct a nonnormal bivariate distribution with normal marginals. [See Exercise 4.8.]) Yet many types of nonnormality are often reflected in the marginal distributions and scatter plots" Moreover, for most practical work, onedimensional and twodimensional investigations are ordinarily sufficient. Fortunately, pathological data sets that are normal in lower dimensional representations, but nonnormal in higher dimensions, are not frequently encountered in practice.
Evaluating the Normality of the Univariate Marginal Distributions Dot diagrams for smaller n and histograms for n > 25 or so help reveal situations where one tail of a univariate distribution is much longer than the other. If the histogram for a variable Xi appears reasonably symmetric, we can check further by counting the number of observations in certain intervals. A univariate normal distribution assigns probability .683 to the interval (J.Li  YU;";, J.Li + YU;";) and probability .954 to the interval (J.Li  2YU;";, J.Li + 2yu;";). Consequently, with a large sample size n, we expect the observed proportion Pi 1 of the observations lying in the
Vs;";) to be about .683. Similarly, the observed proportion 2Vs;";, Xi + 2~) should be about .954. Using the normal approximation to the sampling distribution of Pi (see [9]), we observe that either (.683)(.317) 1.396 I Pi!  .683 I > 3 n Vii
would indicate departures from an assumed normal distribution for the ith characteristic. When the observed proportions are too small, parent distributions with thicker tails than the normal are suggested. Plots are always useful devices in any data analysis. Special plots caIled QQ plots can be used to assess the assumption of normality. These plots can be made for the marginal distributions of the sample observations on each variable. They are, in effect, plots of the sample quantile versus the quantile one would expect to observe if the observations actually were normally distributed. When the points lie very nearly along a straight line, the normality assumption remains tenable. Normality is suspect if the points deviate from a straight line. Moreover, the pattern of the deviations can provide clues about the nature of the nonnormality. Once the reasons for the nonnormality are identified, corrective action is often possible. (See Section 4.8.) To simplify notation, let Xl, Xz, ... , XII represent n observations on any single characteristic Xi' Let x(1) ~ x(z) ~ .. , ~ x(n) represent these observations after they are ordered according to magnitude. For example, x(z) is the second smallest observation and x(n) is the largest observation. The x(j)'s are the sample quantiles. When the x(j) are distinct, exactly j observati~ns are less than or ~qual to xU). (~is is theoretically always true when the observahons are of the contmuous type, which we usually assume.) The proportion j I n of the sample at or to the left of xU) is often approximated by (j  !)In for analytical convenience.' For a standard normal distribution, the quantiles %) are defined by the relation P[ Z ~ q(j)]
1 v17iez2/2dz = .65. [See (430).]
Let us now construct the QQ plot and comment on its appearance. The QQ plot for th.e forego.ing data,.whi.ch is a plot of the ordered data xu) against the normal quanbles qV)' IS ~hown m Figure 4.5. The pairs of points (%), x(j» lie very nearly along a straight lme, and we would not reject the notion that these data are normally distributedparticularly with a sample size as small as n = 10. x{j)
(See Table 1 in the appendix). Here PU) is the probability of getting a value less than or equal to q( ') in a single drawing from a standard normal population. The idea is to look at the pairs of quantiles (qU), xU» with the same associated cumulative probability (j  Din. If the data arise from a normal populati~n, the pairs (%), x(j) will be approximately linearly related, since U%) + IL is nearly the expected sample quantile. 2 lThe! in the numerator of (j 
Din is a "continuity" correction. Some authors (see [5) and [10))
have suggested replacing (j  !)In by (j  n/( n + ~). 2 A better procedure is to plot (mU)' x(j))' where m(j) = E(z(j)) is the expected value of the jthorder statistic in a sample of size n from a standard normal distribution. (See [13) for further discussion.)
Figure 4.S A QQ plot for the data in Example 4.9.
The calculations required fo'r QQ plots are easily programmed for electronic computers. Many statistical programs available commercially are capable of producing such plots. , The steps leading to a QQ plot are as follows:
1. Order the original observations to get x(1),
x(2), . .. , x(n)
probability values (1 1)ln, (2 1)ln, ... , (n 1)ln; 2. Calculate the standard normal quantiles q(l), q(2)"'" q(n); and 3. ~lot th~pair.s of observations (q(l), X(I»' (q(2), X(2», .•• , (q(n), x(n», and examme the straightness" of the outcome.
Q_Q plots are not particularly informative unless the sample size is.mode rate to largef or instance , n ;::: 20. There can be quite a bit of variabili ty in the straightn ess of the Q_Q plot for small samples, even when the observat ions are known to come from a normal populati on.
Example 4.10 (A Q_Q plot for radiation data) The qualitycontrol departm ent of a manufa cturer of microwave ovens is required by the federal governm eI:1t to monitor the amount of radiatio n emitted when the doors of the ovens are closed. Observa tions of the radiatio n emitted through closed doors of n = 42 random ly selected ovens were made. The data are listed in Table 4.1.
. 10
Table 4.1 Radiatio n Data (Door Closed) Oven no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Oven no. 16 17 18 19 20 21
Oven no.
Source: Data courtesy of 1. D. Cryer.
In order to determin e the probability of exceeding a prespeci fied toleranc e level, a probabi lity distribution for the radiation emitted was needed. Can we regard the observa tions here as being normally distributed? A comput er was used to assemble the pairs (q(j)' x(j» and construc t the QQ plot, pictured in Figure 4.6 on page 181. It appears from the plot that the data as a whole are not normally distributed. The points indicated by the circled location s in the figure are outliers values that are too large relative to the rest of the observa tions. For the radiatio n data, several observations are equal. When this occurs, those observa tions with like values are associated with the same normal quantile . This quantile is calculat ed using the average of the quantiles the tied observa tions would have if they all differed slightly.
the radiation data (door closed) from Exampl e 4.10. (The integers in the plot indicate the number of points occupying the same location.)
The straightness of the QQ plot can be . efficient ofthe points in the plot Th I ' measured. by calculatm g the correlati on co. e corre atIOn coefficIe nt for the QQ plot is defined by 11
and a powerfu l test of normali ty can be ba d . we reject the hypothe sis of normali ty at 1~~e~n/t .. (S~ [5], [lO],.and [12].) Formally, appropr iate value in Table 4.2. 0 sIgn lcance a If rQ falls below the
Table 4.~ Critical Points for the QQ Plot CorrelatIOn Coefficient Test for Normali ty Sample size n 5 10 15 ,20 25 30 35 40 45 50 55 60 75 100 150 200 300
Significance levels a .01 .8299 .8801 .9126 .9269 .9410 .9479 .9538 .9599 .9632 .9671 .9695 .9720 .9771 .9822 .9879 .9905 .9935
.05 .8788 .9198 .9389 .9508 .9591 .9652 .9682 .9726 .9749 .9768 .9787 .9801 .9838 .9873 .9913 .9931 .9953
.10 .9032 .9351 .9503 .9604 .9665 .9715 .9740 .9771 .9792 .9809 .9822 .9836 .9866 .9895 .9928 .9942 .9960
Chapter 4 The Multivariate Normal Distribution Example 4.11 (A correlation coefficient test for normality) Let us calculate the cor
has probability .5. Thus, we should expect rou hi the sa 0 sample observations to lie in the ellipse given b; y me percentage, 50 Yo, of
for normality. Using the information from Example 4.9, we have 10
where I~e have re~lac~d JL by its estimate norma 1ty assumptlOn 1S suspect.
x and l;1 by its estimate Sl. If not
!::~~: 4.~: t (Che~king bivariate ~ormality) Although not a random sample, data compani;s in t~: ~~~~do: ~~~~r~a~lOEns (Xl. = sales, x2 = profits) for the 10 largest r 1S e m xerC1se lA. These data give
x = [155.60J A test of normality at the 10% level of significance is provided by referring rQ = .994 to the entry in Table 4.2 corresponding to n = 10 and a = .10. This entry is .9351. Since 'Q > .9351, we do not reject the hypothesis of normality. • Instead of rQ' some software packages evaluate the original statistic proposed by Shapiro and Wilk [12]. Its correlation form corresponds to replacing %) by a function of the expected value of standard normalorder statistics and their covariances. We prefer rQ because it corresponds directly to the points in the normalscores plOt. For large sample sizes, the two statistics are nearly the same (see [13]), so either can be used to judge lack of fit. Linear combinations of more than one characteristic can be investigated. Many
Frt~mf Table 3 in the appendix, rz(.5) = 1.39. Thus, any observation x'  [x
ejXj where Se1 = A1e 1 in which A1 is the largest eigenvalue of S. Here xj = [xi!' Xj2,···, Xjp] is the jth observation on the p variables Xl' X 2 , •• ·, Xp. The linear combination e~Xj corresponding to the smallest eigenvalue is also frequently singled out for inspection. (See Chapter 8 and [6] for further details.)
the• estimated 50O/C0 con t our. OtherW1se . the observation is outside this is on or inside • ~~~~~::~~e first pa1r of observations in Exercise lA is [Xl> X2]' = (108.28,17.05J. 108.28  155.60J' [ .000253 [ 17.05  14.70  .002930
Evaluating Bivariate Normality We would like to check on the assumption of normality for all distributions of 2,3, ... , p dimensions. However, as we have pointed out, for practical work it is usually sufficient to investigate the univariate and bivariate distributions. We considered univariate marginal distributions earlier. It is now of interest to examine the bivariate case. In Chapter 1, we described scatter plots for pairs of characteristics. If the observations were generated from a multivariate normal distribution, each bivariate distribution would be normal, and the contours of constant density would be ellipses. The scatter plot should conform to this structure by exhibiting an overall pattern
and this point falls outside the 50% t Th ... P . alized distances from x of .30,.62 1~~~ ~~~ 4 ;8re1~~n~nff3 1l11n7e1 omts have generf th d. ' , . , . , . , . , . , and 1.16 respectively Since fo less 1.39, a proportion, 040, of data falls would expect about half ~. f th e observat~o~s w~re normally distributed, we . . . ,o.r ,0 t em to be Wlthm th1S contour. This difference in ~~~~~~~~~~:;~~rO~dmanlY rO~ide evid~nce for rejecting the notion of bivariate also Exa~ple 4.13.)' ur samp e SlZe of 10 1S too small to reach this conclusion. (See
that is nearly elliptical. Moreover, by Result 4.7, the set of bivariate outcomes x such that ing
Y compar~o~r:t:~; ~:!r:~~~:~ ;~~~:~~:~si:i~h~::f~~n~outr anthder sUbjecthivel , u ra roug , procedure.
184 Chapter 4 The Multivariate Normal Distribution Assessing the Assumption of Normality
A somewhat more formal method for judging the joint normality of a data set is based on the squared generalized distances 5
j = 1,2, ... , n 4.5
where XI, Xz, .. ' , l:n are the sample observationl'. The procedure we are about to describe is not limited to the bivariate case; it can be used for all p ~ 2. When the parent population is multivariate normal and both nand n  pare greater than 25 or 30, each of the squared distances di, d~, ... , d~ should behave like a chisquare random variable. [See Result 4.7 and Equations (426) and (427).] Although these distances are not independent or exactly chisquare distributed, it is helpful to plot them as if they were. The resulting plot is called a chisquare plot or gamma plot, because the chisquare distribution is a special case of the more general gamma distribution. (See [6].) To construct the chisquare plot, 1. Order the squared distances in (432) from smallest to largest as d71) :s d7z) :s ... :S d[n). 2. Graph the pairs (qcj(j  Dln),d7j)), where qc,A(j  !)In) is the 100(j  Din quantile of the chisquare distribution with p degrees of freedom.
Figure 4.7 A chisquare plot of the ordered distances in Example 4.13.
u~ g:;rh of the pairs (qc.z( (j  !)/1O), dfj)) is shown in Figure 4.7. The points in
~re reasona?ly straight. Given the small sample size it is difficult to
~eJect blvanate ~ormalIty on the evidence in this graph. If further analysis of the ata were req~lre~, it might be reasonable to transform them to observations ms ort~ ne a rl y blvanate normal. Appropriate transformations are discussed ec
4 . 8.
. ~n addition ~o inspecting univariate plots and scatter plots, we should check multlvanate normalIty by constructing a chisquared or d Z plot. Figure 4.8 contains dZ
Example 4.13 (Constructing.a chi~square plot) Let us construct a c~isquare plot of the generalized distances given I~ Example 4,12, The ordered. dlsta~ces and the corresponding chisquare percentIles for p = 2 and n = 10 are lIsted III the following table:
chisquared distribution. In particular, qc,p( (j  Din) = x~( (n  j + Din). The plot should resemble a straight line thro~gh the origin hav~ng slope 1. A systematic curved pattern suggests lack of normalIty. One or two POlllts far above the line indicate large distances, or outlying observations, that merit further attention.
Quantiles are specified in terms of proportions, whereas percentiles are speci. fied in terms of percentages. The quantiles qc) (j  !)In) . are related to the upper percentiles of a
plots based on two computergenerated samples of 30 fourvariate normal random vectors. As expected, the plots have a straightline pattern, but the top two or three ordered squared distances are quite variable. . The next example contains a real data set comparable to the sImulated data set that produced !he plots in Figure 4.8. Example 4.14 (Evaluating multivariate normality for a fourvariable data set) The data in Table 4.3 were obtained by taking four different measures of stiffness, x 1, x 2" X3 and x 4, of each of n = 30 boards. The first measurement involves sending . . a shock wave down the board, the second measurement IS determined while vibrating the board, and the last tw_o ,m_~asuren:ents are obtained fr~m static tests. The squared distances dj = (Xj  x) S (Xj  x) are also presented In the table. .
Observation no.
Observation no.
1778 .60 2197 5.48 2222 7.62 1533 5.21 1883 1040 1546 2.22 1671 4.99 1874 1.49 2581 12.26 1508 .77 1667 1.93 1898 .46 1741 2.70 1678 .13 1714 1.08
1281 16.85 1176 3.50 1308 3.99 1755 1.36 1646 1.46 2111 9.90 1477 5.06 1516 .80 2037 2.54 1533 4.58 1469 3.40 1834 2.38 1597 3.00 2234 6.28 1284 2.58
Source: Data courtesy ofWilliam Galligan.
The marginal distributions appear quite normal (see Exercise 4.33), with the . possible exception of specimen (~oard) 9. . To further evaluate mu/tivanate normalIty, we constructed the chIsquare plot shown in Figure 4.9. The two specimens with the largest squared distances are clearly removed from the straightline pattern. Together, with the next largest point or two, they make the plot appear curved at the upper end. We will return to a discus• sion of this plot in Example 4.15. We have discussed some rather simple techniques for checking the multivariate j = 1,2, ... , n normality assumption. Specifically, we advocate calculating the [see Equation' (432)] and comparing the results with .i quantiles. For example, pvariate normality is indicated if
1. Roughly half of the dy are less than or equal to qc,p( .50).
.
Figure 4.9 A chisquare plot for the data in Example 4.14.
:,.r~or)~:"O(~'~lfU~'::, (~~l), :~;,:ti:::y,: .:,;1:',,::: line having slope 1 and that passes through the origin.
(See [6] for a more complete exposition of methods for assessing normality.) We close this section by noting that all measures of goodness offit suffer the same serious drawback, When the sample size is small, only the most aberrant behavior will be identified as lack of fit. On the other hand, very large samples invariably produce statistically significant lack of fit. Yet the departure from the specified distribution may be very small and technically unimportant to the inferential conclusions.
4.7 Detecting Outliers and Cleaning Data Most data sets contain one or a few unusual observations that do not seem to belong to the pattern of variability produced by the other observations. With data on a single characteristic, unusual observations are those that are either very large or very small relative to the others. The situation can be more complicated with multivariate data, Before we address the issue of identifying these outliers, we must emphasize that not all outliers are wrong numbers, They may, justifiably, be part of the group and may lead to a better understanding of the phenomena being studied.
188 Chapter 4 The Multivariate Normal Distribution OutIiers are best detected visually whenever this is possible. When the number of observations n is large, dot plots are not feasible. When the number of characteristics p is large, the large number of scatter plots p(p  1)/2 may prevent viewing them all. Even so, we suggest first visually inspecting the data whenever possible. What should we look for? For a single random variable, the problem is one dimensional, and"we look for observations that are far from the others. For instance, the dot diagram
•• • •••• .... . ....... ..... . ..
@ I .. x
reveals a single large observation which is circled. In the bivariate case, the situation is more complicated. Figure 4.10 shows a situation with two unusual observations. The data point circled in the upper right corner of the figure is detached from the pattern, and its second coordinate is large relative to the rest of the X2
.
...
Figure 4.10 Two outliers; one univariate and one bivariate.
measurements, as shown by the vertical dot diagram. The second outIier, also circled, is far from the elliptical pattern of the rest of the points, but, separately, each of its components has a typical value. This outlier cannot be detected by inspecting the marginal dot diagrams. In higher dimensions, there can be outliers that cannot· be detected from the univariate plots or even the bivariate scatter plots. Here a large value of (Xj  X)'Sl(Xj  x) will suggest an unusual observation, even though it cannot be seen visually.
Steps for Detecting Outliers 1. Make a dot plot for each variable. 2. Make a scatter plot for each pair of variables. 3. Calculate the standardized values Zjk = (Xjk  Xk)/YS;;; for j = 1,2, ... , n and each column k = 1,2, ... , p. Examine these standardized values for large or small values. 4. Calculate the generalized squared distances (Xj  X)'SI(Xj  x). Examine these distances for unusually large values. In a chisquare plot, these would be the points farthest from the origin. In step 3, "large" must be interpreted relative to the sample size and number of variables. There are n X p standardized values. When n = 100 and p = 5, there are 500 values. You expect 1 or 2 of these to exceed 3 or be less than 3, even if the data came from a multivariate distribution that is exactly normal. As a guideline, 3.5 might be considered large for moderate sample sizes. In step 4, "large" is measured by an appropriate percentile of the chisquare distribution with p degrees of freedom. If the sample size is n = 100, we would expect 5 observations to have values of that exceed the upper fifth percentile of the chisquare distribution. A more extreme percentile must serve to determine observations that do not fit the pattern of the remaining data . The data we presented in Table 4.3 concerning lumber have already been cleaned up somewhat. Similar data sets from tl!e same study also contained data on Xs = tensile strength. Nine observation vectors, out of the total of 112, are given as rows in the following table, along with their standardized values.
The standardized values are based on the sample mean and variance, calculated from al1112 observations. There are two extreme standardized values. Both are too large with standardized values over 4.5. During their investigation, the research ers recorded measurements by hand in a logbook and then performed calculations that produce d the values given in the table. When they checked their records regardin g the values pinpointed by this analysis, errors were discovered. The value X5 = 2791 was correcte d to 1241, andx4 = 2746 was corrected to 1670. Incorrect readings on an individu al variable are quickly detected by locating a large leading digit for the standard ized value. The next example returns to the data on lumber discussed in Exampl e 4.14.
Example 4.15 (Detecting outliers in the data on lumber) Table 4.4 contains the data in Table 4.3, along with the standardized observations. These data consist of four different measures of stiffness Xl, X2, X3, and X4, on each of n = 30 boards. ReCall that the first measurement involves sending a shock wave down the board, the second measurement is determined while vibrating the board, and the last two measure ments are obtained from static tests. The standardized measurements are
I I I I _r'. J.....L ....l
1889 2403 2119 1645 1976 1712 1943 2104 2983 1745 1710 2046 1840 1867 1859 1954 1325 1419 1828 1725 2276 1899 1633 2061 1856 1727 2168 1655 2326 1490
1651 2048 1700 1627 1916 1712 1685 1820 2794 1600 1591 1907 1841 1685 1649 2149 1170 1371 1634 1594 2189 1614 1513 1867 1493 1412 1896 1675 2301 1382
1561 2087 1815 1110 1614 1439 1271 1717 2412 1384 1518 1627 1595 1493 1389 1180 1002 1252 1602 1313 1547 1422 1290 1646 1356 1238 1701 1414 2065 1214
1778 2197 2222 1533 1883 1546 1671 1874 2581 1508 1667 1898 1741 1678 1714 1281 1176 1308 1755 1646 2111 1477 1516 2037 1533 1469 1834 1597 2234 1284
Observation no. 1 2 3 4 5 6 7 8 9
.2 .8 .7 3.0 .4 .0 .4 .3 .1 .4 1.1 1.7 .8 .3 .6 .1 .3 .7 .5 .5 .9 .6 .3 1.8 1.0
..1 1 1 J.L.L.. \ I 1 .L.l... I I 1200 r.lL.... I I
Figure 4.11 Scatter plots for the lumber stiffness data with specimens 9 and 16 plotted k = 1,2,3,4 ;
as solid dots.
j = 1,2, ... ,30
and the squares of the distance s are d?J = (x·J  )'Sl( X x·  x) . Theast I column in Table reveals th . J . . SIDce x~(.OO5) = 14.86' yet all4.4 of th . d' .;t speCImen 16 IS. a. multIva nate outlier, respecti ve univaria te s~atters Spe . e ID 9IVI I uaIhmea suremen ts are Th . . clmen a so as a large d 2 value well within their e two speclffiens (9 and 16) with lar . . differen t from the rest of the I t ' . g~ squared distance s stand out as clearly removed , the remainin g patter: a er; ID Igure 4.9. Once these two points are Scatter plots for the lumber stiffn con orms to the. expected straightline relation e~s measure ments are given in Figure 4.11 above..
The solid dots in these figures correspond to specimens 9 and 16. Although the dot for specimen 16 stands out in all the plots, the dot for specimen 9 is "hidden" in the scatter plot of X3 versus X4 and nearly hidden in that of Xl versus ~3. However, s~ecimen 9 is clearly identified as a multivariate outlier when all four vanables are considered. Scientists specializing in the properties of wood conjectured that specimen 9 was unusually cH~ar and therefore very stiff and strong. It would also appear that specimen 16 is a bit unusual, since both of its dynamic measurements are above average and the two static measurements are low. Unf?rtunately, it was not possible to investigate this specimen further because the matenal was no longer available. • If outliers are identified, they should be examIned for content, as was done in the case of the data on lumber stiffness in Example 4.15. Depending upon the nature of the outliers and the objectives of the investigation, outIiers may be deleted or appropriately "weighted" in a subsequent analysis. Even though many statistical techniques assume normal populations, those based on the sample mean vectors usually will not be disturbed by a few moderate outliers. Hawkins [7] gives an extensive treatment of the subject of outliers.
In ma~y ~nstances, ~he choice of a transformation to improve the approximation to normaht~ IS not obvIOus. For such cases, it is ~onvenient to let the data suggest a transformatIOn. A useful family of transformations for this purpose is the family of power transformations. Power transformations are defined only for positive variables. However, this is not as restrictive as it seems, because a single constant can be added to each observation in the data set ifsome of the values are negative. . . Let X represent an arbitrary observation. The power family of transformations IS mdexed by a parameter A. A given value for A implies a particular transformation. For example, consider XA with A = 1. Since XI = l/x, this choice of A corresponds to the recip~ocal transformation. We can trace the family of transformations as A ranges from negative to positive powers of x. For A = 0, we define XO = In x. A sequence of possible transformations is ... ,X
4.8 Transformations to Near Normality If normality is not a viable assumption, what is the next step? One alternative is to
ignore the findings of. a ~ormality check and p:ocee~ as if t~e data w~re normally distributed. This practice IS not recommended, smce, m many mstances, It could lead to incorrect conclusions. A second alternative is to make nonnormal data more "normal looking" by considering transformations of the data. Normaltheory analyses can then be carried out with the suitably transformed data. 1Tansformations are nothing more than a reexpression of the data in different units. For example, when a histogram of positive observations exhibits a long righthand tail, transforming the observations by taking their logarithms or square roots will often markedly improve the symmetry about the mean and the approximation to a normal distribution. It frequently happens that the new units provide more natural expressions of the characteristics being studied. Appropriate transformations are suggested by (1) theoretical considerations or (2) the data themselves (or both). It has been shown theoretically that data that are counts can often be made more normal by taking their square roots. Similarly, the logit transformation applied to proportions and Fisher's ztransformation applied to correlation coefficients yield quantities that are approximately normally distributed.
1. Counts,y 2. Proportions, jJ 3. Correlations, r
To select a power transformation, an investigator looks at the marginal oot diagram or histogram and decides whether large values have to be "pulled in" or "pushed out" to improve the symmetry about the mean. Trialanderror calculations . ~ith a fe~ of the foregoing transformations should produce an improvement. The fmal chOIce should always be examined by a QQ plot or other checks to see whether the tentative normal assumption is satisfactory. The transformations we have been discussing are data based in the sense that it is ?nly the appear~nce of the data themselves that influences the choice of an appropnate trans~ormatlOn. There are no external considerations involved, although the tr~nsformatlOn actually used is often determined by some mix of information supphed by the d~ta and extradata factors, such as simplicity or ease of interpretation. A convement analytical method is available for choosing a power transformation. We begin by focusing our attention on the univariate case. Box and Cox (3) consider the slightly modified family of power transformations X(A) =
which is continuous in A for x > O. (See [8].) Given the observations Xl, X2, .. . , X n , the BoxCox solution for the choice of an appropriate power A is the solution that maximizes the expression
...
=.!. n
±xy) = .!. ±(xt  1)
194 Chapter 4 The Multivariate Normal Distribution Transformations to Near Normality
is the arithmetic average of the transformed observations. The first term in (435) is, apart from a constant, the logarithm of a normal likelihood function, after maximizing it with respect to the population mean and variance parameters. The calculation of e( A) for many values of Ais an easy task for a computer. It is helpful to have a graph of eCA) versus A, as. well as a tabular displflY of the pairs (A, e(A)), in orderto study the be~avior near the maxim~zing value A. For instance, if either A = 0 (logarithm) or A = 2 (square root) is near A, one of these may be preferred because of its simplicity. Rather than program the calculation of (435), some statisticians recommend the equivalent procedure of fixing A, creating the new variable j
= 1, ... , n
and then calculating the sample variance. The minimum of the variance occurs at the same Athat maximizes (435). Comment. It is now understood that the transformation obtained by maximizing e(A) usually improves the approximation to normality. However, there is no guarantee that even the best choice of A will produce a transformed set of values that adequately conform to a normal distribution. The outcomes produced by a transformation selected according to (435) should always be carefully examined for possible violations of the tentative assumption of normality. This warning applies with equal force to transformations selected by any other technique. Example 4.16 (Determining a power transformation for univariate data) We gave readings of the microwave radiation emitted through the closed doors of n = 42 ovens in Example 4.10. The QQ plot of these data in Figure 4.6 indicates that the observations deviate from what would be expected if they were normally distributed. Since all the observations are positive, let us perform a power transformation of the data which, we hope, will produce results that are more nearly normal. Restricting our attention to the family of transformations in (434), we must find that value of A maximizing the function e(A) in (435). The pairs (A, e(A» are listed in the following table for several values of A:
70.52 75.65 80.46 84.94 89.06 92.79 96.10 98.97 101.39 103.35 104.83 105.84 106.39 106.51)
Figure 4.12 Plot of C(A) versus A for radiation data (door closed).
Th~ cFiurve of e(A) versus A that allows the more exact determination sh own In Igure 4.12.
28 is .
~t ~s eVi~e(nt) from both th~ table and the plot !hat a value of Aaround .30 maXImIzes A. For convemence, we choose A = 25 The d t reexpressed as .. a a Xj were (1/4) x}l4  1 Xi = :1:
j = 1,2, ... ,42
~n;.a Q~ ot was constructed from the transformed quantities. This plot is shown Igure. on page 196. The quantile pairs fall very close to a straight line and we . ' would conclude from this evidence that the x(I/4) j are approxImately normal. In
Transforming Multivariate Observations ith Wh m~lbtII'variate observations, a power transformation must be selected for each of t e vana es. Let A A b h e power transformations for the measured . . 1, 2,···, Apet charactenstIcs. Each Ak can be selected by maximizing P ek(A) =
196 Chapter 4 The Multivariate Normal Distribution The procedure just described is equivalent to making each marginal distribution approximately normal. Although normal marginals are not sufficient to ensure that the joint distribution is normal, in practical applications this may be good enough. If not, we could start with the values AI, A2 , ... , Ap obtained from the preceding transformations and iterate toward the set of values A' = (A'I, A2, ... , Ap], which collectively maximizes
+ ... + (A p  1) '" In X·JP £.J.
.
j = 1,2, ... , n
re 4 13 A QQ plot of the transformed radiat~on data (d?or closed).
flgu.. . the plot indicate the number of pomts occupymg the same (The mtegers III location.)
are the n observations on the kth variable, k = 1, 2, ... , p. Xlk> X2b""
. . e of the transformed observations. The jth transformed mulis the anthmetlc averag tivariate observation is
; are the values that individually maximize (438).
where AI, "2,' .. , "p
Maximizing (440) not only is substantially more difficult than maximizing the individual expressions in (438), but also is unlikely to yield remarkably better results. The selection method based on Equation (440) is equivalent to maximizing a muItivariate likelihood over ft, 1: and A, whereas the method based on (438) corresponds to maximizing the kth univariate likelihood over JLb akk, and Ak' The latter likelihood is generated by pretending there is some Ak for which the observations (x;~  1)/Ak , j = 1, 2, ... , n have a normal distribution. See [3] and [2] for detailed discussions of the univariate and multivariate cases, respectively. (Also, see [8].)
Example 4.17 (Determining power transformations for bivariate data) Radiation measurements were also recorded through the open doors of the n = 42 microwave ovens introduced in Example 4.10. The amount of radiation emitted through the open doors of these ovens is listed in Table 4.5. In accordance with the procedure outlined in Example 4.16, a power transformation for these data was selected by maximizing £(A) in (435). The approximate maximizing value was A= .30. Figure 4.14 on page 199 shows QQ plots of the untransformed and transformed dooropen radiation data. (These data were actually
Table 4.S Radiation Data (Door Open) Oven no. 1 2 3 4 5 6 7 8 9
Oven no.
Oven no.
''_..L. _ _...l_ _L _ _1 _ _ ..
Source: Data courtesy of 1. D. Cryer.
transformed by taking the fourth root, as in Example 4.16.) It is clear from the figure that the transformed data are more nearly normal, although the normal approximation is not as good as it was for the doorclosed data. Let us denote the doorclosed data by XII ,X2b"" x42,1 and the dooropen data by X12, X22," . , X42,2' Choosing a power transformation for each set by maximizing the expression in (435) is equivalent to maximizing fk(A) in (438) with k = 1,2. Thus, using th~ outcomes from Example 4.16 and the foregoing results, we have Al = .30 and A2 = .30. These powers were determined for the marginal distributions of Xl and X2' We can consider the joint distribution of Xl and X2 and simultaneously determine the pair of powers (Ab A2) that makes this joint distribution approximately bivariate normal. To do this, we must maximize f(Al' A2) in (440) with respect to both Al and A2· We computed f(AJ, A2) for a grid of Ab A2 values covering 0 :S Al :S .50 and o :S A2 :;; .50, and we constructed the contour pl<2t s~hown in Figure 4.15 on page 200. We see that the maxirilUm occurs at about (AI' A2) = (.16, .16). The "best" power transformations for this bivariate case do not differ substantially from those obtained by considering each marginal distribution. As we saw in Example 4.17, making each marginal distribution approximately normal is roughly equivalent to addressing the bivariate distribution directly and making it approximately normal. It is generally easier to select appropriate transformations for the marginal distributions than for the joint distributions.
Figure 4.14 QQ plots of (a) the original and (b) the transformed radiation data (with door open). (The integers in the plot indicate the number of points occupying the same location.)
(b) Write out the squared generalized distance expression (x  p.)'II(x _ p.) as a function of xI and X2' 222
(c) Determine (and sketch) the. constantdensity contour that contains 50% of the probability.
4.3. Let X be N 3 (p., I) with p.' = [3,1,4) and 0.4
Which of the following random variables are independent? Explain. (a) X 1 and X 2 (b) X 2 and X3 (c) (X1 ,X2 ) and X3 Xl + X 2 (d) 2 and X3
225 9 .
X2 If the data includes some large negative values and have a single .l~ng tail, a more general transformation (see Yeo and Johnson [14]) should be apphe .
2X2 + X .
3 (b) Relabelthe variables if necessary, and find a 2 x 1 vector a such that X and
Figure 4.1 5 Contour plot of C( AI' A2 ) for the radiation data.
are independent.
4.5. Specify each of the following. (a) The conditional distribution of XI> given that X 2 = Exercise 4.2.
(b) The conditional distribution of X 2 , given that XI = xI and X3 tribution in Exercise 4.3.
(c) The conditional distribution of X 3 , given that XI tribution in Exercise 4.4.
4.6. Let X be distributed asN3 (p.,I), wherep.' = [1, 1,2) and
Exercises (1"1 = 2, (1"22 = 1 and ILl = 1,IL2  3, 1 .8. . (a) Write out the bivariate normal density. . ( x  p. )'II(xp.)asaqua(b) Write out the squared statistical distance expresslOn dratic function of XI and X2'
. WI'th 4.1· Consider a bivariate normal distributlOn
· WI'th 4.2. Consider a bivariate normal popu Iabon PI2 = .5. . (a) Write out the bivariate normal density.
0,.2 11.
Which of the following random variables are independent? Explain. (a) XI andX2 (b) X 1 and X3 ' (c) X 2 and X3 (d) (X1' X 3 ) and X 2 (e) XI and XI + 3X2  2X3
Refer to Exercise 4.6 and specify each of the following. (a) The conditional distribution of Xl, g~ven that X 3 = x3' _ (b) The conditional distribution of Xl, gtven that X 2 = X2 and X 3  .X3' onnonna l bivariate distribut ion with normal margmal s.) Let XI be 4.8. (ExampIe 0 f a n N(O, 1), and let~
4.1.
AIel 1 = 1. Now use the result in Part a.
4.1 I. Show that, if A is square,
Show each of the following. (a) X also has an N(O, 1) distribution. .,. 2 (b) XI and X do not have a bivariate normal dlstnbutl On. 2
. is N(O 1) P[1 < XI S x] = P[x S XI < 1 ) for any x. Wh en (a) Smce XI< 1 P[X ~ x) = P[X S 1) + P[l <X S X2] = P[XI S 1) 2 2 1 <xI2<_ X' <x2) =2p [X s1) + P[X2S X
p[lXII> 1] = .3174. . . but modify the construc tion by replacin g the break pomt 1 by Refer to E xerclse 48 ., c so that XI ifc S XI S C X  { 2XI elsewhe re osen so that Cov (XI X 2 ) = 0 but that the two random variables Show that c can be ch " are not independent.
0, evaluate Cov (Xl' X 2) = E[ X:I (XI)] For c very large, evaluate Cov (XI' X 2 ) = E [XI (  XI)]'
4.12. Show that, for A symmetr ic,
Thus, (A\1  A 12 A 2iA 2l )1 is the upper lefthand block of AI. Hint: Premult iply the expressi on in the hint to Exercise 4.11 by I [ 0'
[A~A21 ~ J'. Take inverses of the res~lting expression.
4.13. Show the followin g if X is Np(IL, I) with / I I # O. (a) Check that /I/ = IInllIl 1  I 12 I iI J/. (Note that /I/ 2 2 can be factored into the product of contribu tions from the margina l and conditio nal distribut ions.) (b) Check that
4.10. ShoW each of the following. (a)
Take determin ants on both sides of this equality. Use Exercise 4.10 for the first and third determin ants on the left and for the determin ant on the right. The second equality for / A / follows by consider ing
0 0 \\ I 0 \. Expandi ng the determin ant \ I, 0 \ by the first roW . (a) 0' B  0' I 0' B O B . ee Definition 2A.24) gives 1 times a determin ant of the sam: form,. wIth t~e ?rder (s d db one This procedur e is repeated until 1 X IB lIS obtamed . SlffitlarIy, ofIre uce Y . \A expanding the determin ant \ by the lastrow gives 0' I = IA I·
+ (X2  ILdI2~(X2  IL2) (Thus, the joint density exponen t can be written as the sum of two terms correspo nding to contribu tions from the conditio nal and margina l distribut ions.) (c) Given the results in Parts a and b, identify the margina l distribut ion of X 2 and the conditio nal distribut ion of XI f X 2 = X2'
Chapter 4 The Multivariate Normal Distribution and Hint: (a) Apply Exercise 4.11. _ (b) Note from Exercise 4.12 that we can write (x  IL)'!, I (x  p.) as l XI  P.IJ' 0J [(!,II  !,!2,!,i"!!,2It J [ X2  P.2  !,22!,21 I 0 22
.
4.18. Find the maximum likelihood estimates of the 2 x 1 mean vector p. and the 2 x 2 covariance matrix!' based on the random sample
Xl  X 2 + X3  X 4 + Xs in terms of p. and !'. Also, obtain the covariance between the two linear combinations of random vectors.
the result follows. · d' 'b t d N (11. !,) with I!' I#'O show that the joint density can be written 4.. . . ' 14 If X IS Istn u e as p"' as the product of marginal denslttes for , XI
4.20. For the random variables XI, X 2, ... , X 20 in Exercise 4.19, specify the distribution of B(19S)B' in each case.
0] [XI  P.I] Ii"! X2  P.2 = (XI  p.1)'!,ll(xI  ILl) + (X2  P.2)'!,i"1( X2  P.2)
Note that I!' I = I!,IIII !,221 from Exercise 4.1O(a). Now factor the joint density.
~ ( )( 11.)' and ~ (x  I" )(x·  x)' are both p X P matrices of 4.15. Show that £.J Xj  X X  ,.~} . j=1 } zeros. Here xi = [Xjl, Xj2,"" Xj pl, j = 1,2, ... , n, and 1
4.16. Let Xj, X 2, X 3, and X 4 be independent Np(p., I) random vectors. (a) Find the marginal distributions for each of the random vectors VI = 4I Xl  4IX 2 + 4IX 3  4IX 4 and
(b) Find the joint density of the random vectors VI and V2 defined in (a). 4 17 Le X
from a bivariate normal population.
4.19. Let XI> X 2, ... , X 20 be a random sample of size n = 20 from an N6(P.,!') population. Specify each of the following completely. (a) The distribution of (XI  p.),!,I(X I  p.) (b) The distributions of X and vIl(X  p.) (c) The distribution of (n  1) S
X X X and X 5 be independent and identically distributed random vectors . th I> 2, t3, 4'and cov ariance matrix!' Find the mean vector and covariance maWIt mean vec or p. . .' . trices for each of the two linear combtna tlOns of random vectors IX !X !X I ~XI+5X2+5 3+5 4+55
4.21. Let X I, ... , X 60 be a random sample of size 60 from a fourvariate normal distribution having mean p. and covariance !'. Specify each of the following completely. (a) The distribution ofK: (b) The distribution of (XI  p. )'!,I(XI  p.) (c) Thedistributionofn(X  p.)'!,I(X  p.) (d) The approximate distribution of n(X  p. },SI(X  p.) 4.22. Let XI, X 2, ... , X 75 be a random sample from a population distribution with mean p. and covariance matrix !'. What is the approximate distribution of each of the following? . (a) X (b) n(X  p. ),Sl(X  p.) 4.23. Consider the annual rates of return (including dividends) on the DowJones industrial average for the years 19962005. These data, multiplied by 100, are 0.6 3.1 25.3 16.8 7.1 6.2 25.2 22.6 26.0. , Use these 10 observations to complete the following. (a) Construct a QQ plot. Do the data seem to be normally distributed? Explain. (b) Carry out a test of normality based on the correlation coefficient 'Q. [See (431).] Let the significance level be er = .10.
4.24. Exercise 1.4 contains data on three variables for the world's 10 largest companies as of April 2005. For the sales (XI) and profits (X2) data: (a) Construct QQ plots. Do these data appear to be normally distributed? Explain.
Chapter 4 The Multivariate Normal Distribution t t of normality based on the correlation coefficient rQ. [See (431).] I I at a = 10 Do the results ofthese tests corroborate the re(b) Carry o~t a.f.es Set the slgm Icance eve ., suits in Part a? th world's 10 largest companies in Exercise 1.4. Construct a chi4 25 Refer to the data for e . '1 . . . II three variables. The chisquare quanti es are square plot uslO.g a 0.3518 0.7978 1.2125 1.6416 2.1095 2.6430 3.2831 4.1083 5.3170 7.8147 . h x measured in years as well as the selling price X2, measured 4.26. Exercise 1.2 glVeds tll e agfe ~ = 10 used cars. Th'ese data are reproduced as follows: in thousands of
.
4.31. Examine the marginal normality of the observations on variables XI, X 2 , • •• , Xs for the multiplesclerosis data in Table 1.6. Treat the nonmultiplesclerosis and multiplesclerosis groups separately. Use whatever methodology, including transformations, you feel is appropriate.
4.32. Examine the marginal normality of the observations on variables Xl, X 2 , ••• , X6 for the radiotherapy data in Table 1.7. Use whatever methodology, including transformations, you feel is appropriate.
4.33. Examine the marginal and bivariate normality of the observations on variables XI' X 2 , X 3 , and X 4 for the data in Table 4.3.
4.34, Examine the data on bone mineral content in Table 1.8 for marginal and bivariate nor5
mality.
4.35. Examine the data on paperquality measurements in Table 1.2 for marginal and multivariate normality.
4.36. Examine the data on women's national track records in Table 1.9 for marginal and mulxercise 1 2 to calculate the squared statistical distances . ,  [ ] (a) Use the resU Its 0 f E (x  X),S1 (Xj  x), j = 1,2, ... ,10, where Xj  Xj~' Xj2 • •• I . . Part a determine the proportIOn of the observatIOns falhng the distances m , . . d' 'b . ( b) Us'ng .I _ . d 500"; probability contour of a blvanate normal Istn utlOn. wlthlO the estimate ° distances in Part a and construct a chisquare plot. (c) 0 r d er th e b" I? . P rts band c are these data approximately Ivanate norma. (d) Given the resu Its m a , Explain. . . ( data (with door closed) in Example 4.10. Construct a QQ plot 4.27. ConSider the radla I?~ of these data [Note that the natural logarithm transformation for the naturall~:r~~h:s A = 0 in (434).] Do the natural logarithms appe~r to be ?orcorres~nd.sbtot d? Compare your results with Figure 4.13. Does the chOice A = 4, or .,? mally dlstn u e . A = 0 make much difference III thiS case. The following exercises may require a computer. . . _ ollution data given in Table 1.5. Construct a QQ plot for the s~lar 4.28. ConsIder the an p d arry out a test for normality based on the correlation d' r measurements an c . 0 . ra la.l?n [ (431)] Let a = .05 and use the entry correspond 109 to n = 4 ID coeffIcient rQ see . Table 4.2. _ I . ollution data in Table 1.5, examine the pairs Xs = N0 2 and X6 = 0 3 for 4.29. GIven t le alfp bivariate nonnality. , 1 _ • . . I d'stances (x  x) S (x  x), ] = 1,2, ... ,42, where I I I (a) Calculate statlstlca x'·= [XjS,Xj6]' . f 11' I . e the ro ortion of observations xj = [XjS,Xj6], ] = 1,2, ... '.42: a .lOg (b) DetermlO p. p te 500"; probability contour of a bivariate normal dlstnbutlOn. ° within the approxlma (c) Construct a chisquare plot of the ordered distances in Part a.
4 30. Consider the usedcar data in Exercise 4.26., .
. . th power transformation AI that makes the XI values approxImately e d ( a) Determllle nstruct a QQ plot for the transforme data. norma.I C0 , . t I . th power transfonnations A2 that makes the X2 values approxlll1a e y (b) Determme e ct a QQ plot for the transform ed data. norma.I C0 nstru , " ] I . th wer transfonnations A' = [AI,A2] that make the [XIoX2 vaues (c) Deterrnmnna\ee p? (440) Compare the results with those obtained in Parts a and b. jointly no usmg  .
tivariate normality.
4.37. Refer to Exercise 1.18. Convert the women's track records in Table 1.9 to speeds measured in meters per second. Examine the data on speeds for marginal and multivariate normality. .
4.38. Examine the data on bulls in Table 1.10 for marginal and multivariate normality. Consider only the variables YrHgt, FtFrBody, PrctFFB, BkFat, SaleHt, and SaleWt
4.39. The data in Table 4.6 (see the psychological profile data: www.prenhall.comlstatistics) consist of 130 observations generated by scores on a psychological test administered to Peruvian teenagers (ages 15, 16, and 17). For each of these teenagers the gender (male = 1, female = 2) and socioeconomic status (low = 1, medium = 2) were also recorded The scores were accumulated into five subscale scores labeled independence (indep), support (supp), benevolence (benev), conformity (conform), and leadership (leader).
Source: Dala courtesy of C. SOlO.
(a) Examine each of the variables independence, support, benevolence, conformity and leadership for marginal normality. (b) Using all five variables, check for multivariate normality. (c) Refer to part (a). For those variables that are nonnormal, determine the transformation that makes them more nearly nonnal.
4.40. Consider the data on national parks in Exercise 1.27. (a) Comment on any possible outliers in a scatter plot of the original variables. (b) Determine the power transformation Al the makes the Xl values approximately • normal. Construct a QQ plot of the transformed observations. (c) Determine the power transformation A2 the makes the X2 values approximately normal. Construct a QQ plot of the transformed observations. . (d) DetermiQe the power transformation for approximate bivariate normality (440).
4.41. Consider the data on snow removal in Exercise 3.20 .. (a) Comment on any possible outliers in a scatter plot of the original variables. (b) Determine the power transformation Al the makes the Xl values approximately normal. Construct a QQ plot of the transformed observations. (c) Determine the power transformation A2 the makes the X2 values approximately normal. Construct a Q Q plot of the transformed observations. (d) Determine the power transformation for approximate bivariate normality (440).
References 1. Anderson, T. W. An lntroductionto Multivariate Statistical Analysis (3rd ed.). New York: John WHey, 2003. 2. Andrews, D. E, R. Gnanadesikan, and J. L. Warner. "Transformations of Multivariate Data." Biometrics, 27, no. 4 (1971),825840. 3. Box, G. E. P., and D. R. Cox. "An Analysis of Transformations" (with discussion). Journal of the Royal Statistical Society (B), 26, no. 2 (1964),211252. 4. Daniel, C. and E S. Wood, Fitting Equations to Data: Computer Analysis of Multifactor Data. New York: John Wiley, 1980. 5. Filliben, 1. 1. "The Probability Plot Correlation Coefficient Test for Normality." Technometrics, 17, no. 1 (1975),111117. 6. Gnanadesikan, R. Methods for Statistical Data AnalysL~ of Multivariate Observations (2nd ed.). New York: WileyInterscience, 1977. 7. Hawkins, D. M. Identification of Outliers. London, UK: Chapman and Hall, 1980. 8. Hernandez, E, and R. A. Johnson. "The LargeSample Behavior of Transformations to Normality." Journal of the American Statistical Association, 75, no. 372 (1980), 85586l. 9. Hogg, R. v., Craig. A. T. and 1. W. Mckean Introduction to Mathematical Statistics (6th ed.). Upper Saddle River, N.1.: Prentice Hall, 2004. . 10. Looney, S. w., and T. R. Gulledge, Jr. "Use of the Correlation Coefficient with Normal Probability Plots." The American Statistician, 39, no. 1 (1985),7579. 11. Mardia, K. v., Kent, 1. T. and 1. M. Bibby. Multivariate Analysis (Paperback). London: Academic Press, 2003. 12. Shapiro, S. S., and M. B. Wilk. "An Analysis of Variance Test for Normality (Complete Samples)." Biometrika, 52, no. 4 (1965),591611. ..
Exercises 209 13. Viern, '11 S., and R. A. Johnson "Tabl d CensoredData Correlation Sta~istic £es ~n . LargeSample Distribution Theory for Statistical ASSOciation, 83, no. 404 (19~)~~19;~~~~7~ormality." Journal of the American 14. Yeo, I. and R. A. Johnson "A New R '1 ity or Symmetry." Biometrika, 87, n~.~l (~~~~~~~~~~sformations to Improve Normal. 15. Zehna, P. "Invariance of Maximu L" Statistics, 37, no. 3 (1966),744. m lkehhood Estimators." Annals of Mathematical
This test statistic has a student's tdistribution with n  1 degrees of freedom (d.f.). We reject Ho, that Mo is a plausible value of M, if the observed It I exceeds a specified percentage point of a tdistribution with n  1 d.t Rejecting Ho when It I is large is equivalent to rejecting Ho if its square, 
is large. The variable t 2 in (51) is the square of the distance from the sample mean
X to the test value /lQ. The units of distance are expressed in terms of s/Yn, or estimated standard deviations of X. Once X and S2 are observed, the test becomes:
INFERENCES ABOUT A MEAN VECfOR 5.1 Introduction This chapter is the first of the methodological sections of the book. We shall now use the concepts and results set forth in Chapters 1 through 4 to develop techniques for analyzing data. A large part of any analysis is concerned with inferencethat is, reaching valid conclusions concerning a population on the basis of information from a sample. . At this point, we shall concentrate on inferences about a populatIOn mean vector and its component parts. Although we introduce statistical inference through initial discussions of tests of hypotheses, our ultimate aim is to present a full statistical analysis of the component means based on simultaneous confidence statements. One of the central messages of multivariate analysis is that p correlated variables must be analyzed jointly. This principle is exemplified by the methods presented in this chapter.
where t,,_1(a/2) denotes the upper lOO(a/2)th percentile of the tdistribution with n  1 dJ. If Ho is not rejected, we conclude that /lQ is a plausible value for the normal population mean. Are there other values of M which are also consistent with the data? The answer is yes! In fact, there is always a set of plausible values for a normal population mean. From the well"known correspondence between acceptance regions for tests of Ho: JL = /lQ versus HI: JL * /lQ and confidence intervals for M, we have {Do not reject Ho: M = Moat level a}
Population Mean Let us start by recalling the univariate theory for determining whether a specific value /lQ is a plausible value for the population mean M. From the point of view of hypothesis testing, this problem can be formulated as a test of the competing hypotheses
Ho: M = Mo and HI: M * Mo Here Ho is the null hypothesis and HI is the (twosided) alternative hypothesis. If Xl, X 2 , ... , Xn denote a random sample from a normal population, the appropriate test statistic is (X  Jko) 1 n 1 n 2 t where X =  ~ XI' and s2 =  (Xj X) = s/Yn ' n~ n  1 j=l
The confidence interval consists of all those values Jko that would not be rejected by the level a test of Ho: JL = /lQ. Before the sample is selected, the 100(1  a)% confidence interval in (53) is a random interval because the endpoints depend upon the random variables X and s. The probability that the interval contains JL is 1  a; among large numbers of such independent intervals, approximately 100(1  a)% of them will contain JL. Consider now the problem of determining whether a given p x 1 vector /Lo is a plausible value for the mean of a multivariate normal distribution. We shall proceed by analogy to the univariate development just presented. A natural generalization of the squared distance in (51) is its multivariate analog
which combines a normal, Np(O, 1:), random vector and a Wishart W _ (1:) random , matrix in the form ' p,n 1
: .
Wishart random matrix ( d.f.
The statistic T2 is called Hotelling's T2 in honor of Harold Hotelling, a pioneer in multivariate analysis, who first obtained its sampling distribution. Here (1/ n)S is the estimated covariance matrix of X. (See Result 3.1.) If the observed statistical distance T2 is too largethat is, if i is "too far" from pothe hypothesis Ho: IL = Po is rejected. It turns out that special tables of T2 percentage points are not required for formal tests of hypotheses. This is true because T
normal. ) random varIable (
scaled) Chisquare)l random variable ( normal ) random variable d.f.
Example.S.1 .(Evaluating T2) Let the data matrix for a random sample of size n = 3 from a blvanate normal population be
Evaluate the observed T2 for Po = [9,5]. What is the sampling distribution of T2 in this case? We find .
It is informative to discuss the nature of the r 2distribution briefly and its correspondence with the univariate test statistic. In Section 4.4, we described the manner in which the Wishart distribution generalizes the chisquare distribution. We can write
for the univariate case. Since the multivariate normal and Wishart random variables are indepen~ently distributed [see (423)], their joint density function is the product of the margmal normal and Wish art distributions. Using calculus, the distribution (55) of T2 as given previously can be derived from this joint distribution and the representation (58). It is rare, in multivariate situations, to be content with a test of Ho: IL = ILo, whe~e a~l o~ t~e mean vector components are specified under the null hypothesis. Ordmanly, It IS preferable to find regions of p values that are plausible in light of the observed data. We shall return to this issue in Section 5.4.
whatever the true p and 1:. Here Fp,llp(a) is the upper (l00a)th percentjle of the Fp,np distribution. Statement (56) leads immediately to a test of the hypothesis Ho: p = Po versus HI: pPo. At the a level of significance, we reject Ho in favor of HI if the observed
Let Xl, X 2, ... , X" be a random sample from an Np(p, 1:) population. Then
where Fp• n  p denotes a random variable with an Fdistribution with p and n  p d.f. To summarize, we have the following:
Before the sample is selected, T2 has the distribution of a (3  1)2 (3  2) F2,3Z
random variable.
The next example illustrates a test of the hypothesis Ho: f.L = f.Lo ~sing. data collected as part of a search for new diagnostic techniques at the Umverslty of Wisconsin Medical School. Example 5.2 (Testing a multivariate mean vector with T2) Perspiration fro~ 20 healthy females was analyzed. Three components, XI = sweat rate, XZ.= sodIUm content, and X3 = potassium content, were measured, and the results, whIch we call the sweat data, are presented in Table 5.1. Test the hypothesis Ho: f.L' = [4,50,10) against HI: f.L' "* [4,50,10) at level of
significance a = .10. Computer calculations provide
SI .
48.5 65.1 47.2 53.2 55.5 36.1 24.8 33.1 47.4 54.1 36.9 58.8 27.8 40.2 13.5 56.4 71.6 52.8 44.1 40.9
9.3 8.0 10.9 12.0 9.7 7.9 14.0 7.6 8.5 11.3 12.7 12.3 9.8 8.4 10.1 7.1 8.2 10.9 11.2 9.4
Source: Courtesy of Dr. Gerald Bargman.
.586 .022 .258J [ 4.640  4 J 45.400  50 .022 .006 .002 [ .402 9.965  10 .258 .002
we see that T Z = 9.74 > 8.18, and consequently, we reject Ho at the 10% level of significance. We note that Ho will be rejected if one or more of the component means, or some combination of means, differs too much from the hypothesized values [4,50, 10). At this point, we have no idea which of these hypothesized values may not be supported by the data . We have assumed that the sweat data are multivariate normal. The QQ plots constructed from the marginal distributions of XI' X z , and X3 all approximate straight lines. Moreover, scatter plots for pairs of observations have approximate elliptical shapes, and we conclude that the normality assumption was reasonable in this case. (See Exercise 5.4.) • One feature of tl1e TZstatistic is that it is invariant (unchanged) under changes in the units of measurements for X of the form
A transformation of the observations of this kind arises when a constant b; is . · subtracted from the ith variable to form Xi  b i and the result is· < by a constant a; > 0 to get ai(Xi  b;). Premultiplication of the f:en!ter,''/ scaled quantities a;(X;  b;) by any nonsingular matrix will yield Equation As an example, the operations involved in changing X; to a;(X;  b;) cor
The mean 110 is now fixed, but :t can be varied to find the value that is "most likely" to have led, with #Lo fixed, to the observed sample. This value is obtained by maximizing L(IIo, :t) with respect to :to Following the steps in (413), the exponent in L(IIo,:t) may be written as
.!.
Therefore, T2 computed with the y's and a hypothesized value IIy.o = CIIo + d is
.!. 2
The last expression is recognized as the value of rZ computed with the x's.
Todetermine whether 110 is a plausible value of 11, the maximum of L(IIo,:t) is compared with the unrestricted maximum of L(II, :t). The resulting ratio is called the likelihood ratio statistic. Using Equations (510) and (511), we get
5.3 Hotelling's T2 and Likelihood Ratio Tests We introduced the TZstatistic by analogy with the univariate squared distance t 2• There is a general principle for constructing test procedures called the likelihood ratio method, and the TZstatistic can be derived as the likelihood ratio test of Ho: 11 = 110' The general theory of likelihood ratio tests is beyond the scope of this book. (See [3] for a treatment of the topic.) Likelihood ratio tests have several optimal properties for reasonably large samples, and they are particularly convenient for hypotheses formulated in terms of multivariate normal parameters. We know from (418) that the maximum of the multivariate normal likelihood as 11 and :t are varied over their possible values is given by (510)
..
.
The equivalent statIstIc A 2/n = Ii III io I is called Wilks' lambda. If the observed value of this likelihood ratio is too small, the hypothesis Ho: 11 = 110 is unlikely to be true and is, therefore, rejected. Specifically, the likelihood ratio test of Ho: 11 = lIoagainstH1:11 110 rejects Ho if
are the maximum likelihood estimates. Recall that P and i are those choices for fL and :t that best explain the observed values of the random sample.
where Ca is the lower (l00a)th percentile of the distribution of A. (Note that the likelihood ratio test statistic is a power of the ratio of generalized variances.) Fortunately, because of the following relation between T Z and A, we do not need the distribution of the latter to carry out the test.
Result 5.1. Let XI' X 2 , ••. , X" be a random sample from an Np(/L, 'i,) population. Then the test in (57) based on T2 is equivalent to the likelihood ratio test Ho: /L = /Lo versus HI: /L #' /Lo because A 2/" =
Incidentally, relation (514) shows that T2 may be calculated from two determinants, thus avoiding the computation of Sl. Solving (514) for T2, we have
[.~~.d. ~!.~.J A21
Likelihood ratio tests are common in multivariate analysis. Their optimal large sample properties hold in very general contexts, as we shall indicate shortly. They are well suited for the testing situations considered in this book. Likelihood ratio methods yield test statistics that reduce to the familiar F and tstatistics in univariate situations.
By Exercise 4.11, IAI = IA22I1All  A12A2"1A2d = IAldIA22  A21AIIAI21, from which we obtain
(n  1) Proof. Let the (p
We shall now consider the general likelihood ratio method. Let 8 be a vector consisting of all the unknown population parameters, and let L( 8) be the likelihood function obtained by evaluating the joint density of X I, X 2 , ... ,X n at their observed values x), X2,"" XI!" The parameter vector 8 takes its value in the parameter set 9. For example, in the pdimensional multivariate normal case, 8' = [,ul,"" ,up, O"ll"",O"lp, 0"22"",0"2p"'" O"pI,P'O"PP) and e consists of the pdimensional space, where  00 <,ul < 00, ... ,  00 <,up < 00 combined with the [p(p + 1)/2]dimensional space of variances and covariances such that 'i, is positive definite. Therefore, 9 has dimension v = p + p(p + 1 )/2. Under the null hypothesis Ho: 8 = 8 0 ,8 is restricted to lie in a subset 9 0 of 9. For the multivariate normal situation with /L = /Lo and 'i, unspecified, 8 0 = {,ul = ,u10,,u2 = .uzo,···,,up = ,upo; O"I!o' .. , O"lp, 0"22,"" 0"2p"'" 0" p_l,p> 0" pp with 'i, positive definite}, so 8 0 has dimension 1'0 = 0 + p(p + 1 )/2 = p(p + 1)/2. A likelihood ratio test of Ho: 8 E 8 0 rejects Ho in favor of HI: 8 fl eo if max L(8) A =
Here Ho is rejected for small values of A 2/" or, equivalently, large values of T2. The critical values of T2 are determined by (56). •
where c is a suitably chosen constant. Intuitively, we reject Ho if the maximum of the likelihood obtained by allowing (J to vary over the set 8 0 is much smaller than the maximum of the likelihood obtained by varying (J over all values in e. When the maximum in the numerator of expression (516) is much smaller than the maximum in the denominator, 8 0 does not contain plausible values for (J. In each application of the likelihood ratio method, we must obtain the sampling distribution of the likelihoodratio test statistic A. Then c can be selected to produce a test with a specified significance level u. However, when the sample size is large and certain regularity conditions are satisfied, the sampling distribution of 2ln A is well approximated by a chisquare distribution. This attractive feature accounts, in part, for the popularity of likelihood ratio procedures.
wI~hm .the space of all possible parameter values. In this case, the region will be an ellipsOid centered at X. This ellipsoid is the 100(1  a)% confidence region for p..
~ 1~(1.  ~)% co~fidence region for the mean of a pdimensional normal dlstnbutlOn IS the ellipsoid determined by all p. such that n(x  p.)'SI(X  p.) s pen  1) F
1 n 1 n where i =  ~ x' S = ~ ( _ ) ( n II ~ I' (n _ 1) 1=1 £.i Xj x Xj the sample observations.
~o determine whether any P.o lies within the confidence region (is a pl~uslble ;a~~e_ for p.), we need to compute the generalized squared distance n(x  p.o~ S (x. p.o) and compare it with [pen  l)/(n  p)]Fp,n_p(a). If the squared distance IS larger than [p(n l)/(n  p)]Fp,n _p (a) , .0 " is not in the confid . S' .. ence regIOn. mce thiS IS analogous to testing Ho: P. = P.o versus HI: p. '" P.o [see (57)],2 we see that the confidence region of (518) consists of all P.o vectors for which the T test would not reject Ho in favor of HI at significance level a. For p 2:: 4, we cannot graph the joint confidence region for p.. However, we can calculate the axes of the confidence ellipsoid and their relative lengths. These are ~etermined from the eigenvalues Ai and eigenvectors ei of S. As in (47), the directions and lengths of the axes of
5.4 Confidence Regions and Simultaneous Comparisons of Component Means To obtain our primary method for making inferences from a sample, we need to extend the concept of a univariate confidence interval to a multivariate confidence region. Let 8 be a vector of unknown population parameters and e be th~ set ?f ~ possible values of 8. A confidence region is a region of likely 8 values. This regIOn IS determined by the data, and for the moment, we shall denote it by R(X), where X = [Xl> X2 ,· •. , XnJ' is the data matrix. The region R(X) is said to be a 100(1  a)% confidence region if, before the sample is selected,
units along the eigenvectors ei' Beginning at the center x the axes of the confidence ellipsoid are '
This probability is calculated under the true, but unknown, value of 8. ., The confidence region for the mean p. of a pdimensional normal populatIOn IS available from (56). Before the sample is selected,
1,2, ... , P
The ratios of the A;,s will help identify relative amounts of elongation along pairs of axes.
whatever the values of the unknown p. and ~. In words, X will be within
Ex:ample 5.3 (Constructing a confidence ellipse for p.) Data for radiation from microwave ovens were introduced in Examples 4.10 and 4.17. Let
[en  l)pFp,n_p(a)/(n  p)j1f2 of p., with probability 1  a, provided that distance is defined in ~erm~ of nS~I. ,For a particular sample, x and S can be computed, and the mequality
The 95 % confidence ellipse for IL consists of all values (ILl, IL2) satisfying 42[ .564  ILl,
Figure 5.1 A 95% confidence ellipse for IL based on microwaveradiation data.
The length of the major axis is 3.6 times the length of the minor axis.
42(203,018) (.564  ILd 2 + 42(200.228) (.603  ILzf  84( 163.391) (.564  ILl) (.603  IL2) To see whether IL'
42(203.018) (.564  .562)2 + 42(200.228) (.603  .589f  84(163.391) (.564  .562)(.603  .589) We conclude that IL'
[.562, .589] is in the region. Equivalently, a test of Ho:
d' f [.562J h 05 Ievel .562J . [ .589 would not be reJecte III avor of HI: IL if:. .589 at tea =.
,of significance. The joint confidence ellipsoid is plotted in Figure 5.1. The center is at X' = [.564, .603], and the halflengths of the major and minor axes are given by
Simultaneous Confidence Statements While the confidence region n(x  IL )'SI(X  IL) :s; c2 , for c a constant, correctly assesses the joint knowledge concerning plausible values of IL, any summary of conclusions ordinarily includes confidence statements about the individual component means. In so doing, we adopt the attitude that all of the separate confidence statements should hold simultaneously with a specified high probability. It is the guarantee of a specified probability against any statement being incorrect that motivates the term simultaneous confidence intervals. We begin by considering simultaneous confidence statements which are intimately related to the joint confidence region based on the T 2statistic. Let X have an Np(lL, l:) distribution and form the linear combination
Z = alXI + a2X2 + ... + apXp = a'X
respectively. The axes lie along et = [.704, .710] and e2 = [.710, .704] when these vectors are plotted with xas the origin. An indication of the elongation of the confidence ellipse is provided by the ratio of the lengths of the major and minor axes. This ratio is vx; /p(n  1) 2 AI\j n(n _ p) Fp,n_p(a) \lA;" .161 ;::==:======== =  =  = 3.6 / p(n  1) \IX; .045 2v%\j n(n _ p) Fp,np(a)
Moreover, by Result 4.2, Z has an N(a' IL, a'l:a) distribution. If a random sample Xl, X 2,··., Xn from the Np(lL, l:) popUlation is available, a corresponding sample of Z's can be created by taking linear combinations. Thus, j = 1,2, ... , n
.... 224
ConSidering the values of a for which t 2 s; c2, we are naturally led to the determination of
where x and S are the sample mean vector and covariance matrix of the xls, respectively. . . Simultaneous confidence intervals can be developed from a conslderatlOn of confidence intervals for a' p. for various choices of a. The argument proceeds as follows. For a fixed and u~ unknown, a 100(1  0')% confidence interval for /Lz = a'p. is based on student's tratio
2 n(a'(i  p.))2 max t = max ''=..:...:• a'Sa Using the maximization lemma (250) with X = a, d = (x  p.), and B = S, we get m,:u
with the maximum occurring for a proportional to Sl(i _ p.). Result 5.3. Let Xl, Xl,"" Xn be a random sample from an N (p., 1:) population with J: positive definite. Then, simultaneously for all a, the inter:al
a'p.
will contain a' p. with probability 1  a. (521) Proof. From (523),
where tn_;(0'/2) is the upper 100(0'/2)th percentile of a (distribution with n  1 dJ. Inequality (521) can be interpreted as a statement about the components of the mean vector p.. For example, with a' = [1,0, ... ,0), a' p. = /L1, and.(52~) becomes the usual confidence interval for a normal population mean. (Note, m this case, that a'Sa = Sll') Clearly, we could make se~eral confid~~ce statements abou~ the ~om ponents of p. each with associated confidence coeffiCient 1  a, by choos1Og different coefficie~t vectors a. However, the confidence associated with all of the statements taken together is not 1  a. . Intuitively, it would be desirable to associate a "collective" confidence ~oeffi. t of 1  a with the confidence intervals that can be generated Clen . by all chOIces fof a. However, a price must be paid for the convenience of a large slI~ultaneous con 1dence coefficient: intervals that are wider (less precise) than the 10terval of (521) for a specific choice of a. . . . Given a data set Xl, X2, ... , Xn and a particular a, the confidence 10terval m (521) is that set<>f a' p. values for which
a' p.
for every a. Choosing c = pen  l)Fp ,,._p(a)/(n  p) [see (56)] gives intervals that will contain a' p. for all a, with probability 1  a = P[T 2 5 c2). • It is convenient to refer to the simultaneous intervals of Result 5.3 as Tlintervals, since the coverage probability is determined by the di~tribution of T2, The successive choices a' = [1,0, .. ,,0], a' = [0,1, ... ,0), and so on through a' = [0,0, ... ,1) for the T 2intervals allow us to conclude that
A simultaneous confidence region is given by the set of a' p. values such that t 2 is relatively small for all choices of a. It seems reasonable to expect that the constant t~_1(0'/2) in (522) will be replaced by a larger value, c 2 , when statements are developed for many choices of a.
all hold simultaneously with confidence coefficient 1  a. Note that without modifying the coefficient 1  a, we can make statements about the diffe~ences /L'  /Lk d' , [ , correspon mg to a = 0, ... ,0, ai, 0, ... ,0, ab 0, ... ,0], where ai = 1 and
ak = 1. In this case a'Sa
The simultaneous T2 confidence intervals are ideal for "data snooping." The confidence coefficient 1  a remains unchanged for any choice of a, so linear combinations of the components ILi that merit inspection based upon an examination of the data can be estimated. In addition, according to the results in Supplement 5A, we can include the state. ments about (ILi, ILd belonging to the sample meancentered ellipses
and still maintain the confidence coefficient (1  ex) for the whole set of statements. The simultaneous T2 confidence intervals for the individual components of a mean vector are just the shadows, or projections, of the confidence ellipsoid on the component axes. This connection between the shadows of the ellipsoid and the simultaneous confidence intervals given by (524) is illustrated in the next example.
In Example 5.3, we obtained the 95% confidence ellipse for the means of the fourth roots of the doorclosed and dooropen microwave radiation measurements. The 95% simultaneous T2 intervals for the two component means are, from (524),
confidence ellipse on the axesmicrowave radiation data.
Ip(n  1) fSll _ Ip(n  1) ~) p ( XI  \j (n _ p) Fp,n_p(·05) \j~' Xl + \j (n _ p) F .n p(·05) \j~
Example 5.4 (Simultaneous confidence intervals as shadows of the confidence ellipsoid)
ll.
In Figure 5.2, we have redrawn the 95% confidence ellipse from Example 5.3. The 95% simultaneous intervals are shown as shadows, or projections, of this ellipse on the axes of the component means. _ Example 5.5 (Constructing simultaneous confidence intervals and ellipses) The scores obtained by n = 87 college students on the College Level Examination Program (CLEP) subtest Xl and the College Qualification Test (CQT) subtests X 2 and X3 are given in Table 5.2 on page 228 for Xl = social science and history, X 2 = verbal, and X3 = science. These data give
.
228 Chapter 5 Inferences about a Mean Vector or X2 X3 Xl (Social science and (Verbal) (Science) history) Individual
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Source:
468 428 514 547 614 501 421 527 527 620 587 541 561 468 614 527 507 580 507 521 574 587 488 488 587 421 481 428 640 574 547 580 494 554 647 507 454 427 521 468 587 507 574 507
Data courtesy of Richard W. Johnson.
26 26 21 33 27 29 22 23 19 32 31 19 26 20 28 21 27 21 21 23 25 31 27 18 26 16 26 19 25 28 27 28 26 21 23 23 28 21 26 14 30 31 31 23
45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
494 541 362 408 594 501 687 633 647 647 614 633 448 408 441 435 501 507 620 415 554 348 468 507 527 527 435 660 733 507 527 428 481 507 527 488 607 561 614 527 474 441 607
24 25 17 17 23 26 33 31 29 34 25 28 24 19 22 20 21 24 36 20 30 18 25 26 31 26 28 25 33 28 29 19 23 19 23 28 28 34 23 30 16 26 32
With the possible exception of the verbal scores, the marginal QQ plots and twodimensional scatter plots do not reveal any serious departures from normality for the college qualification test data. (See Exercise 5.18.) Moreover, the sample size is large enough to justify the methodology, even though the data are not quite n~)fmally distributed. (See Section 5.5.) The simultaneous T 2intervals above are wider than univariate intervals because all three must hold with 95% confidence. They may also be wider than necessary, be.cause, with the same confidence, we can make statements about differences. For instance, with a' = [0, 1, 1], the interval for IL2  IL3 has endpoints ( _ ) ± )p(n  1) F (05»)S22 X2 X3 (n _ p) p,np'
so (26.44,32.68) is a 95% confidence interval for IL2  IL3' Simultaneous intervals can also be constructed for the other differences. Finally, we can construct confidence ellipses for pairs of means, and the same 95% confidence holds. For example, for the pair (IL2, IL3)' we have
This ellipse is shown in Figure 5.3 on page 230, along with the 95 % confidence ellipses for the other two pairs of means. The projections or shadows of these ellipses on the axes are also indicated, and these projections are the T 2intervals.
A Comparison of Simultaneous Confidence Intervals with OneataTime Intervals An alternative approach to the construction of confidence intervals is to consider the components ILi one at a time, as suggested by (521) with a' = [0, ... ,0, ai, 0, ... ,0] where ai = 1. This approach ignores the covariance structure of the P variables and leads to the intervals
To guarantee a probability of 1  a that' all of the statements about the component means hold simultaneously, the individual intervals must be wider than the separate tintervals;just how much wider depends on both p and n, as well as on 1  a. For 1  a = .95, n = 15, and p = 4, the multipliers of ~ in (524) and (527) are 1) 4(14) (n _ p) Fp,np(.05) = 11 (3.36) = 4.14
and tnI(.025) = 2.145, respectively. Consequently, in this case the simultaneous intervals are lOD( 4.14  2.145)/2.145 = 93% wider than those derived from the oneatatime t method. Table 5.3 gives some critical distance multipliers for oneatatime tintervals computed according to (521), as well as the corresponding simultaneous T 2intervals. In general, the width of the T 2intervals, relative to the tintervals, increases for fixed n as p increases and decreases for fixed p as n increases.
Table ·S.3 Critical Distance Multipliers for OneataTime t Intervals and T 2 Intervals for Selected nand p (1  a = .95) )(n  l)p (n _ p) Fp,n_p(.05)
Figure S.3 95 % confidence ellipses for pairs of means and the simultaneous T 2 intervalscollege test data.
Although prior to sampling, the ith interval has probabili~~ 1  a o.f covering lLi, we do not know what to assert, in general, about the probability of all mtervals containing their respective IL/S. As we have pointed out, this probability is not 1  a. To shed some light on the problem, consider the special case where the observations have a joint normal distribution and
Since the observations on the first variable are independent of those on the second variable, and so on, the product rule for independent events can be applied. Before the sample is selected, P[allt_intervalsin(527)containthelL;'S) = (1  a)(l a)···(l  a) =
= .95 and p = 6, this probability is (.95)6 = .74.
The comparison implied by Table 5.3 is a bit unfair, since the confidence level associated with any collection of T 2intervals, for fixed nand p, is .95, and the overall confidence associated with a collection of individual t intervals, for the same n, can, as we have seen, be much less than .95. The oneatatime t intervals are too short to maintain an overall confidence level for separate statements about, say, all p means. Nevertheless, we sometimes look at them as the best possible information concerning a mean, if this is the only inference to be made. Moreover, if the oneatatime intervals are calculated only when the T 2 test rejects the null hypothesis, some researchers think they may more accurately represent the information about the means than the T 2intervals do. The T 2intervals are too wide if they are applied only to the p component means. To see why, consider the confidence ellipse and the simultaneous intervals shown in Figure 5.2. If ILl lies in its T 2interval and 1L2lies in its T 2interval, then (ILl, IL2) lies in the rectangle formed by these two intervals. This rectangle contains the confidence ellipse and more. The confidence ellipse is smaller but has probability .95 of covering the mean vector IL with its component means ILl and IL2' Consequently, the probability of covering the two individual means ILl and f.L2 will be larger than .95 for the rectangle formed by the T 2intervals. This result leads us to consider a second approach to making· multiple comparisons known as the Bonferroni method.
The Bonferroni Method of Multiple Comparisons .' . small number of individual confidence statements. Often, attentIOn IS rest~lcted t?bla d better than the simultan eous intervals of 't fons it IS pOSSI e to 0 . In t h ese SI ua I T d component means ILi or linear corn b" Result 5.3. If th: number m of spe~lle 11 simultaneous confidence interval matIons s can be '+aJ .L2+ ···+ a J.Llssm a, 3 ,... = alJ.LI ' 2 ( P ecise) than the simultan eous T 2mterval s. develop ed that are shorter ~~r~ pr mparis ons is called the Bonferroni method, _ The alte~n?tive method for ~u ~~b:~i~itY inequality carrying that name. because It IS develop.ed from !llectio n of data, confidence stateme nts about m linSuppose that, pnor to the . , requI'red Let C. denote a confiden ce state.' " 3 J.L are . I earcomb matlOnS 311L,32 /L';",.m [C ] = 1 a· i = 1,2, ... ,m. Now (see ment about the value of aiIL WIth P i t r u e" . Exercis e 5.6), P[ all C true] = 1  P[ at least one Ci false] m i
The stateme nts in (529) can be compar ed with those in (524). The percent age point tn_l(a/2p ) replaces V(n  l)pFp.n _p(a)/(n  p), but otherwi se the intervals are of the same structur e. Example S.6 (Constructing Bonferroni simultaneous confidence interval s and comparing them with T2 intervals) Let us return to the microwa ve oven radiatio n data in Exampl es 5.3 and 5.4. We shall obtain the simultan eous 95% Bonferr oni confidence interval s for the means, ILl and ILz, of the fourth roots of the doorclo sed and doorop en measure ments with Cli = .05/2, i = 1,2. We make use of the results in Exampl e 5.3, noting that n = 42 and 1 1(.05/2(2» = t41(.0125) = 2.327, to get 4
1  (al + a2 + ... + am) . f the Bonferroni inequality, allows an investiInequality (528), a special case 0 + + .,. + a regardless of the correlagator to control the. overall erro~ ~ate al stat::nents. The;; is also the flexibility of tion structure behmd the confl ence of important statements and balancin g it by controll ing the error rate for a group . f th I ss important ~atements. . . another chOice or .e e interval estimates for the restricted set consIstm g Let us develop slmultaneou~ . fonnation on the relative importa nce of these of the components J.Lj of J.L. Lackmg ID. I =
Figure 5.4 shows the 95% T2 simultan eous confiden ce intervals for ILl, IL2 from Figure 5.2, along with the correspo nding 95% Bonferr oni intervals . For each component mean, the Bonferr oni interval falls within the T 2interval . Consequ ently, the rectangu lar Goint) region formed by the two Bonferr oni interval s is contain ed in the rectangu lar region formed by the two T 2intervals. If we are interest ed only in the compon ent means, the Bonferr oni interval s provide more precise estimate s than
. with a· = a/m. SIDce P[X.I ± t111 (a/2m)~ i = 1,2:... , m, we have, from (528),
p[x. I
+ .,. + : )
.h Therefo re, Wit an overall confidence .level greater than or equal to 1  a, we can make the following m = p statements.
tnI(~) f¥:$ J.Ll:$ XI + ttlI(2~) fij _ t (!:...) fs2i.:$ J.L2 :$ X2 + tnI(;p ) rs2j 2p '1; . : \j;
.
Figure S.4 The 95% T2 and 95% Bonferroni simultaneous confiden ce intervals for the component meansm icrowav e radiation data.
the T 2intervals. On the other hand, the 95% confidence region for IL gives the plausible values for the pairs (ILl, 1L2) when the correlation between the measured variables is taken into account. • The Bonferroni intervals for linear combinations a' IL and the T 2intervals (recall Result 5.3) have the same general form:
which does not depend on the random quantities Xand S.As we have pointed out, for a small number m of specified parametric functions a' IL, the Bonferroni intervals will always be shorter. How much shorter is indicated in Table 5.4 for selected nand p. Table S.4 (Length of Bonferroni Interval)/(Length of T 2Interval) for 1  Cl = .95 and Cli = .05/m m=p
IL and positive definite covariance matrix :to When n  p is large, the hypothesis Ho: fL = lLa is rejected in favor of HI: IL ,p lLa, at a level of significance approxi
Large Sample Inferences about a Population Mean Vector When the sample size is large, tests of hypotheses and confidence regions for IL can be constructed without the assumption of a normal population. As illustrated in Exercises 5.15,5.16, and 5.17, for large n, we are able to make inferences about the population mean even though the parent distribution is discrete. In fact, serious departures from a normal population can be overcome by large sample sizes. Both tests of hypotheses and simultaneous confidence statements will then possess (approximately) their nominal levels. The advantages associated with large samples may be partially offset by a loss in sample information caused by using only the summary statistics X, and S. On the other hand, since (x, S) is a sufficient summary for normal populations [see (421)],
Here X~( a) is the upper (100a )th percentile of a chisquare distribution with p dJ.
Comparing the test in Result 5.4 with the corresponding normal theory test in (57), we see that the test statistics have the same structure, but the critical values are different. A closer examination, however, reveals that both tests yield essentially the same result in situations where the x2test of Result 5.4 is appropriate. This follows directly from the fact that (n  l)pFp,n_p(a)/(n  p) and x~(a) are approximately equal for n large relative to p. (See Tables 3 and 4 in the appendix.)
Result 5.5. Let XI, X 2, ... , Xn be a random sample from a population with mean IL and positive definite covariance
a'X ± We see from Table 5.4 that the Bonferroni method provides shorter intervals when m = p. Because they are easy to apply and provide the relatively short confidence intervals needed for inference, we will often apply simultaneous tintervals based on the Bonferroni method.
where x~(a) is the upper (l00a)th percentile of the x~distribution. Equation (531) immediately leads to large sample tests of hypotheses and simultaneous confidence regions. These procedures are summarized in Results 5.4 and 5.5.
Consequently, in every instance where Cli = Cl/ rn,. Length of Bonferroni interval =
the closer the underlying population is to multivariate normal, the more efficiently the sample information will be utilized in making inferences. All largesample inferences about IL are based on a ,idistribution. From (428), we know that (X  1L)'(n1Srl(X  fL) = n(X  IL)'SI(X  IL) is approximately X2 with p d.f., and thus,
will contain a' IL, for every a, with probability approximately 1  a. Consequently, we can make the 100(1  a)% simultaneous confidence statements
contains ILp and, in addition, for all pairs (lLi, ILk)' i, k = 1,2, ... , p, the sample meancentered ellipses
Proof. The first part follows from Result 5A.1, with c2 = x~(a). The probability level is a consequence of (531). The statements for the f.Li are obtained by the special choices a' = [0., ... ,0., ai, 0., ... ,0], where ai = 1, i = 1,2, ... , p. The ellipsoids for pairs of means follow from Result 5A.2 with c2 = X~( a). The overall confidence. level of approximately 1  a for all statements is, once again, a result of the large • sample distribtltion theory summarized in (531). The question of what is a large sample size is not easy to answer. In one or two dimensions, sample sizes in the range 3D to 50. can usually be considered large. As the number characteristics bec9mes large, certainly larger sample sizes are required for the asymptotic distributions to provide good approximations to the true distributions of various test statistics. Lacking definitive studies, we simply state that f'I  P must be large and realize that the true case is more complicated. An application with p = 2 and sample size 50. is much different than an application with p = 52 and sample size 100 although both have n  p = 48. It is good statistical practice to subject these large sample inference procedures to the same checks required of the normaltheory methods. Although small to moderate departures from normality do not cause any difficulties for n large, extreme deviations could cause problems. Specifically, the true error rate may be far removed from the nominal level a. If, on the basis of QQ plots and other investigative devices outliers and other forms of extreme departures are indicated (see, for example, [2b, appropriate corrective actions, including transformations, are desirable. Methods for testing mean vectors of symmetric multivariate distributions that are relatively insensitive to departures from normality are discussed in [11]. In some instances, Results 5.4 and 5.5 are useful only for very large samples. The next example allows us to illustrate the construction of large sample simultaneous statements for all single mean components. >
Example S.7 (Constructing large sample simultaneous confidence intervals) A music educator tested thousands of FInnish students on their native musical ability in order to set national norms in Finland. Summary statistics for part of the data setare given in Table 5.5. These statistics are based on a sample of n = 96 Finnish 12th graders.
Let us construct 90.% simultaneous confidence intervals for the individual mean components f.Li' i = 1,2, ... ,7. From Result 5.5, simultaneous 90.% confidence limits are given by Xi
i = 1,2, ... ,7, where
X~(.lO) = 12.0.2.
22.0. ± Y12.D2
Based, perhaps, upon thousands of American students, the investigator could hypothesize the musical aptitude profile to be 1'0 = [31,27,34,31,23,22,22]
We see from the simultaneous statements above that the melody, tempo, and meter components of 1'0 do not appear to be plausible values for the corresponding means of Finnish scores. ' .
12thGrade Finnish Students Participating in a Standardization Program Raw score Variable
28.1 26.6 35.4 34.2 23.6 22.0. 22.7
Source: Data courtesy ofY. Sell.
When the sample size is large, the oneatatime confidence intervals for individual means are
i = 1,2, ... ,p
where z(a/2) is the upper l00(a/2)th percentile of the standard normal distribution. The Bonferroni simultaneous confidence intervals for the m = p statements about the individual means take the same form, but use the modified percentile z( a/2p) to give
i = 1,2, ... , P
Table 5.6 gives the individual, Bonferroni, and chisquarebased (or shadow of the confidence ellipsoid) intervals for the musical aptitude data in Example 5.7. Table 5.6 The Large Sample 95% Individual, Bonferroni, and T 2Intervals for
the Musical Ap..titude Data The oneatatime confidence intervals use z(.025) = 1.96. The simultaneous Bonferroni intervals use z( .025/7) = 2.69. The simultaneous T2, or shadows of the ellipsoid, use .0(.05) = 14.07. Variable
Oneatatime Bonferroni Intervals Shadow of Ellipsoid Lower Upper Upper Lower Upper Lower
Xl = melody X 2 = harmony X3 = tempo X4 = meter Xs = phrasing X6 = balance X 7 = style
Although the sample size may be large, some statisticians prefer to retain the F and tbased percentiles rather than use the chisquare or standard normalbased percentiles. The latter constants are the infinite sample size limits of the· former constants. The F and t percentiles produce larger intervals and, hence, are more conservative. Table 5.7 gives the individual, Bonferroni, and Fbased, or shadow of the confidence ellipsoid, intervals for the musical aptitude data. Comparing Table 5.7 with Table 5.6, we see that all of the intervals in Table 5.7 are larger. However, with the relatively large sample size n = 96, the differences are typically in the third, or tenths, digit. Table 5.7 The 95% Individual, Bonferroni, and T2IntervaIs for the
Musical Aptitude Data The oneatatime confidence intervals use t95(.025) = 1.99. The simultaneous Bonferroni intervals use t95(.025/7) = 2.75. The simultaneous T2, or shadows of the ellipsoid, use F7,89(.05)
= 2.11.
Oneatatime Bonferroni Intervals Shadow of Ellipsoid Lower Upper Lower Upper Lower Upper
Xl = melody X 2 = harmony X3 = tempo X 4 = meter Xs = phrasing X6 = balance X 7 = style
5.6 Multivariate Quality Control Charts To improve the quality of goods and services, data need to be examined for causes of variation. When a manufacturing process is continuously producing items or when we are monitoring activities of a service, data should be collected to evaluate the capabilities and stability of the process. When a process is stable, the variation is produced by common causes that are always present, and no one cause is a major source of variation. The purpose of any control chart is to identify occurrences of special causes of variation that come from outside of the usual process. These causes of variation often indicate a need for a timely repair, but they can also suggest improvements to the process. Control charts make the variation visible and allow one to distinguish common from special causes of variation. A control chart typically consists of data plotted in time order and horizontal lines, called control limits, that indicate the amount of variation due to common causes. One useful control chart is the X chart (read Xbar chart). To create an X chart, 1. Plot the individual observations or sample means in time order. 2. Create and plot the centerline X, the sample mean of all of the observations. 3. Calculate and plot the controllirnits given by Upper control limit (UCL)
The standard deviation in the control limits is the estimated standard deviation of the observations being plotted. For single observations, it is often the sample standard deviation. If the means of subs am pies of size m are plotted, then the standard deviation is the sample standard deviation divided by Fm. The control limits of plus and minus three standard deviations are chosen so that there is a very small chance, assuming normally distributed data, of falsely signaling an outofcontrol observationthat is, an observation suggesting a special cause of variation.
Example 5.8 (Creating a univariate control chart) The Madison, Wisconsin, police department regularly monitors many of its activities as part of an ongoing quality improvement program. Table 5.8 gives the data on five different kinds of overtime hours. Each observation represents a total for 12 pay periods, or about half a year. We examine the stability of the legal appearances overtime hours. A computer calculation gives Xl = 3558. Since individual values will be plotted, Xl is the same as Xl' Also, the sample standard deviation is ~ = 607, and the controllirnits are
Multivariate Quality Control Charts 241 240 Chapter 5 Inferences about a Mean Vector Table 5.8 Five'lYpes 0
.
14,861 11,367 13,329 12,328 12,847 13,979 13,528 12,699 13,534 11,609 14,189 15,052 12,236 15,482 14,900 15
3387 3109 2670 3125 3469 3120 3671 4531 3678 3238 3135 5217 3728 3506 3824 3516 1 Compensatory
875 957 1758 868 398 1603 523 2034 1136 5326 1658 1945 344 807 1223 overtime allowed.
The data, along with the center1me an Figure 5.5.
The legal appearances overtime hours are stable over the period in which the data were collected. The variation in overtime hours appears to be due to common _ causes, so no specialcause variation is indicated. With more than one important characteristic, a multivariate approach should be used to monitor process stability. Such an approach can account for correlations between characteristics and will control the overall probability of falsely signaling a special cause of variation when one is not present. High correlations among the variables can make it impossible to assess the overall error rate that is implied by a large number of univariate charts. The two most common multivariate charts are (i) the ellipse format chart and (ii) the T 2chart. Two cases that arise in practice need to be treated differently:
1. Monitoring the stability of a given sample of multivariate observations 2. Setting a control region for future observations Initially, we consider the use of multivariate control procedures for a sample of multivariate observations Xl, X2,"" X". Later, we discuss these procedures when the observations are subgroup means.
Charts for Monitoring a Sample of Individual Multivariate Observations for Stability We assume that XI, X 2 , .•• , X" are independently distributed as Np(p" !,). By Result 4.8,
 '!'X'_ n } I  .!.X. n J+ 1  .. , _.!.X n n
x\ = 3558
;>
§ 3500 ~
~
_ = (1 ;;1)2
Cov(Xj  X)
!,
(nl)
+ (n  l)n2 !, = n!'
Each X j  X has a normal distribution but, X j  X is not independent of the sample covariance matrix S. However to set control limits, we approximate that (Xj  X)'SI(Xj  X) has a chisquare distribution.
2500 LCL = 1737
1500 15
o
Ellipse Format Chart. The ellipse format chart for a bivariate control region is the
more intuitive of the charts, but its approach is limited to two variables. The two characteristics on the jth unit are plotted as a pair (Xjl, Xj2)' The 95% quality ellipse consists of all X that satisfy
ObserVation Number
• X  h rt for Figure S.S The c a
X
1
'"
legal appearances overtime hours.
(x  i)'SI(X  x)
s;
¥Z(05)
(532)
Multivariate Quality Control Charts 243
242 Chapter 5 Inferences about a Mean Vector
Extraordinary Event Hours Example 5.9 (An ellipse format chart for overtime hours) Let us refer to Example 5.8 and create a quality ellipse for the pair of overtime characteristics (legal appear
6000
ances, extraordinary event) hours. A computer calculation gives
5000
~_
x
=
[3558J 1478 and
S
=
[ 367,884.7 72,093.8J 72,093.8 1,399,053.1
We illustrate the quality ellipse format chart using the 99% ellipse, which consists of all x that satisfy
4000
"::>
3000
>
2000
Oi
Oi ::>
~
1000
.s
'6
Here p = 2, so X~(.01) = 9.21, and the ellipse becomes
xd _ 2s12 (Xl  xd (X2 
(Xl '.:'"':....
Slls22 SllS22 
SI2
Sll
X2)
+ (X2
1000
xd)
S22
SllS22
2000
LCL =  2071
3000
o
(367844.7 X 1399053.1)
= 367844.7
X
1399053.1  (72093.8)2
5 10 Observation Number
15
Figure 5.7 TheX'" chart for X2 = extraordinary event hours. 3558)2 (XI  3558) (X2  1478) 367844.7  2( 72093.8) 367844.7 X 1399053.1
Xl 
X (
1478)2) < 1399053.1  9.21
(X2 
+
This ellipse format chart is graphed, along with the pairs of data, in Figure 5.6.
.§ " Of!e ,." 0 c " & ~
'"
••• • +. •• • • • • • •
~
"
~
S
Notice that one point, indicated with an arrow, is definitely outside of the ellipse. When a point is out of the control region, individual X charts are constructed. TheX'" chart for XI was given in Figure 5.5; that for X2 is given in Figure 5.7. When the lower control limit is less than zero for data that must be nonnegative, it is generally set to zero. The LCL = 0 limit is shown by the dashed line in Figure 5.7 . Was there a special cause of the single point for extraordinary event overtime that is outside the upper control limit in Figure 5.?? During this period, the United States bombed a foreign capital, and students at Madison were protesting. A majority of the extraordinary overtime was used in that fourweek period. Although, by its very definition, extraordinary overtime occurs only when special events occur and is therefore unpredictable, it still has a certain stability. • T 2Chart. A T 2chart can be applied to a large number of characteristics. Unlike the ellipse format, it is not limited to two variables. Moreover, the points are displayed in time order rather than as a scatter plot, and this makes patterns and trends visible. For the jth point, we calculate the T 2statistic
•
(533)
tI\
We then plot the T 2values on a time axis. The lower control limit is zero and we use the upper control limit '
~
ueL =
I
1500
2500
3500
4500
Appearances Overtime
Figure 5.6 The quality control 5500 99% ellipse for legal
appearances and extraordinary event overtime.
x7,(.05)
or, sometimes, x7,( .01). There is no centerline in the T 2 chart. Notice that the T 2 statistic is the same as the quantity used to test normality in Section 4.6.
dJ
244
Multivariate Quality Control Charts 245
Chapter 5 Inferences about a Mean Vector
hour~)
Example 5.10 (A T 2 chart for overtime Using the police departm ent data· . Exampl e 5.8, we construc t a T2plot based on the two variable s Xl = legal . ances hours and X = extraordinary event hours. T 2charts with more than 2 variable s are conside red in Exercise 5.26. We take a = .01 to be consiste nt the ellipse format chart in Example 5.9. . The T2chart in Figure 5.8 reveals that the pair (legal appeara nces, "'Ylrr<..,,~"': nary event) hours for period 11 is out of control. Further investigation, as in pie 5.9, confirm s that this is due to the large value of extraord inary event OV"rh~'" during that period. 12
• 10
      
 
Table 5.9 gives the values of these variable s at fivesecond intervals . Table 5.9 Welder Data
Case 1 2 3 4 5 6 7 8 9 10
11 h
6
4
•
•
2
•
0 0
• 2
• 4
•
• 6
10
8
12
14
16
Period
Figure 5.8 The T 2 chart for legal appearances hours and extraordinary event hours,
a = .01.
When the multivariate T 2chart signals that the jth unit is out of control, it should be determi ned which variables are responsible. A modified region based on Bonferroni intervals is frequent ly chosen for this purpose. The kth variable is out of control if Xjk does not lie in the interval (Xk  tn_I(.OO5/p)~, Xk + tn_l(.005Ip)~) where p is the total number of measured variables. Example 5.11 (Control of robotic welders more than T2 needed) The assembly of a drivesha ft for an automob ile requires the circle welding of tube yokes to a tube. The inputs to the automat ed welding machines must be controlled to be within certain operatin g limits where a machine produces welds of good quality. In order to control the process, one process engineer measure d four critical variables : Xl = Voltage (volts) X2 = Current (amps) X3 = Feed speed(in /min) X 4 = (inert) Gas flow (cfm)
12 1314 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Voltage (Xt> 23.0 22.0 22.8 22.1 22.5 22.2 22.0 22.1 22.5 22.5 22.3 21.8 22.3 22.2 22.1 22.1 21.8 22.6 22.3 23.0 22.9 21.3 21.8 22.0 22.8 22.0 22.5 22.2 22.6 21.7 21.9 22.3 22.2 22.3 22.0 22.8 22.0 22.7 22.6 22.7
Current (X2 ) 276 281 270 278 275 273 275 268 277 278 269 274 270 273 274 277 277 276 278 266 271 274 280 268 269 264 273 269 273 283 273 264 263 . 266 263 272 217 272 274 270
Source: Data courtesy of Mark Abbotoy.
Feed speed (X3 ) 289.6 289.0 288.2 288.0 288.0 288.0 290.0 289.0 289.0 289.0 287.0 287.6 288.4 290.2 286.0 287.0 287.0 290.0 287.0 289.1 288.3 289.0 290.0 288.3 288.7 290.0 288.6 288.2 286.0 290.0 288.7 287.0 288.0 288.6 288.0 289;0 287.7 289.0 287.2 290.0
Gas flow (X4 ) 51.0 51.7 51.3 52.3 53.0 51.0 53.0 54.0 52.0 52.0 54.0 52.0 51.0 51.3 51.0 52.0 51.0 51.0 51.7 51.0 51.0 52.0 52.0 51.0 52.0 51.0 52.0 52.0 52.0 52.7 55.3 52.0 52.0 51.7 51.7 52.3 53.3 52.0 52.7 51.0
246
Multivariate Quality Control Charts 247
Chapter 5 Inferences about a Mean Vector The normal assumption is reasonable for most variables, but we take the natur_ al logarithm of gas flow. In addition, there is no appreciable serial correlation for. successive observations on each variable. A T 2chart for the four welding variables is given in Figure 5.9. The dotted line is the 95% limit and the solid line is the 99% limit. Using the 99% limit, no points are out of contf6l, but case 31 is outside the 95% limit. What do the quality control ellipses (ellipse format charts) show for two variables? Most of the variables are in control. However, the 99% quality ellipse for gas flow and voltage, shown in Figure 5.10, reveals that case 31 is out of ~ntrol and this is due to an unusually large volume of gas flow. The univariate X chart for· In(gas flow), in Figure 5.11, shows that this point is outside the three sigma limits. . It appears that gas flow was reset at the target for case 32. All the other univariate X charts have all points within their three sigma control limits.
14
~
__________________________~99~~~o~L~irrn~'~t
10
8 6 4 2
1:><
3.95
Mean = 3.951
3.90
LCL= 3.896
o
30
40
Figure S.II The univariate
Case
X chart for In(gas flow).
95% Limit
Control Regions for Future Individual Observations
• • • • • • • • • • • • •• • • • •• • • •• • •• ••• •••• •• •• ••
0l,rr,,J
o
30
20
10
40
Case
Figure S.9 The T2~chart for the welding data with 95% and 99% limits.
The goal now is to use data Xl, X2,"" Xn , collected when a process is stable, to set a control region for a future observation Xor future observations. The region in which a future observation is expected to lie is called a forecast, or prediction, region. If the process is stable, we take the observations to be independently distributed as Np(/L, 1;). Because these regions are of more general importance than just for monitoring quality, we give the basic distribution theory as Result 5.6. Result S.6. Let Xl, X 2, ... , Xn be independently distributed as Np(/L, 1;), and let X be a future observation from the same distribution. Then T
4.05
••
•• •••1.•••••
.....
3.95
..s
n
,
= 1 (X  X) n+
sI (X 

X) is distributed as
. )'Sl( ) (x  x x X
•
~
0 <0::::
2
(n  1)p Fp np np ,
and a 100(1  a)% pdimensional prediction ellipsoid is given by all X satisfying
•
4.00
~ ~
20
10
In this example, a shift in a single variable was masked with 99% limits, or almost masked (with 95% limits), by being combined into a single T2value. •
•
12
UCL=4.005
4.00
2
:5
(n  1)p F () n(n _ p) p,np a
Proof. We first note that X  X has mean O. Since X is a future observation, X and
•••••
X are independent, so _ Cov(X  X)
3.90
= Cov(X)' +
_ Cov(X)
1 n
= 1; + 1; =
(n
+ 1) n
1;
and, by Result 4.8, v'nj(n + 1) (X  X) is distributed as N p (O,1;). Now,
3.85 20.5
21.0
21.5
22.0
22.5
Voltage
23.0
23.5
24.0
Figure S.IO The 99% quality control ellipse for In(gas flow) and voltage.
)
n (X  X),Sl n+1
J
n (X  X) n+1
248 Chapter 5 Inferences about a Mean Vector
Multivariate Quality Control Charts 249
which combines a multivariate normal, Np(O, I), random vector and an independent Wishart, Wp ,III (I), random matrix in the form (
mUltiVariate normal)' (Wishart random matrix)I (multivariate normal) random vector dJ, random vector
has the scaled r distribution claimed according to (58) and the discussion on page 213. The constant for the ellipsoid follows from (56),
•
§
••
N
•
.~
•
!t
Note that the prediction region in Result 5,6 for a future observed value x is an ellipsoid, It is centered at the initial sample mean X, and its axes are determined by the eigenvectors of S, Since
 , _]
:5
1:1
~
~
•
§
]
(n  l)p ] n(n _ p) Fp,lI_p(ex) = 1  ex
~
before any new observations are taken, the probability that X will fall in the prediction ellipse is 1  ex. Keep in mind that the current observations must be stable before they can be used to determine control regions for future observations. Based on Result 5.6, we obtain the two charts for future observations.
8
"
III
•
• •
'"
8
'"I 1500
Control Ellipse for Future Observations
e.r
• • • •
. ",
2

P [ (X  X) S (X  X)
2500
3500
4500
5500
Appearances Overtime
With P = 2, the 95% prediction ellipse in Result 5.6 specializes to
Figure S.12 The 95% control ellipse for future legal appearances and extraordinary event overtime.
2
 1)2 F ( 05) (x  x)'Sl( x  x) < (n n(n _ 2) 2.112'
(534)
in time order. Set LCL
=
0, and take
Any future observation x is declared to be out of control if it falls out of the control ellipse.
(n  l)p VCL = ( n  p ) Fp llp(.05) ' .
Example S.12 CA control ellipse for future overtime hours) In Example 5.9, we checked the stability of legal appearances and extraordinary event overtime hours. Let's use these data to determine a control region for future pairs of values. From Example 5.9 and Figure 5.6, we find that the pair of values for period 11 were out of control. We removed this point and determined the new 99% ellipse. All of the points are then in control, so they can serve to determine the 95% prediction region just defined for p = 2. This control ellipse is shown in Figure 5.12 along with the initial 15 stable observations. Any future observation falling in the ellipse is regarded as stable or in control. An observation outside of the ellipse represents a potential outofcontrol observa_ tion or specialcause variation.
Points above the upper control limit represent potential special cause variation and suggest that the process in question should be examined to determine whether immediate corrective action is warranted. See [9] for discussion of other procedures.
T2 Chart for Future Observations For each new observation x, plot T2 =
_n_ (x  x)'Sl(x  x) n +1
Control Charts Based on Subsample Means It is ass~m~d that each random vector of observations from the process is independent~~ dIstnbuted as Np(O, I). We proceed differently when the sampling procedure specIfies that m > 1 units be selected, at the same time, from the process. From the first sample, we determine its sample mean XI and covariance matrix SI' When the population is normal, these two ra~o~ qua!!,!ities are independent. For a general subsample mean X j , Xj  X has a normal distribution with mean oand = Cov(Xj  X)
=
(
1)2
1 n
_.+ n  1CoV (X
Cov(Xj)
2
n
1
)
= (n  1) ~ nm

Inferences about Mean Vectors When Some Observations Are Missing 251
250 Chapter 5 Inferences about a Mean Vector
Control Regions for Future Subsample Observations
where
_ X
1~
=  4J
n
Xj
j=1
As will be .described in Section 6.4, the sample covariances from the n samples can be combined to give a single estimate (called Spooled in Chapter 6) of the. common covariance :to This pooled estimate is .
Once data are collected from the stable operation of a process, they can be used to set control limits for future observed subsample means. If X is a future subsample mean, then X  X has a multivariate normal distribution with mean 0 and .
_
=
_
Cov(X  X)
=
Cov(X)
1 n
_
+  Cov(X I )
+ 1)
(n =
nm
:t
Consequently, Here (nm  n)S is independent of each Xj and, then~for~, of their mean X. Further, (nm  n)S is distributed as a Wishart random matrIX with nm  n degrees. of freedom. Notice that we are estimating I internally from the. data collected in any given period. These estimators are combined to give a single estimator with a large number of degrees of freedom. Consequently,
is distributed as
(nm  n)p (nm  n  p + 1) Fp,nmnp+1 Control Ellipse for Future Subsample Means. The prediction ellipse for a future subsample mean for p = 2 characteristics is defined by the set of an X such that _ =, 1 _ = (n + l)(m  1)2 (x  x) S (x  x):5 m ( nm  n  1) F2 ' nmnl('OS)
is distributed as
(nm  n)p Fp,nmnp+1 nmnp + 1) (
where, again, the righthand side is usually approximated as
Ellipse Format Chart. In an analogous fashion to our. discussion on individu~ multivariate observations, the ellipse format chart for paIrs of subsample means IS _ _ = (n  1)(m  1)2 (X  x)'Sl(x  x) ~ m(nm  n  1) F2.nmnl('OS)
(S36)
although the righthand side is usually approxi~ated as X~(·OS)/m .. Subsamples corresponding to points outside of the c~ntrol elhpse. s~ould .be carefully checked for changes in the behavior of the qu~h.ty cha~acter~st1cs bemg measured. The interested reader is referred to [10] for additIonal diSCUSSion. T2Chart. To construct a T 2chart with subsample data and p characteristics, we plot the quantity
TJ =
=, 1  = m(Xj  X) S (Xj  X)
for j = 1, 2, ... , n, where the VCL
(n  1)(m  1)p ) Fp,nmnp+1('OS)  n  p +1
= (nm
The VCL is often approximated as x;,(.OS) when n is large. Values of T~ that exceed the VCL correspond to potentially outofcontrol or special cause va~iation, which should be checked. (See [10].)
T2Cbart for Future Subsample Means. control limit and plot the quantity
T2 = m(X 
(S37)
x1( .OS)/m.
As before, we bring n/(n + 1) into the
X)'SI(X  X)
for future sample means in chronological order. The upper control limit is then (n + 1) (m  l)p VCL = (nmnp + 1). Fp,nmnp+l(.OS)
The VCL is often approximated as X~( .OS) when n is large. Points outside of the prediction ellipse or above the VCL suggest that the current values of the quality characteristics are different in some way from those of the previous stable process. This may be good or bad, but almost certainiy warrants a careful search for the reasons for the change.
S.7 Inferences about Mean Vectors When Some Observations Are Missing Often, some components of a vector observation are unavailable. This may occur because of a breakdown in the recording equipment or because of the unwillingness of a respondent to answer a particular item on a survey questionnaire. The best way to handle incomplete observations, or missing values, depends, to a large extent, on the

252 Chapter 5 Inferences about a Mean Vector Inferences about Mean Vectors When Some Observations Are Missing 253 experimental context. If the pattern of missing values is closely tied to the value of the response, such as people with extremely high incomes who refuse to respond in a survey on salaries, subsequent inferences may be seriously biased. To date, no statisti_ cal techniques have been developed for these cases. However, we are able to treat situations where data are missing at randomthat is, cases in which the chance mechanism responsible for the missing values is not influenced by the value of the variables. A general approach for computing maximum likelihood estimates from incomplete data is given by Dempster, Laird, and Rubin [5]. Their technique, called the EM algorithm, consists of an iterative calculation involving two steps. We call them the prediction and estimation steps:
1. Prediction step. Given some estimate (j of the unknown parameters, predict the contribution of any missing observation to the (completedata) sufficient statistics. 2. Estimation step. Use the predicted sufficient statistics to compute a revised estimate of the parameters. The calculation cycles from one step to the other, until the revised estimates do not differ appreciably from the estimate obtained in the previous iteration. When the observations Xl, X 2 , ... , Xn are a random sample from a pvariate normal population, the predictionestimation algorithm is based on the completedata sufficient statistics [see (421)]
and ~
XP>X (2), = j
~
I J'
E(X,P)X(2)' x(2). ,
ii,~) =
( x/)x)2)'
The contributions in (538) and (5 39) nents. The results are combined with t~ are1summed o~er ag Xi wit£ missing com~ e samp e data to Yield TI and T2 . Estimation step. Compute the revised m' '. . _ ax:Jmum likelihood estImates (see Result 4.11):  _ Tl IL  ;:;,

1
~ = ;; T2 
ii'ji'
(540)
We illustrate the computational as e t . . in Example 5.13. p c s of the predIctIonestimation algorithm . .Example 5.13 (Illustrating the EM algorithm IL and covariance ~ using the incom I t d) EstImate the normal population mean pe e ata set
n
Tl =
2: Xj = nX
Here n = 4 , P = 3 , and part s 0 f 0 b servatlOn . t We obtain the initial sample averages vec ors XI and
i=l
and
_ ILl
n
T2
=
2: XiX; = (n 
1)S + nXX'
j=1
In this case, the algorithm proceeds as follows: We assume that the population mean and varianceIL and ~, respectivelyare unknown and must be estimated. Prediction step. For each vector Xj with missing values, let xjI) denote the missing components and denote those components which are available. Thus,
x?)
, _ [(I)'
Xi 
Xi
(2),]
,xi
~(I)
_ E(X(I) I (2).

;
~ ~)
,IL,~
Xj
x?)
(i)(l), _ E(X(l)X(I)' I
1 If all L" .
~

0+2+1 p.,2 = = 1, 3
i
i
(2).
Xi
~ ~)
,IL,~

~(I)
IL
+
~
~I(
~12~22
(2)
Xi
~(2»
 IL
(538)
Uu = (6  6)2 + (7  6)2 + (5  6)2 + (6 
1
U22=2'
_
~
~11
the components Xj are missing, set Xj = j1. and
_
~ ~l~
·..... 12~22 .....21
x/x; = I
+ j1.j1.'.
+ ~(I)~(I)' Xi Xi
(539)
_ U 33 =
= 3+6+2+5 = 4
_
4 3
U23 =
4'
6)2
1
2
5
2
Ul2 = (6  6)(0  1) + (7  6)(2  1) + (5
1
xlI) to T2 is
p.,3
are missing.
4 from the available observations. Substitutin so that XII = 6, for example, we can obt .g ~h~s.e averag~s for any missing values, construct these estimates using th d" alllblllltIal covanance estimates. We shall ' e IVlsor n ecause the I 'th d uces the maximum likelihood estimate i Thus, a gon m eventually pro4
estimates.the contribution of to T I . Next, the predicted contribution of Xi Xi
+5
.
Given estimates ii and ~ from the estimation step, use the mean of the conditional normal distribution of x(l), given x(2), to estimate the missing values. That is,! Xi
7
=  2  = 6,
X4
6)(1. 1) + (6
6)(1
1)
4
The prediction step consists of usin th . . . . _ _ contributions of the missing values to t~ e :~I~Ial estlll~at.es IL and ~ to predict the and (539).J e su Clent statIstIcs Tl and T 2. [See (538)
Inferences about Mean Vectors When Some Observations Are Missing 255 254
Chapter 5 Inferences about a Mean Vector
The first component of Xl is missing, so we partition ii and ~ as
are the contributions to T2 • Thus, the predicted completedata sufficient statistics are X21 + X31 + ~41] [5.73 + 7 + 5 + 6.4] [24.13] + X22 + X32 + X42 = 0 + 2 + 1 + 1.3 = 4.30 + X23 + x33 + X43 3+6 +2 +5 16.00
Xll + ==
\ \
and predict
~xlI ~
XII
=
~
= ILl
~ ~1 [X!2 + I12I22 X13 ~ ~
IL2J  6 + [1 ~

2
U11  ~12I2~I21 + Xli
1]
4'
f.L3
1
3  4
[13Jl [lJ
1"2 54 ~ 2
[1
2  4' 1
=
[1i i~Jl [0 1J 14
= 32.99
~[X12' ~) =Xll[XI2, X13) =5.73[0, 3) = [0, 17.18)
~ [~lJ = [1i(1)] ;':;(2)'
IL =
f.L2 .;.:;'" f.L3
.
~ [~11 ~12 \ ~13J [~!.d ~.I.~]
I
f.L
=
0"12
0"22: 0"23
0"13
0"23
·~········~····1··~····
i
0"33
=
~:
~
=
32.99 + 72 + 52 + 41.06 = 0 + 7(2) + 5(1) + 8.27 [ 17.18 + 7(6) + 5(2) + 32
=
148.05 27.27 27.27 6.97 [ 101.18 20.50
02 + 22 + 12 + 1.97 0(3) + 2(6) + 1(2) + 6.5
101.18] 20.50 74.00
I21 i In '
This completes one prediction step. The next esti!llation step, using (540), provides the revised estimates 2
and predict
[8
X13
= 5.73
+ (5.73)2
For the two missing components of X4, we partition ii and ~ as
[
X12
E([~:J \
X43
=
5;ii,I)
=
[~J + I!2~2Hx43
1 Ii = ;;1\ = ~ [24.13] 4.30 =
1L3)
16.00
[~J + [nm (5  4) [~::J
[6.03] 1.08 4.00
1
=
=
for the contribution to T1. Also, from (539),
and
_ ! [148.05
27.27 101.18

4
=
.61 .33 [ 1.17
27.27 101.18] [6.03] 6.97 20.50  1.08 [6.03 20.50 74.00 4.00
1.08 4.00]
.33 1.17] .59 .83 .83 2.50
Note that U11 = .61 and U22 = .59 are larger than the corresponding initial estimates obtained by replacing the missing observations on the first and second variables by the sample means of the remaining values. The third variance estimate U33 remains unchanged, because it is not affected by the missing components. The iteration between the prediction and estimation steps continues until the elements of Ii and ~ remain essentially unchanged. Calculations of this sort are easily handled with a computer. _ 2The final entries in I are exact to two decimal places.
256
Chapter 5 Inferences about a Mean Vector Difficulties Due to Time Dependence in Multivariate Observations 257 Once final estimates jL and i are obtained and relatively few missing compo_ nents occur in X, it seems reasonable to treat allpsuchthatn(jL  p)'iI(it  p):5 x~(a)
As shown in Johnson and Langeland [8], 1 n * S =. n _ 1 ~ (X t  X)(Xt  X)' ~ Ix
(541)
as an approximate 100(1  a)% confidence ellipsoid. The simultaneous confidence·. statements would then follow as in Section 5.5, but with x replaced by jL and S replaced by I.
where the arrow above indicates convergence in probability, and (543)
Caution. The predictionestimation algorithm we discussed is developed on the. basis that component observations are missing at random. If missing values are related to the response levels, then handling the missing values as suggested may introduce serious biases into the estimation procedures; 'TYpically, missing values are related to the responses being measured. Consequently, we must be dubious of any computational scheme that fills in values as if they were lost at random. When more than a few values are missing, it is imperative that the investigator search for the systematic causes that created them.
Moreover, for large n, Vn (X  JL) is approximately normal with mean 0 and covariance matrix given by (543). To make the calculat~ons easy, suppose the underlying process has = cpI where Icp I < 1. Now consIder the large sample nominal 95% confidence ellipsoid for JL. {all JL such that n(X  JL )'SI(X  JL)
:5
x~(.05)}
This ellipsoid has large sample coverage probability .95 if the observations are inde
pe~de~t.1f the observations are related by our autoregressive model, however, this
5.8 Difficulties Due to Time Dependence in Multivariate
ellIpsOId has large sample coverage probability
Observations
P[x~
For the methods described in this chapter, we have assumed that the multivariate observations Xl, X 2,.··, Xn constitute a random sample; that is, they are independent of one another. If the observations are collected over time, this assumption may not be valid. The presence of even a moderate amount of time dependence among the observations can cause serious difficulties for tests, confidence regions, and simultaneous confidence intervals, which are all constructed assuming that independence holds. We will illustrate the nature of the difficulty when the time dependence can be represented as a multivariate first order autoregressive [AR(l)] model. Let the p X 1 random vector X t follow the multivariate AR(l) model
Xt  P
=
(X t  I  p)
+ et
00
Ix =
L
'IEct>'j
j=O
The AR(l) model (542) relates the observation at time t, to the observation at time t  1, through the coefficient matrix <1>. Further, the autoregressive model says the observations are independent, under multivariate normality, if all the entries in the coefficient matrix are o. The name autoregressive model comes from the fact that (542) looks like a multivariate version of a regression with X t as the dependent variable and the previous value X t  I as the independent variable.
(1  CP)(l +
Table 5.10 shows how the coverage probability is related to the coefficient cp and the number of variables p. According to Table 5.10, the coverage probability can drop very low to 632 even for the bivariate case. ' . , . The independ:nce a.ssuI?ption is crucial, and the results based on this assumptIOn can be very mlsleadmg If the observations are, in fact, dependent.
Ta~le 5: I 0 Coverage Probability of the Nominal 95% Confidence EllIpSOId
(542)
where the et are independent and identically distributed with E [et] = 0 and Cov (et) = lE and all of the eigenvalues of the coefficient matrix are between 1 and 1. Under this model Cov (Xt' X t,) = <1>'1. where
:5
P
1 2 5 10 15
cp
.25
.25
.5
.989 .993 .998 .999 1.000
.950 .950 .950 .950 .950
.871 .834 .751 .641 .548
.742 .632 .405 .193 .090
p Simultaneous Confidence Intervals and Ellipses as Shadows of the p·Dimensional Ellipsoids 259
Supplement
elli~soi~ o~is cVu' ~u ~/u'u, and its length is cVu' Au/u'u. With the unit vector
eu
u/ v u'u, the proJectlOn extends

The projection of the ellipsoid also extends the same length in the direction u.
•
Result SA.2. Suppose that the ellipsoid {z' z' Alz < c2 } " d IS given an that U = [UI i U2] is arbitrary but of rank two. Then'
SIMULTANEOUS CONFIDENCE INTERVALS AND ELLIPSES AS SHADOWS OF THE p DIMENSIONAL ELLIPSOIDS
zin the ellipsoid } 2 { based on AI and c
implies that
{fO II U U' .. h . .} r a , z IS 1U t ~ ellIpSOId based on (U' AU) 1 and c2
or for all U We fjr2st establish a basic inequality. Set P = AI/2U(U' AU)lU' AI/2 where A. = A/_~1/2. Nlote that P = P' and p2 = P, so (I  P)P' = P _ p2 = 0' d A I/2' Next, usmg A = A /2A I/2, we write z' Alz = (A 1/2z)' (A1/2 ) = PA l /2z + (I  P)A I/2z. Then z an z
. Proof.
We begin this supplementary section by establishing the general result concerning the projection (shadow) of an ellipsoid onto a line.
z' Alz
Result SA. I. Let the constant c > 0 and positive definite p x p matrix A determine the ellipsoid {z: z' AIz ::s c2 }. For a given vector u 0, and z belonging to the ellipsoid, the
l2
*'
(
Projection (shadow) Of) {z'A1z::sc 2 }onu
=
c Vu'Au u u'u
= (z'u)2::s
(z'K1z)(u'Au) :;; c2u' Au
for all z: z' A1z ::s c2
The choice z = cAul Vu' Au yields equalities and thus gives the maximum shadow, besides belonging to the boundary of the ellipsoid. That is, z' Alz = cZu' Au/u' Au = c2 for this z that provides the longest shadow. Consequently, the projection of the 258
1
+ (I  P)K I/ 2Z)'(PAl/2z + (I _ P)KI/2z) I2
= (PA /2Z), (PA / Z)
S'
Proof. By Definition 2A.12, the projection of any z on u is given by (z'u) u/u'u. Its squared length is (z'u//u'u. We want to maximize this shadow over all z with z' AIz ::s c2• The extended CauchySchwarz inequality in (249) states that (b'd)2::s (b'Bd) (d'B1d), with equality when b = kB1d. Setting b = z, d = u, and B = AI, we obtain
(u'u) (length of projection?
= (PA / z
2:
",hich extends from 0 along u with length cVu' Au/u'u. When u is a unit vector, the shadow extends cVu'Au units, so Iz'ul:;; cVu'Au. The shadow also extends cVu' Au units in the u direction.
(AI/2z)' (Al/2z )
=
'I
12 z'A / p'PA l/2z
mce z A Z::S
C
2
+ ((I  P)Al/2z)' «I  P)Kl/2Z)
= z'A 1/2PA I/2z = z'U(U'AUrIU'z
(SAI)
and U was arbitrary, the result follows.
•
Our next . . · result . establishes ' . the twodimensional confidence ell'Ipse as a proJectlOn o f the p d lIDenslOnal ellipsoId. (See Figure 5.13.) 3
"'2 UU'z
Figure 5.13 The shadow of the ellipsoid z' AI z ::s c2 on the UI, u2 plane is an ellipse.
260 Chapter 5 Inferences about a Mean Vector
Exercises 261
Projection on a plane is simplest when the two vectors UI and Uz determi ning the plane are first convert ed to perpendicular vectors of unit length. (See Result 2A.3.)

Exercises 5.1.
(a) Evaluate y2, for testing Ho: p.' = [7,
Result SA.3. Given the ellipsoid {z: z' AIz :s; CZ } and two perpend icular unit vectors UI and Uz, the projection (or shadow) of {z'A1z::;;; CZ} on the u1o U2 2 plane results in the twodimensional ellipse {(U'z)' (V' AVrl (V'z) ::;;; c }, where V = [UI ! U2]'
11], using the data
2 X = 8 12] 9
r
6 9 8 10
(b) Specify the distribution of T2 for the situation in (a). (c) Using (a) and (b), test Ho at the Cl! = .05Ieve!. What conclusion do you reach?
Proof. By Result 2A.3, the projecti on of a vector z on the Ul, U2 plane is
5.2.
The projection of the ellipsoid {z: z' AIz :s; c2 } consists of all VV'z with z' AIz :s; c2. Consider the two coordin ates V'z of the projection V(V'z). Let z belong to the set {z: z' A1z ::;;; cz} so that VV'z belongs to the shadow of the ellipsoid. By Result SA.2,
~:~n!: t~~ 2~~~~si~e~I:~:f~y 5C1~j~e~~r~hat T Z remains unchanged if each obser~ation
Note that the observations
(V'z)' (V' AVrl (U'z) ::;;; c 2 yield the data matrix
so the ellipse {(V'z)' (V' AVrl (V'z) ::;;; c 2 } contains the coefficient vectors for the shadow of the ellipsoid. Let Va be a vector in the UI, U2 plane whose coefficients a belong to the ellipse {a'(U' AVrla ::;;; CZ}. If we set z = AV(V' AVrla, it follows that V'z = V' AV(V' AUrla
=
(6  9) [ (6+9)
(8  3)J' (8+3)
5.3. (a) Use expression (515) to evaluate y2 for the data in Exercise 5.1. (b) Use the data in Exercise 5.1 to evaluate A in (513). Also, evaluate Wilks' lambda. 5.4. Use the sweat data in Table 5.1. (See Example 5.2.) (a) ~:~:r:::s~ the axes of the 90% confidence ellipsoid for p. Determi ne the lengths of
a
and
(b) Thus, U'z belongs to the coefficient vector ellipse, and z belongs to the ellipsoid z' AIz :s; c2 . Consequently, the ellipse contains only coefficient vectors from the projection of {z: z' AIz ::;;; c 2 } onto the UI, U2 plane. Remark. Projecting the ellipsoid z' AIz :s; c2 first to the UI, U2 plane and then to the line UJ is the same as projecting it directly to the line determi ned by UI' In the context of confidence ellipsoids, the shadows of the twodimensional ellipses give the single compon ent intervals. Remark. Results SA.2 and SA.3 remain valid if V = 2 < q :s; p linearly indepen dent columns.
(10  6) (10+6 )
[Ub""
u q ] consists of
Const~uct
rate sodium content a ~~~~:~~~ ~i~:~e6; ::~~~tiv elj~.co~ struct the three possibl~ scatter plots for'pa~~ case?
QQ plots for the observations on sweat
Commen~.
mu Ivanate normal assumption seem justified in this
5.5. The quantities X, S, and SI are give i E f radiation data. Conduct a test of the ~ul~ hyxpa:::heI~is53 'H ~r ~h~ tran sf0rmed microwavelev lof' T I o·  [. 55 " 6O] atthe Cl! = 05 tur:d in s:.fgn~rleca5n1c~·Es Ylo~rresult consistent with the 95%P confiden ce ellipse for p ~ic.. xpam. .
5.6.
V~rify the Bonferroni inequality in (528) for m = 3 Hmt: A Venn diagram .for the three events CI, C2, a'nd C h I 3 may e p. 5.7. dence Use the sweat data in Table 51 (S E I interval f . e~ xamp e 5.2.) Find simultaneous 95% y2 confivals using (5_2~)0~::r;p~2re' atnhd tP3 usmg Rf~Sult 5.3. Construct the 95% Bonferroni intei. e wo se t s 0 mtervals.
262
Chapter 5 Inferences about a Mean Vector
Exercises 263
k that rZ is equal to the largest squared univaria te tvalue 5.8. From (523), we nOewlinear combination a'xj with a = stcx  ILo), Using the construc ted from th 3 d th H, in Exercise 5.5 evaluate a for the transform ed· It . Example 5. an e o' . h" resu s ID I Z .' d microwaveradiatIOn ata. ¥ en'fy that the tZvalue'computed with t IS a IS equa to T , in Exercise 5.5. ~ I' t < the Alaska Fish and Game departm ent, studies grizzly a natura IS l o r . 5.9. H arry.R oberts e ' oal of maintaining a healthY population. ~easurements on n = 61 bears bear~ wldthhth fgllOwing summary statistics (see also ExerCise 8.23): prOVide t e O · Variable
Sample mean x
Weight (kg)
95.52
Body length (cm) 164.38
Neck (cm)
55.69
Girth (cm)
93.39
Head length (cm) 17.98
(d) Refer to Parts a and b. Constru ct the 95% Bonferro ni confiden ce intervals for the set consisting of four mean lengths and three successive yearly increase s in mean length. (e) Refer to Parts c and d. Compar e the 95% Bonferr oni confiden ce rectangl e for the mean increase in length from 2 to 3 years and the mean increase in length from 4 to 5 years with the confiden ce ellipse produce d by the T 2procedu re. 5.1 1. A physical anthropo logist perform ed a mineral analysis of nine ancient Peruvian hairs. The results for the chromiu m (xd and strontium (X2) levels, in parts per million (ppm), were as follows:
Head width (cm) 31.13
X2(St)
.48
40.53
12.57
73.68
2.19
.55
.74
.66
.93
.37
.22
11.13 20.03 20.29 .78 4.64 .43 1.08 Source: Benfer and others, "Mineral Analysis of Ancient Peruvian Hair," American
Journal of Physical Anthropo logy, 48, no. 3 (1978),277282.
Covariance matrix
S=
3266.46 1343.97 731.54 1343.97 721.91 324.25 731.54 324.25 179.28 1175.50 537.35 281.17 80.17 39.15 162.68 238.37 117.73 56.80
1175.50 162.68 238.37 537.35 80.17 117.73 56.80 281.17 39.15 94.85 474.98 63.73 13.88 9.95 63.73 94.85 13.88 21.26
I (a) Obtain the large samp e 95°;(° simultaneous confidence intervals for the six population mean body measurements. . I (b) Obtain the large samp e 95°;(° simultaneous confidence ellipse for mean weight and mean girth. . . P t , h 950' Bonferroni confidence intervals for the SIX means ID ar a. (c) ObtaID t e 10 ' t th 95°;' Bonferrom. confidence rectangIe for t he mean (d) Refer to Part b. Co?struc. e =° Compare this rectangle with the confidence 6 weight and mean girth usmg m . ellipse in Part b. . . h 950/. Bonferroni confidence mterval for (e) Obtam t e, ° mean head width  mean head length . _ 6 1 = 7 to alloW for this statement as well as statemen ts about each usmg m  + individual mean. . th data in Example 1.10 (see Table 1.4). Restrict your attention to 5.10. Refer to the bear grow the measurements oflength. . s . h 950;' rZ simultaneous confidence intervals for the four populatIO n mean (a) Obtam t e ° , for length. ' f h th ee , Obt' the 950/. T Z simultaneous confidence .mterva1 sort e r am . ° (b) Refer to Part a. h . e yearly increases m mean lengt . succeSSlV . . I th from 2 to 3 . h 950/. T Z confidence ellipse for the mean mcrease ID eng (c) Obtam td~he r:ean increase in length from 4 to 5 years. years an
It is known that low levels (less than or equal 'to .100 ppm) of chromiu m suggest the presence of diabetes, while strontiu m is an indication of animal protein intake. (a) Constru ct and plot a 90% joint confidence ellipse for the populati on mean vector IL' = [ILl' ILZ], assuming that these nine Peruvian hairs represen t a random sample from individuals belonging to a particula r ancient Peruvian culture. (b) Obtain the individual simultan eous 90% confiden ce intervals for ILl and ILz by"projecting" the ellipse construc ted in Part a on each coordina te axis. (Alterna tively, we could use Result 5.3.) Does it appear as if this Peruvian culture has a mean strontiu m level of 10? That is, are any of the points (ILl arbitrary, 10) in the confiden ce regions? Is [.30, 10]' a plausible value for IL? Discuss. (c) Do these data appear to be bivariate normal? Discuss their status with referenc e to QQ plots and a scatter diagram. If the data are not bivariate normal, what implications does this have for the results in Parts a and b? (d) Repeat the analysis with the obvious "outlying" observat ion removed . Do the inferences change? Commen t.
5.12. Given the data
with missing components, use the predictio nestima tion algorithm of Section 5.7 to estimate IL and I. Determi ne the initial estimates, and iterate to find the first revised estimates. 5.13. Determi ne the approxim ate distribut ion of n In( I i Table 5.1. (See Result 5.2.)
1/1 io i)
for the sweat data in
5.14. Create a table similar to Table 5.4 using the entries (length of oneata time tinterva l)/ (length of Bonferro ni tinterval).
Exercises 265
264 Chapter 5 Inferences about a Mean Vector and
Exercises 5.15, 5.16, and 5.17 refer to the following information:
Frequently, some or all of the population characteristics of interest are in the form of attributes. Each individual in the population may then be described in terms of the attributes it possesses. For convenience, attributes are usually numerically coded with respect to their presence or absence. If we let the variable X pertain to a specific attribute, then we can distinguish between the presence or absence of this attribute by defining X =
{I
o
if attribute present if attribute absent
1
2
k
q
q + 1
1 0 0
0 1 0
0 0 0
0 0 0
0 1 0
Outcome (value)
Probability (proportion)
0 1 0
PI
P2
Pk
Pq
p
=
:
[ , . Pq+1
l
1
=
n
2: Xj
n j=1
[
.:
C7'I,q+l
(T2,q+1
l
(T q+:,q+1
vn(p 
p)
Pq+1 = 1
E(p) = P = [
~: Pq+1
l
N(O,I)
is approximately
where the elements of I are (Tkk = Pk(l  Pk) and (Tik = PiPk' The normal approximation remains valid when (Tkk is estimated by Ukk = Pk(l  Pk) and (Tik is estimated k. by Uik = P;Pb i Since each individual must belong to exactly one category, Xq+I,j = 1  (Xlj + X 2j + ... + X qj ), so Pq+1 = 1  (PI + Pz + ... + Pq), and as a result, i has rank q. The usual inverse of i does not exist, but it is still possible to develop simultaneous 100(1  a)% confidence intervals for all linear combinations a'p.
*
Result. Let XI, X 2 , ... , Xn be a random sample from a q + 1 category multinoinial distribution with P[Xjk = 1] = Pt. k = 1,2,.,., q + 1, j = 1,2, ... , n. Approximate simultaneous 100(1  a)% confidence regions for all linear combinations a'p = alPl + a2P2 + .,. + aq+IPq+1 are given by the observed values of
n
2: Xj' and i
=
{uid is a (q + 1) x
(q
+ 1)
*
matrix with Ukk = k. Also, x~(a) is the upper (100a )th percentile of the chisquare distribution with q d.t •
0 1
.
WIth
(TI,q+1 (T2,q+1
(T21
j=1 Pk(1  Pk) and Uik = PiPt, i
q
P2 PI
= n1
provided that n  q is large, Here p = (l/n)
2: Pi ;=1
Let Xj,j = 1,2, ... , n, be a random sample of size n from the multinomial distribution. The kth component, Xj k, of Xj is 1 if the observation (individual) is from category k and is 0 otherwise. The random sample X I, X 2 , ... , Xn can be converted to a sample proportion vector, which, given the nature of the preceding observations, is a sample mean vector. Thus,
'
1 = I n
For large n, the approximate sampling distribution of p is provided by the central limit theorem. We have
In this way, we can assign numerical values to qualitative characteristics. When attributes are numerically coded as 01 variables, a random sample from the population of interest results in statistics that consist of the counts of the number of sample items that have each distinct set of characteristics. If the sample counts are large, methods for producing simultaneous confidence statements can be easily adapted to situations involving proportions. We consider the situation where an individual with a particular combination of attributes can be classified into one of q + 1 mutually exclusive and exhaustive categories. The corresponding probabilities are denoted by PI, P2, ... , Pq, Pq+I' Since the categories include all possibilities, we take Pq+1 = 1  (PI + P2 + .,. + Pq ). An individual from category k will be assigned the «( q + 1) Xl) vector value [0, ... , 0, 1,0, ... , O)'with 1 in the kth position. The probability distribution for an observation from the population of individuals in q + 1 mutually exclusive and exhaustive categories is known as the multinomial distribution. It has the following structure: Category
(TII
,1 Cov(p) = Cov(X) n )
In this result, the requirement that n  q is large is interpreted to mean npk is about 20 or more for each category. We have only touched on the possibilities for the analysis of categorical data. Complete discussions of categorical data analysis are available in [1] and [4J. 5.15. Le,t X ji and X jk be the ith and kth components, respectively, of Xj'
and (Tjj = Var(X j ;) = p;(l  p;), i = 1,2, ... , p. (b) Show that (Tik = Cov(Xji,Xjk ) = PiPbi k. Why must this covariance neceSsarily be negative?
(a) Show that JLi
= E(Xji)
= Pi
*
5.16. As part of a larger marketing research project, a consultant for the Bank of Shorewood wants to know the proportion of savers that uses the bank's facilities as their primary vehicle for saving. The consultant would also like to know the proportions of savers who use the three major competitors: Bank B, Bank C, and Bank D. Each individual contacted in a survey responded to the following question:
Exercises 7,67 266
C
hapter 5 Inferences about a Mean Vector Construct 95% simultaneous confidence intervals for the three proportions PI, P2' and P3 = 1  (PI + P2)'
Which bank is your primary savings bank?
\
Response:
\
A sample of n = 355 people with savings accounts produced.the follo~ing . when asked to indicate their primary savings banks (the people with no savmgs Will ignored in the comparison of savers, so there are five categories):
\\ \\
I I I I I
Bank (category)
Bank of Shorewood
BankB BankC BankD
Observed number
105
119
56
25
populatio~
PI
P2
P3
P4
, _ 105 355
PI 
=
30 .
P2
= .33 P3 =.16 P4
=
5.18. Use the college test data in Table 5.2. (See Example 5.5.) (a) Test the null hypothesis Ho: P' = [500,50, 30J versus HI: P' [500,50, 30J at the a = .05 level of significance. Suppose [500,50,30 J' represent average scores for thousands of college students over the last 10 years. Is there reason to believe that the group of students represented by the scores in Table 5.2 is scoring differently? Explain. .
*'
Another bank
(b) Determine the lengths and directions for the axes of the 95% confidence ellipsoid for p. (c) Construct QQ plots from the marginal distributions of social science and history, verbal, and science scores. Also, construct the three possible scatter diagrams from the pairs of observations on different variables. Do these data appear to be normally distributed? Discuss.
50
5.19. Measurements of Xl = stiffness and X2 = bending strength for a sample of n = 30 pieces of a particular grade of lumber are given in Thble 5.11. The units are pounds/(inches)2. Using the data in the table,
proportIOn Observed .sample proportIOn
The following exercises may require a computer.
Bank of Another No Shorewood Bank B Bank C Bank D Bank Savings
.D7
P5 = .14
Table 5.11 Lumber Data Xl
Let the population proportions be PI = proportion of savers at Bank of Shorewood P2 = proportion of savers at Bank B P3
=
proportion of savers at Bank C
P4 = proportion of savers at Bank D
1  (PI + P2 + P3 + P4) = proportion of savers at other banks (a) Construct simultaneous 95% confidence intervals for PI , P2, ... , P5' • ()"f • • I th at aIlows a comparison of the .. (b) Construct a simultaneous 95/0 confidence mterva Bank of Shorewood with its major competitor, Bank B. Interpret thiS mterval. b h' h school students in a S.I 7. In order to assess the prevalence of a drug pro lem among I~ , ive hi h schools articular city a random sample of 200 students from the city s f g P , . h onding responses are were surveyed. One of the survey questIOns and t e corresp as follows:
1232 1115 2205 1897 1932 1612 1598 1804 1752 2067 2365 1646 1579 1880 1773
X2
Xl
Xz
(Bending strength)
(Stiffness: . modulus of elasticity)
(Bending strength)
4175 6652 7612 10,914 10,850 7627 6954 8365 9469 6410 10,327 7320 8196 9709 10,370
1712 1932 1820 1900 2426 1558 1470 1858 1587 2208 1487 2206 2332 2540 2322
7749 6818 9307 6457 10,102 7414 7556 7833 8309 9559 6255 10,723 5430 12,090 10,072
Source: Data courtesy of U.S. Forest Products Laboratory.
What is your typical weekly marijuana usage? Category
Number of responses
(Stiffness: modulus of elasticity)
Heavy
None
Moderate (13 joints)
(4 or more joints)
117
62
21
(a) Construct and sketch a 95% confidence ellipse for the pair [ILl> IL2J', where ILl = E(X I ) and IL2 = E(X2)' (b) Suppose ILIO = 2000 and IL20 = lO,DOO represent "typical" values for stiffness and bending strength, respectively. Given the result in (a), are the data in Table 5.11 consistent with thesevalues? Explain.
268 Chapter 5 Inferences about a Mean Vector
Exercises 269
(c) Is the bivariate normal distributio n a viable population model? Exp lain with refer . ence to Q_Q plots and a scatter diagr am. . 5.20: A wildlife ecologist measured XI = taillength (in millim:ters) and X2 = wing. length (in millimeters) for a sample of n = 45 fema le hookbilled kites. These data are displ ayed in Tabl e 5.12. Usi~g the data in the table ,
Xl
X2
(Tai l leng th)
(Wing length)
284 191 285 197 288 208 273 180 275 180 280 188 283 210 288 196 271 191 257 179 289 208 285 202 272 200 282 192 280 199 Source: Data courtesy of S. Temple.
Xl
X2
Xl
x2
. (Tail length)
(Wing length)
(Tail leng th)
(Wing leng th)
186 197 201 190 209 187 207 178 202 205 190 189 211 216 189
266 285 295 282 305 285 297 268 271 285 280 277 310 305 274
173 194 198 180 190 191 196 207 209 179 186 174 181 189 188
271 280 300 272 292 286 285 286 303 261 262 245 250 262 258
(a) Find and sketch the 95% confidenc e ellipse for the population mea ns ILl and Suppose it is known that iLl = 190 mm and iL2 = 275 mm for male hook IL2' billed kites. Are these plausible values for the mean tail length and mea n wing leng th for the female birds? Explain. (b) Construct the simultane ous 95% T2_intervals for ILl and IL2 and the 95% Bonferroni intervals for iLl and iL2' Compare the two sets of intervals. Wha t advantage, if any, do the T2_intervals have over the Bonferron i intervals? (c) Is the bivariate normal distributio n a viable popu latio n model? Exp lain with reference to QQ plots and a scatter diagr am. 5.21. Usin g the data on bone mineral roni conte intervals for the individual means. nt in Table 1.8, construct the 95% Bon Also, find the 95% simultaneous 2 fer T intervals. Com pare the two sets of intervals. 5.22 . A portion of the data contained in Table The se data represent various costs assoc 6.10 in Chapter 6 is repr oduc ed in Table 5.13. iated with transporting milk from farm s to dairy plan ts for gasoline trucks. Only the first 25 multivariate observations for gaso line trucks are given. Observations 9 and 21 have been identified as outliers from the full data set of 36 observations. (See [2].)

Table 5.13 Milk Tran spor tatio nCo st Dat a
Fue l (xd '16.44 7.19 9.92 4.24 11.20 14.25 13.50 13.32 29.11 12.68 7.51 9.90 10.25 11.11 12.17 10.24 10.18 8.88 12.34 8.51 26.16 12.95 16.93 14.70 10.32
Rep air (xz) 12.43 2.70 1.35 5.78 5.05 5.78 10.98 14.27 15.09 7.61 5.80 3.63 5.07 6.15 14.26 2.59 6.05 2.70 7.73 14.02 17.44 8.24 13.37 10.78 5.16
Cap ital (X3) 11.23 3.92 9.75 7.78 10.67 9.88 10.60 . 9.45 3.28 10.23 8.13 9.13 10.17 7.61 14.39 6.09 12.14 12.23 11.68 12.01 16.89 7.18 17.59 14.58 17.00
(a) Construct QQ pIo tsof t h e marg Inal . distributio ~lso, construct the three possible scatt . d' ns of fuel, repair, and capi tal costs. d~fferent va~iables. Are the outliers ev~~ e~:~ rams from the pairs of obse rvat ions on dlagran;ts ~Ith, the appa rent outliers rem ov' :ze at the QQ plots and mally dlstn bute d? Discuss. the scat ter e. 0 the data now appe ar to be nor(b) Constr~ct 95% Bonferroni inter vals for t 95% T intervals. Com pare the two . .. t mdlvldual cost means. Also find se S 0 f~e Inter the vals. ' 5.23 . Tabl Con side r the 30 obse rvations on male E e 6.13 on page 349. gyph.an skulls for the first time peri od given in (a) Con struc t QQ plots of the mar inal . . . basl~ngt~ and nasheight varia bYes. ~~s~nbuhons of the ~axbreat h, bash eigh t, mul hvan ate obse rvat ions Do th d ' cons truc t Exp lain. quare plot of the . ese ata appe ar to abechis normally distr ibut ed? (b) Con struc t 95% Bon ferro ni inter Also, find the 95% TZintervals Cvals for .. . the IndlV 5 2" ldual skull dimension variables. . omp are the two sets of intervals. . 4. !:!smg the Madison, Wisconsin Polic X char ts .fo! X3 = hold over hour e D t s and e.!'a~men t data in Table 5.8, cons truct indi vidu al char acte nshc s seem to be in cont ro\? (Tb 4 . COA hours. Do these indiv . a t IS, are they stab le?) Comment. idual proc ess
• 270
Exercises
Chapter 5 Inferences about a Mean Vector 5.25. Refer to Exercise 5.24. Using the data on the holdover and COA overtime hours, construct a quality ellipse and a r 2chart.. Does the process represented by the bivariate observa tions appear to be in control? (That is, is it stable?) Commen t. Do you somethi ng from the multivar iate control charts that was not apparent in the'
I
X charts? 5.26. Construc t a r 2 chart using the data on Xl = legal appearances overtime X2 = extraord inary event overtime hours, and X3 = holdover overtime Table 5.8. Compar e this chart with the chart in Figure 5.8 of Example 5.10. Does r2 with an additional characteristic change your conclusion about process Explain. 5.27. Using the data on X3 = holdove r hours and X4 = COA hours from Table 5.8, a predictio n ellipse for a future observation x' = (X3' X4)' Rememb er, a ellipse should be calculate d from a stable process. Interpret the result. As part of a study of its sheet metal assembly process, a major automob ile manufacturer 5.28 uses sensors that record the deviation from the nominal thickness (miJIimeters) at six 10cations on a car. The first four are measured when the car body is complete and the two are measure d on the underbo dy at an earlier stage of assembly. Data on 50 cars are given in Table 5.14. (a) The process seems stable for the first 30 cases. Use these cases to estimate Sand i. Then construc t a r 2chart using all of the variables. Include all 50 cases. (b) Which individual locations seem to show a cause for concern? Refer to the car body data in Exercise 5.28. These are all measured as deviations from 5.29 target value so it is appropr iate to test the null hypothesis that the mean vector is zero. Using the first 30 cases, test Ho: JL = 0 at ll' = .05 Refer to the data on energy consumption in Exercise 3.18. 5.30 (a) Obtain the large sample 95% Bonferroni confidence intervals for the mean con· sumptio n of each of the four types, the total of the four, and the differenc e, petroleurn minus natural gas. (b) Obtain the large sample 95% simultaneous intervals for the mean consump of each of the four types, the total of the four, and the difference, petroleum tion minus natural gas. Compar e with your results for Part a.
\ \
\
r
5.31 Refer to the data on snow storms in Exercise 3.20. (a) Find a 95% confidence region for the mean vector after taking an appropri
ate trans
formation. (b) On the same scale, find the 95% Bonferroni confidence intervals for the two component means. ~
..
~
l "1
k"
~71
TABLE 5.14 Car Body Assemb ly Data Index
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Xl
0.12 0.60 0.13
X2
0040
0.36 0.35 0.05 0.37 0.24 0.16 0.24 0.05 0.16 0.24 0.83 0.30 0.10 0.06 0.35 0.30 0.35 0.85 0.34 0.36 0.59 0.50 0.20 0.30 0.35 0.36 0.35 0.25 0.25 0.16 0.12
0.60
0040
0046 0046 0046 0046 0.13 0.31 0.37 1.08
0042 0.31 0.14 0.61 0.61 0.84 0.96 0.90
0046 0.90 0.61 0.61
0046 0.60 0.60 0.31 0.60 0.31 0.36
0047 0046 0044
0.16 0.18 ....:0.12
0.90 0.50 0.38 0.60 0.11 0.05 0.85 0.37 0.11 0.60 0.84
0040
0046 0.56 0.56 0.25
0.35 0.08 0.35 0.24 0.12 0.65 0.10 0.24 0.24 0.59 0.16 0.35 0.16 0.12
Source: Data Courtesy of Darek Ceglarek.
X3
0040 0.04 0.84 0.30 0.37 0.Q7 0.13 0.01 0.20 0.37 0.81 0.37 0.24 0.18 0.24 0.20 0.14 0.19 0.78 0.24 0.13 0.34 0.58 0.10
0045 0.34
0045 0042 0.34 0.15
0048 0.20 0.34 0.16 0.20 0.75 0.84 0.55 0.35 0.15 0.85 0.50 0.10 0.75 0.13 0.05 0.37 0.10 0.37 0.05
X4
0.25 0.28 0.61 0.00 0.13 0.10 0.02 0.09 0.23 0.21 0.05 0.58 0.24 0.50 0.75 0.21 0.22 0.18 0.15 0.58 0.13 0.58 0.20 0.10 0.37 0.11 0.10 0.28 0.24 0.38 0.34 0.32 0.31 0.01
0048 0.31 0.52 0.15 0.34 0.40 0.55 0.35 0.58 0.10 0.84 0.61 0.15 0.75 0.25 0.20
X5
1.37 0.25 1.45 0.12 0.78 1.15 0.26 0.15 0.65 1.15 0.21 0.00 0.65 1.25 0.15 0.50 1.65 1.00 0.25 0.15 0.60 0.95 1.10 0.75 1.18 1.68 1.00 0.75 0.65 1.18 0.30 0.50 0.85 0.60
1040 0.60 0.35 0.80 0.60 0.00 1.65 0.80 1.85 0.65 0.85 1.00 0.68
0045 1.05 1.21
X6
0.13 0.15 0.25 0.25 0.15 0.18 0.20 0.18 0.15 0.05 0.00
0045 0.35 0.05 0.20 0.25 0.05 0.08 0.25 0.25 0.08 0.08 0.00 0.10 0.30 0.32 0.25 0.10 0.10 0.10 0.20 0.10 0.60 0.35 0.10 0.10 0.75 0.10 0.85 0.10 0.10 0.21 0.11 0.10 0.15 0.20 0.25 0.20 0.15 0.10
272 Chapter 5 Inferences about a Mean Vector
References 1 A sti A. Categorical Data Analysis (2nd ed.), New York: John WHey, 2~. . WK F "A New Graphical Method for Detectmg Smgle . gre , 2. BaconSone , J:, an~ U· : ~nt· and Multivariate Data." Applied Statistics, 36, no. 2 Multiple Outh~rs m mvana e (1987),153162. 0 k Mathematical Statistics: Basic Ideas and Selected Topics, 3. Bickel, P. J., and K. A. 0 sum. . . H 11 2000 Vo!. I (2nd ed.), Upper Saddle River, NI: PrentIce a, . . . ' .. . band P.W Holland B' h Y M M S E Fem erg, .. . Discrete Multlvanate AnalysIS. Theory 4. a~~ ~~~c;ice' (p~p~rb~ck). Cambridge, MA: The MIt Press, 1977. M L . d nd D B Rubin. "Maximum Likelihood from Incomplete 5. Demps~er, A. P., N. . .ahlr ,(a 'th Di~cussion)." Journal of the Royal Statistical Society Data via the EM Algont m Wl (B) 39 no. 1 (1977),138. . ". '. , , . L'k rhood Estimation from Incomplete Data. BIOmetriCS, 14 6. Hartley, H. O. "MaXimum I e 1 (1958) 174194. " B' . 27 ' R H k' "The Analysis of Incomplete Data. IOmetrrcs, 7. Hartley, H. 0., and R. . oc mg. (1971),783808. . . . S . IC L l d "A Linear CombmatlOns Test for Detectmg ena or8. Iohnson, R. A. a~d a~ "Topics in Statistical Dependence. (1991) Institute of . relation in MultIvanate amp es. I 299 313 M thematical Statistics Monograph, Eds. Block, H. et a ., . a d R L' "Multivariate Statistical Process Control Schemes for Control9. Johnson, R.A. an .' I H db k of Engineering Statistics (2006), H. Pham, Ed. ling a Mean." Sprmger an 00 Springer Berlin. v k J h WI 's . t' I Methods for Quality Improvement (2nd ed.). New .or : 0 n Iey, 10. Ryan, T. P. tafts Ica ' . 2000. . . t M S' h "Robust Statistics for Testing Mean Vectors 0 f M uI'tlvana e 11. Tiku, M. L., and . mg... . StatisticsTheory and Methods, 11, no. 9 (1982), Distributions." CommunIcatIOns In
'f:
9851001.
ant
COMPARISONS OF SEVERAL MULTIVARIATEMEANS 6.1
Introduction The ideas developed in Chapter 5 can be extended to handle problems involving the comparison of several mean vectors. The theory is a little more complicated and rests on an assumption of multivariate normal distributions or large sample sizes. Similarly, the notation becomes a bit cumbersome. To circumvent these problems, we shall often review univariate procedures for comparing several means and then generalize to the corresponding multivariate cases by analogy. The numerical examples we present will help cement the concepts. Because comparisons of means frequently (and should) emanate from designed experiments, we take the opportunity to discuss some of the tenets of good experimental practice. A repeated measures design, useful in behavioral studies, is explicitly considered, along with modifications required to analyze growth curves. We begin by considering pairs of mean vectors. In later sections, we discuss several comparisons among mean vectors arranged according to treatment levels. The corresponding test statistics depend upon a partitioning of the total variation into pieces of variation attributable to the treatment sources and error. This partitioning is known as the multivariate analysis o/variance (MANOVA).
6.2 Paired Comparisons and a Repeated Measures Design , Paired Comparisons Measurements are often recorded under different sets of experimental conditions to see whether the responses differ significantly over these sets. For example, the efficacy of a new drug or of a saturation advertising campaign may be determined by comparing measurements before the "treatment" (drug or advertising) with those 273
Paired Comparisons and a Repeated Measures Design 275
274 Chapter 6 Comparisons of Several Multivariate Means after the treatment. In other situations, two or more treatments can be aOInm:istelrl'j to the same or similar experimental units, and responses can be compared to the effects of the treatments. One rational approach to comparing two treatments, or the presence and sence of a single treatment, is to assign both treatments to the same or identical (individuals, stores, plots of land, and so forth). The paired responses may then analyzed by computing their differences, thereby eliminating much of the of extraneous unittounit variation. In the single response (univariate) case, let X jI denote the response treatment 1 (or the response before treatment), and let X jZ denote the response treatment 2 (or the response after treatment) for the jth trial. That is, (Xjl, are measurements recorded on the jth unit or jth pair of like units. By design, n differences . j = 1,2, ... , n should reflect only the differential effects of the treatments. Given that the differences Dj in (61) represent independent observations an N (0, u~) distribution, the variable l5  8
and the p paireddifference random variables become
Let Dj =
where
_
1
2:n Dj
D = 
versus
=
_ D
2:
d  t,,_I(a/2)
Vn
:5
8
:5
Sd
d + fll I(a/2) Yn
(64)
(For example, see [11].) Additional notation is required for the multivariate extension of the pairedcomparison procedure. It is necessary to distinguish between p responses, two treatments, and n experimental units. We label the p responses within the jth unit as Xli! = variable 1 under treatment 1 Xl j2
= variable 2 under treatment 1
X lj p =
variab!.~.~.~.~.~~.~.e~~~~.~~.~....
X;;~';;;'~~;:f~ble 1 under treatment 2
X 2jZ = variable 2 under treatment 2 X 2j p = variable p under treatment 2

X 2jp
and assume, for j = 1,2, ... , n, that
(67)
1
1
Il
2: Dj n J=I
=
and
Sd
n
= _ 2: n
1
j=I
(Dj  D)(Dj  D)'
(68)
=
n(D  8)'Sd I (D  8)
is distributed as an [( n  1 )p/ (n  p) )Fp.np random variable, whatever the true 8 and l:d' .
*

Djp = X ljp Djp),
TZ
HI: 0 0 may be conducted by comparing It I with tll _l(a/2)the upper l00(a/2)th percentile of a tdistribution with n  1 dJ. A 100(1  a) % confidence interval for the mean difference 0 = E( Xi!  X j2 ) is provided the statement Sd
D jz , ••• ,
Result 6.1. Let the differences Db Oz, ... , Dn be a random sample from an Np ( 8, l:d) population. Then
1 j=l
(zerome~ndifferencefortreatments)
_
fDjI ,
(65)
where
has a tdistribution with n  1 dJ. Consequently, an alevel test of
Ho: 0
X 2j2
If, in addition, D I , D 2 , ... , Dn are independent N p ( 8, l:d) random vectors, inferences about the vector of mean differences 8 can be based upon a TZstatistic. Specificall y,
Yn n
X ZiI

T Z = n(D  8)'S;?(D  8)
1 " and s~ =  _ (Dj _l5)z
n j=I

= X lj2
(66)
t=Sd/
Dj~ = X lj1 Dj2
If nand n  p are both large, T Z is approximately distributed as a ~ random variable, regardless of the form of the underlying population of difference~. Proof. The exact distribution of T2 is a restatement of the summary in (56), with vectors of differences for the observation vectors. The approximate distribution of TZ, for n andn  p large, follows from (428). •
The condition 8 = 0 is equivalent to "no average difference between the two treatments." For the ith variable, 0; > 0 implies that treatment 1 is larger, on average, than treatment 2. In general, inferences about 8 can be made using Result 6.1. Given the observed differences dj = [djI , dj2 , .•• , d j p), j = 1,2, ... , n, corresponding to the random variables in (65), an alevel test of Ho: 8 = 0 versus HI: 8 0 for an N p ( 8, l:d) population rejects Ho if the observed
*
TZ = nd'SId > (n  l)p F () d (n _ p) ~np a where Fp,n_p(a) is tEe upper (l00a)th percentile of an Fdistribution with p and n  p dJ. Here d and Sd are given by (68).
276
Paired Comparisons and a Repeated Measures Design ~77
Chapter 6 Comparisons of Several Multivariate Means
A lOD( 1  a)% confidence region for B consists of all B such that _
(n1)p
,t
( d  B) Sd (d  B) ~ n( n  p ) Fp,lI_p(a) .
(69)
Also, 100( 1  ~a)% simultaneous confidence intervals for the individual mean [Ji are given by 1)p (610) (n _ p) Fp,np(a) \j;
differences
g
en 
where di is the ith element of ii.and S~i is the ith diagon~l e~ement of Sd' , For n  p large, [en  l)p/(n  p)JFp,lI_p(a) = Xp(a) and normalIty need not be assumed. . ' The Bonferroni 100(1  a)% simultaneous confidence mtervals for the individual mean differences are
ai : di ± tnI(2~) ~
Do the two laboratories' chemical analyses agree? If differences exist, what is their nature? The T 2 statistic for testing Ho: 8' = [01, a2 ) = [O,OJ is constructed from the differences of paired observations: dj! =
Xljl 
X2jl
d j2 =
Xlj2 
X2j2
19 22 18 27
10
12
42
15
4 10
14
11
4
1
17
9
4 19
60 2
10
7
Here
d=
[~IJ = d 2
s
[9.36J 13.27 '
d
= [199.26 88.38
88.38J 418.61
and
(610a)
T2 = l1[ 9.36
where t _t(a/2p) is the upper 100(a/2p)th percentile of a tdistribution with n
,
13.27J [
.0055 .0012
.0012J [9.36J .0026 13.27
=
13.
6
n  1 dJ.
Checking for a mean difference with paired observations) Municipal Examp Ie 6 . I ( . h' d' h . treatment plants are required by law to momtor t elr lSC arges mto t was t ewa er . b'l' fd t f
rivers and streams on a regular basis. Concern about the rella 1 Ity 0 a a rom one of these selfmonitoring programs led to a study in whi~h samples of effluent were divided and sent to two laboratories for testing. Onehalf of each sample ,:"as sent to the Wisconsin State Laboratory of Hygiene, and onehalf was sent to a prIvate co~ merciallaboratory routinely used in the monitoring pr~gram. Measuremen~s of biOchemical oxygen demand (BOD) and suspended solIds (SS~ were o?tamed, for n = 11 sample splits, from the two laboratories. The data are displayed 111 Table 6.1.
Taking a = .05, we find that [pen 1)/(n  p»)Fp.n_p(.05) = [2(1O)/9)F2 ,9(·05) = 9.47. Since T2 = 13.6 > 9.47, we reject Ho and conclude that there is a nonzero mean difference between the measurements of the two laboratories. It appears, from inspection of the data, that the commercial lab tends to produce lower BOD measurements and higher SS measurements than the State Lab of Hygiene. The 95% simultaneous confidence intervals for the mean differences a1 and 02 can be computed using (610). These intervals are

01: d] ±
~(n1)p J?j;~J ( ) Fp np(a) np'
n
= 9.36
± V9.47
J199.26 .11 or
Table 6.1 Effluent Data Commercial lab Xlj2 (SS) Xljl (BOD) Samplej 27 6 1 23 6 2 64 lR 3 44 8 4 30 11 5 75 34 6 26 28 7 124 71 8 54 43 9 30 33 10 14 20 11 Source: Data courtesy of S. Weber.
State lab of hygiene X2j2 (SS) X2jl (BOD)
25 28 36 35 15 44 42 54 34 29 39
15 13 22 29 31 64
30 64 56 20 21
[J2:
13.27 ± V9.47
)418.61 11
or
(22.46,3.74)
(5.71,32.25)
The 95% simultaneous confidence intervals include zero, yet the hypothesis Ho: iJ = 0 was rejected at the 5% level. What are we to conclude? The evideQ.ce points toward real differences. The point iJ = 0 falls outside the 95% confidence region for li (see Exercise 6.1), and this result is consistent with the T 2test. The 95% simultaneous confidence coefficient applies to the entire set of intervals that could be constructed for all possible linear combinations of the form al01 + a202' The particular intervals corresponding to the choices (al = 1, a2 '" 0) and (aJ = 0, a2 = 1) contain zero. Other choices of a1 and a2 will produce siIl1ultaneous intervals that do not contain zero. (If the hypothesis Ho: li '" 0 were not rejected, then all simultaneous intervals would include zero.) The Bonferroni simultaneous intervals also cover zero. (See Exercise 6.2.)
278
Chapter 6 Comparisons of Several Multivariate Means Paired Comparisons and a Repeated Measures Design
Our analysis assumed a normal distribution for the Dj. In fact, the situation further complicated by the presence of one or, possibly, two outliers. (See 6.3.) These data can be transformed to data more nearly normal, but with small sample, it is difficult to remove the effects of the outlier(s). (See Exercise The numerical results of this example illustrate an unusual circumstance can occur when.making inferences. The experimenter in Example 6.1 actually divided a sample by first shaking it then pouring it rapidly back and forth into two bottles for chemical analysis. This prudent because a simple division of the sample into two pieces obtained by the top half into one bottle and the remainder into another bottle might result in suspended solids in the lower half due to setting. The two laboratories would then be working with the same, or even like, experimental units, and the conclusions not pertain to laboratory competence, measuring techniques, and so forth. Whenever an investigator can control the aSSignment of treatments to experimental units, an appropriate pairing of units and a randomized assignment of ments can' enhance the statistical analysis. Differences, if any, between supposedly identical units must be identified and mostalike units paired. Further, a random assignment of treatment 1 to one unit and treatment 2 to the other unit will help eliminate the systematic effects of uncontrolled sources of variation. Randomization can be implemented by flipping a coin to determine whether the first unit in a pair receives treatment 1 (heads) or treatment 2 (tails). The remaining treatment is then assigned to the other unit. A separate independent randomization is conducted for each pair. One can conceive of the process as follows: Experimental Design for Paired Comparisons
Like pairs of experimental units
3
2
{6
D ••• 0 D ···0
D D
t
t
Treatments I and 2 assigned at random
n
Treatments I and2 assigned at random
•••
Treatments I and2 assigned at random
[XII, X12,"" Xl p' X2l> Xn,·.·, X2p]
and S is the 2p x 2p matrix of sample variances and covariances arranged as
S ==
th . .I~~ ar y, 22 contaIns the sample variances and covariances computed or .e p vana es on treatment 2. Finally, S12 = Sh are the matrices of sample cov.arbIances computed from Observations on pairs of treatment 1 and treatment 2 vana les. Defining the matrix
r
.
e =
(px2p)
0 0
1
1
0 1
1
~
(613)
j (p + 1 )st column
we can verify (see Exercise 6.9) that j =
d = ex
and
1,2, ... , n
Sd =
esc'
(614)
Thus, (615) d 0 th th and it .is. not necessary first to calculate the differences d d hand t . t I I 1, 2"", n' n eo er , ~ IS WIse 0 ca cu ate these differences in order to check normality and the assumptIOn of a random sample. Each row eI of . the . m a t' . a contrast vector because its elements nx e'In (6  13) IS sum t 0 zero. A ttention IS usually t d ' Ea h . . cen ere on contrasts when comparing treatments. c contrast IS perpendIcular to the vector l' = [1 1 1]' '1  0 Th com t 1" , "", smce Ci  . e ·p?neT~ Xj, rep~ese~tmg the overall treatment sum, is ignored by the test t s a IShc presented m thIS section.
t
A Repeated Measures Design for Comparing Treatments
We conclude our discussion of paired comparisons by noting that d and Sd, and hence T2, may be calculated from the fullsample quantities x and S. Here x is the 2p x 1 vector of sample averages for the p variables on the two treatments given by
x' ==
~:t:~~~~ SS~ c~nt~in~ the sample variances and covariances for the p variables on
f
t
t
Treatments I and 2 assigned at random
Atnothter generalization of the univariate paired tstatistic arises in situations where q rea ments are compared with res tt . I or . I" pec 0 a smg e response variable. Each subject e~Pthenbmenta .Ulll~ receIves each treatment once over successive periods of time Th eJ 0 servatlOn IS .
(611) j = 1,2, ... ,n
[(~;~) (~~~)] S21 (pXp)
522 (pxp)
279
where X ji is the response to the ith treatment on the ,'th unl't The d m as t fr . name repeate e ures s ems om the fact that all treatments are administered to each unit.
280
Paired Comparisons and a Repeated Measures Design 281
Chapter 6 Comparisons of Several Multivariate Means For comparative purposes, we consider contrasts of the components IL = E(X j ). These could be 1 0 0 1 ILl : IL3 = ~ . ..
['
r~J ~.
1
ILl  ILq
or
jJm~c,p
: ] l~ ~ : ~ . . .~ ~ll~~J l :~ ~
=
ILq  ILql
0 0
. A co~fidence region for contrasts CIL, with IL the mean of a normal population, IS determmed by the set of all CIL such that n(Cx  CIL),(CSCT\Cx  CIL)
(617)
c'x ±
)(n 
1)(q  1) F ( ) (n  q + 1) q1.nq+1 a
)CIsc n
(618)
Example .6.2 (Testing for equal treatments in a repeated measures design) Improved
1 1J ILq
anesthetIcs are often developed by first studying their effects on animals. In one 19 dogs were initially given the drug pentobarbitol. Each dog was then admIlllstered carbon dioxide CO 2 at each of two pressure levels. Next halothane (H) was added, and the administration of CO 2 was repeated. The respon~e, milliseconds between heartbeats, was measured for the four treatment combinations: st~~y,
Both Cl and C are called contrast matrices, because their q  1 rows are linearly' 2 independent and each is a contrast vector. The nature of the design eliminates much of the influence of unittounit variation on treatment comparisons. Of course, . experimenter should randomize the order in which the treatments are presented to
Present
each subject. When the treatment means are equal, C1IL = C 2IL = O. In general, the hypothesis that there are no differences in treatments (equal treatment means) becomes CIL = 0 for any choice of the contrast matrix C. Consequently, based on the contrasts CXj in the observations, we have means 2 C x and covariance matrix CSC', and we test CIL = 0 using the T statistic T2 =
(n  1)(q  1) F ( ) (n  q + 1) ql,nq+1 ex
whe~e x an~ S are as defined in (616). Consequently, simultaneous 100(1  a)% c?nfIdence mtervals for single contrasts c' IL for any contrast vectors of interest are gIven by (see Result 5A.1)
C'IL:
= C 21L
:5
Halothane Absent Low
High
C02 pressure
n(Cx),(CSCTlCX
Table 6.2 contains the four measurements for each of the 19 dogs, where
Test for Equality of Treatments in a Repeated Measures Design Consider an N q ( IL, l:) population, and let C be a contrast matrix. An alevel test of Ho: CIL = 0 (equal treatment means) versus HI: CIL * 0 is as follows: Reject Ho if (n  1)(q  1) (616) T2 = n(Cx)'(CSCTICX > (n _ q + 1) FqI.nq+l(a) where F 1.nq+l(a) is the upper (lOOa)th percentile of an Fdistribution wit~ q q _ 1 and n  q + 1 dJ. Here x and S are the sample mean vector and covanance matrix defined, respectively, by
x=
1 ~ LJ
n
j=1
Xj
and S =
1 LJ ~ (Xj =1
n
x) ( Xj

x)'
Treatment 1 Treatment 2 Treatment 3 Treatment 4
l
I Any pair of contrast matrices Cl and C2 must be related by Cl = BC2, with B nonsingular. This follows because each C has the largest possible number, q  1. of linearly independent rows, all perpendicular to the vector 1. Then (BC2),(BC2SCiBTI(BC2) = CiB'(BTI(C2SCirIB~IBC2 = Q(C Sq)I C2 • so T2 computed with C2 orC I = BC2gives the same result. 2
= Iow CO 2 pressure without H = high CO2 pressure with H = Iow CO2 pressure with H
. We shall analyze the anesthetizing effects of CO 2 pressure and halothane from thIS repeatedmeasures design. There are three treatment contrasts that might be of interest in the experiment. Let ILl , IL~' IL3, and IL4 correspond to the mean responses for treatments 1,2,3, and 4, respectIvely. Then Halothane contrast representing the) difference between the presence and ( absence of halothane
(IL3
+ 1L4)
 (ILl
+
IL2) =
(ILl
+ IL3)
 (IL2
+
IL4) = (C0 2 contrast. representing the difference)
+ IL4)
 (IL2
+
IL3) =
j=1
It can be shown that T2 does not depend on the particular choice of C.
= high CO 2 pressure without H
(ILl
between hIgh and Iow CO 2 pressure Contrast representing the influence ) of halothane on CO 2 pressure differences ( (H C02 pressure "interaction")
282
Paired Comparisons and a Repeate d Measure s Design
Chapter 6 Compari sons of Several Multivariate Means
With a = .05,
Table 6.2 SleepingDog Data
Treatment Dog 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
~
1
2
3
4
426 253 359 432 405 324 310 326 375 286 349 429 348 412 347 434 364 420 397
609 236 433 431 426 438 312 326 447 286 382 410 377 473 326 458 367 395 556
556 392 349 522 513 507 410 350 547 403 473 488 447 472 455 637 432 508 645
600 395 357 600 513 539 456 504 548 422 497 547 514 446 468 524 469 531 625
18(3) 18(3) (n  l)(q  1) (3.24) = 10.94 (n  q + 1) Fq I ,Il_q+l(a ) = ~ F3,16(·05) = 16 nt effects). From (616), rZ = 116> 10.94, and we reject Ho: Cp. =: 0 (no treatme HQ, we construc t of n rejectio the for ble responsi are s contrast the of which see To (618), the 95% simulta neous confide nce intervals for these contrasts. From contrast cip. = (IL3 + IL4)  (J.LI + J.L2)
=:
halotha ne influence
is estimate d by the interval
(X3 + X4)  (XI + X2) ±
18(3) F ,16(.05) )CiSCl ~= 16" 3
d by where ci is the first row of C. Similarly, the remaining contrasts are estimate CO2 pressure influence = (J.Ll + J.L3)  (J.Lz + J.L4):  60.05 ± VlO.94
=
[P.l, ILz, IL3, IL4j, the contrast matrix C is
C=
1
i =
f
and
502.89
S=
2819.29 3568.42 7963.14 2943.49 5303.98 6851.32 2295.35 4065.44 4499.63
f
It can be verified that
209.31] Cx = 60.05 ; [ 12.79
CSC'
9432.32 1098.92 927.62] 1098.92 5195.84 914.54 = [ 927.62 914.54 7557.44
and
rZ
 12.79 ± VlO.94
)7557.44 = 12.79 ± 65.97 19
1
The data (see Table 6.2) give 368.21J 404.63 479.26
4 )5195.8   = 60.05 ± 54.70 19
HC02 pressure "interac tion" = (J.Ll + J.L4)  (J.L2 + J.L3):
[~1 =~ ~ ~] 1
2 ~ )9432.3 19
. 209.31 ± v 10.94
= 209.31 ± 73.70
Source: Data courtesy of Dr. 1. Atlee.
With p.'
283
= n(Cx)'( CSCTl (Ci) = 19(6.11) = 116
The presThe first confidence interval implies that there is a halotha ne effect. at both occurs This ts. heartbea between times longer s produce ence of halotha ne , contrast ion interact levels of CO2 pressure , since the HC0 2 pressure third the (See zero. from t differen ntly significa (J.LI + J.L4)  (li2  J.L3), is not there is an confidence interval.) The second confidence interval indicate s that between times longer s produce pressure CO 2 effect due to CO2 pressure : The lower heartbeats. the Some caution must be exercised in our interpre tation of the results because due be may t Heffec t apparen The without. those follow must ne trials with halotha determi ned at to a time trend. (Ideally, the time order of all treatme nts should be _ random.) (X) = l:, The test in (616) is appropr iate when the covariance matrix, Cov that l: assume to ble reasona is cannot be assumed to have any special structure. If it higher have mind in e structur this with designed tests has a particular structure, e (814), see power than the one in (616). (For l: with the equal correlation structur [22).) or (17J in design block" ized a discussion of the "random
284
Comparing Mean Vectors from l\vo Populations
Chapter 6 Comparisons of Several Multivariate Means'
285
Further Assumptions When nl and n2 'Are Small
6.3 Comparing Mean Vectors from Two Populations A TZstatistic for testing the equality of vector means from two multivariate tions can be developed by analogy with the univariate procedure. (See [l1J for cussion of the univariate case.) This T 2 statistic is appropriate for <Ulnn,.r... ;;;' responses from oneset of experimental settings (population 1) with independent sponses from another set of experimental settings (population 2). The COlnD,ari~:nn. can be made without explicitly controlling for unittounit variability, as in pairedcomparison case. If possible, the experimental units should be randomly assigned to the sets experimental conditions. Randomlzation will, to some extent, mitigate the of unit"tounit variability in a subsequent comparison of treatments. Although precision is lost relative to paired comparisons, the inferences in the tW'O~)oDluhlti('ln case are, ordinarily, applicable to a more general collection of experimental units simply because unit homogeneity is not required. . Consider a random sample of size nl from population 1 and a sample of', size n2 from population 2. The observations on p variables can be arranged as follows:
1. Both populations are muItivariate normal. 2. Also, ~I = ~z (same covariance matrix).
(620)
The second assumption, that ~I = ~z, is much stronger than its univariate counterpart. Here we are assuming that several pairs of variances and covariances are nearly equal. When ~I
= ~2 = ~,
n2
n1
L
j=1
(xlj  XI) (Xlj  xd is an estimate of (n}  1)~ and
L(X2j  X2)(X2j  xz)'isanestimateof(n2  1)~.Consequently,wecanpoolthe j=1 information in both samples in order to estimate the common covariance ~. We set
(621) Sample
Summary statistics
~
Since
(Population 1) XII,xI2"",XlnJ
~
XI) (xlj  xd has
1 dJ. and
nl 
j=1
L (X2j 
X2) (X2j  xz)' has
j=1
1 dJ., the divisor (nl  1) + (nz  1) in (621) is obtained by combining the two component degrees of freedom. [See (424).J Additional support for the pooling procedure comes from consideration of the multivariate normal likelihood. (See Exercise 6.11.) To test the hypothesis that ILl  IL2 = 8 0 , a specified vector, we consider the squared statistical distance from XI  Xz to 8 0 , Now, n2 
(Population 2) X21, XZ2, ... , X2n2 In this notation, the first subscriptl or 2denotes the population. We want to make inferences about (mean vector of population 1)  (mean vector of population 2) =
L (Xlj 
£(XI  X 2) ILl  ILz.
For instance, we shall want to answer the question, Is ILl = IL2 (or, equivalently, is O)? Also, if ILl  IL2 * 0, which component means are different? With a few tentative assumptions, we are able to provide answers to these questions.
ILl  IL2 =
= £(XI)
  Xz) COV(XI
Assumptions Concerning the Structure of the Data
We shall see later that, for large samples, this structure is sufficient for making inferences about the p X 1 vector ILl  IL2' However, when the sample sizes nl and n2 are small, more assumptions are needed.
= ILl
 ILz
Since the independence assumption in (619) implies that Xl and X 2 are independent and thus Cov (Xl, Xz) = 0 (see Result 4.5), by (39), it follows that = Cov(Xd +
 ) Cov(X z
Because Spooled estimates ~, we see that
1. The sample XII, X I2 ,.·., X ln1 , is a random sample of size nl from a pvariate population with mean vector ILl and covariance matrix ~I' 2. The sample X 21 , X 2Z , ... , X 2n2 , is a random sample of size n2 from a pvariate population with mean vector IL2 and covariance matrix ~2' (619) 3. Also, XII, X IZ ,"" XlnJ' are independent ofX2!,X zz "", X 2n2 .
 £(X 2)
(:1
+
1 = ~ + nl
:J
1 ~ nz
= (1  + 1) ~ nl
nz
(622)
Spooled
is an estimator of Cov (X I  X 2). The likelihood ratio test of
Ho: ILl

ILz = 8 0
is based on the square of the statistical distance, T2, and is given by (see [1]). Reject Ho if
T = (XI  X2  ( 0)' [ (:1 + Z
:JSPooled JI (XI 
X2  ( 0) >
CZ (623)
Comparing Mean Vectors from Two Populations 287
Chapter P Comparisons of Several Multivariate Means
286
where the critical distance cZ is determined from the distribution of the twosample T 2.statistic. Result 6.2. IfX ll , X 12 ' ... , XlIII is a random sample of size nl from Np(llj, I) X 2 1> X 22, ••. ' X 21lZ is an independent random sample of size nz from N p (1l2, I), 2



T = [Xl  Xz  (Ill  Ilz)]
, [(
is distributed as
1
1)
nl + nz Spooled
nz  2)p
(n! + ( nl + nz 
P  1)
Jl ( [XI  X z  III  Ilz)j
We are primarily interested in confidence regions for III  1l2' From (624), we conclude that all III  112 within squared statistical distance CZof Xl  xz constitute the confidence region. This region is an ellipsoid centered at the observed difference Xl  Xz and whose axes are determined by the eigenvalues and eigenvectors of Spooled (or S;;';oled)' Example 6.3 (Constructing a confidence region for the difference of two mean vectors)
Fifty bars of soap are manufactured in each of two ways. Two characteristics, Xl = lather and X z = mildness, are measured. The summary statistics for bars produced by methods 1 and 2 are
Fp.",+I7,pl
XI = [8.3J 4.1'
SI =
X = [1O.2J 2 3.9'
Sz =
U!J [~ !J
Consequently, P
[
1   Xz  (Ill  Ilz» , [ ( III (Xl
1 ) Spooled + nz
JI (Xl  X 2 
zJ
(Ill  1l2» s c
= 1  er .
(624) where
Obtain a 95% confidence region for III  1l2' We first note that SI and S2 are approximately equal, so that it is reasonable to pool them. Hence, from (621),
49 SI + 98 49 Sz = [21 51J Spooled = 98 Proof. We first note that
_ 1 X  X =  X ll 1
2
n1
Also,
1 1 IX 1X IX + n1 X I2 + '" +  XI  21  22  '"  2 nl "I n2 nZ nZ "2
is distributed as
= .. , =
C'"
= llnl
and C",+I
= C"I+2 = .. , =
(n1  1 )SI is distributed as w,'I l (I) and (nz  1)Sz as W1l2 
j
C"'+"2
=
T2 =
(

nl
+
_  Xz )1 /2(Xl
(Ill 
, Ilz» S~ooled 1
nZ
(
1
nl
1
+
)l/Z(Xl   X z 
dJ.
nl
i = 1,2
.290J el = [ .957
and ez =
[
.957J
_ .290
By Result 6.2, 1 1) 2 (1 ( nl + n2 C = 50
random vector
= N (0, I)' [Wn l +nr P
+9
are
(Ill  IlZ)
nZ
= (multivariate normal)' (Wishart random matrix)I (multivariate normal)
random vector
0= ISpooled  All = /2  A I / = A2  7A 15 A
so A = (7 ± y49  36)/2. Consequently, Al = 5.303 and A2 = 1.697, and the corresponding eigenvectors, el and ez, determined from
Cl)
By assumption, the X1/s and the X 2/s are independent, so (nl  l)SI and (nz  1 )Sz are also independent. From (424), Cnl  1 )Sj + (nz  1 )Sz is then distributed as Wnl+nzz(I). Therefore, 1
X2 = [1.9J .2
so the confidence ellipse is centered at [ 1.9, .2)'. The eigenvalues and eigenvectors of Spooled are obtained from the equation
by Result 4.8, with Cl = C2 l/nz. According to (423),
1
 Xl
2(I)J1 N (0, I) + nz  2 P
which is the TZ·distribution specified in (58), with n replaced by nl (55). for the relation to F.]
1 ) (98)(2)
+ 50 (97) F2•97 (·05)
= .25
since F2,97(.05) = 3.1. The confidence ellipse extends
+ n2

1. [See •
v'A;
1(1.. + 1..) c
\j
nl
n2
2
=
v'A; v'25
..
288 Chapter 6 Comparisons of Several Multivariate Means
Comparing Mean Vectors from lWo Populations 289 are both estimators of a'1:a, the common popUlation variance of the linear combinations a'XI and a'Xz' Pooling these estimators, we obtain
2.0
S~, pooled
(111  I)Sf,a
+ (I1Z 
l)s~,a
== ':":~'':'"'(nl + 112  2) == a' [111 ';
~ ~ 2 SI + 111 '; ~ ~ 2 S2 J a
(625)
== a'Spooled a To test Ho: a' (ILl  ILz) == a' 00, on the basis of the a'X lj and a'X Zj , we can form the square of the univariate twosample 'statistic
[a'(X I  X 2  (ILl ~ ILz»]z
Figure 6.1 95% confidence ellipse forlLl  IL2'
1.0
units along the eigenvector ei, or 1.15 units in the el direction and .65 units in the ez direction. The 95% confidence ellipse is shown in Figure 6.1. Clearly, ILl  ILz == 0 is not in the ellipse, and we conclude that the two methods of manufacturing soap produce different results. It appears as if the two processes produce bars of soap with about the same mildness (Xz), but lhose from the second process have more lather (Xd. •
It is possible to derive simultaneous confidence intervals for the components of the vector ILl  ILz· These confidence intervals are developed from a consideration of all possible linear combinations of the differences in the mean vectors. It is assumed that the parent multivariate populations are normal with a common covariance 1:.
Result 6.3. Let cZ == probability 1  a.
[(111
+
I1Z 
2)p/(nl +
I1Z 
will cover a'(ILI  ILz) for all a. In particular ILli

(~ + ~) Sii,pooled 111
P  1)]Fp.l1l+n2pI(a). With
ILZi
will be covered by
for i == 1,2, ... , p
112
Proof. Consider univariate linear combinations of the observations
XII,XIZ,,,,,X1nl
According to the maximization lemma with d = (XI  X 2 B == (1/111 + 1/11z)Spooled in (250),
z (XI   X z  (ILl  ILz» , [(1 ta:s:
11.1
and X21,X22"",XZn2
given by a'X lj == alXljl + a ZX lj2 + ., . + apXljp and a'X Zj == alXZjl '+ azXZjz + ... + ap X 2jp ' These linear combinations have~ample me~s and covariances a'X1 , a'Sla and a'Xz, a'S2a, respectively, where Xl> SI, and X 2 , Sz are the mean and covariance statistics for the two original samples, (See Result 3.5.) When both parent populations have the same covariance matrix, sf.a == a'Sla and s~,a == a'Sza
+ 1 ) Spooled I1.z

(ILl  IL2»
and
JI (XI
== T Z for all a # O. Thus, (1  a) == P[Tz:s: c Z] = P[t;:s: cZ,
Simultaneous Confidence Intervals
(626)
a ,( 1 + 1 ) Spooleda 111 I1Z
==p[la'(X I
~ Xz) 
for all a]
a'(ILI  ILz)1 :s: c
a ,( 1 + 1 ) Spooleda nl I1Z
where cZ is selected according to Result 6,2.
for all
a] •
Remark. For testing Ho: ILl  ILz == 0, the linear combination a'(X1  xz), with coefficient vector a ex S~60Icd(Xl  xz), quantifies the largest popUlation difference, That is, if T Z rejects Ho, then a'(xI  Xz) will have a nonzero mean. Frequently, we try to interpret the components of this linear combination for both subject matter and statistical importance. Example 6.4 (Calculating simultaneous confidence intervals for the differences in mean components) Samples of sizes 111 == 45 and I1Z == 55 were taken of Wisconsin homeowners with and without air conditioning, respectively, (Data courtesy of Statistical Laboratory, University of Wisconsin,) Two measurements of electrical usage (in kilowatt hours) were considered, The first is a measure of total onpeak consumption (XI) during July, and the second is a measure of total offpeak consumption (Xz) during July. The resulting summary statistics are XI
=
[204.4J 556.6'
[130.0J Xz == 355.0'
. [13825.3 SI == 23823.4
Sz ==
23823.4J 73107.4 '
[8632,0 19616.7J 19616.7 55964.5 '
nz == 55
290
Comparing Mean Vectors from TWo PopuJations
Chapter 6 Comparisons of Several Multivariate Means
(The offpeak consumption is higher than the onpeak consumption because there are more offpeak hours in a month.) Let us find 95% simultaneous confidence intervals for the differences in the mean components. Although there appears to be somewhat of a discrepancy in the sample variances, for illustrative purposes we proceed to a calculation of the pooled sample covariance matrix. Here nl  1
Spooled
= nl
+
n2 
n2 
2 SI + nl +
1
n2 
2 S2
~
[10963.7 21505.5J 21505.5 63661.3
291
300
200
100
and
o
=
(2.02)(3.1)
JL2l:
Figure 6.2 95% confidence ellipse for
JLI  JL2
= (f.L]]
 f.L2], f.L12 
f.L22)·
= 6.26
With ILl  IL2 = [JLll  JL2!> JL12  JL22), the 95% simultaneous confidence intervals for the population differences are JLlI 
L1'002....t00~ P"  P21
(204.4  130.0) ± v'6.26
+ ~) 10963.7 (~ 45 55
The coefficient vector for the linear combination most responsible for rejection 
isproportionaltoSp~oled(xl  X2)' (See Exercise 6.7.)
The Bonferroni 100(1  a)% simultaneous confidence intervals for the p population mean differences are
or
:s: 127.1
(onpeak)
JL22: (556.6  355.0) ± V6.26
J(4~ + 515)63661.3
21.7 :s: JL12 
JLlI 
JL2l
where nl
or 74.7 :s:
JL12 
JL22
:s: 328.5
and
(~l + ~J c2= v'3301.5 )
U5 +
tnJ +nz2( a/2p)
n2 
is the upper 100 (a/2p )th percentile of a tdistribution with
2 dJ.
(offpeak)
We conclude that there is a difference in electrical consumption between those with airconditioning and those without. This difference is evident in both onpeak and offpeak consumption. The 95% confidence ellipse for JLI  IL2 is determined from the eigenvalueeigenvector pairs Al = 71323.5, e; = [.336, .942) and ,1.2 = 3301.5, e2 = [.942, .336). Since
vx; )
+
;5) 6.26
= 28.9
we obtain the 95% confidence ellipse for ILl  IL2 sketched in Figure 6.2 on page 291. Because the confidence ellipse for the difference in means does not cover 0' = [0,0), the T 2statistic will reject Ho: JLl  ILz = 0 at the 5% level.
The TwoSample Situation When 1: 1 =F 1:2 When II *" I 2 . we are unable to find a "distance" measure like T2, whose distribution does not depend on the unknowns II and I 2 • Bartlett's test [3] is used to test the equality of II and I2 in terms of generalized variances. Unfortunately, the conclusions can be seriously misleading when the populations are nonnormal. Nonnormality and unequal covariances cannot be separated with Bartlett's test. (See also Section 6.6.) A method of testing the equality of two covariance matrices that is less sensitive to the assumption of multivariate normality has been proposed by Tiku and Balakrishnan [23]. However, more practical experience is needed with this test before we can recommend it unconditionally. We suggest, without much factual support, that any discrepancy of the order eTI,ii = 4eT2,ii, or vice versa, is probably serious. This is true in the univariate case. The size of the discrepancies that are critical in the multivariate situation probably depends, to a large extent, on the number of variables p. A transformation may improve things when the marginal variances are quite different. However, for nl and n2 large, we can avoid the complexities due to unequal covariaI1ce matrices.
292
Comparing Mean Vectors from Two Populations
Chapter 6 Comparisons of Several Multivariate Means Result 6.4. Let the sample sizes be such that 11)  P and 112  P are large. Then,
approximate 100(1  a)% confidence ellipsoid for 1'1 satisfying
[x\  Xz  (PI  I'z)]'
[~S) + ~SzJ) [x) 111

1'2 is given by all 1'1
xz  (I')  I'z)]
112
$

Example 6 .• S (Large sample procedures for inferences about the difference in means)
We shall analyze the electricalconsumption data discussed in Example 6.4 using the large sample approach. We first calculate
~(a)
1 S 111
1
+
1 S I1Z
where ~ (a) is the upper (l00a }th percentile of a chisquare distribution with p d.f. Also, 100(1  a)% simultaneous confidence intervals for all linear combinations a'(I')  I'z) are provided by a'(I')  1'2)
belongs to a'(x)  Xz) :;I:
V~(a) \j;la'
(l..8 + I1r
1
293
1 [13825.3 2 = 45 23823.4 464.17 = [ 886.08
23823.4J 1 [ 8632.0 19616.7J 73107.4 + 55 19616.7 55964.5
886.08J 2642.15
The 95% simultaneous confidence intervals for the linear combinations l..sz)a 112 a '( 1')  ILz )
= [0][1'11 1,
a '( ILl  ILz )
=
 I'ZIJ
1')2  I'Z2
= 1'1)
 I'ZI
= 1'12

and
Proof. From (622) and (39),
£(Xl  Xz) = 1'1  I'z
and
[0,1 ] [1'))  1'21] 1'12  1'22
1'2Z
are (see Result 6.4)
By the central limit theorem, X)  Xz is nearly Np[l')  ILz, I1~Il ~ 11Z I z]· If Il and I2 were known, the square of the statistical distance from Xl  X 2 to 1')  I'z would be I
1'))  I'ZI:
74.4 ± v'5.99 v'464.17
or
(21.7,127.1)
J.L12  J.L2Z:
201.6 ± \15.99 \12642.15
or
(75.8,327.4)
Notice that these intervals differ negligibly from the intervals in Example 6.4, where the pooling procedure was employed. The T 2statistic for testing Ho: ILl  ILz = 0 is T Z = [XI 
This squared distance has an approximate x7,distribution, by Result 4.7. When /11 and /12 are large, with high probability, S) will be close to I) and 8 z will be close to I z· Consequently, the approximation holds with SI and S2 in place of I) and I 2, respectively. The results concerning the simultaneous confidence intervals follow from Result 5 A.1. • Remark. If 11)
= I1Z = 11, then (11
1 1  SI +  S2 /1)
112

1 /1
1)/(11
=  (SI + S2) = =
SpoOJedG
+
(11 
11 
2)
= 1/2, so
1) SI + (11  1) 82 (1 1)  +11 + n  2 11 n
+;)
With equal sample sizes, the large sample procedure is essentially the same as the procedure based on the pooled covariance matrix. (See Result 6.2.) In one dimension, it is well known that the effect of unequal variances is least when 11) = I1Z and greatest when /11 is much less than I1Z or vice versa.
Jl
1 xz]' [ 181 + 8 2 11)
I1Z
204.4. 130.0J' [464.17 886.08
= [ 556.6  355.0
= [74.4 For er
= .05,
201.6] (104) [
[XI  X2] 886.08JI [204.4  130.0J 2642.15 556.6  355.0
59.874 20.080
the critical value is X~(.05)
20.080J [ 74.4J 10.519 201.6
= 5.99
and, since T Z
= 1566 .
= 15.66 >
x~(.05)
= 5.99, we reject Ho.
The most critical linear combination leading to the rejection of Ho has coefficient vector
a ex:
(l..8
/11 I
+ l..8
)1 ( _)
/12 2
= (104) [
Xl
Xz
59.874 20.080
20.080J [ 74.4J 10.519 201.6
= [.041J
.063
The difference in offpeak electrical consumption between those with air conditioning and those without contributes more than the corresponding difference in onpeak consumption to the rejection of Ho: ILl  ILz = O. •
294 Chapter 6 Comparisons of Several Multivariate Means
Comparing Mean Vectors fromTho Populations 295
A statistic similar to T2 that is less sensitive to outlying observations for and moderately sized samples has been developed byTiku and Singh [24]. lOvvev'~rE if the sample size is moderate to large, Hotelling's T2 is remarkably unaffected slight departures from normality and/or the presence of a few outliers.
An Approximation to the Distribution of r2 for Normal Populations When Sample Sizes Are Not Large "
One can test Ho: ILl  IL2 = .a when the population covariance matrices are unequal even if the two sample sizes are not large, provided the two populations are multivariate normal. This situation is often called the multivariate BehrensFisher problem. The result requires that both sample sizes nl and n2 are greater than p, the number of variables. The approach depends on an approximation to the distribution of the statistic
For normal populations, the approximation to the distribution of T2 given by (628) and (629) usually gives reasonable results.
Example 6.6 (The approximate T2 distribution when l:. #= l:2) Although the sample sizes are rather large for the electrical consumption data in Example 6.4, we use these data and the calculations in Example 6.5 to illustrate the computations leading to the approximate distribution of T Z when the population covariance matrices are unequal. We first calculate
~S  ~ [13825.2 23823.4J nl
I 
1 nz S2 =
45 23823.4
= [307.227
529.409J 529.409 1624.609
73107.4
1 [8632.0 19616.7] 55 19616.7 55964.5
=
[156.945 356.667] 356.667 1017.536
and using a result from Example 6.5, which is identical to the large sample statistic in Result 6.4. However, instead of using the chisquare approximation to obtain the critical value for testing Ho the recommended approximation for smaller samples (see [15] and [19]) is given by
vp
2 _
T  vp
+1
+ ~Sz]I = (104) [ 59.874 ( ~SI nl n2 20.080
20.080] 10.519
F
P.vp+1
Consequently,
where the d!,!grees of freedom v are estimated from the sample covariance matrices using the relation
(629) [ where min(nJ> n2) =:; v =:; nl + n2' This approximation reduces to the usual Welch solution to the BehrensFisher problem in the univariate (p = 1) case. With moderate sample sizes and two normal populations, the approximate level a test for equality of means rejects Ho: IL I  ""2 = 0 if
1
1
(XI  Xz  (ILl  IL2»' [ SI + S2 nl n2
J (Xl  Xz I


(ILl  ILz»
>
v
_ vp + 1 Fp.v_p+l(a)
307.227 529.409
529.409] (104) [ 59.874 1624.609 20.080
20.080] = [ .776 10.519 .092
.060J .646
and
[~Sl + ~Sz]I)Z = [ .776 ( ~SI nl nl nz .092
.060][ .776 .646 .092
.060] .646
= [ .608 .085] .131
.423
Further,
p
where the degrees of freedom v are given by (629). This procedure is consistent with the large samples procedure in Result 6.4 except that the critical value x~(a) is vp replaced by the larger constant v _ p + 1 Fp.v_p+l(a). Similarly, the approximate 100(1  a)% confidence region is given by all #LI  ILz such that
]1 (Xl_  Xz_ IL2»' nl SI + n2 Sz 1
(XI  X2  (PI 
[
1
vp
(""1  ""2»
=:; v _ p
+ 1 Fp, vp+l(a) (630)
[
156.945 356.667](104)[ 59.874 356.667 1017.536 20.080
20.080] = [.224 10.519 .092
 .060] .354
and
+ l...sz]I)Z = ( ~S2[~SI n2 nl n2
[
.224 .060][ .224 .060] [.055 .092 .354 .092 .354 = .053
.035] .131

296
Comparing Several Multivariate Population Means (Oneway MANOVA) 297
Chapter 6 Comparisons of Several Multivariate Means
2. AIl populations have a common covariance matrix I.
Then
3. Each population is multivariate normal. Condition 3 can be relaxed by appealing to the central limit theorem (Result 4.13) when the sample sizes ne are large. A review of the univariate analysis of variance (ANOVA) will facilitate our discussion of the multivariate assumptions and solution methods.
1
= 55 {(.055
+ .131) + (.224 + .354f} =
Using (629), the estimated degrees of freedom v is
2 + 2z v and the a
=
=
.0678 + .0095
= 77.6
.05 critical value is
77.6 X 2 0' ,·_p+I(.05) = 77. 6  2 + 1 F?776,+l05) v  p + . . 1 . vp
155.2
= 6 76. 3.12 = 6.32
From Example 6.5, the observed value of the test statistic is rZ = 15.66 so hypothesis Ho: ILl  ILz = 0 is rejected at the. 5% level. This is the same cOUlclu:sioIi reached with the large sample procedure described in Example 6.5.
A Summary of Univariate ANOVA In the univariate situation, the ass~mptions are that XCI, Xez, ... , XCne is a random sample from an N(/Le, a 2 ) population, e = 1,2, ... , g, and that the random samples are independent. Although the nuIl hypothesis of equality of means could be formulated as /L1 = /L2 = ... = /Lg, it is customary to regard /Lc as the sum of an overalI mean component, such as /L, and a component due to the specific population. For instance, we can write /Le = /L + (/Le  IL) or /Lc = /L + TC where Te = /Le  /L. Populations usually correspond to different sets of experimental conditions, and therefore, it is convenient to investigate the deviations Te associated with the eth population (treatment). The reparameterization
(
As was the case in Example 6.6, the Fp • v  p + 1 distribution can be defined noninteger degrees of freedom. A slightly more conservative approach is to use integer part of v.
6.4 Comparing Several Multivariate Population Means (OneWay MANOVA)
eth pOPUlation) mean
Xll,XI2, ... ,Xlnl
Population 2: X ZI , X zz , ... , X2",
Te
eth population ) ( ( treatment) effect
OVerall) ( mean
(632)
leads to a restatement of the hypothesis of equality of means. The null hypothesis becomes Ho: Tt = T2 = ... = Tg = 0 The response Xc;, distributed as N(JL form
XC;
+ Te, a 2 ), can be expressed in the suggestive +
/L
=
Often, more than two populations need to be compared. Random samples, "V'.n..",,,u.,,,,,,, from each of g populations, are arranged as Population 1:
+
ILe
Te
(overall mean)
(
+
treatment) effect
ec;
(random) (633) error
where the et; are independent N(O, a 2 ) random variables. To define uniquely the model parameters and their least squares estimates, it is customary to impose the constraint
±
nfTf
= O.
t=1
Population g: X gI , Xgb ... , Xgn g MANOVA is used first to investigate whether the population mean vectors are the same and, if not, which mean components differ significantly.
Assumptions about the Structure of the Data for OneWay L XCI, X C2 ,"" Xcne,is a random sample of size ne from a population with mean e = 1, 2, ... , g. The random samples from different populations are
Motivated by the decomposition in (633), the analysis of variance is based upon an analogous decomposition of the observations, XCj
x
(observation)
overall ) ( sample mean
+
(XC  x) estimated ) ( treatment effect
+ (xe;  xc) (634)
(residual)
where x is an estimate of /L, Te = (xc  x) is an estimate of TC, and (xCi  xc) is an estimate of the error eej.
198
Chapter 6 Comparisons of Several Multivariate Means
Comparing Several Multivariate Population Means (Oneway MANOV A) 199 "
Example 6.1 (The sum of squares decomposition for univaria te ANOVA ) Consider
the following independent samples. Population 1: 9,6,9 population 2: 0,2 Population 3: 3, I, 2 Since, for example, X3 = (3 + 1 + 2)/3 = 2 and x = (9 + 6 + 9 +0 +2 3 + 1 + 2)/8 = 4, wefind that 3 = X31 = ~ + (X3  x) + (.~31  X3) = 4 + (2  4) + (3  2) = 4 + (2) + 1
The sum of squares decomp osition illustrat ed numerically in Exampl e 6.7 is so basic that the algebrai c equivale nt will now be develop ed. Subtrac ting x from both sides of (634) and squaring gives (XCi  X)2 =
We can sum both sides over j, note that ~
2./=1
observation (xCi)
4 4 4 mean
(x)
+
2 2 2 treatment effect (xe  x)
+
1 1 0 residual (xCi  XC)
Th uestion of equality of means is answered by assessing whether the 'be t~ relative to the residuals. (Our esticont n u IOn 0 f the treatment array is large g ~    x of Te always satisfy ~ neTe = O. Under Ho, each Tc is an ma t es Te  Xe ~ estimate of zero.) If the treatment contribution is large, Ho should. be rejected . The size of an array is quantified by stringing the ~ows of the array out mto a vector and calculating its squared length. This quantity IS, called the sum of squares (SS). For the observations, we construct the vector y = [9,6,9,0 ,2,3,1, 2J. Its squared length is
Similarly, SS
;~n
Ir
= 42 + 42 + 42 + 42 + 4 2 + 4 2 + 42 + 4 2 = 8(4 2) = 128
42 + 42 + 42 + (_3)2 + (3f + (2)2 + (_2)2 + (_2)2 = 3(4 2) + 2(3f + 3(2j2 = 78
=
and the residual sum of squares is SSre. = 12 + (_2)2 + 12 + (If + 12 + 12 + (1)2 + 02 = 10 The sums of squares satisfy the same decomposition, (634), as the observat ions. Consequently, SSobs = SSmean + SSlr + SSre. or 216 = 128 + 78 + 10. The breakup into sums of sq~ares apportio ns variability in the combined samples into mean, treatmen t, and re~ldu~1 (error) compon ents. An analysis of variance proceeds by comparing the relative SIzes of S~lr and SSres· If Ho is true, variances computed from SSlr and SSre. should be approxImately equal. 
xd + 2(xt 

.t
j:1
x)(xej  xc)
(XCi  xel = 0, and obtain
Z
(XCi  x) = n(xc  x/
~ + 2.
(Xti  xel
z
j:]
Next, summing both sides over
e we get
±~ ± ±i; co,~~~::;;~ro) ~ (:"we
Repe(a~T~)operatio(n:,,:' :)07'("::' w~tru:)fu' ~a(y,_: _~ ') 3 1 2
(xc  x/ + (xCj
(XCi  x)2 =
ncCxc  x)2
+
SS }
(XCj  xe)2
+ (Wifuin
or g
~
"i'
2: x7i
(n]
(:1 j:1
+ n2 + ... + n g )x2 +
g
2: nc(xc 
x)2
+
c:]
g
~ {:I
(635)
(~;~1")
~
SS)
2. (XCj  xc)
2
j:1
(SSobs)
(SSme.n) + (SSres) (636) + In the course of establishing (636), we have verified that the arrays represen ting the mean, treatme nt effects, and residuals are orthogonal. That is, these arrays, conside red as vectors, are perpend icular whateve r the observa tion vector y' = [XlI, .. ·, XI,,!, X2I'···' xz Il2 ' . · . , Xgll ]. Consequently, we could obtain SSre. by subtract ion, without having to calculate' the individual residuals , because SS res = SSobS  SSme.n  SSlr' Howeve r, this is false econom y because plots of the residuals provide checks on the assumpt ions of the model. The vector represen tations of the arrays involved in the decomp~sition (634) also have geometr ic interpre tations that provide the degrees of freedom . For an arbitrar~ set of observatio~s, let [XII,' .. : Xl "l' Xz j, .•. , X21l2' ... , XgngJ. = Y". The observatIOn vector y can he anywhe re m n = nl + n2 + ... + n climensIOns; the mean vector xl = [x" .. , x]' must lie along the equiang ular line ~f I, and the treatment effect vector 1
(XI  x)
1 0
}n, + (X2  x)
0 1
}
+ ... + (x, 
x)
0 0
n2
0 0
1 0
0 1
}n,
1 = (Xl  X)UI + (X2  x)uz + .. , + (Xg  x)u g
Comparing Several Multivariate Population Means (Oneway MANOVA) 301
300 Chapter 6 Comparisons of Several Multivariate Means lies in the hyperplane of linear combinations of the g vectors 1I1, U2,"" u g • Since 1 = Ul + U2 + ." + ug , the mean vector also lies in this hyperplane, and it is always perpendicular to the treatment vector. (See Exercise 6.10.) Thus, the mean vector has the freedom to lie anywhere along the onedimensional equiangular line and the treatment vector has the freedom to lie anywhere in the other g  1 di~> mensions. The residual vector,e = y  (Xl)  [(Xl  X)Ul + .. , + (x g  x)u g ] is perpendicular to both the mean vector and the treatment effect vector and has the freedom to lie anywhere in the subspace of dimension n  (g  1) , 1 = n that is perpendicular to their hyperplane. To summarize, we attribute 1 d.f. to SSmean,g .1 d.f. to SSt" and n  g '" (nl + n2 + ... + ng)  g dJ. to SS,es' The total number of degrees of freedom is n = n~ + n2 + .. , + n g • Alternatively, by appealing to the univariate distribution theory, we find that these are the degrees of freedom for the chisquare distributions' associated with the corresponding sums of squares. The calculations of the sums of squares and the associated degrees of freedom are conveniently summarized by an ANOVA table.
ANOVA Table for Comparing Univariate Population Means Source of variation
Degrees of freedom (d.f.)
Sum of squares (SS) SSt,
=
2: ne(xc 
g1
C=1 g
Residual (error)
x)2
SS,es
=
ne
2: 2: (XCj 
XC)2
neatments
SStr
Residual
Degrees of freedom
g1=31=2
78
=
±
ne  g = (3 + 2 + 3)  3 = 5
SS,es = 10
(=1
g
Total (corrected)
SScor
=
L nc  1 = 7
88
C=1
Consequently,
F
=
SSt,/(g  1) SSres/(l;nc  g)
= 78/2 = 195 10/5
Since F = 19.5 > F2 ,s(.01) = 13.27, we reject Ho: effect) at the 1% level of significance.
71
.
= 72 = 73 = 0 (no treatment _
Lne  g C=1
MANOVA Model For Comparing g Population Mean Vectors
±
Total (corrected for the mean)
Sum of squares
Paralleling the univariate reparameterization, we specify the MANOVA model:
g
f=l j=1
Source of variation
Multivariate Analysis of Variance (MANOVA)
g
neatments
Example 6.8 CA univariate ANOVA table and Ftest for treatment effects) Using the
information in Example 6.7, we have thefoIlowingANOVA table:
X Cj
ne 1
C=1
=,."
+
Te
+
eCj,
j
= 1,2, ... ,nc
and
e = 1,2, ... ,g
(638)
~here IS
the eCj are independent Np(O, l;) variables. Here the parameter vector,." an overall mean (level), and TC represents the eth treatment effect with
g
The usual Ftest rejects Ho: 71 =
72
L neTc = C=1
= ... = 7 g = 0 at level a if
O.
According to the model in (638), each component of the observation vector XC' satisfies the univariate model (633). The errors for the components of Xc' are c~rrelated, but the covariance matrix l; is the same for all populations. ] A vector of observations may be decomposed as suggested by the model. Thus,
SSt,/(g  1)
'2:ri
1 1 + SSt, /SS,es
SS,es SS,es + SSt,
(637)
(observation)
+
x
XCj
where F 1 :2:n _g(O') is the upper (I00O')th percentile of the Fdistribution with g _ 1 a~d c  g degrees of freedom. This is equivalent to rejecting Ho for large values of SSt,/SS,es or for large values of 1 + SSt,/S5,.es· The statistic appropriate for a multivariate generalization rejects Ho for small values of the reciprocal
(
overall sa~Ple) mean,."
(xe  x) estimated) treatment ( effectTc
+
(XCj 
Xe)
(res~dual) _
(639)
eCj
The decomposition in (6~39) leads to the muItivariate analog of the univariate sum of squares breakup in (635). First we note that the product (XCj 
x)(XCj 
x)'
302
Chap ter 6 Com paris ons of Several Multivariate Means
can be written as (XCj  x)(XCj  x)'
Com~aring Seve ral MuIt ivari ate Popu latio n Mea ns (One way MAN OVA )
= [(x!,j 
xc) + (Xt  x)] [(XCj  ic) + (xc x)J' = (XCj  ic)(xCj  i )' + (Xt; c  xc) (xc  x)' + (Xt  X)(Xtj  xc)' + (Xe  X)(Xc  i)' The sum over j of the middle two expressions is the zero matrix, ~ (xc;  it) = O. Hence, summing the cross product over e and j yields
303
This tabl e is exactly the sam e form , com pon ent by com pon ent, as the ANO VA table, exce pt that squa res of scal ars are repl aced by thei r vect or coun terp arts. For exam ple, (xc  x? beco mes (xc  x)(x c  x)'. The degr ees of free dom corr espo nd to the univ ariat e geom etry and also to som e mul tiva riate distr ibut ion theo ry involving Wis hart densities. (See [1].) One test of Ho: TI = TZ = '" = Tg = 0 involves generalized variances. We reject Ho if the ratio of gene raliz ed variances
~I
~ ~ (x. "'' ~ (/
C=1 /=1
x) (xc'  i)' = /
.
±
nc(xc  x){xc  x)' + c=)
1: ~ (xc; 
(=1 /=1
A* =
xc) (XCj  xc)'
(residual (Wi thin ) sum ) of squares and cross prod ucts
C=I j=1
(640)
The within sum of squares and cross prod ucts matrix can be expressed as g
W=
"I
2: L (xej 
Xe)(Xfj  xc)'
C=I j=1
= (n)
 1)SI
+ (n2 
1)~
+ ... + (ng 
(641)
I)Sg
I±
.s(X t;  x)(XCj 
.
treatment <_Between») sum of squares and ( cross products
tota l (correcte sum ») of squares an dd cross ( products /
Iwl IB+wl
where Se is the sample covariance matr ix for the fth sam~le. This matr ix is a gener. . f the (n + n2  2) S ) d matrix enco a}Izat) on 0 ) untered III the twosample case. It e plays a dominant role in testing poo for the presence of t~eatment effects. Analogous to the univariate result, the hypotheSIS of no trea tme nt effects, Ho: T) = T2 = ... =Tg = 0 . t ted by considering the relative sizes of the treatment and residual Ise s sums of squares and crosS products. Equivalen tly, we may conS.Ider the re I" atlve SlZes 0 fth e residual and total (corrected) sum of squares and cross products. Formally, we summarize the calculations leading to the test statistic in a MAN OVA table.
is too small. The quan tity A * = IWill B + w I, prop osed orig inally by Wilks (see [25]), corr espo nds to the equi vale nt form (637) of the Ftest of Ho: no trea tmen t effects in the univ ariat e case . Wilk s' lamb da has the virtu e of bein g conv enie nt and rela ted to the likel ihoo d ratio z criterion. The exac t distributIon of A * can be deri ved for the special cases liste d in Table 6.3. For othe r cases and larg e sam ple sizes, a modification of A* due to Bart lett (see [4]) can be used to test Ho. Table 6.3 Dist ribu tion ofW ilks ' Lam bda, A* = Iwl/lB + wl No. of No. of variables grou ps Sam plin g distr ibut ion for mul tiva riate norm al data p =
1
p= 2
p;;: :1
g;;: :2
g;;::2
g= 2
(Ln c g 1
g)
(Ln c  g g 1 (Ln e  P P
MANOVA Table for Comparing Popu lation
Mean Vectors Matrix of sum of squares and cross products (SSP)
Source of variation
Deg rees of free dom (dJ.)
x)'1
(642)
p;;:: 1
g= 3
(Ln e  p p
e
A* ) A*
~
e
FgI, 'I:.ne g
1) VAVA* *) 1) (~) ~ 2) eVAVA* *) A*
~
FZ(g I),Z ('I:.n erl)
Fp,'I :.nep 1
~
F Zp, Z('I:. n,p 2)
g
Treatment
B=
2: ne(xe (=1
g
Residual (Error)
W=
g
g1 g
"f
L 2: (xc; t=1 j=1
Total (corrected for the mean)
x) (ic  x)'
ic) (XCj  xc)'
2: ne 
g
C=I nl
B + W = ~ ~ (xc;  x)(XCj  x)' (=1 j=1
g
~ ne 1 e=1
2Wilks' lambda can also be expressed as a function of the eigenvalues of Ab A2 , .•• , As of W1B as
A'=llC~J
where s = min (p, g  1), the rank of B. Othe r statistics for checking the equality of se~eral multivariate means, such as Pillai's statistic, the LawleyHotelling statistic, and Roy' s largest root statistic can also be written as particular functions ofthe eigenvalues ofW 1B. For large samp les, all of these statistics are, essentially equivalent. (See the addit ional discussion on page 336.)
Comparing Several Multivariate Population Means (Oneway MANOVA)
304 Chapter 6 Comparisons of Several Multivariate Means
Bartlett (see [4]) has shown that if Ho is true and 2
and
Ln( = n is large,
(n1(P+g»)lnA*=(n1(P+g»)ln( 2
IWI) IB+ WI
(p + g») 2
In
(
SSobs = SSmean + SStr
(643)
has approximately a chisquare distribution with peg  1) dJ. Consequently, for Lne = n large, we reject Ho at significance level a if  (n  1 
305
) IB Iwl + wl > x7,(gl)(a)
(644)
where x;,(gl)(a) is the upper (l00a)th percentile of a chisquare distribution with peg  1) dJ.
+
SSres
272 = 200 + 48 + 24 Total SS (corrected)
= SSobs
 SSmean = 272  200 = 72
These two singlecomponent analyses must be augmented with the sum of entrybyentry cross products in order to complete the entries in the MANOVA table. Proceeding row by row in the arrays for the two variables, we obtain the cross product contributions: Mean: 4(5) + 4(5) + '" + 4(5) = 8(4)(5) = 160 Treatment: 3(4)(1) + 2(3)(3) + 3(2)(3) = 12
Example 6.9 CA MANOVA table and Wilks' lambda for testing the equality of three mean vectors) Suppose an additional variable is observed along with the variable introduced in Example 6.7, The sample sizes are nl = 3, n2 = 2, and n3 = 3. Arranging the observation pairs Xij in rows, we obtain
[~] [~] [~] [~] [~] [~] [~] [~J
WithXl = andx =
[!l
x2 =
[~l
X3 =
(observation)
G::)
+
[~].
Total (corrected) cross product = total cross product  mean cross product
Thus, the MANOVA table takes the following form:
[:J
Source of variation
(=~ =~ J
+
treatment) ( effect
(mean)
Total: 9(3) + 6(2) + 9(7) + 0(4) + ... + 2(7) = 149
= 149  160 = 11
We have already expressed the observations on the first variable as the sum of an overall mean, treatment effect, and residual in our discussion of univariate ANOVA. We found that
(P:)
Residual: 1(1) + (2)(2) + 1(3) + (1)(2) + ... + 0(1) = 1
(:
~:
Matrix of sum of squares and cross products
Treatment
[
78 12
Residual
[
Total (corrected)
[
12J 48
10
1
2!J
88 11
l1J
Degrees of freedom
3  1= 2
3+2+33=5
:)
(residual)
and
72
7
Equation (640) is verified by noting that
SSobs = SSmean + SStr + SSres
216 = 128 + 78 + 10 Total SS (corrected) = SSobs  SSmean
= 216
 128 = 88
Repeating this operation for the obs,ervations on the second variable, we have
(! ~ 7) 8 9 7
(observation)
(~~ 5)
+
(=~ =~ 1)
5 5 5 (mean)
3 (
3
3
treatment) effect
+
(~ =~ 3) 0
11
Using (642), we get
. A*
(residual)
10
IWI 11 = IB + WI =
11
24
I 111 88 11
72
10(24)  (1)2 239 =   = .0385 88(72)  (11? 6215
_~,,c..::
Comparing Several Multivariate Population Means (Oneway MANOVA)
306 Chapter 6 Comparisons of Several Multivariate Means
Since p = 2 and g = 3, Table 6.3 indicates that an exact test (assuming normal_ ity and equal group covariance matrices) of Ho: 1'1 = 1'2 = 1'3 = 0 (no treatment effects) versus HI: at least one Te 0 is available. To carry out the test, we compare the test statistic
Sample covariance matrices
*
\f.0385) (8 3 3 1 1) = 8..19 v'Av'*A*) (Lne(g  g1)' 1) = (1 V.0385
1(
with a percentage point of an Fdistribution having Vi = 2(g  1) == 4 == 2( Lne  g  1) == 8 dJ. Since 8.19 > F4,8(.01) = 7.01, we reject Ho at a = .01 level and conclude that tI:eatment differences exist.
SI =
l·291 .001 .002
S3 =
.030 l~l .003
V2
eej == Xej  Xf
.011
.000 .001 .003 .000 .010
.018 When the number of variables, p, is large, the MANOVA table is usually not constructed. Still, it is good practice to have the computer print the matrices Band W so that especially large entries can be located. Also, the residual vectors
.017 .000 .006
.004 .001
Group
Number of observations
n2 = 138
_ XI
=
2.066] .480 .082; l .360
e = 3 (government)
lS61 .011 .001 .037
.025 .004 . .005 .007 .002
.J
oJ
Since the Se's seem to be reasonably compatible,3 they were pooled [see (641)] to obtain
W = (ni  l)SI
+ (n2  1)S2 + (n3  I)S3
182.962 4.408 8.200 1.695 .633 9.581 2.428 l
] . 1.484 .394 6.538
Also,
and
B 
~
£.; C=1
nc(Xe  ) X (Xc  x)' =
l~:;~~
.821 .584
1.225 .453 .235 .610 .230
Sample mean vectors
e = 1 (private) e = 2 (nonprofit)
oJ
S = 2
Source: Data courtesy of State of Wisconsin Department of Health and SociatServices.
should be examined for normality and the presence of outhers using the techniques discussed in Sections 4.6. and 4.7 of Chapter 4. Example 6.10 CA multivariate analysis of Wisconsin nursing home data) The Wisconsin Department of Health and Social Services reimburses nursing homes in the state for the services provided. The department develops a set of formulas for rates for each facility, based on factors such as level of care, mean wage rate, and average wage rate in the state. Nursing homes can be classified on the basis of ownership (private party, nonprofit organization, and government) and certification (skilled nursing facility, intermediate care facility, or a combination of the two). One purpose of a recent study was to investigate the effects of ownership Or certification (or both) on costs. Four costs, computed on a perpatientday basis and measured in hours per patient day, were selected for analysis: XI == cost of nursing labor,X2 = cost of dietary labor,X3 = cost of plant operation and maintenance labor, and X 4 = cost of housekeeping and laundry labor. A total of n = 516 observations on each of the p == 4 cost variables were initially separated according to ownership. Summary statistics for each of the g == 3 groups are given in the following table.
307
l2.167] _ .596 x2 = .124; .418
_ X3
=
l2.273] .521 .125 .383
To test Ho: 1'1 = 1'2 = 1'3 (no ownership effects or, equivalently, no difference in average costs among the three types of ownersprivate, nonprofit, and government), we can use the result in Table 6.3 for g = 3. Computerbased calculations give
A*
=
IB
IWI + WI = .7714
3
:2:: ne =
e=1
516
3However, a normaltheory test of Ho: I1 = I2 = I3 would reject Ho at any reasonable significance level because ofthe large sample sizes (see Example 6.12).
Simultaneous Confidence Intervals for Treatment Effects 309
308 Chapter 6 Comparisons of Several Multivariate Means
It remains to apportion the error rate over the numerous confidence state
and 2:. n e  p (
v'Av'*A*)
2) (1 
p
=
v:77I4)
(516  4  2) (1 4 v.7714
~ents. Relation (528) still applies. There are p variables and g(g  1)/2 pairwise
differences, so each twosample tinterval will employ the critical value tn g ( a/2m), where
= 17.67
m = pg(g  1)/2 Let a = .01, so that F2(4),i(51O)(.01) == /s(.01)/8 = 2.51. Since 17.6? > F8•1020( .01) == 2.51, we reject Ho at the 1% level and conclude that average costs differ, depending on type of ownership. ." " . It is informative to compare the results based on this exact test With those obtained using the largesample procedure summarized in (643) and (644). For the present example, 2:.nr = n = 516 is large, and Ho can be tested at the a = .01 level by comparing
en  1 
(p + g)/2)
InCBI:~I) = 511.5 In (.7714) = 132.76
with X~(gl)(.01) = X§(·01) =: 20.09 .. Since 1~2.76 > X§(·Ol) = 20.09, we reject .Ho at the 1 % level. This result IS consistent With the result based on the foregomg Fstatistic.
•
6.S Simultaneous Confidence Intervals for Treatment Effects When the hypothesis of equal treatment effects is rejected, those effects that led to the rejection of the hypothesis are of interest. For pairwise. comparisons, th~ Bonferroni approach (see Section 5.4) can be used to construct sImultaneous confI~ence intervals for the components of the differences Tk  Te (or ILk  lLe)· These mtervals are shorter than those obtained for all contrasts, and they require critical values only for the univariate tstatistic. . .. • _ _ Let Tki be the ith component of Tk· Smce Tk IS estimated by Tk = Xk  X
is the number of simultaneous confidence statements.

nk.
+
ne
(l  a),
belongs to
___ _  (1 1)
Var(Xki

Xe;) =

nk
+
where Wji is the ith diagonal element of Wand n
ne
Wii
n  g
= n l + ... + n g •
xki 
Xc; ± t n  g (
a
pg(g  1)
for all components i = 1, ... , p and all differences ith diagonal element of W.
e<
)
J~ (1. + 1.) n  g nk ne
k == 1, ... , g. Here Wii is the
We shall illustrate the construction of simultaneous interval estimates for the pairwise differences in treatment means using the nursinghome data introduced in Example 6.10. Example 6.11 (Simultaneous intervals for treatment differencesnursing homes) We saw in Example 6.10 that average costs for nursing homes differ, depending on the type of ownership. We can use Result 6.5 to estimate the magnitudes of the differences. A comparison of the variable X 3 , costs of plant operation and maintenance labor, between privately owned nursing homes and governmentowned nursing homes can be made by estimating T13  T33. Using (639) and the information in Example 6.10, we have
•
_
.D70j .039 , [ .020 .020
_
71=(X1 X)=
182.962 W = 4.408 8.200 [ 1.695 .633 1.484 9.581 2.428 .394
Uii
where U·· is the ith diagonal element of:t. As suggested by (641), Var (Xki  X ei) is estim~~ed by dividing the corresponding element of W by its degrees of freedom. That is,
nk. For the model in (638), with confidence at least
k=I
and Tki  Tfi = XA;  XCi is the difference between two independent sample means. The twosample (based confidence interval is valid with an appropriately modified a. Notice that
_ _ (1 1)
f
Result 6.S. Let n =
(645)
Var(Tki  Te;) = Var(Xki  Xli) =
(646)
Consequently,
T13 and n = 271
+ 138 + 107
J(
1
n1
+
1)
n3
733
• 73
_
_
= (X3  x) =
.137j .002 [ .023 .003
.J
= .020  .023 = .043
= 516, so that W33
n  g
=
~( 2711
1) 1.484
+ 107 516  3 = .00614
310 Chapter 6 Comparisons of Several Multivariate Means
•
_
Testing for Equality of Covarian ce Matrices
== 3 for 95% simultan eous confidence stat~ments we require
Box's test is based on his X 2 approxi mation to the samplin g distribu tion of  2 In A (see Result 5.2). Setting 21n A = M (Box's M statistic ) gives
~~:~~5~(~~:~~2~ == 2:87. (See Appendix, Table 1.) The 95% SImultaneous confidence statement is
J( 1+ 1)
belongs to. T13  T33 ± t513(.00208)
nl
n3
M =
W33 n  g
mainten ance and labor cost for governm entown ed We ~onclude th~t h~ehave~age025 to .061 hour per patient day than for privately nursmg homes IS Ig er y. . th t . h mes With the same 95% confIden ce, we can say a owne d nursmg 0 . _ ~ belongs to the interval (.058, .026) 'T13 • 23 7"23
_ ~ •
33
[2:(n e  1)]ln I Spooled I  2:[(ne  l)ln ISell e e
== .043 ± 2.87(.00614) ==  .043 ± .018, or (  .061,  .025)
and
311
belongs to the interval ( .021, .019)
. . th's cost exists between private and nonprofit nursing homes, Thus a difference m I . h 'ff' 's observed between nonprof it and government nursmg omes. but no dI erence 1
(650)
If the null hypothe sis is true, the individual sample covarian ce matrices are not expecte d to differ too much and, consequently, do not differ too much from the pooled covarian ce matrix. In this case, the ratio of the determi nants in (648) will all be close to 1, A will be near 1 and Box's M statistic will be small. If the null hypothesis is false, the sample covarian ce matrices can differ more and the differen ces in their determi nants will be more pronoun ced. In this case A will be small and M will be relatively large. To illustrat e, note that the determi nant of the pooled covarian ce matrix, I Spooled I, will lie somewh ere near the "middle " of the determi nants ISe I's of the individual group covarian ce matrices. As the latter quantiti es become more disparat e, the product of the ratios in (644) will get closer to O. In fact, as the ISf I's increase in spread, IS(1) I1I Spooled I reduces the product proporti onally than IS(g) I1I Spooled I increases it, where IS(l) I and IS(g) I are the minimu m andmore maximu m determi nant values, respectively.
,
Box's Test for Equality of Covariance Matrices
6.6 Testing for Equality of Covariance Matrices
Set
. d when compari ng two or more multivar iate mean vecOne of the assumptI~ns ma et' of the potentia lly different populati ons are the tors is that the cova~lanc~ ma nces . m' Chapter 11 when we discuss discrimina(Th' umptlon wIll appear agam s~me. d IS ass'fi f n) Before pooling the variatio n across samples to fo~m a tlOn an clas.sl ca 10 ~ . hen compari ng mean vectors, it can be worthwhile to pooled covanl~:ce f~:enp:pwulation covariance matrices. One common ly employed test the equa I y 0 test for equal covariance matrices is Box'~ M. test ([8] , [9]) . With g populations, the null hypothesIs IS Ho: 'i. == 'i.2 = ... = 'i. g = ' i . ( 6  4 7 ) 1
. r" ance matrix for the eth population, e ~ 1, 2, ... , g, and I is where Ie IS the cova 1 . the presumed common covanance ma trix. The alternative hypothesis is that at least . e matrices are not equal. two of the covanan~. f ons a likelihood ratio statistic for testAssuming multlvanate normaI popu Ia I, ing (&47) is given by (see [1])
A=
ne ( I
I Se I Spooled
)(n
CI)12
(648)
I
Here ne is the sample size for the eth group,.Se is the e~h ~roup sample covariance . matnx an d Spooled 'IS the pooled sample covanan ce matnx given by Spooled ==
1 ~(ne  1) t
{(nl _ l)SI + (nz  1)S2 + ... + (ng  l)Sg}
(649)
u 
[2:
1 1 e (ne  1) ~(ne _ 1)
J[
2p2 + 3p  1 ] 6(p + l)(g  1)
(651)
where p is the number of variable s and g is the number of groups. Then
C = (1  u)M = (1  u){[
~(ne l)Jtn ISpooled I  ~[(ne l)ln I Se IJ}(652)
has an approxi mate X2 distribu tion with
1 1 1 + 1)  Zp(p + 1) = Zp(p
v = gzp(p
degrees of freedom . At significance level
(1',
reject
Ho
+ 1)(g
 1)
(653)
if C > ~(p+l)(gI)I2«I').
K
Box's approxi mation works well if each ne exceeds 20 and if p and g do not exceed 5. In situations where these conditions do not hold, Box ([7J, [8]) has provide d a more precise F approxi mation to the samplin g distribu tion of M.
Example 6.12 (Testing equality of covariance matrice snursin g homes) We introduced the Wisconsin nursing home data in Exampl e 6.10. In that example the sample covarian ce matrices for p = 4 cost variables associat ed with g = 3 groups of nursing homes are displayed. Assumi ng multiva riate normal data, we test the hypothe sis HO::I1 = :I2 = :I3 = 'i..
312
Chapter 6 Comparisons of Several Multivariate Means
lWoWay Mu/tivariate Analysis of Variance 313
Using the information in Example 6.10, we have nl = 271, n2 == 138, 8 8 X 10 ,1 s21 = 89.539 X 10 ,1 s31 = 14.579 X 108 , and 1Spooled 1 = 17.398 X 108. Taking the natural logarithms of the determinants gives In 1SI 1= 17.397, In 1Sz 1= 13.926, In 1s31 = 15.741 and In 1Spooled 1= 15.564. We calculate
n3
= 107 and 1SI 1= 2.783
If
u = [ 270
1
+ 137 + 106
1
 270
+ 137 + 106
e
,nations of levels. Denoting the rth observation at level of factor 1 and level k of factor 2 by X fkr , we specify the univariate twoway model as
X ekr = JL
][2W) + 3(4) 
1] 6(4 + 1)(3 _ 1) = .0133
= [270 +137 + 106)(15.564)  [270(17.397) + 137(13.926) + 106( 15.741) J = 289.3 and C = (1 .0133)289.3 = 285.5. Referring C to a i table with v = 4(4 + 1)(3 1)12 M
= 20 degrees of freedom, it is clear that Ho is rejected at any reasonable level of significance. We conclude that the covariance matrices of the cost variables associated with the three populations of nursing homes are not the same. _
Box's Mtest is routinely calculated in many statistical computer packages that do MANOVA and other procedures requiring equal covariance matrices. It is known that the Mtest is sensitive to some forms of nonnormality. More broadly, in the presence of nonnormality, normal theory tests on covariances are influenced by the kurtosis of the parent populations (see [16]). However, with reasonably large samples, the MANOVA tests of means or treatment effects are rather robust to nonnormality. Thus the Mtest may reject Ho in some nonnormal cases where it is not damaging to the MANOVA tests. Moreover, with equal sample sizes, some differences in covariance matrices have little effect on the MANOVA tests. To summarize, we may decide to continue with the usual MANOVA tests even though the Mtest leads to rejection of Ho.
f3k + 'Yek 1,2, ... ,g k = 1,2, ... , b
+ eekr (654)
r = 1,2, ... ,n b
g
where
b
g
2: Te = k=1 2: f3k = e=1 2: 'Yek = k=1 2: 'Yek = 0 e=1
and the elkr are independent
N(O, (T2) random variables. Here JL represents an overall level, Te represents the fixed effect of factor 1, f3 k represents the fixed effect of factor 2, and 'Ye k is the interaction between factor 1 and factor 2. The expected response at the eth level of factor 1 and the kth level of factor 2 is thus
mean) ( response
JL
+
Tt
+
f3k
( overall) level
+
( effect Of) factor 1
+
( effect Of) factor 2
e=I,2, ... ,g,
k = 1,2, ... , b
+
'Yek
2) + (fa~tOr1fa~tor InteractIOn (655)
The presence of interaction, 'Yek> implies that the factor effects are not additive and complicates the interpretation of the results. Figures 6.3(a) and (b) show
Level I offactor I Level 3 offactor I Level 2 offactor I
6.7 TwoWay Multivariate Analysis of Variance Following our approach to tile oneway MANOVA, we shall briefly review the analysis for a univariate twoway fixedeffects model and then simply generalize to the multivariate case by analogy.
+ Te +
e=
2
3
4
Level of factor 2
Univariate TwoWay FixedEffects Model with Interaction
(a)
Level 3 of factor I
We assume that measurements are recorded at various levels of two factors. In some cases, these experimental conditions represent levels of a single treatment arranged within several blocks. The particular experimental design employed will not concern us in this book. (See (10) and (17) for discussions of experimental design.) We shall, however, assume that observations at different combinations of experimental conditions are independent of one another. Let the two sets of experimental conditions be the levels of, for instance, factor 1 and factor 2, respectively.4 Suppose there are g levels of factor 1 and b levels of factor 2, and that n independent observations can be observed at each of the gb combi
Level I offactor I Level 2 offactor I
3
2
4The use of the tenn "factor" to indicate an experimental condition is convenient. The factors discussed here should not be confused with the unobservable factors considered in Chapter 9 in the context of factor analysis.
Level of factor 2 (b)
4
Figure 6.3 Curves for expected responses (a) with interaction and (b) without interaction.
TwoWay Mu/tivariate Analysis of Variance 315
314 Chapter 6 Comparisons of Several Multivariate Means
expected responses as a function of the factor levels with and without interaction, respectively. The absense of interaction means 'Yek = 0 for all .and k. In a manner analogous to (655), each observation can be decomposed as
The Fratios of the mean squares, SSfact/(g  1), SSfaczl(b  1), and SSintl (g  1)( b  1) to the mean square, SS,es I (gb( n  1» can be used to test for the effects of factor 1, factor 2, and factor Ifactor 2 interaction, respectively. (See [11] for a discussion of univariate twoway analysis of variance.)
where x is the overall average, Xf· is the average for the eth level of factor 1, x'k is the average for the kth level of factor 2, and Xlk is the average for the eth level factor 1 and the kth level of factor 2. Squaring and summing the deviations (XCkr  x) gives
Multivariate TwoWay FixedEffects Model with Interaction
e
g
b
n
2: bn(xf· 
x)2 =
(=1 k=1 ,=1
X)2
+
f=1
2: gn(x'k 
X)2
e=
k=1
g
+
X ekr = po + 'Te + Ih + 'Ytk + eCk,
b
g
2: 2: 2: (Xtkr 
Proceeding by analogy, we specify the twoway fixedeffects model for a vector response consisting ofp components [see (654)]
b
2: 2: n(Xfk 
1,2, ... ,g
(659)
k = 1,2, ... ,b
Xc 
X'k
+ X)2
r = 1,2, ... ,n
f=1 k=1 g
where
Q
b
g
2: 'T C = k=1 2: Ih = C=I 2: 'Y Ck = k=1 2: 'Ye k =
O. The vectors are all of order p X 1,
f~1
and the eCkr are independent Np(O,::£) random vectors. Thus, tbe responses consist of p measurements replicated n times at each of the possible combinations of levels of factors 1 and 2. Following (656), we can decompose the observation vectors xtk, as
or SSco, = SSfacl
+
SSfac2 + SSint
+ SSres
The corresponding degrees of freedom associated with the sums of squares in the breakup in (657) are gbn  1 = (g  1)
+ (b  1) + (g  1) (b  1) + gb(n  1)
XCkr = X + (xe·  x)
ANOVA Table for Comparing Effects of Two Factors and Their Interaction Degrees of freedom (d.f.)
Sum of squares (SS)
g
b
SSfac1 =
2: bn(xe. 
x)2
g1
i)(XCk'  x)' =
2: bn(ic· C=I
Interaction
SSfac2 = SSint
=
2: gn(x'k 
x)2
b  1
k=1 g
b
C=I
k=1
f=1
k=l r=1
2: 2: n(xCk 
±2: 2: ±2: 2: =
Residual (Error)
SSres =
Total (corrected)
SScor
b
b
"
n
C=1 k=! ,=1
Xc·  X'k
+ X)2
XCk)
(660)
i)(xe·  i)'
b
(=1
+
2: gn(i' k k=l
+
2: 2: n(itk t=1 k=l
b
Factor 2
+ (XCkr 
g
n
2: 2: 2: (XCkr (=1 k=1 r=1
g
Factor 1
i' k + i)
where i is the overall average of the observation vectors, ic. is the average of the observation vectors at the etb level of factor 1, i' k is the average of the observation vectors at the kth level of factor 2, and ie k is the average of the observation vectors at the eth level of factor 1 and the kth level of factor 2. Straightforward generalizations of (657) and (658) give the breakups of the sum of squares and cross products and degrees of freedom:
(658)
TheANOVA table takes the following form:
Source of variation
+ (X'k  x) + (XCk  xc· 
g
i)(i' k

i)'
b
Xc·  i' k + i) (iek  Xt·  i' k + i)'
(g  1)(b  1)
(661)
(XCkr  fed
gb(n  1)
(Xek'  x)2
gbn  1
gbn  1 = (g  1)
+
(b  1)
+
(g  1)(b  1)
+ gb(n
 1)
(662)
Again, the generalization from the univariate to the multivariate analysis consists simply of replacing a scalar such as (xe.  x)2 with the corresponding matrix
(i e·  i)(xc.  i)'.
316
Chapter 6 Comparisons of Several Multivariate Means
'!WoWay Multivariate Analysis of Variance 3,17
The MANOVA table is the following: Factors and Their Interaction
MANOVA Table for
Matrix of sum of squares and cross products (SSP)
Source of variation
g
SSPtacl =
Factor 1
2: bn(xe· 
SSPtac2 =
Interaction
SSPint =
2: gri(X'k 
±±
b 1
1: ±:±
(=]
g
SSPcor =
(XCkr 
XCk)(XCkr 
Reject Ho: 'Tl
xcd
gb(n1)[
n
2: 2: r=1 2:
(Xtkr 
X)(Xfkr  x)'
A test (the likelihood ratio test)5 of
= 1'12 = ... = 1'gb = 0
versus
HI: Atleast one 1't k
(no interaction effects)
/SSPres / + SSPres /
:"'""=,
/SSPfac2
1
)(b l)JInA* >
xTgI)(bl)p(a)
where A * is given by (664) and xfgI)(bl)p(a) is the upper (lOOa)th percentile chisquare distribution with (g  .1)(?  l!p d.f. Ordinarily the test for interactIOn IS earned out before the tests for fects. If interadtion effects exist, the factor effects do not hav.e a clear in.t4.erpallret8Itl( From a practical standpoint, it is not advisable to proceed WIth the addltich0n . . variatetests. Instead,p umvanate twoway analyses 0 f variance . (onee for res eanses are often conducted to see whether the interaction appears m som po . that p 5The likelihood test procedures reqwre (with probability 1).
:5
(667)
(668)
are consistent with HI' Once again, for large samples and using Bartlett's correction: Reject Ho: PI = P2 = ... = Pb = 0 (no factor 2 effects) at level a if
For large samples, Wilks' lambda, A *, can be referred. to a .chisquar~ . n Using Bartlett's multiplier (see [6]) to improve th~ chIsquare approxlmatto , reject Ho: I'll = 1'12 = '" = l' go = 0 at the a level if
i[..
InA*>xfg_l)p(a)
*"
A* =
ISSPresl  ''"'',  ISSPint + SSPres I
[gb(n  1)  P + 1  (g2
2
where A * is given by (666) and Xtgl)p(a) is the upper (l00a)th percentile of a Chisquare distribution with (g  l)p d.f. In a similar manner, factor 2 effects are tested by considering Ho: PI = P 2 = ... = Pb = 0 and HI: at least one Pk O. Small values of
*" 0
is conducted by rejecting Ho for small values of the ratio A*
P+1(g1)]
gbn 1
(=1 k=1
Ho: 1'11
(666)
= 'T2 = ... = 'Tg = 0 (no factor 1 effects) at level a if
k=1 r=1
b
/SSPresl I SSPtacl + SSPres I
'':':0.=.:._ _
so that small values of A * are consistent with HI' Using Bartlett's correction, the likelihood ratio test is as follows:
n(Xtk  it·  X'k + x) (Xlk  I.e·  X'k + x)'
SSPres =
Total (corrected)
A* =
.
x) (X'k  x)'
k=l
e=1 k=1
Residual (Error)
*"
e=1 b
Factor 2
gl
x) (I.e·  x)'
others. Those responses without interaction may be interpreted in terms of additive factor 1 and 2 effects, provided that the latter effects exist. In any event, interaction plots similar to Figure 6.3, but with treatinent sample means replacing expected values, best clarify the relative magnitudes of the main and interaction effects. In the multivariate model, we test for factor 1 and factor 2 main effects as follows. First, consider the hypotheses Ho: 'Tl = 'T2 = ... = 'Tg = 0 and HI: at least one 'Tt O. These hypotheses specify no factor 1 effects and some factor 1 effects, respectively. Let
go(n  1), so that SSPres will be positive
 [ gb(n  1) 
p
+ 1  (b  l)J 2
In A* > XtbI)p(a)
(669)
where A * is given by (668) and XTbI)p( a) is the upper (100a)th percentile of a chisquare distribution witlt (b  1) P degrees of freedom. Simultaneous confidence intervals for contrasts in the model parameters can provide insights into the nature of the·factor effects. Results comparable to Result 6.5 are available for the twoway model. When interaction effects are negligible, we may concentrate on contrasts in the factor 1 and factor 2 main . effects. The Bonferroni approach applies to the components of the differences 'Tt  'Tm of the factor 1 effects and the components of Pk  Pq of the factor 2 effects, respectively. The 100(1  a)% simultaneous confidence intervals for 'Tei  'Tm; are Tti  Tm;
belongs to
(Xt.; 
~m'i)
± tv Cg(ga_ l»));i b~
(670)
where v = gb(n  1), Ei; is the ith diagonal element of E = SSPres , and xe.;  Xm.i is the ith component of I.e.  xm •• I
L
318
TwoWay Multivariate Analysis of Variance
Chapter 6 Comparisons of Several Multivariate Means
Similarly, the 100(1  a) percent simultaneous confidence intervals for f3ki  f3 qi are f3ki  f3 qi
where
jJ
belongsto
(i·ki  i·qi)
a) ~;g;;. fE::2 ± tv (pb(b 1)
Source of variation
[1.7405
1 change in rate ractor : of extrusion
1.5045 1.3005
[7~
and Eiiare as just defined and i·ki  i·qiis the ith component ofx·k  x. q • n 2 amountof ractor : additive
Comment. We have considered the multivariate twoway model with replications. That is, the model allows for n replications of the responses at each combination of factor levels. This enables us to examine the "interaction" of the factors. If only one observation vector i~ available at each combination of factor levels, the twoway model does not allow for the possibility oca general interaction term 'Yek· The corresponding MANOVA table includes only factor 1, factor 2, and residual sources of variation as components of the total variation. (See Exercise 6.13.)
d.f.
SSP
n
(671)
Interaction
Residual
.6825 .6125
319
.8555 ]
.7395 .4205
1.9305] 1.7325
1
1
4.9005
[
.0165 .5445
r7~
.D200 2.6280
0445]
1
3.0700] .5520
16
1.4685 3.9605
64.9240 Example 6.13 (A twoway multivariate analysis of variance of plastic film data) The optimum conditions for extruding plastic film have been examined using a technique called Evolutionary Operation. (See [9].) In the course of the study that was done, three responsesXl = tear resistance, Xz = gloss, and X3 = opacitywere measured at two levels of the factors, rate of extrusion and amount of an additive. The measurements were repeated n = 5 times at each combination of the factor levels. The data are displayed in Table 6.4. Table 6.4 Plastic Film Data Xl = tear resistance, X2 = gloss, and X3 = opacity Factor 2: Amount of additive
Low (1.0%) ~
Factor 1: Change
[6.5 [6.2 Low (10)% [5.8 [6.5 [6.5
in rate of extrusion
High (10%)
~
~
9.5 9.9 9.6 9.6 9.2
4.4] 6.4] 3.0] 4.1] 0,8]
~
Xz
X3
[6.7 [6.6 [7.2 [7.1 [6.8
9.1 9.3 8.3 8.4 8.5
2.8] 4.1] 3.8] 1.6] 3.4]
High (1.5%) ~
X2
X2
2395] 1.9095
19
PANEL 6.1
SAS ANALYSIS FOR EXAMPLE 6.13 USING PROC GLM
title 'MANOVA'; data film; infile 'T64.dat'; input xl x2 x3 factorl factor2; proc glm data =film; class factorl factor2; model xl x2 x3 =factorl factor2 factorl *factor2/ss3; manova h =factorl factor2 factorl *factor2/printe; means factorl factor2;
PROGRAM COMMANDS
X3
X3
[7.1 9.2 8.4] [7.0 8.8 5.2] [7.2 9.7 6.9] [7.5 10.1 2.7] [7.6 9.2 1.9]
The matrices of the appropriate sum of squares and cross products were calcu6 lated (see the SAS statistical software output in Panel 6.1 ), leading to the following MANOVA table: 6Additional SAS programs for MANOVA and other procedures discussed in this chapter are available in [13].
.7855 5.0855
74.2055
[6.9 9.1 5.7] [7.2 10.0 2.0] [6.9 9.9 3.9] [6.1 9.5 1.9] [6.3 9.4 5.7] ~
[42655
Total (corrected)
General linear Models Procedure Class Level Information
LrleR~!l~~ri~ ~~rillbt~:~1 I Source Model Error Corrected Total
Source
Class Levels FACTOR 1 2 FACTOR2 2 Number of observations in OF 3 16 19
Sum of Squares 2.50150000 1.76400000 4.26550000
Mean Square 0.83383333 0.11025000
RSquare 0.586449
C.V.
4.893724
Root MSE 0.332039
OF
OUTPUT
Values 0 1 0 1 data set =20 F Value 7.56
Pr> F 0.0023
Xl Mean 6.78500000
Mean Square
F Value
Pr> F
1.74050000 0.76050000 0.00050000
15.79 6.90 0.00
0.0011 0.0183 0.9471
(continues on next page)
320
Two·Way Multivariate Analysis of Variance
Chapter 6 Comparisons of Several Multivariate Means
PANEL 6.1
321
(continued)
(continued)
Manova Test Criteria and Exact F Statistics for the
i
Sum of Squares 2.45750000 2.62800000 5.08550000
Mean Square 0.81916667 0.16425000
R·Square 0.483237
C.V. 4.350807
Root M5E ·0.405278
OF
Type /11 SS
Mean Square
F Value
1.300$0000 0.612soOOo 0.54450000
1.30050000 0.61250000 0.54450000
7.92 3.73 3.32
source Model Error corrected Total
\\ source
F Value 0.76
OF 3 16 19
Sum of Squares 9.28150000 64.92400000 74.20550000
Mean Square 3.09383333 4.05775000
R·Square 0.125078
C.V. 51.19151
RootMSE 2.014386
OF
Type /11 SS
Mean Square
F Value
0A20SOOOO 4.90050000 3.960SOOOO
0.42050000 4.90050000 3.96050000
0.10 1.21 0.98
Source
I.
1.764 0.02 3.07
Pillai's Trace HotellingLawley Trace Roy's Greatest Root
0.7517 0.2881 0.3379
0.61814162 1.61877188 1.61877188
7.5543 7.5543 7.5543
.F 1.3385
. Numb!' 3
DenDF 14
Pr> F 0.3018
0.22289424 0.28682614 0.28682614
1.3385 1.3385 1.3385
3 3 3
14 14 14
0.3018 0.3018 0.3018
o
Mean ·6.49000000 7.08000000
X3
o
3.07 0.552 64.924
1
Level of FACTOR2
N 10 10
N 10 10
SO 0.42018514 0.32249031
X2Mean 9.57000000 9.06000000
X3Mean SO 3.79000000 4.08000000
1.85379491 2.18214981
X2
Mean 6.59000000 6.98000000
Mean 9.14000000 9.49000000
Level of FACTOR2
N 10 10
SO 0.40674863 0.47328638
X3Mean 3.44000000 4.43000000
SO 1.55077042 2.30123155
To test for interaction, we compute 3 3
SO . 0.29832868 0.57580861
Xl
o
H = Type'" SS&CP Matrix for FACTORl S= 1 M =0.5
Pillai's Trace HotellingLawley Trace ROy's Greatest Root
Value 0.77710.576
N 10 10
Manova Test Criteria and Exact F Statistics for
1HYpOthi!sis. of no Overall fACTOR1 Effect 1
0.0247 0.0247 0.0247
E = Error SS&CP Matrix
Xl
Level of FACTOR 1
o the
14 14 14
3 3 3
Hypothl!sis of no Qverall FAcrOR1~.FAcrOR2 Effect
Level of FACTOR 1 X2 0.02 2.628 0.552
4.2556 4.2556 4.2556
H = Type III SS&CP Matrix for FACTOR 1*FACTOR2 S = ·1 M = 0.5 N=6
Pr> F 0.5315
E= Error SS&CP M'!trix Xl
Xl X2 X3
I
Manova Test Criteria and Exact F Statistics for
X3.1
Source Model Error Corrected Total
Hypothesis of no ()ve~a"FACTOR2 Effect
0.47696510 0.91191832 0.91191832
pillai's Trace HotellingLawley Trace Roy's Greatest Root
the [ Dependi!li~Varlal:i'e;
I
F Value 4.99
OF 3 16 19
A* =
/SSPres / /SSPint + SSPres /
275.7098 354.7906 = .7771
SO 0.56015871 0.42804465
322
Profile Analysis
Chapter 6 Comparisons of Several Multivariate Means For
(g  1)(b  1) = 1, F =
1A*) (I (g (A*
(gb(n 1)  p + 1)/2 l)(b  1)  pi + 1)/2
has an exact Fdistribution with VI = I(g  l)(b  1) gb(n 1)  p + 1d.f.(See[1].)Forourexample.
F
pi + 1
From before, F3 ,14('OS) = 3.34. We have FI = 7.5S > F3,14('OS) = 3.34, and therefore, we reject Ho: 'TI = 'T2 = 0 (no factor 1 effects) at the S% level. Similarly, Fz = 4.26 > F3,14( .OS) = 3.34, and we reject Ho: PI = pz = 0 (no factor 2 effects) at the S% level. We conclude that both the change in rate of extrusion and the amount of additive affect the responses, and they do so in an additive manner. The nature of the effects of factors 1 and 2 on the responses is explored in Exercise 6.1S. In that exercise, simultaneous confidence intervals for contrasts in the components of 'T e and Pk are considered. _
= (1  .7771) (2(2)(4)  3 + 1)/2 = 1 .7771 (11(1) .31 + 1)/2 34
6.8 Profile Analysis
VI =
(11(1)  31 + 1)
V2 =
(2(2)(4)  3 + 1) = 14
=
3
and F3 ,14( .OS) = 3.34. Since F = 1.34 < F3,14('OS) = 3.34, we do not reject hypothesis Ho: 'Y11 = 'YIZ = 'Y21 = 'Y22 = 0 (no interaction effects). Note that the approximate chisquare statistic for this test is (3 + 1  1(1»/2] In(.7771) = 3.66, from (665). Since x1(.05) = 7.81, we reach the same conclusion as provided by the exact Ftest. To test for factor 1 and factor 2 effects (see page 317), we calculate
A~
=
ISSPres I = 27S.7098 = ISSPfac1 + SSPres I 722.0212
.3819
and
A; = For both g  1
323
ISSPres I = 275.7098 = .5230 ISSPfacZ + SSP,es I 527.1347
= 1 and b
 1
= 1,
_(1 A~ Pi 
A~) (gb(n  1)  P + 1)/2
and
(I (g 
A;)
_ (1 Fz A;
1) 
pi + 1)/2
(gb(n  1)  p + 1)/2 (i (b  1)  pi + 1)/2
Profile analysis pertains to situations in which a battery of p treatments (tests, questions, and so forth) are administered to two or more groups of subjects. All responses must be expressed in similar units. Further, it is assumed that the responses for the different groups are independent of one another. Ordinarily, we might pose the question, are the population mean vectors the same? In profile analysis, the question of equality of mean vectors is divided into several specific possibilities. Consider the population means /L 1= [JLII, JLI2 , JLI3 , JL14] representing the average responses to four treatments for the first group. A plot of these means, connected by straight lines, is shown in Figure 6.4.1bis brokenline graph is the profile for population 1. Profiles can be constructed for each population (group). We shall concentrate on two groups. Let 1'1 = [JLll, JLl2,"" JLlp] and 1'2 = [JLz!> JL22,"" JL2p] be the mean responses to p treatments for populations 1 and 2, respectively. The hypothesis Ho: 1'1 = 1'2 implies that the treatments have the same (average) effect on the two populations. In terms of the population profiles, we can formulate the question of equality in a stepwise fashion.
1. Are the profiles parallel? Equivalently: Is H01 :JLli  JLlil
= JLzi  JLzil, i = 2,3, ... ,p, acceptable? 2. Assuming that the profiles are parallel, are the profiles coincident? 7 Equivalently: Is H 02 : JLli = JLZi, i = 1,2, ... , p, acceptable?
Mean response
have Fdistributions with degrees of freedom VI = I(g  1)  pi + 1, gb (n  1)  P + 1 and VI = I (b  1)  pi + 1, V2 = gb(n  1)  p + 1, tively. (See [1].) In our case, = (1  .3819) (16  3 + 1)/2 = 7.55
FI
~:
F2
~<
l
~: '·J···· ..
.3819 = (
(11 31+ 1)/2
1  .5230) (16  3 + 1)/2 .5230 (11  31 + 1)/2 = 4.26
and VI
= 11  31
+1
= 3
V2
= (16  3 + 1) = 14
L..._ _L  _   l_ _l_ _l._
2
3
4
_+
Variable
Figure 6.4 The population profile p = 4.
7The question, "Assuming that the profiles are parallel, are the profiles linear?" is considered in Exercise 6.12. The null hypothesis of parallel linear profIles can be written Ho: (/Lli + iL2i)  (/Llil + /L2H) = (/Llil + iL2H)  (/Lli2 + iL2i2), i = 3, ... , p. Although this hypothesis may be of interest in a particular situation, in practice the question of whether two parallel profIles are the same (coincident), whatever their nature, is usually of greater interest.
324 Chapter 6 Comparisons of Several Multivariate Means
Profile Analysis 325
3. Assuming that the profiles are coincident, are the profiles level? That is, are all the means equal to the same constant? Equivalently: Is H03: iLl I = iL12 = ... = iLlp = JL21 = JL22 = ... = iL2p acceptable?
Test for Coincident Profiles. Given That Profiles Are Parallel
The null hypothesis in stage 1 can be written
where C is the contrast matrix
1
C
1 0 0 1 1 0 0
=
((pI)Xp)
(672)
~
[
o
0 0
For independent samples of sizes nl and n2 from the two popu]ations, the null hypothesis can be tested by constructing the transformed observations CXI;,
j=1,2, ... ,nl
CX2j,
j = 1,2, ... ,n2
For coincident profiles, xu. X12,'·" Xl nl and XZI> xzz, ... , xZ n2 are all observations from the same normal popUlation? The next step is to see whether all variables have the same mean, so that the common profile is level. When HOI and Hoz are tenable, the common mean vector #' is estimated, using all nl + n2 observations, by
_=  1  ( "+ "I ""2) =
and
x
These have sample mean vectors CXI and CX2, respectively, and pooled covariance matrix CSpooledC" Since the two sets of transformed observations have Np1(C#'1, Cl:C:) and NpI(CiL2, CIC') distributions, respectively, an application of Result 6.2 provides a test for parallel profiles.
nl
+
nz
£.; Xl'
;=1
)
£.; X2'
. j=l
)
nl (nl
+
_
n2)
Xl
+
(nl
nz_ X2 + n2)
If the common profile is level, then iLl = iL2 = .. , = iLp' and the null hypothesis at stage 3 can be written as H03: C#' = 0
where C is given by (672). Consequently, we have the following test.
Test for level Profiles. Given That Profiles Are Coincident Test for Parallel Profiles for Two Normal Populations Reject HoI : C#'l
For two normal populations: Reject H03: C#' = 0 (profiles level) at level a if I (nl + n2)x'C'[CSCT Cx > c 2 (675)
= C#'2 (parallel profiles) at level a if
T2 = (Xl  X2)'C{
(~I + ~JCSpooledC' Jl C(Xl 
X2) > c
2
(673)
where S is the sample covariance matrix based on all nl + n2 observations and c 2 = (nl + n2  l)(p  1) ( ) (nl + n2  P + 1) Fpcl,nl+nzP+l et
where
When the profiles are parallel, the first is either above the second (iLli > JL2j, for all i), or vice versa. Under this condition, the profiles will be coincident only if the total heights iLl 1 + iL12 + ... + iLlp = l' #'1 and IL21 + iL22 + ... + iL2p = 1'1'"2 are equal. Therefore, the null hypothesis at stage 2 can be written in the equivalent form
H02 : I' #'1
=
Example 6.14 CA profile analysis of love and marriage data) As part of a larger study of love and marriage, E. Hatfield, a sociologist, surveyed adults with respect to their marriage "contributions" and "outcomes" and their levels of "passionate" and "companionate" love. Receqtly married males and females were asked to respond to the following questions, using the 8point scale in the figure below.
I' #'2
We can then test H02 with the usual twosample tstatistic based on the univariate observations i'xli' j = 1,2, ... , nI, and l'X2;, j = 1,2, ... , n2'
2
3
4
5
6
7
8
326
Chapter 6 Comparisons of Several Multivariate Means
Profile Analysis 327
1. All things considered, how would you describe your contributions to the marriage? 2. All things considered, how would you describe your outcomes from themarriage? SubjeGts were also asked to respond to the following questions, using the 5point scale shown.
Sample mean response 'i (i
6
3. What is the level of passionate love that you feel for your partner? 4. What is the level of companionate love that you feel for your partner?
 d
t..o~
4 None at all
I
Very little
Some
A great deal
Tremendous amount
4
5
X
Key:
x  x Males
I
0 oFemales
2
2 L     _ L _ _ _L _ _ _L_ _ _L_ _+_
Let Xl
= an 8point scale response to Question 1
X2 =
an 8point scale response to Question 2
X3 =
a 5point scale response to Question 3
X4
= a 5point scale response to Question 4
2
3
CSpOoJedC'
[ 1 = ~
and the two populations be defined as Population 1 Population 2
= married men = married women
The population means are the average responses to the p = 4 questions for the populations of males and females. Assuming a common covariance matrix I, it is of interest to see whether the profiles of males and females are the same. A sample of nl = 30 males and n2 = 30 females gave the sample mean vectors
Xl
=
r;:n 4.700J
(males)
_ X2 =
l
6.633j 7.000 4.000 4.533
(females)
and pooled covariance matrix
SpooJed =
Figure 6.S Sample profiles for marriagelove responses.
Variable
4
[ =
1 1 0
.719  .268
.125
0 1 1
.268 1.101 .751
~}~~r ~
0 1 1 0
fj
125]
.751 1.058
and
Thus, .719
T2 = [.167, .066, .200J (k +
ktl [ .268 .125
.268 1.101 .751
.125]1 [.167] .751 .066 1.058 .200
= 15(.067) = 1.005
l
·606 .262 .066 .262 .637 .173 .066 .173 .810 .161 .143 .029
.161j .143 .029 .306
The sample mean vectors are plotted as sample profiles in Figure 6.5 on page 327. Since the sample sizes are reasonably large, we shall use the normal theory methodology, even though the data, which are integers, are clearly nonnormal. To test for parallelism (HOl: CILl =CIL2), we compute
Moreover, with a= .05, c 2 = [(30+302)(41)/(30+30 4)JF3,56(.05) = 3.11(2.8) = 8.7. Since T2 = 1.005 < 8.7, we conclude that the hypothesis of parallel profiles for men and women is tenable. Given the plot in Figure 6.5, this finding is not surprising. Assuming that the profiles are parallel, we can test for coincident profiles. To test H 02 : l'ILl = l' IL2 (profiles coincident), we need Sum of elements in (Xl  X2) = l' (Xl  X2) = .367 Sum of elements in Spooled
= I'Spooled1 = 4.207
Repeated Measures Designs and Growth Curves 329
328 Chapter 6 Comparisons of Several Multivariate Means Using (674), we obtain
Table 6_S Calcium Measurements on the Dominant Ulna; Control Group T2 = (
.367
V(~ + ~)4.027
)2 = .501
With er = .05, F1,;8(.05) = 4.0, and T2 = .501 < F1,58(.05) = 4.0, we cannot reject the hypothesis that the profiles are coincident. That is, the responses of men and women to the four questions posed appear to be the same. We could now test for level profiles; however, it does not make sense to carry out this test for our example, since Que'stions 1 and i were measured on a scale of 18, while Questions 3 and 4 were measured on a scale of 15. The incompatibility of these scales makes the test for level profiles meaningless and illustrates the need for similar measurements in order to carry out a complete profIle analysis. _ When the sample sizes are small, a profile analysis will depend on the normality assumption. This assumption can be checked, using methods discussed in Chapter 4, with the original observations Xej or the contrast observations CXej' The analysis of profiles for several populations proceeds in much the same fashion as that for two populations. In fact, the general measures of comparison are analogous to those just discussed. (See [13), [18).)
6.9 Repeated Measures Designs and Growth Curves
Subject
Initial
1 year
2 year
3 year
1 2 3 4 5 6 7 8 9 10
87.3 59.0 76.7 70.6 54.9 78.2 73.7 61.8 85.3 82.3 68.6 67.8 66.2 81.0 72.3
86.9 60.2 76.5 76.1 55.1 75.3 70.8 68.7 84.4 86.9 65.4 69.2 67.0 82.3 74.6
86.7 60.0 75.7 72.1 57.2 69.1 71.8 68.2 79.2 79.4 72.3 66.3 67.0 86.8 75.3
75.5 53.6 69.5 65.3 49.0 67.6 74.6 57.4 67.0 77.4 60.8 57.9 56.2 73.9 66.1
72.38
73.29
72.47
64.79
11 12 13 14 15 Mean
Source: Data courtesy of Everett Smith.
When the p measurements on all subjects are taken at times tl> t2,"" tp, the PotthoffRoy model for quadratic growth becomes
As we said earlier, the term "repeated measures" refers to situations where the same characteristic is observed, at different times or locations, on the same subject. (a) The observations on a subject may correspond to different treatments as in Example 6.2 where the time between heartbeats was measured under the 2 X 2 treatment combinations applied to each dog. The treatments need to be compared when the responses on the same subject are correlated. (b) A single treatment may be applied to each subject and a single characteristic observed over a period of time. For instance, we could measure the weight of a puppy at birth and then once a month. It is the curve traced by a typical dog that must be modeled. In this context, we refer to the curve as a growth curve. When some subjects receive one treatment and others another treatment, the growth curves for the treatments need to be compared.
To illustrate the growth curve model introduced by Potthoff and Roy [21), we consider calcium measurements of the dominant ulna bone in older women. Besides an initial reading, Table 6.5 gives readings after one year, two years, and three years for the control group. Readings obtained by photon absorptiometry from the same subject are correlated but those from different subjects should be independent. The model assumes that the same covariance matrix 1: holds for each subject. Unlike univariate approaches, this model does not require the four measurements to have equal variances.A profile, constructed from the four sample means (Xl, X2, X3, X4), summarizes the growth which here is a loss of calcium over time. Can the growth pattern be adequately represented by a polynomial in time?
where the ith mean ILi is the quadratic expression evaluated at ti • Usually groups need to be compared. Table 6.6 gives the calcium measurements for a second set of women, the treatment group, that received special help with diet and a regular exercise program. When a study involves several treatment groups, an extra subscript is needed as in the oneway MANOVA model. Let X{1, X{2,"" Xene be ~he ne vectors of measurements on the ne subjects in group e, for e = 1, ... , g.
Assumptions. All of the X ej are independent and have the same covariance matrix 1:. Under the quadratic growth model, the mean vectors are
330 Chapter 6 Comparisons of Several Multivariate Means
Repeated Measures Designs and Growth Curves 331 g
Table 6.6 Calcium Measurements on the Dominant Ulna; Treatment
with N =
Group
ne, is the pooled estimator of the common covariance matrix l:. The
e=l
Subject 1 2 3 4 5 6 7 8 9
L
Initial
1 year
2 year
3 year
83.8 65.3 81.2 75.4 55.3 70.3 76.5 66.0 76.7 77.2 67.3 50.3 57.7 74.3 74.0 57.3 69.29
85.5 66.9 79.5 76.7 58.3 72.3 79.9 70.9 79.0 74.0 70.7 51.4 57.0 77.7 74.7 56.0 70.66
86.2 67.0 84.5 74.3 59.1 70.6 80.4 70.3 76.9 77.8 68.9 53.6 57.5 72.6 74.5 64.7 71.18
81.2 60.6 75.2 66.7 54.2 68.6 71.6 64.1 70.3 67.9 65.9 48.0 51.5 68.0 65.7 53.0 64.53
,
10
11 12 13 14 15 16 Mean
estimated covariances of the maximum likelihood estimators are 
k,
A
Wq =
g
~
e=1
j=l
L
tl t~t1]
[ f3eo ]
tz
and
f
A
tl
t'{
t2
t5.
B=
~;~
Pe =
(676)
(680)
=
IWI IWql
(681)
Pe
q
[Pr. pzJ (677)
=
=
Under the assumption of multivariate normality, the maximum likelihood estimators of the Pe are (678) where
=N
1 _ gW
73.0701 3.6444 [ 2.0274
70.1387] 4.0900 1.8534
so the estimated growth curves are
f3eq
1 Spooled = (N _ g) «nl  I)SI + ... + (ng  I)Sg)
(682)
Example 6.IS (Fitting a quadratic growth curve to calcium loss) Refer to the data in
Control group: tp
xrpql)g(a)
Tables 6.5 and 6.6. Fit the model for quadratic growth. A computer calculation gives
f3eo f3n and
tp
A
~(p  q+ g») In A * >
If a qthorder polynomial is fit to the growth data, then
1
(679)
Under the polynomial growth model, there are q + 1 terms instead of the p means for each of the groups. Thus there are (p  q  l)g fewer parameters. For large sample sizes, the null hypothesis that the polynomial is adequate is rejected if
~ t~ t~
1 1
f = 1,2, ... , g
~ (X ej  BPe) (Xej  Bpe)'
A*
( N 
=
for
has ng  g + p  q  1 degrees of freedom. The likelihood ratio test of the null hypothesis that the qorder polynomial is adequate can be based on Wilks' lambda
where
B
1
where k =IN  ¥) (N  g  l)j(N  g  p + q)(N  g  p + q + 1). Also, Pe and Ph are independent, for f # h, so their covariance is O. We can formally test that a qthorder polynomial is adequate. The model is fit without restrictions, the error sum of squares and cross products matrix is just the within groups W that has N  g degrees of freedom. Under a qthorder polynomial, the error sum of squares and cross products
Source: Data courtesy of Everett Smith.
1l
1
Cov(Pe) =  (B SpooledB) ne
73.07 + 3.64t  2.03(2 (2.58) (.83) (.28) .
Treatment group: 70.14 (2.50)
+ 4.09t  1.85t2 (.80)
(.27)
where
(B'Sp601edBr1 =
93.1744 5.8368 [ 0.2184
5.8368 9.5699 3.0240
0.2184] 3.0240 1.1051
and, by (679), the standard errors given below the parameter estimates were obtained by dividing the diagonal elements by ne and taking the square root.
Perspect ives and a Strategy for Analyzing Multivar iate Models 333
332 Chapter6 Comparisons of Several Multivar iate Means
Examination of the estimates and the standard errors reveals that the (2 terms are needed. Loss of calcium is predicte d after 3 years for both groups. Further, there o s not seem to be any substantial difference between the two g~oups. . d e . th sis that the quadratic growth model IS Wilks' lambda for testIng e nu1I hypothe ~. adequate becomes 2660.749 2660.749 2756.009 2343.514 2327~961 2369.308 2343.514 2301.714 2098.544 2335.912 23?7.961 2098.544· 2277.452 = .7627 2698.589 2363.228 2698.589 2832.430 2331.235 2381..160 2363.228 2331.235 2303.687 2089.996 2362.253 2381.160 2089.996 2314.485 Since, with a _( N _
=
~ (p 
r62~
2369308 2335.91]
l'781.O17
~~~31
.01, q + g»)tn A *
=
(31 
i
(4  2 + 2») In .7627
= 7.86 <
_
xt42l)2( .01)  9.21
;ea~~~~~ ~r:c:s~~~:~:~~,as~:! :~~d~~:~r;~~~ f:~:~:a~r~t!~ ~:~: ~~I~~:r~::i~ We could, without restr!cti ng to ~uadratIc growth, test for par dent calcium loss using profile analYSIS.
_ .
owth curve model holds for more general designs than The Potthoff and Roy gr , I . b (6 78) and the expresNOVA Howeve r the fJ( are no onger gIven y oneway. MA . ' . b' ore complic ated than (679). We refer the sion for Its covanance matnx ecomes m reader to [14] for moretheexrammop~~~c:~~~~r:!~~!e:~'del treated here. They include the There are many 0 following: (a) Dropping the restriction to. pol~nomial growth. Use nonline ar parametric models or even nonpara metnc sphnes. . .al f such as equally correlated (b) Restricting the covariance matriX to a specl onn responses on the same individual. . .. . bl f (c) Observing more than one respon~e vana e, over Ime, on the same IndIVIdual. This results in a multivariate verSIOn of the growth curve model.
6.10 Perspectives and a Strategy for Analyzing Multivariate Models We emphasize that with several characteristics, it is ~port~nt to co~trol the ~~~:~
probability of making any incorrect decision. This IS partIcularl~ ~p~~~nc hapter testing for the equality of two or more treatme nts as the exarnp es In
indicate. A single multivariate test, with its associated. single pvalue, is preferab le to performing a large number of univariate tests. The outcom e tells us whether or not it is worthwhile to look closer on a variable by variable and group by group analysis. A single multivariate test is recomm ended over, say,p univariate tests because, as the next example demonstrates, univariate tests ignore importa nt informa tion ·and can give misleading results. Example 6.16 (Comparing multivariate and univariate tests for the differences in means) Suppose we collect measure ments on two variables Xl and X 2 for ten randomly selected experimental units from each of two groups. The hypothetical data are noted here and displayed as scatter plots and marginal dot diagrams in Figure 6.6 on page 334.
X2
Group
5.0 4.5 6.0 6.0 6.2 6.9 6.8 5.3 6.6
3.0 1 3.2 1 3.5 1 4.6 1 5.6 1 5.2 1 6.0 1 5.5 1 7.3 1 ___?} ___________________________f?:_~______________________________ .!___ _ 4.6 4.9 2 4.9 5.9 2 4.0 4.1 2 3.8 5.4 2 6.2 6.1 2 5.0 7.0 2 5.3 4.7 2 7.1 6.6 2 5.8 7.8 2 6.8 8.0 2 It is clear from the horizontal marginal dot diagram that there is conside rable overlap in the Xl values for the two groups. Similarly, the vertical margina l dot diagram shows there is considerable overlap in the X2 values for the two groups. The scatter plots suggest that there is fairly strong positive correlat ion between the two variables for each group, and that, although there is some overlap, the group 1 measurements are generally to the southea st of the group 2 measurements. Let PI = [PlI, J.l.12J be the populat ion mean vector for the first group, and let /Lz = [J.l.2l, /L22J be the populat ion mean vector for the second group. Using the Xl observations, a univariate analysis of variance gives F = 2.46 with III = 1 and 112 = 18 degrees of freedom . Consequently, we cannot reject Ho: J.l.1I = J.l.2l at any reasonable significance level (F1.18(.10) = 3.01). Using the X2 observa tions, a univariate analysis of variance gives F = 2.68 with III = 1 and 112 = 18 degrees of freedom. Again, we cannot reject Ho: J.l.12 = J.l.22 at any reasonable significa nce level.
Perspectives and a Strategy for Analyzing Multivariate Model~ 335 334 Chapter 6 Comparisons of Several Multivariate Means Table 6.7 Lizard Data for Two Genera
C
fjgure 6.6 Scatter plots and marginal dot diagrams for the data from two groups.
The univariate tests suggest there is no difference between the component means for the two groups, and hence we cannot discredit 111 = 112' On the other hand, if we use Hotelling's T2 to test for the equality of the mean vectors, we find
Mass
SVL
Mass
SVL
7.513 5.032 5.867 11.088 2.419 13.610 18.247 16.832 15.910 17.035 16.526 4.530 7.230 5.200 13.450 14.080 14.665 6.092 5.264 16.902
74.0 69.5 72.0 80.0 56.0 94.0 95.5 99.5 97.0 90.5 91.0 67.0 75.0 69.5 91.5 91.0 90.0 73.0 69.5 94.0
13.911 5.236 37.331 41.781 31.995 3.962 4.367 3.048 4.838 6.525 22.610 13.342 4.109 12.369 7.120 21.077 42.989 27.201 38.901 19.747
77.0 62.0 108.0 115.0 106.0 56.0 60.5 52.0 60.0 64.0 96.0 79.5 55.5 75.0 64.5 87.5 109.0 96.0 111.0 84.5
14.666 4.790 5.020 5.220 5.690 6.763 9.977 8.831 9.493 7.811 6.685 11.980 16.520 13.630 13.700 10.350 7.900 9.103 13.216 9.787
80.0 62.0 61.5 62.0 64.0 63.0 71.0 69.5 67.5 66.0 64.5 79.0 84.0 81.0 82.5 74.0 68.5 70.0 77.5 70.0
SVL = snoutvent length. Source: Data courtesy of Kevin E. Bonine. 4~~ ~c 800
and we reject Ho: 111 = 112 at the 1% level. The multivariate test takes into account the positive correlation between the two measurements for each groupinforma2 tion that is unfortunately ignored by the univariate tests. This T test is equivalent to the MANOVA test (642). •
'. nl = 20
S:
K 1
nz
= 40
= [2.240J 4.394
2.368J K2 = [ 4.308
s = [0.35305 1
S2
0.09417J 0.09417 0.02595 0.50684 0.14539J 0.04255
= [ 0.14539
°°
00
3
°
°
S
,Rn° , ..'••
o<e~· of.P
Qi0cY tit
2
Example 6.11 (Data on lizards that require a bivariate test to establish a difference in means) A zoologist collected lizards in the southwestern United States. Among other variables, he measured mass (in grams) and the snoutvent length (in millimeters). Because the tails sometimes break off in the wild, the snoutvent length is a more representative measure of length. The data for the lizards from two genera, Cnemidophorus (C) and Sceloporus (S), collected in 1997 and 1999 are given in Table 6.7. Notice that there are nl = 20 measurements for C lizards and n2 = 40
C
S
SVL
(18)(2) T2 = 17.29 > c 2 = ~ F2,17('01) = 2.118 X 6.11 = 12.94
measurements for S lizards. After taking natural logarithms, the summary statistics are
S
Mass
<e, 1
3.9
?f o •• #
° • 4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
In(SVL)
Figure 6.7
Scatter plot of In(Mass) versus In(SVL) for the lizard data in Table 6.7.
!" plot of ~ass (Mass) versus snoutvent length (SVL), after taking natural logarithms, IS. shown ~
Figure 6.7. The large sample individual 95% confidence intervals for the difference m In(Mass) means and the difference in In(SVL) means both cover O. In (Mass ): In(SVL):
ILll  IL21: IL12  IL22:
( 0.476,0.220) (0.011,0.183)
336
Exercises
Chapter 6 Comparisons of Several Multivariate Means
merit further study but, with the current data, cannot be taken as conclusive evidence for the existence of differences. We summarize the procedure developed in this chapter for comparing treatments. The first step is to check the data for outliers using visual displays and other calculations.
The corresponding univariate Student's ttest statistics for test.ing for no difference in the individual means have pvalues of .46 and .08, respectlvely. Clearly, from a univariate perspective, we cannot detect a diff~ence in mass means or a difference in snoutvent length means for the two genera of lizards. However, consistent with the scatter diagram in Figure 6.7, a bivariate analysis strongly supports a difference in size between the two groups of lizards. Using ReSUlt 6.4 (also see Example 6.5), the T2statistic has an approximate X~ distribution. For this example, T2 = 225.4 with a pvalue less than .0001. A multivariate method is essential in this case. •
A Strategy for the Multivariate Comparison of Treatments 1. Try to identify outliers. Check the data group by group for outliers. Also check the collection of residual vectors from any fitted model for outliers. Be aware of any outliers so calculations can be performed with and without them. 2. Perform a multivariate test of hypothesis. Our choice is the likelihood ratio test, which is equivalent to Wilks' lambda test.
Examples 6.16 and 6.17 demonstrate the efficacy of ~ m~ltiv~riate. test relative to its univariate counterparts. We encountered exactly this SituatIOn with the efflllent data in Example 6.1. In the context of random samples from several populations (recall the oneway MANOVA in Section 6.4), multivariate tests are based on the matrices W
=
±~
e=1
3. Calculate the Bonferroni simultaneous confidence intervals. If the multivariate test reveals a difference, then proceed to calculate the Bonferroni confidence intervals for all pairs of groups or treatments, and all characteristics. If no differences are significant, try looking at Bonferroni intervals for the larger set of responses that includes the differences and sums of pairs of responses.
(xej  xe)(xcj  xe)' and B = ±ne(xe  x)(xe  x)' e=1
j=!
Throughout this chapter, we have used Wilks'lambdastatisticA*
=
IBI:~I
We must issue one caution concerning the proposed strategy. It may be the case that differences would appear in only one of the many characteristics and, further, the differences hold for only a few treatment combinations. Then, these few active differences may become lost among all the inactive ones. That is, the overall test may not show significance whereas a univariate test restricted to the specific active variable would detect the difference. The best preventative is a good experimental design. To design an effective experiment when one specific variable is expected to produce differences, do not include too many other variables that are not expected to show differences among the treatments.
which is equivalent to the likelihood ratio test. Three other multivariate test statistics are regularly included in the output of statistical packages. LawleyHotelling trace = tr[BWI ] Pillai trace
= tr[B(B + W)IJ
Roy's largest root
=
maximum eigenvalue of W (B
+ W)I
All four of these tests appear to be nearly equivalent for extremely large samples. For moderate sample sizes, all comparisons are based on what is necessarily a limited number of cases studied by simulation. From the simulations reported to date the first three tests have similar power, while the last, Roy's test, behaves differe~tly.lts power is best only when there is a single nonzero eigenvalue and, at the same time, the power is large. This may approximate situations where a large difference exists in just one characteristic and it is between one group and all of the others. There is also some suggestion that Pillai's trace is slightly more robust against nonnormality. However, we suggest trying transformations on the original data when the residuals are nonnormal. All four statistics apply in the twoway setting and in even more complicated MANOVA. More discussion is given in terms of the multivariate regression model in Chapter 7. When, and only when, the multivariate tests signals a difference, or de~arture from the null hypothesis, do we probe deeper. We recommend calculatmg the Bonferonni intervals for all pairs of groups and all characteristics. The simultaneous confidence statements determined from the shadows of the confidence ellipse are, typically, too large. The oneatatime intervals may be suggestive of differences that
337
Exercises 6.1.
Construct and sketch a joint 95% confidence region for the mean difference vector I) using the effluent data and results in Example 6.1. Note that the point I) = 0 falls outside the 95% contour. Is this result consistent with the test of Ho: I) = 0 considered in Example 6.1? Explain. 6.2. Using the information in Example 6.1. construct the 95% Bonferroni simultaneous intervals for the components of the mean difference vector I). Compare the lengths of these intervals with those of the simultaneous intervals constructed in the example. 6.3. The data corresponding to sample 8 in Thble 6.1 seem unusually large. Remove sample 8. Construct a joint 95% confidence region for the mean difference vector I) and the 95% Bonferroni simultaneous intervals for the components of the mean difference vector. Are the results consistent with a test of Ho: I) = O? Discuss. Does the "outlier" make a difference in the analysis of these data?
338 Chapter 6 Comparisons of Several Multivariate Means
Exercises JJ9
6.4. Refer to Example 6.l. (a) Redo the analysis in Example 6.1 after transforming the pairs of observations to In(BOD) and In (SS). (b) Construct the 95% Bonferroni simultaneous intervals for the components of the mean vector B of transformed variables. (c) Discuss any possible violation of the assumption of a bivariate normal distribution for the difference vectors of transformed observations. 6.S. A researcher considered three indices measuring the severity of heart attacks. The values of these indices for n = 40 heartattack patients arriving at a hospital emergency room produced the summary statistics .
x=
46.1] 57.3 and S [ 50.4·
=
(a) Calculate Spooled' (b) Test Ho: ILz  IL3 = 0 employing a twosample approach with a = .Ol. (c) Construct 99% simultaneous confidence intervals for the differences J.tZi  J.t3i, i = 1,2. 6.1. Using the summary statistics for the electricitydemand data given in Example 6.4, compute T Z and test the hypothesis Ho: J.tl  J.t2 = 0, assuming that 11 = 1 2, Set a = .05. Also, determine the linear combination of mean components most responsible for the rejection of Ho. 6.8. Observations on two responses are collected for three treatments. The obser
[:~J are Treatmentl:
Treatment 2:
Treatment 3:
[!J DJ [~J [~l DJ DJ UJ [~l [~J
[~l
GJ
1 1 0
}n,
UI =
,°2
0 1
0 0
=
0 0
1 0
, ... ,Dg
= 0 1 1
}n,
,
1
I = + [(nl  1) SI + (n2  1) S2] = (nl + n2 nl n2 nl + n2
2)
Spooled
Hint: Use (416) and the maximization Result 4.10. 6.12. (Test for
linear prOfiles, given that the profiles are parallel.) Let ILl [J.tI1,J.tIZ,··· ,J.tlp] and 1'2 = [J.tZI,J.t22,.·· ,J.tz p ] be the mean responses to p treatments for populations 1 and 2, respectively. Assume that the profiles given by the two mean vectors are parallel.
(a) ShowthatthehypofuesisthattheprofilesarelinearcanbewrittenasHo:(J.t li + J.t2i)(J.tliI + J.tzid = (J.tliI + J.tzid  (J.tliZ + J.tZiZ), i = 3, ... , P or as Ho: C(I'I + 1'2) =0, where the (p  2) X P matrix 2
o o
o
000
1
~ ~
1
(a) Break up the observations into mean, treatment, and residual components, as in (639). Construct the corresponding arrays for each variable. (See Example 6.9.) (b) Using the information in Part a, construct the oneway MAN OVA table. (c) Evaluate Wilks' lambda, A *, and use Table 6.3 to test for treatment effects. Set a = .01. Repeat the test using the chisquare approximation with Bartlett's correction. [See (643).] Compare the conclusions.
}n,
6.1 I. A likelihood argument provides additional support for pooling the two independent sample covariance matrices to estimate a common covariance matrix in the case of two normal populations. Give the likelihood function, L(ILI, IL2' I), for two independent samples of sizes nl and n2 from Np(ILI' I) and N p(IL2' I) populations, respectively. Show that this likelihood is maximized by the choices ill = XI, il2 = X2 and
2
[~J
d = Cx, and
6.10. Consider the univariate oneway decomposition of the observation xc' given by (634). Show that the mean vector x 1 is always perpendicular to the treat~ent effect vector (XI  X)UI + (xz  X)U2 + ... + (Xg  x)u g where
[101.3 63.0 71.0] 63.0 80.2 55.6 71.0 55.6 97.4
(a) All three indices are evaluated for each patient. Test for the equality of mean indices using (616) with a = .05. (b) Judge the differences in pairs of mean indices using 95% simultaneous confidence intervals. [See (618).] 6.6. Use the data for treatments 2 and 3 in Exercise 6.8.
vation vectors
6.9. Using the contrast matrix C in (613), verify the relationships d· = Cx·, Sd = CSC' in (614). ) )
0J0
(b) Following an argument similar to the one leading to (673), we reject Ho: C (1'1 + 1'2) = 0 at level a if Z
T = (XI + X2)'C[ where
(~I + ~JCSpooledC'JIC(XI + X2) > c Z
340
Exercises 341
Chapter 6 Comparisons of Several Multivariate Means Let nl
Hint: This MANOVA table is consistent with the twoway MANOVA table for comparing factors and their interactions where n = 1. Note that, with n = 1, SSPre , in the general twoway MANOVA table is a zero matrix with zero degrees of freedom. The matrix of interaction sum of squares and cross products now becomes the residual sum of squares and cross products matrix. (d) Given the summary in Part c, test for factor 1 and factor 2 main effects at the a = .05 level. Hint: Use the results in (667) and (669) with gb(n  1) replaced by (g  1)(b  1). Note: The tests require that p :5 (g  1) (b  1) so that SSPre , will be positive definite (with probability 1).
= 30, n2 = 30, xi = [6.4,6.8,7.3, 7.0],i2 = [4.3,4.9,5.3,5.1], and
SpooJed =
l
·61 .26 .07 .161 .26 .64 .17 .14 .07 .17 .81 .03 .16 .14 .03 .31
Test for linear profiles, assuming that the profiles are parallel. Use a
= .05.
6.13. (Twoway MANOVA without replications.) Consider the observations on two responses, XI and X2, displayed in the form of the following twoway table (note that there is a single observation vector at each combination of factor levels):
6.14. A replicate of the experiment in Exercise 6.13 yields the following data: Factor 2
Factor 2
Level 1 Factor 1
Level 2 Level 3
2
[~J [~J
[~J
[~J
[=:]
Level 1
Level 4
Level 3
Level
Level 1
[l~J [~J [~ J
[:]
Level 1 Factor 1
Level 3
With no replications, the twoway MANOVA model is g
b
f=1
k=1
2: 'rf = 2: Ih =
+ (X'k
 x)
+ (XCk
 xe·  X.k
+ x)
+ SSfac I + SSfac2 + SSre,
and sums of cross products SCPtot = SCPmean + SCPt• cl + SCPf•c2
+ SCPre,
Consequently, obtain the matrices SSPcop SSPf•cl , SSPfac2 , and SSPre, with degrees of freedom gb  1, g  1, b  1, and (g  1)(b  1), respectively. (c) Summarize the calculations in Part b in a MANOVA table.
Level 3
Level 4
[1:J [~J [~J [~!J DJ L~J [1~J [~J [~J [~J [1~J [~J
xek = x + (xe.  x) + (X.k  x) + (Xfk  xe.  x.k + x)
similar to the arrays in Example 6.9. For each response, this decomposition will result in several 3 X 4 matrices. Here x is the overall average, xc. is the average for the lth level of factor 1, and X'k is the average for the kth level of factor 2. (b) Regard the rows of the matrices in Part a as strung out in a single "long" vector, and compute the sums of squares SStot = SSme.n
2
(a) Use these data to decompose each of the two measurements in the observation vector as
where the eek are independent Np(O,!) random vectors. (a) Decompose the observations for each of the two variables as Xek = X + (xc.  x)
Level 2
Level
6.1 s.
where x is the overall average, xe. is the average for the lth level of factor 1, and X'k is the average for the kth level of factor 2. Form the corresponding arrays for each of the two responses. (b) Combine the preceding data with the data in Exercise 6.13 and carry out the necessary calculations to complete the general twoway MANOVA table. (c) Given the results in Part b, test for interactions, and if the interactions do not exist, test for factor 1 and factor 2 main effects. Use the likelihood ratio test with a = .05. (d) If main effects, but no interactions, exist, examine the natur~ of the main effects by constructing Bonferroni simultaneous 95% confidence intervals for differences of the components of the factor effect parameters. Refer to Example 6.13. (a) Carry out approximate chisquare (likelihood ratio) tests for the factor 1 and factor 2 effects. Set a =.05. Compare these results with the results for the exact Ftests given in the example. Explain any differences. (b) Using (670), construct simultaneous 95% confidence intervals for differences in the factor 1 effect parameters for pairs of the three responses. Interpret these intervals. Repeat these calculations for factor 2 effect parameters.
Exercises 343
342 Chapter 6 Comparisons of Several Multivariate Means The following exercises may require the use of a computer.
6.16. Four measures of the response stiffness on .each of 30 boards are listed in Table 4.3 (see ' Example 4.14). The measures, on a given board, are repeated in ~he sense ~hat they were made one after another. Assuming that the measures of stiffness anse from four treatments test for the equality of treatments in a repeated measures design context. Set a = .05. Construct a 95% (simultaneous) confidence interval for a co~trast in the mean levels representing a comparison of the dynamic measurements WIth the static measurements. 6.1,7. The data in Table 6.8 were collected to test two psychological models of numerical , cognition. Does the processfng oLnumbers d~pend on the w~y the numbers ar~ presented (words, Arabic digits)? Thirtytwo subjects were requued to make a senes of Table 6.8 Number Parity Data (Median Times in Milliseconds) ArabicSame ArabicDiff WordSame WordDiff
(Xl) 869.0 995.0 1056.0 1126.0 1044.0 925.0 1172.5 1408.5 1028.0 1011.0 726.0 982.0 1225.0 731.0 975.5 1130.5 945.0 747.0 656.5 919.0 751.0 774.0 941.0 751.0 767.0 813.5 1289.5 1096.5 1083.0 1114.0 708.0 1201.0
(X2) 860.5 875.0 930.5 954.0 909.0 856.5 896.5 1311.0 887.0 863.0 674.0 894.0 1179.0 662.0 872.5 811.0 909.0 752.5 ' 659.5 833.0 744.0 735.0 931.0 785.0 737.5 750.5 1140.0 1009.0 958.0 1046.0 669.0 925.0
Source: Data courtesy of J. Carr.
(X3)
691.0 678.0 833.0 888.0 865.0 1059.5 926.0 854.0 915.0 761.0 663.0 831.0 1037.0 662.5 814.0 843.0 867.5 777.0 572.0 752.0 683.0 671.0 901.5 789.0 724.0 711.0 904.5 1076.0 918.0 1081.0 657.0 1004.5
(X4)
601.0 659.0 826.0 728.0 839.0 797.0 766.0 986.0 735.0 657.0 583.0 640.0 905.5 624.0 735.0 657.0 754.0 687.5 539.0 611.0 553.0 612.0 700.0 735.0 639.0 625.0 7~4.5
983.0 746.5 796.0 572.5 673.5
quick numerical judgments about two numbers presented as either two number words ("two," "four") or two single Arabic digits ("2," "4"). The subjects were asked to respond "same" if the two numbers had the same numerical parity (both even or both odd) and "different" if the two numbers had a different parity (one even, one odd). Half of the subjects were assigned a block of Arabic digit trials, followed by a block of number word trials, and half of the subjects received the blocks of trials in the reverse order. Within each block, the order of "same" and "different" parity trials was randomized for each subject. For each of the four combinations of parity and format, the median reaction times for correct responses were recorded for each subject. Here ' Xl = median reaction time for word formatdifferent parity combination X z = median reaction time for word formatsame parity combination X3 == median reaction time for Arabic formatdifferent parity combination X 4 = median reaction time for Arabic formatsame parity combination
(a) Test for treatment effects using a repeated measures design. Set a = .05. (b) Construct 95% (simultaneous) confidence intervals for the contrasts representing the number format effect, the parity type effect and the interaction effect. Interpret the resulting intervals. (c) The absence of interaction supports the M model of numerical cognition, while the presence of interaction supports the C and C model of numerical cognition. Which model is supported in this experiment? (d) For each subject, construct three difference scores corresponding to the number format contrast, the parity type contrast, and the interaction contrast. Is a multivariate normal distribution a reasonable population model for these data? Explain. 6.18. 10licoeur and Mosimann [12] studied the relationship of size and shape for painted turtles. Table 6.9 contains their measurements on the carapaces of 24 female and 24 male turtles. (a) Test for equality of the two population mean vectors using a = .05. (b) If the hypothesis in Part a is rejected, find the linear combination of mean components most responsible for rejecting Ho. (c) Find simultaneous confidence intervals for the component mean differences. Compare with the Bonferroni intervals. Hint: You may wish to consider logarithmic transformations of the observations. 6.19. In the first phase of a study of the cost of transporting milk from fanns to dairy plants, a survey was taken of finns engaged in milk transportation. Cost data on X I == fuel, X 2 = repair, and X3 = capital, all measured on a permile basis, are presented in Table 6.10 on page 345 for nl = 36 gasoline and n2 = 23 diesel trucks. (a) Test for differences in the mean cost vectors. Set a = .01. (b) If the hypothesis of equal cost vectors is rejected in Part a, find the linear combination of mean components most responsible for the rejection. (c) Construct 99% simultaneous confidence intervals for the pairs of mean components. Which costs, if any, appear to be quite different? (d) Comment on the validity of the assumptions used in your analysis. Note in particular that observations 9 and 21 for gasoline trucks have been identified as multivariate outIiers. (See Exercise 5.22 and [2].) Repeat Part a with these observations deleted. Comment on the results.
Exercises 345
344 Chapter 6 Comparisons of Several Multivariate Means
Table 6.10 Milk TransportationCost Data
Table 6.9 Carapace Measurements (in Millimeters) for Painted Thrtles
Gasoline trucks
Male
Female
Width
Height
Width
Height
Length
(Xl) 
(X2)
(X3)
(Xl)
(X2)
(X3)
98 103 103 105 109 123 123 133 133 133 134 136 138 138 141 147 149 153 155 155 158 159 162 177
81 84 86 86 88 92 95 99 102 102 100 102 98 99 105 108 107 107 115 117 115 118 124 132
38 38 42 42 44 50 46 51 51 51 48 49 51 51 53 57 55 56 63 60 62 63 61 67
93 94 96 101 102 103 104 106 107 112 113 114 116 117 117 119 120 120 121 125 127 128 131 135
74 78 80 84 85 81 83 83 82 89 88 86 90 90 91 93 89 93 95 93 96 95 95 106
37 35 35 39 38 37 39 39 38 40 40 40 43 41 41 41 40 44 42 45 45 45 46 47
Length
6.20. The tail lengths in millimeters (xll and wing lengths in rniIlimeters (X2) for 45 male hookbilled kites are given in Table 6.11 on page 346. Similar measurements for female hookbilled kites were given in Table 5.12. (a) Plot the male hookbilled kite data as a scatter diagram, and (visually) check for outliers. (Note, in particular, observation 31 with Xl = 284.) (b) Test for equality of mean vectors for the populations of male and female hookbilled kites. Set a = .05. If Ho: ILl  ILz = 0 is rejected, find the linear combination most responsible for the rejection of Ho. (You may want to eliminate any out/iers found in Part a for the male hookbilled kite data before conducting this test. Alternatively, you may want to interpret XJ = 284 for observation 31 as it misprint and conduct the test with XI = 184 for this observation. Does it make any difference in this case how observation 31 for the male hookbilled kite data is treated?) (c) Determine the 95% confidence region for ILl  IL2 and 95% simultaneous confidence intervals for the components of ILl  IL2' (d) Are male or female birds generally larger?
Diesel trucks
Xl
X2
X3
Xl
X2
X3
16.44 7.19 9.92 4.24 11.20 14.25 13.50 13.32 29.11 12.68 7.51 9.90 10.25 11.11 12.17 10.24 10.18 8.88 12.34 8.51 26.16 12.95 16.93 14.70 10.32 8.98 9.70 12.72 9.49 8.22 13.70 8.21 15.86 9.18 12.49 17.32
12.43 2.70 1.35 5.78 5.05 5.78 10.98 14.27 15.09 7.61 5.80 3.63 5.07 6.15 14.26 2.59 6.05 2.70 7.73 14.02 17.44 8.24 13.37 10.78 5.16 4.49 11.59 8.63 2.16 7.95 11.22 9.85 11.42 9.18 4.67 6.86
11.23 3.92 9.75 7.78 10.67 9.88 10.60 9.45 3.28 10.23 8.13 9.13 10.17 7.61 14.39 6.09 12.14 12.23 11.68 12.01 16.89 7.18 17.59 14.58 17.00 4.26 6.83 5.59 6.23 6.72 4.91 8.17 13.06 9.49 11.94 4.44
8.50 7.42 10.28 10.16 12.79 9.60 6.47 11.35 9.15 9.70 9.77 11.61 9.09 8.53 8.29 15.90 11.94 9.54 10.43 10.87 7.13 11.88 12.03
12.26 5.13 3.32 14.72 4.17 12.72 8.89 9.95 2.94 5.06 17.86 11.75 13.25 10.14 6.22 12.90 5.69 16.77 17.65 21.52 13.22 12.18 9.22
9.11 17.15 11.23 5.99 29.28 11.00 19.00 14.53 13.68 20.84 35.18 17.00 20.66 17.45 16.38 19.09 14.77 22.66 10.66 28.47 19.44 21.20 23.09
Source: Data courtesy of M. KeatoD.
6.21. Using Moody's bond ratings, samples of 20 Aa (middlehigh quality) corporate bonds and 20 Baa (topmedium quality) corporate bonds were selected. For each of the corresponding companies, the ratios Xl = current ratio (a measure of shortterm liquidity) X 2 = longterm interest rate (a measure of interest coverage) X3 = debttoequity ratio (a measure of financial risk or leverage) X 4 = rate of return on equity (a measure of profitability)
346
Exercises 347
Chapter 6 Comparisons of Several Multivariate Means
(c) Calculate the linear combinations of mean components most responsible for rejecting Ho: 1'1  1'2 = 0 in Part b. (d) Bond rating companies are interested in a company's ability to satisfy its outstanding debt obligations as they mature. Does it appear as if one or more of the foregoing financial ratios might be useful in helping to classify a bond as "high" or "medium" quality? Explain. (e) Repeat part (b) assuming normal populations with unequal covariance matices (see (627), (628) and (629». Does your conclusion change?
Table 6.1 1 Male HookBilled Kite Data Xl
X2
Xl
x2
(Tail length)
(Wing length)
(Tail length)
(Wing length)
(Tail length)
(Wing length)
ISO
278 277 308 290 273 284 267 281 287 271 302 254 297 281 284
185 195 183 202 177 177 170 186 177 178 192 204 191 178 177
282 285 276 308 254 268 260 274 272 266 281 276 290 265 275
284 176 185 191 177 197 199 190 180 189 194 186 191 187 186
277 281 287 295 267 310 299 273 278 280 290 287 286 288 275
Xl
Xl
186 206 184 177 177 176 200 191 193 212 181 195 187 190
6.22. Researchers interested in assessing pulmonary function in nonpathological populations asked subjects to run on a treadmill until exhaustion. Samples of air were collected at definite intervals and the gas contents analyzed. The results on 4 measures of oxygen consumption for 25 males and 25 females are given in Table 6.12 on page 348. The variables were XI = resting volume 0 1 (L/min) X 2 = resting volume O 2 (mL/kg/min) X3 = maximum volume O 2 (L/min) X 4 = maximum volume O 2 (mL/kg/min)
(a) Look for gender differences by testing for equality of group means. Use a = .05. If you reject Ho: 1'1  1'2 = 0, find the linear combination most responsible. (b) Construct the 95% simultaneous confidence intervals for each JLli  JL2i, i = 1,2,3,4. Compare with the corresponding Bonferroni intervals. (c) The data in Thble 6.12 were collected from graduatestudent volunteers, and thus they do not represent a random sample. Comment on the possible implications of this infonnation.
Source: Data courtesy of S. Temple. were recorded. The summary statistics are as follows:
Aa bond companies:
nl
= 20, x; = [2.287,12.600, .347, 14.830J, and
6.23. Construct a oneway MANOVA using the width measurements from the iris data in Thble 11.5. Construct 95% simultaneous confidence intervals for differences in mean components for the two responses for each pair of populations. Comment on the validity of the assumption that I,l = I,2 = I,3'
.459 .254 .026 .2441 .254 27.465 .589 .267 SI = .026 .589 .030 .102 [ .244 .267 .102 6.854
Baa bond companies:
6.24. Researchers have suggested that a change in skull size over time is evidence of the interbreeding of a resident population with immigrant populations. Four measurements were made of male Egyptian skulls for three different time periods: period 1 is 4000 B.C., period 2 is 3300 B.c., and period 3 is 1850 B.c. The data are shown in Thble 6.13 on page 349 (see the skull data on the website www.prenhall.com/statistics). The measured variables are
n2 = 20, xi = [2.404,7.155, .524, 12.840J, 944 .089 .002 .719 1 .089 16.432 .400 19.044 S2 .002  .400 .024  .094 [ .719 19.044 .094 61.854 _
and
XI
X3 = basialveolar length of skull (mm)
[.701 Spooled =
= maximum breadth of skull (mm)
Xl = basibregmatic height of skull (mm)
481 .083 .012 9.388 .494 .083 21.949  .0041 . _ .012 .027 .494 .004 34.354 .481 9.388
X 4 = nasalheightofskujl(mm)
(a) Does pooling appear reasonable here? Comment on the pooling procedure in this
0;
case. f th e with (b) Are the financial characteristics of fir~s with A~ bonds different rof. mean Baa bonds? Using the pooled covanance matnx, test for the equa Ity 0 vectors. Set a = .05.
Construct a oneway MANOVA of the Egyptian s~uJl data. Use a = .05. Construct 95 %' simultaneous confidence intervals to determine which mean components differ among the populations represented by the three time periods. Are the usual MANOVA assumptions realistic for these data? Explain. 6.25. Construct a oneway MANOVA of the crudeoil data listed in Table 11.7 on page 662. Construct 95% simultaneous confidence intervals to detennine which mean components differ among the populations. (You may want to consider transformations of the data to make them more closely conform to the usual MANOVA assumptions.)
Exercises 349 Table 6.13 Egyptian Skull Data MaxBreath
~~~~~~~~~~~S~~~~~~~~~8~~~
~~~~~~~~~.~~~~~~~~~~~~~~~~
~~~g~~~~~~~~~~~~~~~g~~~~~ 0000000000000000000000000
(xd
BasHeight (X2)
BasLength (X3)
NasHeight (X4)
Tlffie Period
131 125 131 119 136 138 139 125 131 134
138 131 132 132 143 137 130 136 134 134
89 92 99 96 100 89 108 93 102 99
49 48 50 54 56 48 48 51 51
1 1 1 1 1 1 1 1 1 1
124 133 138 148 126 135 132 133 131 133
138 134 134 129 124 136 145 130 134 125
101 97 98 104 95 98 100 102 96 94
48 48 45 51 45 52 54 48 50 46
2 2 2 2 2 2 2 2 2 2
132 133 138 130 136 134 136 133 138 138
130 131 137 127 133 123 137 131 133 133
91 100 94 99 91 95 101 96 100 91
52 50 51 45 49 52 54 49 55 46
3 3 3 3 3 3 3 3 3 3
:
~~~g~~~~~~~8~~~~~S~~~g~g~ ~~~~~~~~~~~~~~~~~~~~~~~~N
44
:
Source: Data courtesy of 1. Jackson.
~ ~ ~ ~ .~ ~ ~ ~ c:1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ :g ~.~ ~ 0000000000000000000000000
6.26. A project was des.igne~ to investigate how consumers in Green Bay, Wisconsin, would rea~t to an electncal tImeofuse pricing scheme. The cost of electricity during peak penods for some customers w~s s~t a~ eight times the cost of electricity during off~eak hours. Hourly consumptIon (m kIlowatthours) was measured on a hot summer day m Jul~ and compared, for both the test group and the control group with baseline consumptIOn measured on a similar day before the experimental rat~s began. The responses, log( current consumption)  10g(baseJine consumption)
348
350
Chapter 6 Comparisons of Several Multivariate Means
Exercises
for the hours ending 9 A.M.ll A.M. (a peak hour), 1 p.M.,and 3 the following summary statistics:
P.M.
(a peak: hour) produced
Table 6.14 Spouse Data Husban d rating wife
nl = 28,i\ = [.153,. 231,32 2,339]
Test group: Control group:
nz = 58, ii = [.151, .180, .256, 257]
and
Spooled
.804 355 = [ 228 .232
355 .722 .233 .199
.228 .233 .592 .239
.232] .199 .239 .479
Source: Data courtesy of Statistical Laboratory, University of Wisconsin. Perform a profile analysis. Does timeofuse pricing seem to make a differenc e in electrical consumption? What is the nature of this difference, if any? Commen t. (Use a significance level of a = .OS for any statistical tests.) 6.27. As part of the study of love and marriage in Example 6.14, a sample of husband s and wives were asked to respond to these questions: 1. What is the level of passionate love you feel for your partner? 2. What is the level of passionate love that your partner feels for you? 3. What is the level of companionate love that you feel for your partner? 4.
What is the level of companionate love that your partner feels for you? The responses were recorded on the following Spoint scale. None at all I
Very little I
A great deal I
Tremendous
Some I 3
4
5
amount
Thirty husbands and 30 wives gave the response s in Table 6.14, where XI = a Spointscale response to Question 1, X = a Spoints cale response to Questio 2 n 2, X3 = a Spointscale response to Question 3, and X 4 == a Spoints cale response to Question 4. (a) Plot the mean vectors for husbands and wives as sample profiles. (b) Is the husband rating wife profile parallel to the wife rating husband profile? Test for parallel profiles with a = .OS. If the profiles appear to be parallel, test for coincident profiles at the same level of significance. Finally, if the profiles are coincident,test for level profiles with a = .OS. What conclusi on(s) can be drawn from this analysis? 6.28. 1\vo species of biting flies (genus Leptoconops) are so similar morphol ogically, that for many years they were thought to be the same. Biological differenc es such as sex ratios of emerging flies and biting habits were found to exist. Do the taxonom ic data listed in part in Table 6.1S on page 3S2 and on the website www.prenhall.comlstatistics indicate any difference in the two species L. carteri and L. torrens? '!est for the equality of the two population mean vectors using a = .OS. If the hypothes es of equal mean vectors is rejected, determin e the mean compone nts (or linear combina tions of mean compone nts) most responsible for rejecting Ho. Justify your use of normalt heory methods for these data. 6.29. Using the data on bone mineral content in Table 1.8, investiga te equality between the dominan t and nondominant bones.
351
Xl
Xz
2 5 4 4 3 3 3 4 4 4 4 5 4 4 4 3 4 5 5 4 4 4 3 5 5 3 4 3 4 4
3 5 5 3 3 3 4 4 5 4 4 5 4 3 4 3 5 5 5 4 4 4 4 3 5 3 4 3 4 4
. x3
5 4 5 4 5 4 4 5 5 3 5 4 4 5 5 4 4 5 4 4 4 4 5 5 3 4 4 5 3 5
Wife rating husband X4
XI
5 4 5 4 5 5 4 5 5 3 5 ·4 4 5 5 5 4 5 4 4 4 4 5 5 3 4 4 5 3 5
4 4 4 4 4 3 4 3 4 3 4 5 4 4 4 3 5 4 3 5 5 4 2 3 4 4 4 3 4 4
x2
X3
4 5 4 5 4 3 3 4 4 4 5 5 4 4 4 4 5 5 4 3 3 5 5 4 3 4 4 4 4 4
5 5 5 5 5 4 5 5 5 4 5 5 5 4 5 4 5 4 4 4 4 4 5 5 5 4 5 4 5 5
X4
5 5 5 5 5 4 4 5 4 4 5 5 5 4 5 4 5 4 4 4 4 4 5 5 5 4 5 4 4 5
S()urce: Data courtesy of E. Hatfield.
(a) Test using a = .OS. (b) Construc t 9S% simultan eous confiden ce intervals for the mean differenc es. (c) ~onstruc~ the Bonferro ni 9S% simultan eous intervals , and compare these with the mtervals m Part b. 6.30. Table 6.16 on page 3S3 C?ntain~ .the bone mineral contents , for the first 24 subjects in Table 1.8, 1 year after thel~ particIpa tion in an experim ental program . Compar e the data from both tables to determm e whether there has been bone loss. (a) Test using a = .OS. (b) Constru ct 9S% simultan eous confiden ce intervals for the mean differenc es. (c) ~nstruc~ the Bonferr oni 9S% simultan eous intervals , and compare these with the mtervals In Part b.
352 Chapter 6 Comparisons of Several Multivariate Means
Exercises 353 Table 6.16 Mineral Content in Bones (After 1 Year)
Xl
X2
(Wing) (Wing) length width 85 87 94 92 96 91 90 92 91 87 L. torrens
c~rrd) palp
X4
length
palp width
palp length
41 38 44 43 43 44 42 43 41 38
31 32 36 32 35 36 36 36 36 35
13 14 15 17 14 12 16 17 14 11
25 22 27· 28 26 24 26 26 23 24
47 46 44 41 44 45 40 44 40 46 19 40 48 41 43 43 45 43 41 44
38 34 34 35 36 36 35 34 37 37 37 38 39 35 42 40 44 40 42 43
15 14 15 14 13 15 14 15 12 14 11 14 14 12 15 15 14 18 15 16
42 45 44 43. 46 47 47 43 50 47
38 41 35 38 36 38 40 37 40 39
14 17 16 14 15 14 15 14 16 14
:
99 110 99 103 95 101 103 99 105 99
Xs
(Thl'd) (FO_)
:
106 105 103 100 109 104 95 104 90 104 86 94 103 82 103 101 103 100 99 100 L. carteri
X3
Source: Data courtesy of William Atchley.
X6
X7
( Longtb of ) ( Length of antennal antennal segment 12 segment 13
9 13 8 9 10 9 9 9 9 9
:
:
8 13 9 9 10 9 9 9 9 10
26 31 23 24 27 30 23 29 22 30 25 31 33 25 32 25 29 31 31 34
10 10 10 10 11 10 9 9 9 10 9 6 10 9 9 9 11 11 10 10
10 11 10 10 10 10 10 10 10 10 9 7 10 8 9 9 11 10 10 10
33 36 31 32 31 37 32 23 33 34
9 9 10 10 8 11 11 11 12 7
9 10 10 10 8 11 11 10 11 7
:
:
Subject number
Dominant radius
Radius
Dominant humerus
Humerus
Dominant ulna
Ulna
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
1.027 .857 .875 .873 .811 .640 .947 .886 .991 .977 .825 .851 .770 .912 .905 .756 .765 .932 .843 .879 .673 .949 .463 .776
1.051 .817 .880 .698 .813 .734 .865 .806 .923 .925 .826 .765 .730 .875 .826 .727 .764 .914 .782 .906 .537 .900 .637 .743
2.268 1.718 1.953 1.668 1.643 1.396 1.851 1.742 1.931 1.933 1.609 2.352 1.470 1.846 1.842 1.747 1.923 2.190 1.242 2.164 1.573 2.130 1.041 1.442
2.246 1.710 1.756 1.443 1.661 1.378 1.686 1.815 1.776 2.106 1.651 1.980 1.420 1.809 1.579 1.860 1.941 1.997 1.228 1.999 1.330 2.159 1.265 1.411
.869 .602 .765 .761 .551 .753 .708 .687 .844 .869 .654 .692 .670 .823 .746 .656 .693 .883 .577 .802 .540 .804 .570 .585
.964 .689 .738 .698 .619 .515 .787 .715 .656 .789 .726 .526 .580 .773 .729 .506 .740 .785 .627 .769 .498 .779 .634 .640
Source: Data courtesy of Everett Smith.
6.31. Peanuts are an important crop in parts of the southern United States. In an effort to develop improved plants, crop scientists routinely compare varieties with respect to several variables. The data for one twofactor experiment are given in Table 6.17 on page 354. Three varieties (5,6, and 8) were grown at two geographical locations (1,2) and, in this case, the three variables representing yield and the two important gradegrain characteristics were measured. The three variables are
X z
= Yield (plot weight) = Sound mature kernels (weight in gramsmaximum of 250 grams)
X3
= Seed size (weight, in grams, of 100 seeds)
Xl
There were two replications of the experiment. (a) Perform a twofactor MANQVA using the data in Table 6.17. Test for a location effect, a variety effect, and a locationvariety interaction. Use a = .05. (b) Analyze the residuals from Part a. Do the usual MANQVA assumptions appear to be satisfied? Discuss. (c) Using the results in Part a, can we conclude that the location and/or variety effects are additive? If not, does the interaction effect show up for some variables, but not for others? Check by running three separate univariate twofactor ANQVAs.
Exercises 355
354 Chapter 6 Comparisons of Several Multivariate Means
Table 6.17 Peanut Data Factor 1 Location
Factor 2 Variety
Xl
X2
X3
Yield
SdMatKer
SeedSize
1 1 2 2 1 1 2 2 1 1 2 2
5 5 5 5 6 6 6 6 8 8 8 8
195.3 194.3 189.7 180.4 203.0 195.9 202.7 197.6 193.5 187.0 201.5 200.0
153.1 167.7 l39.5 121.1 156.8 166.0 166.l 161.8 164.5 165.1 166.8 173.8
51.4 53.7 55.5 44.4 49.8 45.8 60.4 54.l 57.8 58.6 65.0 67.2
Source: Data courtesy of Yolanda Lopez.
(d) Larger numbers correspond to better yield and gradegrain characteristics. Using cation 2, can we conclude that one variety is better than the other two for each acteristic? Discuss your answer, using 95% Bonferroni simultaneous intervals pairs of varieties. 6.32. In one experiment involving remote sensing, the spectral reflectance of three lyearold seedlings was measured at various wavelengths during the growing The seedlings were grown with two different levels of nutrient: the optimal coded +, and a suboptimal level, coded . The species of seedlings used were spruce (SS), Japanese larch (JL), and 10dgepoJe pine (LP).1\vO of the variables sured were Xl = percent spectral reflectance at wavelength 560 nrn (green) X 2 = percent spectral reflectance at wavelength 720 nrn (near infrared) The cell means (CM) for Julian day 235 for each combination of species and level are as follows. These averages are based on four replications. 560CM
nOCM
10.35 13.41 7.78 10.40 17.78 10.40
25.93 38.63 25.15 24.25 41.45 29.20
Species
Nutrient
SS
+ + +
JL LP
SS JL LP
(a) 'freating the cell means as individual observations, perform a twoway test for a species effect and a nutrient effect. Use a = .05. (b) Construct a twoway ANOVA for the 560CM observations and another ANOVA for the nOCM observations. Are these results consistent MANOVA results in Part a? If not, can you explain any differences?
6.33. Refer to Exercise 6.32. The data in Table 6.18 are measurements on the variables Xl = percent spectral reflectance at wavelength 560 nm (green) X 2 = percent spectral reflectance at wavelength no nm (near infrared) for three species (sitka spruce [SS], Japanese larch [JL), and lodgepole pine [LP]) of lyearold seedlings taken at three different times (Julian day 150 [1], Julian day 235 [2], and Julian day 320 [3]) during the growing season. The seedlings were all grown with the optimal level of nutrient. (a) Perform a twofactor MANOVA using the data in Table 6.18. Test for a species effect, a time effect and speciestime interaction. Use a = .05.
Table 6.18 Spectral Reflectance Data 560 run
720nm
Species
TIme
Replication
9.33 8.74 9.31 8.27 10.22 10.l3 10.42 10.62 15.25 16.22 17.24 12.77 12.07 11.03 12.48 12.12 15.38 14.21 9.69 14.35 38.71 44.74 36.67 37.21 8.73 7.94 8.37 7.86 8.45 6.79 8.34 7.54 14.04 13.51 13.33 12.77
19.14 19.55 19.24 16.37 25.00 25.32 27.12 26.28 38.89 36.67 40.74 67.50 33.03 32.37 31.31 33.33 40.00 40.48 33.90 40.l5 77.14 78.57 71.43 45.00 23.27 20.87 22.16 21.78 26.32 22.73 26.67 24.87 44.44 37.93 37.93 60.87
SS SS SS
1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
SS SS SS SS SS SS SS SS SS JL JL JL JL JL JL JL JL JL JL JL JL LP LP LP LP LP LP LP LP LP LP LP LP
Source: Data courtesy of Mairtin Mac Siurtain.
Exercises :357
356 Chapter 6 Comparisons of Several Multivariate Means (b) Do you think the usual MAN OVA assumptions are satisfied for the these data? cuss with reference to a residual analysis, and the possibility of correlated tions over time. (c) Foresters are particularly interested in the interaction of species and time. teraction show up for one variable but not for the other? Check by running· variate twofactor ANOVA for each of the two responses. . (d) Can you think of another method of analyzing these data (or a different tal design) that would allow for a potential time trend in the spectral numbers? 6.34. Refer to Exampl e 6.15. (a) Plot the profiles, the components of Xl versus time and those of X2 versuS the same graph. Comment on the comparison. (b) Test that linear growth is adequate. Take a = .01. 6.35. Refer to Exampl e 6.15 but treat all 31 subjects as a single group. The maximum hood estimate of the (q + 1) X 1 P is
P= (B'SlB rIB'Sl x

where S is the sample covariance matrix. The estimate d covariances of the maximum likelihood estimators are
CoV(P)
='
(n  l)(n  2) (n  1  P
+ q) (n
 p
+
(B'SIB r
J
q)n
Fit a quadrati c growth curve to this single group and comment on the fit. 6.36. Refer to Example 6.4. Given the summary information on electrical usage in this pie, use Box's Mtest to test the hypothesis Ho: IJ = ~2 =' I. Here Il is the ance matrix for the two measures of usage for the population of Wisconsi n with air conditioning, and ~2 is the electrical usage covariance matrix for the of Wisconsin homeowners without air conditioning. Set a = .05. 6.31. Table 6.9 page 344 contains the carapace measurements for 24 female and 24 male ties. Use Box's Mtest to test Ho: ~l = ~2 = I. where ~1 is the populatio n matrix for carapace measurements for female turtles, and I2 is the populatio n ance matrix for carapace measurements for male turtles. Set a '" .05. 6.38. Table 11.7 page 662 contains the values of three trace elements and two measures of drocarbons for crude oil samples taken from three groupS (zones) of Box's Mtest to test equality of population covariance matrices for the sandstone. three.s: groups. Set a = .05. Here there are p = 5 variables and you may wish to conSIder formations of the measurements on these variables to make them more nearly 6.39. Anacondas are some of the largest snakes in the world. Jesus Ravis and his searchers capture a snake and measure its (i) snout vent length (cm) or the length the snout of the snake to its vent where it evacuates waste and (ii) weight sample of these measurements in shown in Table 6.19. (a) Test for equality of means between males and females using a = .05. large sample statistic. (b) Is it reasonable to pool variances in this case? Explain. (c) Find the 95 % Boneferroni confidence intervals for the mean differenc es males and females on both length and weight.
andlstone;:~
Table 6.19 Anacon da Data
Snout vent Length 271.0 477.0 306.3 365.3 466.0 440.7 315.0 417.5 307.3 319.0 303.9 331.7 435.0 261.3 384.8 360.3 441.4 246.7 365.3 336.8 326.7 312.0 226.7 347.4 280.2 290.7 438.6 377.1
Weight
Gender
18.50 82.50 23.40 33.50 69.00 54.00 24.97 56.75 23.15 29.51 19.98 24.00 70.37 15.50 63.00 39.00 53.00 15.75 44.00 30.00 34.00 25.00 9.25 30.00 15.25 21.50 57.00 61.50
F F F F F F F F F F F F F F F F F F F F F F F F F F F F
Snout vent length 176.7 259.5 258.0 229.8 233.0 237.5 268.3 222.5 186.5 238.8 257.6 172.0 244.7 224.7 231.7 235.9 236.5 247.4 223.0 223.7 212.5 223.2 225.0 228.0 215.6 221.0 236.7 235.3
Weight
Gender
3.00 9.75 10.07 7.50 6.25 9.85 10.00 9.00 3.75 9.75 9.75 3.00 10.00 7.25 9.25 7.50 5.75 7.75 5.75 5.75 7.65 7.75 5.84 7.53 5.75 6.45 6.49 6.00
M M M M M M M M M M M M M M M M M M
M M M M M M M M
M M
Source: Data Courtesy of Jesus Ravis.
6.40. Compare the male national track records in 1: b . records in Table 1.9 using the results for the 1~rr:e2~6 WIth the female national track neat the data as a random sample of siz 64 f h' m, 4OOm, and 1500m races. e 0 t e twelve recordSOOm values. (a) Test for equality of means between males and fema e . .' may be appropriate to analyze differences. I s usmg a  .05. Explam why It (b) Find the 95% Bonferroni confidence in male and females on all of the races. tervals for the mean differences between
6.41. When cell phone relay towers are not worki . . amounts of money so it is importa nt to be a~re~~OKerly, wrreless prov~~ers can lose great toward understanding the problem s' I d ' IX problems expedItiously. A [lISt step ment .involving three factors. A prOb~:;::;e ~s.~ ~olI~ct ~ata from a designed experisimple or complex and the en ineer . as ml a y c assified as low or high severity, expert (guru ).' g a~sJgned was rated as relatively new (novice) or
358 Chapter 6 Comparisons of Several Multivariate Means References 359 Tho times were observed. The time to assess the pr?blem and plan an atta~k the time to implement the solution were each measured In hours. The data are given Table 6.20. . If· rta t Perform a MANOVA including appropriate confidence mterva s or Impo n I"
,.
Problem Severity Level Low Low Low Low Low Low Low Low High High High High High High High High
9. Box, G. E. P., and N. R. Draper. Evolutionary Operation:A Statistical Method for Process Improvement. New York: John Wiley, 1969. 10. Box, G. E. P., W. G. HUnter, and 1. S. Hunter. Statistics for Experimenters (2nd ed.). New York: John Wiley, 2005. 11. Johnson, R. A. and G. K. Bhattacharyya. Statistics: Principles and Methods (5th ed.). New York: John Wiley, 2005.
Problem Complexity Level Simple Simple Simple Simple Complex Complex Complex Complex Simple Simple Simple Simple Complex Complex Complex Complex
Engineer Experience Level Novice Novice Guru Guru Novice Novice Guru Guru Novice Novice Guru Guru Novice Novice Guru Guru
Problem Assessment Tune
Problem Implementation Time
3.0 2.3 1.7 1.2 6.7 7.1 5.6 4.5 4.5 4.7 3.1 3.0 7.9 6.9 5.0 5.3
6.3 5.3 2.1 1.6 12.6 12.8 8.8 9.2 9.5 10.7 6.3 5.6 15.6 14.9 10.4 10.4
Total Resolution Time 9.3 7.6 3.8 2.8 19.3 19.9 14.4 13.7 14.0 15.4 9.4 8.6 23.5 21.8 15.4 15.7
Source: Data courtesy of Dan Porter.
References 1. Anderson, T. W. An Introduction to Multivariate Statistical Analysis (3rd ed.). New York: John Wiley, 2003. . . 2 B acon Shone,., for Detectmg Smgle J and W.K. Fung. "A New Graphical Method " r d S . . 36 noand2 tatlstrcs, , . . Multiple Outliers in Univariate and Multivariate Data. App le (1987),153162. . h R I 3. Bartlett, M. S. "Properties of Sufficiency and Statistical Tests." Proceedmgs of t e oya Society of London (A), 166 (1937), 268282. .". 0 4. Bartlett, M. S. "Further Aspects of the Theory of Multiple RegressIOn. Proceedings f the Cambridge Philosophical Society, 34 (1938),3340. 5. Bartlett, M. S. "Multivariate Analysis." Journal of the Royal Statistical Society Supplement (B), 9 (1947), 176197. . . " . . F:ac torsor f Various X2 ApprOXimations. 6. Bartlett, M. S.• ~ Note on the Multlplymg Journal of the Royal Statistical Society (B), 16 (1954),296298. . ." 7. Box, G. E. P., "A General Distribution Theory for a Class of Likelihood Cntena. Biometrika, 36 (1949),317346. . 6 8. Box, G. E. P., "Problems in the Analysis of Growth and Wear Curves." Biometrics, (1950),362389.
12. Jolicoeur, P., and 1. E. Mosimann. "Size and Shape Variation in the Painted ThrtJe: A Principal Component Analysis." Growth, 24 (1960),339354. 13. Khattree, R. and D. N. Naik, Applied Multivariate Statistics with SAS® Software (2nd ed.). Cary, NC: SAS Institute Inc., 1999. 14. Kshirsagar,A. M., and W. B. Smith, Growth Curves. New York: Marcel Dekker, 1995. 15. Krishnamoorthy, K., and 1. Yu. "Modified Nel and Van der Merwe Test for the Multivariate BehrensFisher Problem." Statistics & Probability Letters, 66 (2004), 161169. 16. Mardia, K. V., "The Effect of Nonnormality on some Multivariate Tests and Robustnes to Nonnormality in the Linear Model." Biometrika, 58 (1971), 105121. 17. Montgomery, D. C. Design and Analysis of Experiments (6th ed.). New York: John Wiley, 2005. 18. Morrison, D. F. Multivariate Statistical Methods (4th ed.). Belmont, CA: Brooks/Cole Thomson Learning, 2005. 19. Nel, D. G., and C. A. Van der Merwe. "A Solution to the Multivariate BehrensFisher Problem." Communications in StatisticsTheory and Methods, 15 (1986), 37193735. 20. Pearson, E. S., and H. O. Hartley, eds. Biometrika Tables for Statisticians. vol. H. Cambridge, England: Cambridge University Press, 1972. 21. Potthoff, R. F. and S. N. Roy. "A Generalized Multivariate Analysis of Variance Model Useful Especially for Growth Curve Problems." Biometrika, 51 (1964),313326. 22. Scheffe, H. The Analysis of Variance. New York: John Wiley, 1959. 23. Tiku, M. L., and N. Balakrishnan. "Testing the Equality of VarianceCovariance Matrices the Robust Way." Communications in StatisticsTheory and Methods, 14, no. 12 (1985), 30333051. 24. Tiku, M. L., and M. Singh. "Robust Statistics for Testing Mean Vectors of Multivariate Distributions." Communications in StatisticsTheory and Methods, 11, no. 9 (1982), 985100l. 25. Wilks, S. S. "Certain Generalizations in the Analysis of Variance." Biometrika, 24 (1932), 471494.
The Classical Linear Regression Model 361
Chapter and Zl
== square feet ofliving area location (indicator for zone of city) = appraised value last year = quality of construction (price per square foot)
Z2 =
Z3 Z4
MULTIVARIATE LINEAR REGRESSION MODELS
The cl~ssicalli~ear regression model states that Y is composed of a mean, which depends m a contmuous manner on the z;'s, and a random error 8, which accounts for measurement error and the effects of other variables not explicitly considered in the mo~eI. Th~ values of the predictor variables recorded from the experiment or set by the mvestigator ~e treated as fixed ..Th~ error (and hence the response) is viewed as a r~dom vanable whose behavlOr IS characterized by a set of distributional assumptIons. Specifically, the linear regression model with a single response takes the form
Y
= 13o
+ 13lZl + ... + 13,z, + 8
[Response] = [mean (depending on
7.1 Introduction Regression analysis is the statistical methodology for predicting values of one or more response (dependent) variables from a collection of predictor (independent) variable values. It can also be used for assessing the effects of the predictor variables· on the responses. Unfortunately, the name regression, culled from the title of the first paper on the sUbject by F. Galton [15], in no way reflects either the importance ..... or breadth of application of this methodology. . In this chapter, we first discuss the multiple regression model for the predic· tion of a single response. This model is then generalized to handle the prediction of several dependent variables. Our treatment must be somewhat terse, as a vast literature exists on the subject. (If you are interested in pursuing regression analysis, see the following books, in ascending order of difficulty: Abraham and Ledolter [1], Bowerman and O'Connell [6], Neter, Wasserman, Kutner, and Nachtsheim [20], Draper and Smith [13], Cook and Weisberg [11], Seber [~3], and Goldberger [16].) Our abbreviated treatment highlights the regressIOn assumptions and their consequences, alternative formulations of the regression model, and the general applicability of regression techniques to seemingly different situations.
Yl = ~ =
130 130
+ 13lZ 11 + 132Z12 + ... + 13rzl r + 81 + 13lZ21 + 132Z22 + ... + 13rZ2r + 82 (71)
Yn =
130
+
Y = current market value of home 360
13lZnl
+ 132Zn2 + ... + 13rZnr + 8 n
where the error terms are assumed to have the following properties: 1. E(8j) = 0;
2. Var(8j) = a2 (constant); and 3. COV(8j,8k) = O,j
(72)
* k.
In matrix notation, (71) becomes Zll
Z12
ZZl
Z22
1
Znl
or Y
(nXl)
=
Z
1. E(e) = 0; and = E(ee')
= a2I.
:
Znr
13r
8n
fJ
(nX(r+l» ((r+l)xl)
and the specifications in (72) become 2. Cov(e)
:
[8 + :
Z2r 131 Zlr] [130]
:::
.
1.2 The Classical linear Regression Model Let Zl, Zz, ... , z, be r predictor variables thought to be related to a response variable Y. For example, with r = 4, we might have
+ [error]
Zl,Z2, ... ,Z,)]
The term "linear" refers to the fact that the mean is a linear function of the unknown pa~ameters 13o, 131>···,13,· The predictor variables mayor may not enter the model as fIrstorder terms. With n independent observations on Yand the associated values of z· the comI' plete model becomes
+ e
(nxl)
82 ]
362
The Classical Linear Regression Model 363
Chapter 7 MuItivariate Linear Regression Models Note that a one in the first column of the design matrix Z is the multiplier of the. constant term 130' It is customary to introduce the artificial variable ZjO = 1, so
I
I
I
130
+ 131Zjl + .,. + 13rzjr =
{3oZjO
The data for this model are contained in the observed response vector y and the design matrix Z, where
+ {3I Zjl + ... + {3r Zj,
Each columnof Z consists of the n values of the corresponding predictor variable· while the jth row of Z contains the values for all predictor variables on the jth trial: Note that we can handle a quadratic expression for the mean response by introducing the term 132z2, with Z2 = zy. The linear regression model for the jth trial in this latter case is
Classic~1 linear Regression Model
E(E) where
13
P+E,
Z
y=
(nX(r+I» ((r+I)XI)
(nXl)
= 0
(nXl)
and Cov(e)
(nXl)
lj =
130
+ 131Zjl + 132zj2 + Sj
lj =
130
+ 13lzjl + 132zJI + Sj
or
= (1"2 I, (nXn)
•
and (1"2 are unknown parameters and the design matrix Z has jth row
[ZjO, Zjb .•• , Zjr]'
Although the errorterm assumptions in (72) are very modest, we shall later need to add the assumption of joint normality for making confidence statements and testing hypotheses. We now provide some examples of the linear regression model.
Example 7.2 (The design matrix for oneway ANOVA as a regression model) Determine the design matrix if the linear regression model is applied to the oneway ANOVA situation in Example 6.6. We create socalled dummy variables to handle the three population means: JLI = JL + 7"1, JL2 = JL + 7"2, and JL3 = JL + 7"3' We set
if the observation is from population 1 otherwise
Example 7.1 (Fitting a straightline regression model) Determine the linear regression model for fitting a straight liiie
Mean response
= E(Y) = f30 +
and
y
o
1
1
4
2 3
3 8
4 9
Yl] [1'~25
,
Z =
.[1~ T ZIl] 'P 1
ZSl
=
= 7"1,132 = 7"2,133 = 7"3' Then lj = 130 + 131 Zjl + 132Zj2 + 133Zj3 + Sj,
130 = JL,131
E' =
Y
ZP + e
(8XI)
where
Y =
[:~J
o
j=1,2, ... ,8
where we arrange the observations from the three populations in sequence. Thus, we obtain the observed response vector and design matrix
Before the responses Y' = [Yi, Yi, ... , Ys] are observed, the errors [el, e2, ... , es] are random, and we can write Y =
{
if the observation is from population 2 otherwise
if the observation is from population 3 otherwise
f3l zl
to the data
I Z2 =
E
=
[SI] ~2 Ss
=
9 6 9 0 2 3 1 2
Z
(8X4)
=
1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1
•
The construction of dummy variables, as in Example 7.2, allows the whole of analysis of variance to be treated within the multiple linear regression framework.
364 Chapter 7 Multivariate Linear Regression Models
Least Squares Estimation ,365
7.3 least Squares Estimation
P
One of the objectives of regression analysis is to develop an equation that will the investigator to predict the response for given values of the predictor Thus it is necessary to "fit" the model in (73) to the observed Yj cOlTes;pollldill2:Jf8: the known values 1, Zjl> ... , Zjr' That is, we must determine the values for regression coefficients fJ and the error variance (}"2 consistent with the available Let b be trial values for fJ. Consider the difference Yj  bo  b1zj1  '" between the observed response Yj and the value bo + b1zj1 + .,. + brzjr that be expected if b were the ·"true" parameter vector. 1)rpicaJly, the Yj  bo  b1zj1  ...  brzjr will not be zero, because the response fluctuates manner characterized by the error term assumptions) about its expected value. method of least squares selects b so as to miI).imize the sum of the squares of differences: S(b) =
Proof. Let = (Z'ZfIZ'y as asserted. Then £ [I  Z(Z'ZfIZ']y. The matrix [I  Z(Z'ZfIZ'] satisfies 1. [I  Z(Z'Zf1z,], = [I  Z(Z'Z)IZ'] l
= I  2Z(Z'Zf z, = [I  Z (Z'Zflz,]
~o  ~IZjl
_
Zp =
(symmetric);
+ Z(Z'Z)IZ'Z(Z'Z)IZ'
(76)
(idempotent);
3. Z'[I  Z(Z'Zflz,] = Z'  Z' = O. Consequently,Z'i = Z'(y  y) = Z'[I  Z(Z'Z)lZ'Jy == O,soY'e = P'Z'£ = O. Additionally, = y'[1  Z(Z'Z)IZ'J[I ~Z(Z'ZfIZ']y = y'[1 _ Z (Z'Z)lZ']Y = y'y  y'ZfJ. To verify the expression for fJ, we write
!'e
bo 
b1zj1 
'"

brz jr )
y  Zb = Y  ZP + ZP  Zb = y  ZP + Z(P  b) so
The coefficients b chosen by the least squares criterion are called least squqres mates of the regression parameters fJ. They will henceforth be denoted by fJ to em~ . phasize their role as e~timates of fJ. . The coefficients fJ are consistent. with the data In the sense that they estimated (fitted) mean responses, ~o + ~IZjl + ... + ~rZj" ~he sum squares of the differences from the observed Yj is as small as possIble. The de\IlatlloriJ:i 
y=y
2. [I  Z(Z'ZfIZ'][I  Z(Z'Z)IZ']
j=l
= Yj

2
2:n (Yj 
= (y  Zb)'(y  Zb)
Sj
=y

.. , 
~rZj"
j
n. l The least squares estimate of fJ in'~
P= (Z'ZfIZ'y Let y = ZfJ~ = Hy denote the fitted values of y, where H "hat" matrix. Then the residuals
= Z (Z'Z)I Z '
is called
+ (P  b),Z'Z(P  b)
+ 2(y  ZP)'Z(P  b) = (y  ZP)'(y  ZP)
Zp
:5
= (y  ZP)'(y  ZP)
= 1,2, ... ,n
are called residuals. The vector of residuals i == y contains the information about the remaining unknown parameter~. (See Result 7.2.) Result 7.1. Let Z have full rank r + 1 (73) is given by
S(b) = (y  Zb)'(y  Zb)
+ (P  b)'Z'Z(P  b)
since (y  ZP)'Z = £'Z = 0'. The first term in S(b) does not depend on b and the' sec~ndisthesquaredlengthofZ(P  b). BecauseZhasfullrank,Z(p  b) '# 0 if fJ '# b, so the minimum sum of squares is unique and Occurs for b = =
P
(Z'Zf1Z'y. Note that (Z'Z)l exists since Z'Z has rank r + 1 :5 n. (If Z'Z is not of full rank, Z'Za = 0 for some a '# 0, but then a'Z'Za = 0 or Za = 0 which con, • tradicts Z having full rank r + 1.)
P
Result 7.1 shows how the least squares estimates and the residuals £ can be obtained from the design matrix Z and responses y by simple matrix operations.
i = y  y = [I  Z(Z'ZrIZ']Y = (I  H)y satisfy Z'
e = 0 and Y' e = O. Also, the
residual sum of squares
=
2:n (Yj 
~o 
~
{3IZjl  '"

~)2 {3rZjr
",'"
=E
E
Example 7.3 (Calculating the least squares estimates, the residuals, and the residual the residuals i, and the resIdual sum of squares for a straightline model
su~ of squares) Calculate the least square estimates
P,
j=l
= y'[1
_ Z(Z'ZrIZ']Y
= y'y
 y'ZfJ fit to the data
IIf Z is not full rank, (Z'Z)l is replaced by (Z'Zr, a generalized inverse of Z'Z. Exercise 7.6.) ,
ZI
o
Y
1
1 4
2 3
3 8
4
9
Least Squares Estimation 367
366 Chapter 7 Multivariate Linear Regression Models
Since the first column of Z is 1, the condition Z'e = 0 includes the requirement
We have
Z'
[~
y
1 ~J 1 1 1 2 3
Consequently,
p=
z'z
[~:J
[1~
m
= (Z'ZrlZ'y =
(Z'Zr
10J 30
[
.6 .2
o=
l
'£L
.2] .1
l'e =
n
n
n
j=l
j=l
j=l
2: ej = 2: Yj  L
Yj' or y =
y.
Subtracting n),2 = n(W from both
sides of the decomposition in (77), we obtain the basic decomposition of the sum of squares about the mean:
[~~J
or n
2: (Yj 
n
n
j=l
j=l
2: (Yj  Y/ + 2: e;
y)2 =
j=l
(78) .
~~!~us;~ ) = (re:~:~~n) + (residu~l (error))
[:~ :~J D~J [~J
( about mean
=
squares
sum 0 squares
The preceding sum of squares decomposition suggests that the quality of the models fit can be measured by the coefficient of determination
and the fitted equation is
n
Y= 1 + 2z R2 = 1 _
The vector of fitted (predicted) values is
11
L e1
2: (Yj 
j=!
j=l
±
(79)
±
(Yj  y)2
j=!
y)2
(Yj _ y/
j=l
The quantity R2 gives the proportion of the total variation in the y/s "explained" by, or attributable to, the predictor variables Zl, Z2,' .. ,Zr' Here R2 (or the multiple correlation coefficient R = + VJi2) equals 1 if the fitted equation passes through all tpe da!a points; s~ that Sj = 0 for all j. At the other extreme, R2 is 0 if (3o = Y and f31 = f32 = ... = f3r = O. In this case, the predictor variables Zl, Z2, ... , Zr have no influence on the response. so
Geometry of least Squares A geometrical interpretation of the least squares technique highlights the nature of the concept. According to the classical linear regression model,
The residual sum of squares is
Mean response vector = E(Y) = ZP =
f30
ll [Zlll [Zlrl [~ + Z~l + ... + Przr f31
1
SumofSquares Decomposition According to Result 7.1, y'i = 0, so the total response sum of squares y'y = satisfies y'y =
(y + Y _ y)'(y + Y _ y)
Znl
ZIIr
Thus, E(Y) is a linear combination of the columns of Z. As P varies, ZP spans the model plane of all linear combinations. Usually, the observation vector y will not lie in the model plane, because of the random error E; that is, y is not (exactly) a linear combination of the columns of Z. Recall that
=
(y + e)'(y + e)
=
y'y + e'e
~yJ
Y
response) ( vector
+
E
error) ( vector
368
Least Squares Estimation
Chapter 7 Multivariate Linear Regression Models
~69
Accordi ng to Result 2A.2 and Definiti on 2A.12, the projecti on of y on a linear com
3
bination of {ql, qz,··· ,qr+l} is
r+l
(r+l
~ (q;y) q; = i~ qjqi) y = Z(Z' Zfl Z 'y
A
=
ZfJ·
Thus, mUltipli cation by Z (Z'Zfl Z ' projects a vector onto the space spanned by the columns of Z.Z Similarly, [I  Z(Z'Zf 1Z'] is the matrix for the projecti on of y on the plane perpend icular to the plane spanned by the columns of Z.
Sampling Properties of Classical Least Squares Estimators The least squares estimato r detailed in the next result.
jJ
and the residual s
i
have the samplin g properti es
Result 7.2. Under the general linear regressi on model in (73), the least squares
estimato r
jJ
= (Z'Zfl Z 'Y has
Figure 7.1 Least squares as a projection for n = 3, r = 1.
E(jJ) = fJ The residual s
· Once the observa t IOns become available' the least squares solution is derived from the deviation vector y _ Zb = (observation vector)  (vector in model plane) ( _ Zb)'(  Zb) is the sum of squares S(b). As illustrat ed in The squared len~th y all as :ssible when b is selected such that Zb is the point in Figure 7.1, S(b) IS as srn ~. oint occurs at the tip of the perpend icular prothe model plane closest tThoy. • I: p th choiceb = Q yA = is the projecti on of . . f on the plane at IS, lor e ,.., JectlOn 0 Y . ti 'of all linear combinations of the columns of Z. The rest'd u.al y on th: plane c,:n.sls ng d' ular to that plane. This geometry holds even when Z IS vector 13 = Y  YIS perpen IC not of full rank. full k the projection operation is expresse d analytic ally as When Z has ran, • . Z(Z'Z)J I Z ' To see this, we use the spectraI d ecompo multipli cation by the matrIX . sition (216) to write Z'Z = Alelel + Azezez + .,. + A'+le'+le~+1
ZP
.,. > A > 0 are the eigenvalues of Z'Z and el, ez,···, er+1 are where Al 2: Az 2: ,+1 . the corresponding eigenvectors.1f Z IS of full rank,
(Z'Z)1 =
. 1
E(i)
=0
and
r+l
s
2
i'i
=n
 (r + 1)
i=1
,+1 ,=1
 Z(Z'Z flZ ']
Y'[I  Z(Z'Z fl Z ']Y nr l
we have
= aZ[1
 H]
Y'[I  H]Y nr l
E(sz) = c? Moreov er,
jJ and i
are uncorre lated.
Proof. (See webpage : www.pr enhall.c om/stati stics)
•
The least squares estimato r jJ possesse s a minimu m variance propert y that was first establis hed by Gauss. The followin g result concern s "best" estimato rs of linear paramet ric function s of the form c' fJ = cof3o + clf31 + ... + c f3r r for any c.
Result 7.3 (Gauss·3 Ieast squares theorem ). Let Y = ZfJ + 13, where E(e) = 0, COY (e) = c? I, and Z has full rank r + 1. For any c, the estimato r "
1,
= ~ qiqj
= aZ[1
Cov(i)
Also,E (i'i) = (n  r  1)c?, so defining
....... .
c' fJ = cof3o
" + clf31 + " . + c,f3,
2If Z is not of full rank. we can use the generalized inverse (Z'Zr =
.=
Z(Z'Z) l z , = ~ Ai1ZejejZ'
Cov(jJ) = c?(Z'Z fl
have the properti es
~elel + ezez + .,. + Aer+le r+1
Al Az ,+1 A:1/2Zej, which is a linear combination of the columns of~. Then qiqk ConsIde r q" 1/2 ' _ 0 if . #0 k or 1 if i = k. That IS, the r + 1 1/2A1/2 'Z'Ze = A· Ak1/2 ejAkek I = Ai k ej k 'e endicular and have unit length. Their linear cornb'IDa~ectors qi ahre mutuallfYaPlll~ear combinations of the columns of Z. Moreov er, tlOns span t e space 0 .
i
and
Al
2: rl+l
= 2:
A2
2: ... 2:
A,,+l
>0
= A,,+2 = ... = A,+l.
rJ+I
2: Ai1eiei.
where
;J
as described in Exercise 7.6. Then Z (Z'Zr Z '
qiq! has rank rl + 1 and generates the unique projection of y on the space spanned by the linearly i=1 independent columns of Z. This is true for any choice of the generalize d inverse. (See [23J.) 3Much later, Markov proved a less general result, which misled many writers into attaching his name to this theorem.
370
Inferences About the Regression Model 371
Chapter7 Multjvariate Linear Regression Models
and is distributed independently of the residuals i = Y 
of c' p has the smallest possible variance among all linear estimators of the form a'Y = all!
I
I
+ a2~ + .. , + anYn
Zp. Further,
na2 =e'i is distributed as O'2rn_r_1
that are unbiased for c' p.
where 0.2 is the maximum likeiihood estimator of (T2.
Proof. For any fixed c, let a'Y be any unbiased estimator of c' p. E(a'Y) = c' p, whatever the value of p. Also, by assumption,. E( E(a'Zp + a'E) = a'Zp. Equating the two exp~cted valu: expressl~ns , a'Zp = c' p or·(c'  a'Z)p = for all p, indudmg the chOIce P = (c  a This implies that c' = a'Z for any unbiased estimator. I Now, C' = c'(Z'Zf'Z'Y = a*'Y with a* = Z(Z'Z) c. Moreover, Result 7.2 E(P) = P, so c' P = a*'Y is an unbiased estimator of c' p. Thus, for a satisfying the unbiased requirement c' = a'Z,
°
A confidence ellipsoid for P is easily constructed. It is expressed in terms of the l estimated covariance matrix s2(Z'Zr , where; = i'i/(n  r  1).
P
Var(a'Y) = Var(a'Zp + a'e) = Var(a'e)
=
Result 7.S. Let Y = ZP + E, where Z has full rank r + 1 and Eis Nn(O, 0.21). Then a 100(1  a) percent confidence region for P is given by
2
a'IO' a
..... , , ' "
(PP) Z Z(PP)
+ a*),(a  a* + a*)  a*)'(a  a*) + a*'a*]
= O' 2 (a  a* = ~[(a
•
Proof. (See webpage: www.prenhall.comlstatistics)
since (a ' a*)'a* = (a  a*)'Z(Z'Zrlc = 0 from the con~ition (: ~ a*)'~ = a'Z  a*'Z = c'  c' = 0'. Because a* is fIxed and (a  a*) (a  ~I) IS posltIye unless a = a*, Var(a'Y) is minimized by the choice a*'Y = c'(Z'Z) Z'Y = c' p.
P
(r
2
+ l)s Fr+l,nrl(a)
where Fr+ I,nrl (a) is the upper (lClOa )th percentile of an Fdistribution with r + 1 and n  r  1 d.f. Also, simultaneous 100(1  a) percent confidence intervals for the f3i are given by
•
This powerful result states that substitution of for p leads to the be,:;t . tor of c' P for any c of interest. In statistical tenninology, the estimator c' P is called the best (minimumvariance) linear unbiased estimator (BLUE) of c' p.
:s;
f3i ± 
"'.
V%(P;) V(r + I)Fr+1,nrl(a) , .
where Var(f3i) IS the diagonal element of s2(Z'Z)
1
i = O,I, ... ,r ,..
corresponding to f3i'
Proof. Consider the symmetric squareroot matrix (Z'Z)I/2. (See (222).J Set 1/2 V = (Z'Z) (P  P) and note that E(V) = 0, A
7.4 Inferences About the Regression Model
Cov(V) = (Z,z//2Cov(p)(Z'Z)I/2 = O'2(Z'Z)I/\Z'Zr 1(Z,z)I/2 = 0'21
We describe inferential procedures based on the classical linear regression model !n (73) with the additional (tentative) assumption that the errors e have a norrr~al distribution. Methods for checking the general adequacy of the model are conSidered in Section 7.6.
Inferences Concerning the Regression Parameters Before we can assess the importance of particular variables in the regression function
E(Y) =
Po + {3,ZI + ... + (3rzr
(710)
P
we must determine the sampling distributions of and the residual sum of squares, i'i. To do so, we shall assume that the errors e have a normal distribution. Result 7.4. Let Y = Zp + E, where Z has full rank r + ~ and E is distributed ~ Nn(O, 0.21). Then the maximum likelihood estimator of P IS the same as the leas
squares estimator
p=
p. Moreover, 2
(Z'ZrIZ'Y is distributed as Nr +l (p,O' (Z'Zr
1 )
and V is normally distributed, since it consists of linear combinations of the f3;'s. Therefore, V'V = (P  P)'(Z'Z)I/2(Z'Z//2(P  P) = (P  P)' (Z'Z)(P ' P) is distributed as U 2 X;+1' By Result 7.4 (n  r  l)s2 = i'i is distributed as U2rn_r_l> independently of and, hence, independently of V. Consequently, [X;+I/(r + 1)l![rnrl/(n  r  I)J = [V'V/(r + l)J;SZ has an Fr+l,ll rl distribution, and the confidence ellipsoid for P follows. Projecting this ellipsoid for P) using Result SA.1 with AI = Z'Z/ s2, c2 = (r + I)Fr+ 1,nrl( a), and u' =
P
(P 
[0, ... ,0,1,0, ... , DJ yields I f3i 
'"
Pd :s; V (r + I)Fr+l,nrl( a) Vv;;r(Pi), where 1
A
Var(f3;) is the diagonal element of s2(Z'Zr corresponding to f3i'
•
The confidence ellipsoid is centered at the maximum likelihood estimate P, and its orientation and size are determined by the eigenvalues and eigenvectors of Z'Z. If an eigenvalue is nearly zero, the confidence ellipsoid will be very long in the direction of the corresponding eigenvector.
372
Inferences About the Regression Model
Chapter 7 Multivariate Linear Regression Models
and
Practitioners often ignore the "simultaneous" confidence property of the interval estimates in Result 7.5. Instead, they replace (r + l)Fr+l.nrl( a) with the oneatatime t value tn  r 1(a/2) and use the intervals
jJ =
y= Example 7.4 (Fitting a regression model to realestate data) The assessment data Table 7.1 were gathered from 20 homes in a Milwaukee, Wisconsin, neighborhood. Fit the regression model =
where Zl = total dwelling size (in hundreds of square feet), Z2 = assessed value (in thousands of dollars), and Y = selling price (in thousands of dollars), to these using the method of least squares. A computer calculation yields 5.1523 ] .2544 .0512 [ .1463 .0172 .0067
PANEL 7.1
30.967
+ 2.634z1 +
(7.88)
(.785)
Total dwelling size (100 ft2)
Assessed value ($1000)
Y Selling price ($1000)
15.31 15.20 16.25 14.33 14.57 17.33 14.48 14.91 15.25 13.89 15.18 14.44 14.87 18.63 15.20 25.76 19.05 15.37 18.06 16.35
57.3 63.8 65.4 57.0 63.8 63.2 60.2 57.7 56.4 55.6 62.6 63.4 60.2 67.2 57.1 89.6 68.6 60.1 66.3 65.8
74.8 74.0 72.9 70.0 74.9 76.0 72.0 73.5 74.5 73.5 71.5 71.0 78.9 86.5 68.0 102.0 84.0 69.0 88.0 76.0
.045z2 (.285)
I
SAS ANALYSIS FOR EXAMPLE 7.4 USING PROC REG.
",OGRAM COMMANOS
=
Table 7.1 RealEstate Data Z2
30.967] 2.634 .045
title 'Regression Analysis'; data estate; infile 'T71.dat'; input zl z2 y; proc reg data estate; model y = zl z2;
~
Zj
[
with s = 3.473. The numbers in parentheses are the estimated standard deviations of the least squares coefficients. Also, R2 = .834, indicating that the data exhibit a strong regression relationship. (See Panel 7.1, which contains the regression analysis of these data using the SAS statistical software package.) If the residuals E pass the diagnostic checks described in Section 7.6, the fitted equation could be used to predict the selling price of another house in the neighborhood from its size
130 + 131 Zj 1 + f32Zj2 + Sj
(Z'Zr1 =
(Z'ZrIZ'y =
Thus, the fitted equation is
when searching for important predictor variables.
Yj
373
OUTPUT
Model: MODEL 1 Dependent Variable: Analysis of Variance
DF 2 17 19
Source Model Error C Total
J Root MSE Deep Mean CV
Sum of Squares 1032_87506 204.99494 1237.87000 3.47254 76.55000 4.53630
Mean Square 516.43753 12.05853
I
f value
42.828
Rsquare
0.8344,1
Adj Rsq
0.8149
Prob > F 0.0001
Parameter Estimates
Variable INTERCEP zl z2
DF 1
Parameter Estimate' 30.966566 ~.~34400
9.045184
Standard Error 7.88220844' 0.78559872 0.28518271
Tfor HO: Parameter 0 3.929 3.353 0.158
=
Prob> ITI 0.0011 0.0038 0.8760
374
Inferences About the Regression Model 375
Chapter 7 Multivariate Linear Regression Models and assessed value. We note that a 95% confidence interval for
132 [see (714)] is
Proof. Given the data and the normal assumption, the likelihood associated with the parameters P and u Z is
given by
~2 ± tl7( .025) VVai (~2) or
L(P,~)
= .045 ± 2.110(.285)
(.556, .647)
Since the confidence interval includes /3z = 0, the variable Z2 might be dropped from the regression model and the analysis repeated with the single predictor variable Zl' Given dwelling size, assessed value seems to add little to the prediction selling price.
=
1
2
(271' t/2u n
1
e(yzp)'(yZP)/2u <:
en/2
 (271')"/20"
with the maxim~~ occurring at p = (Z'ZrIZ'y and oZ Under the restnctlOn of the null hypothesis, Y = ZIP (I)
=
(y  ZP)'(y  Zp)/n.
+ e and
1
max L(p{!),u2 ) = en / 2 2 (271' )R/2of
P(l),U
• where the maximum occurs at
likelihood Ratio Tests for the Regression Parameters Part of regression analysis is concerned with assessing the e~fect~ of particular predictor variables on the response variable. One null hypotheslS of mterest states that certain of the z.'s do not influence the response Y. These predictors will be labeled ' The statement that Zq+l' Zq+2,"" Zr do not influence Y translates Z
p(t) =
(ZjZlr1Ziy. Moreover,
Rejecting Ho: P(2) = 0 for small values of the likelihood ratio
Z q+l' Z q+2,···, ro
into the statistical hypothesis Ho: f3 q +1 = /3q+z where p(Z) = Setting
= ... = /3r = 0
or Ho:
p(Z)
=0
(712) is equivalent to rejecting Ho for large values of (cT}  UZ)/UZ or its scaled version,
[f3 q +1> /3q+2'"'' f3r]·
Z = [Zl
nX(q+1)
n(cT}  UZ)/(r  q) _ (SSres(Zl)  SSres(Z»/(r  q) F nUZ/(n  r  1) S2 
Z2 ],
1
1 nX(rq)
The preceding Fratio has an Fdistribution with r  q and n  r  1 d.f. (See [22] or Result 7.11 with m = 1.) •
we can express the general linear model as y = Zp
+e
=
[Zl
1 Zz] •
Under the null hypothesis Ho: P(2) of Ho is based on the Extra sum of squares
= SSres(ZI)
[/!mJ + p(Z)
E
= ZIP(l)
+ Z2P(2) + e
= 0, Y = ZIP(1) + e. The. likelihood ratio test
 SSres(Z)
(713)
= (y _ zJJ(1»'(Y  ZJJ(1»  (y  Z{J)'(y  Z{J) where
p(!)
= (ZiZt>lZjy.
Result 7.6. Let Z have full rank r + 1 and E be distributed as Nn(O, 0.21). The likelihood ratio test of HO:P(2) = 0 is equivalent ~o,a test of Ho based on the extra sum of squares in (713) and SZ = (y  Zf3) (y  Zp)/(n  r  1). In particular, the likelihood ratio test rejects Ho if (SSres(ZI)  S;es(Z»/(r  q) >
Comment. The likelihood ratio test is implemented as follows. To test whether all coefficients in a subset are zero, fit the model with and without the terms corresponding to these coefficients. The improvement in the residual sum of squares (the • e~tra sum of.squares) is compared to the residual sum of squares for the full model via the FratlO. The same procedure applies even in analysis of variance situations . where Z is not of full rank.4 Mor~ ge~erally, it is possible to formulate null hypotheses concerning r  q linear combmatIons of P of the form Ho: Cp = A Q • Let the (r  q) X (r + 1) matrix. C have full rank, let Ao = 0, and consider
Ho:CP = 0 (This null hypothesis reduces to the previous choice when C =
[0 ii
I ].)
(rq)x(rq)
Frq,nrl(a)
where Frq,nrl(a) is the upper (l00a)thpercentile of anPdistribution with r  q and n  r  1 d.f.
4Jn situations where Z is not of full rank, rank(Z) replaces Result 7.6.
r
+ 1 and rank(ZJ) replaces
q
+ 1 in
376
Inferences About the Regression Model 377
Chapter 7 Multivariate Linear Regression Models
constant
2
Under the full model, Cp is distributed as Nr_q(CP, a C (Z'ZrlC'). We Ho: C P = 0 at level a if 0 does not lie in the 1DO( 1  a) % confidence ellipsoid Cp. Equivalently, we reject Ho: Cp = 0 if (CP)' (C(Z'ZrIC') 1(CP)
,
s2
~
0 0 0 0 0
1 1 1 1 1
100 100
o
1
o
1
010000 010000
1
010 o 1 0 o 1 0 010 010
1 1 1 1 1
0 0 0 0 0
001000 00100 0 001000 001 000 o 0 1 000
1 1
010 010
o o
1
1
000 000
1 1
001 001
1 1
001 001
1
1 1
(See [23]). The next example illustrates how unbalanced experimental designs are handled by the general theory just described. Example 7.S (Testing the importance of additional predictors using the extra squares approach) Male and female patrons rated the service in three establish: ments (locations) of a large restaurant chain. The service ratings were converted into an index. Table 7.2 contains the data for n = 18 customers. Each data point in the table is categorized according to location (1,2, or 3) and gender (male = 0 and female = 1). This categorization has the format of a twoway table with unequal numbers of observations per cell. For instance, the combination of location 1 and male has 5 responses, while the combination of location 2 and female has 2 responses. Introducing three dummy variables to account for location and two dummy variables to account for gender, we can develop a regression model linking the service index Y to location, gender, and their "interaction" using the design matrix
1 1
Z=
inter!lction
~
1 1 1 1 1
1
where S2 = (y  Zp)'(y  Zp)/(n  r  1) and Frq,nrI(a) is the (l00a)th percentile of an Fdistribution with r  q and n  r  1 dJ. The (714) is the likelihood ratio test, and the numerator in the Fratio is the extra sum of squares incurred by fitting the model, subject to the restriction that Cp ==
gender
100 100 100 100 100
1
> (r  q)Frq,llrl(a)
location
~
1 1 1 1
0 000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 000 0
"'pon'"
} 2 responses
0 0
} 2 responses
1 0 1 0
000 0 1 0 000010
} 2 responses
o o
00000 00000
} 2 responses
1 1
1 0 1 0
I'
1 1
The coefficient vector can be set out as {J' = [/30, /3 j, /32, /33, Tj, T2, 1'11, 1'12, 1'21> 1'22, 1'31, 1'32J
Table 7.2 RestaurantService Data Location
Gender
Service (Y)
1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3
0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 1 1
15.2 21.2 27.3 21.2 21.2 36.4 92.4 27.3 15.2 9.1 18.2 50.0 44.0 63.6 15.2 30.3 36.4 40.9
whe:e the /3;'S, (i > 0) represent the effects of the locations on the determination of service, tthehTils re~resent the effects of gender on the service index, and the 'Yik'S represen t e ocatlOngender interaction effects. The design matrix Z is not of full rank. (For instance, column 1 equals the sum of columns 24 or columns 56.) In fact, rank(Z) = 6. For the complete model, results from a computer program give SSres(Z) = 2977.4 and n  rank(Z) = 18  6 = 12. '!he ~odel without the interaction terms has the design matrix Zl consisting of the flTSt sIX columns of Z. We find that SSres(ZI) == 3419.1
== 14 110 test no· 1I • with _n  rank(ZI). == 18  4 . '. 1'11  1'12 1'32  0 (no locatIOngender mteractlOn), we compute
 1'21 = 1'22 = 1'31 =
F == (SSres(Zl)  SSres(Z»/(6  4) _ (SSres(Zl)  SSres(Z»/2 S
2
_ (3419.1  2977.4)/2 2977.4/12 == .89

SSres(Z)/12
378
Chapter 7 Multivar iate Linear Regression Models
point of an The Fratio may be compared with an appropriate percenta ge ble sigFdistrib ution with 2 and 12 d.f. This Fratio is not significant for any reasona depend not does index service the that conclude nificance level a. Consequently, we from the . upon any location gender interaction, and these terms can be dropped model. no differ_ Using the extra sumofsquares approach, we may verify that there is nt; that is ence between location s (no location effect), but that gender is significa ' males and females do not give the same ratings to service. the varia, unequal are counts cell the where s situation nce ofvaria analysis In their interac_ tion in the respons e attributable to different predictor variables and evaluate the To . amounts ent independ into d separate be usually tions cannot y to fit necessar is it case, this in relative influenc es of the predictors on the response iate appropr the compute and question in terms the without the model with and • Ftest statistics.
Inferences from the Estimate d Regression Function 379
, . is distribu ted as X~ _ /(n  r  1) C ation zofJ is combin hnear the ntIy, l onseque . and ;0) O(z'zr 0'2 N(zop, z
(zoP  z(JP)/Y 0"2 z0 (Z'Z)I ZO
Vr=2(~'=(Z='~)~l= YS10'2 Z zo) S Zo . d' t 'b follows. interval ce confiden The IS IS n uted as (nrl'
be used to Once an investig ator is satisfied with the fitted regression model, it can for the values selected be ZOr] ZOl,"" [1, = Zo 4t s. problem on predicti solve two funcon regressi the estimate to (1) used predicto r variables. Then Zo and fJ can be Y response the of value the estimate to (2) and Zo at f3rzor + , .. + tion f30 + f3lz01 at zoo
Estimating the Regression Function at Zo s have values Let Yo denote the value of the response when the predictor variable 000 is value d expecte the (73), in za = [1, zOJ,· . . , ZOr]. According to the model (715) E(Yo I zo) = f30 + f3lZ0l + ... + f3r zor = zofJ Its least squares estimate is
zop.
d linear Result 7.7. For the linear regression model in (73), zoP is the unbiase If the 1zo0'2. zb(Z'Zr estimato r of E(Yolzo) with minimum variance, Var(zoP ) = for interval ce confiden % a) 100(1 a then errors E are normall y distributed, by provided is zofJ E(Yo I zo) =
ution with where t"rl(a /2) is the upper l00(a/2 )th percentile of a tdistrib n  r  1 d.f. so R~sult . Proof. For a fixed Zo, zofJ)s just a lin~ar combination of the f3;'s, (fJ) = . Cov since 0'2 lzo zo(Z'Zr = (fJ)zo Cov Zo = (zofJ) 7.3 applies. Also, Var distriby normall is E ~(Z'Zrl by Result 7.2. UIlder the further assu~l'tion that s2/0'2, which uted, Result 7.4 asserts that fJ is Nr+1(fJ, 0'2(Z'Z) ) indepen dently of
•
Forecasting a New Observation at Zo . . Predicti on of a new observa tion, such as Y, at z' = [1 ,ZOl"", zor]lsm oreunce rtam o. fY, 0, thanest imating theexpe cted I va ue 0 o· Accordm g to the regression model of (73),
Yo = zoP + BO
or
7.S Inferences from the Estimated Regression Function
(' zoP  zoP)
(new respons e Yo) = (expecte d value of Yoat zo) + (new error) distributed as N(O 2) ap d"IS Illdependent of E and hence of a, and S2 is BO where ,0' a d 2 . fl " p . Tb e errors E III uence the est' t Illla ors p an s through the responses Y, but BO does not. . Result 7.S. Given the linear regression model of (7 3 ) , a new observatIOn YcJ has the unbiased predictor
ZoP =
Po + PIZOI + is The variance of the forecast error Yo Var(Yo 
zoP
ZoP) =
... + PrZor
0'2(1 + zb(Z'Z) I ZO )
~7:;i~~: ~;ors E have a normal distribution, a lOD( 1 zoP ± t"_r_1
(~) Ys2(1
a) % prediction interval for
+ ZO(Z'ZrIZO)
w~re
n
f,,r_l(a /2) is the upper lOO(a/2 )th percenti le of a tdistrib ution with r  1 degrees of freedom .
'a h' h . zOP,' W IC estImates E(Yo I zo). By ReSUlt 7.7, zoP has . J) = z'(Z'Z) lz 2. The f orecast error IS E(zofJ) = zofJ and Var(zof then 00" ,0 ,' y, = E(BO) + zoP) E(Yo Thus, ). zo(PP + =.BO zoP BO + zafJ_ EO : ZO~ =, r is unbiase d Since B0 and P are m d epen d ent, ( o( P fJ»  0 so the predIcto, . , ' V (Y, zo(Z'Z rlz ). ar. o.  zofJ) = Var (BO) + Var (zom = 0'2 + zo(Z'Z) I Z00'2 = 0'2(1 + then P °is If It IS f~rt~er assumed that E has a normal distribution, I C z' _ y, tion combina linear the is so normall y, dlstnbu ted, and op· onseque nt y, 0 ,,J (Y,O  z' P)/V, 2 0" (1 + zo(Z Z) ZO) is distribu ted as N(O, 1). Dividing this ratio by V 2 ~o 2 b' /(n  r 1) is distribu ted as YX"rl which , s/ , we 0 taln Proof. We forecast y, by , "
a
.
a
(1'0  ZoP) Ys2(l + zo(Z'Zr Jzo) . . . which IS dIstribu ted as t n"'rI' Th e pred'" IctIon mterval follows immediately.
•
Model Checking and Other Aspects of Regression
380 Chapter 7 Multivariate Linear Regression Models
The prediction interval for Yois wider than the confidence interval for estimating the value of the regression function E(Yo Izo) = zop· The additional uncertainty in forecasting Yo, which is represented by the extra term S2 in the expression s2(1 + zo(Z'Zrlzo), comes from the presence ofthe unknown error term eo· Example 7.6 (Interval estimates for a mean response and a future response) Companies considering the purchase of a computer must first assess their future needs in to determine the proper equipment. A computer scientist collected data from seven similar company sites so that a forecast equation of computerhardware requirements for inventory management could be developed. The data are given in Table 7.3 for ZI
Y
adddelete item count (in thousands) CPU (central processing unit) time (in hours)
=
Since sY1 + zO(Z'ZflZO = (1.204)(1.16071) = 1.40, a 95% prediction interval for the CPU time at a new facility with conditions Zo is
z'oP ± t4(.025)sY1 + zo(Z'Zr1zo = 151.97 ± 2.776(1.40)
•
or (148.08,155.86).
1.6 Model Checking and Other Aspects of Regression Does the Model Fit?
= customer orders (in thousands)
Z2 =
Construct a 95% confidence interval for the mean CPU time, E(Yolzo) '= + fJrzol + f32Z02 at Zo '= [1,130,7.5]. Also, find a 95% prediction interval for a new facility's CPU requirement corresponding to the same zo° A computer program provides the estimated regression function
130
Assuming that the model is "correct," we have used the estimated regression function to make inferences. Of course, it is imperative to examine the adequacy of the model before the estimated function becomes a permanent part of the decisionmaking apparatus. All the sample information on lack of fit is contained in the residuals
81
= Yl 
e2 = Y2  130 
8.17969 [
.08831
en = Yn 
.00052 .00107
,
f31Z21  ... 
~o  ~IZnl
 ... 
f3rZ2r
~rZnr
or
e=
and s = 1.204. Consequently,
zoP = 8.42 + 1.08(130) + .42(7.5) = 151.97 ,:
and s Yzo(Z'Zrlzo = 1.204( .58928) = .71. We have t4( .025) confidence interval for the mean CPU time at Zo is
=
2.776, so the 95%
zoP ± t4(.025)sYzo(Z'Zrlzo = 151.97 ± 2.776(.71) or (150.00,153.94). Table 7.3 Computer Data
Zl (Orders)
(Adddelete items)
Y (CPU time)
123.5 146.1 133.9 128.5 151.5 136.2 92.0
2.108 9.213 1.905 .815 1.061 8.603 1.125
141.5 168.9 154.8 146.5 172.8 160.1 108.5
Z2
~o  ~IZI1  ...  ~rZlr A,
y = 8.42 + 1.08z1 + .42Z2 (Z'ztl = .06411
381
Source: Data taken from H. P. Artis, Forecasting Computer Requirements: A Forecaster's Dilemma (Piscataway, NJ: Bell Laboratories, 1979).
[I  Z(Z'ZfIZ']Y
=
[I  H]y
(716)
If the model is valid, each residual ej is an estimate of the error ej' which is assumed to be a normal random variable with mean zero and variance (1'2. Although the residuals ehaveexpectedvalueO,theircovariancematrix~[1  Z(Z'Zr1Z'] = (1'2[1  H] is not diagonal. Residuals have unequal variances and nonzero correlations. Fortunately, the correlations are often small and the variances are nearly equal. Because the residuals have covariance matrix (1'2 [I  H], the variances of the ej can vary greatly if the diagonal elements of H, the leverages h jj , are substantially different. Consequently, many statisticians prefer graphical diagnostics based on studentized residuals. Using the residual mean square S2 as an estimate of (1'2, we have
e
Va;(ei) = s2(1  kJj),
j = 1,2, ... ,n
(717)
and the studentized residuals are j == 1,2, ... ,n
(718)
We expect the studentized residuals to look, approximately, like independent drawings from an N(0,1) distribution. Some software packages go one step further and studentize ej using the deleteone estimated variance ;(j), which is the residual mean square when the jth observation is dropped from the analysis.
382 Chapter 7 Multivariate Linear Regression Models
Model Checking and Other Aspects of Regression 383
Residuals should be plotted in various ways to detect possible anomalies. For general diagnostic purposes, the following are useful graphs: 1. Plot the residuals Bj against the predicted values Yj = Po + 13) Zjl + ... + P,Zj'" Departures from the assumptions of the model are typically indicated by two' types of pheno1J.1ena: (a) A dependence of the residuals on the predicted value. This is illustrated in Figure 7.2(a). The numerical calculations are incorrect, or a f30 term been omitted from the model. (b) The variance is not constant. The pattern of residuals may be funnel shaped, as in Figure 7.2(bY, so that there is large variability for large Yandsmall variability for small y. If this is the case, the variance of the error .is . not constant, and transformations or a weighted least squares approach (or both) are required. (See Exercise 7.3.) In Figure 7.2( d), the residuals form a horizontal band. This is ideal and indicates equal variances and no dependence on y. 2. Plot the residuals Bj against a predictor variable, such as ZI, or products ofpredictor variables, such as ZI or ZI Zz. A systematic pattern in these plots suggests the need for more terms in the model. This situation is illustrated in Figure 7.2(c). 3. QQ plots and histograms. Do the errors appear to be normally distributed? To answer this question, the residuals Sj or can be examined using the techniques discussed in Section 4.6. The QQ plots, histograms, and dot diagrams help to detect the presence ~f unusual observations or severe departures from normality that may require special attention in the analysis. If n is large, minor departures from normality will not greatly affect inferences about p.
si
4. Plot the residuals versus time. The assumption of independence is crucial, but hard to check. If the data are naturally chronological, a plot of the residuals versus time may reveal a systematic pattern. (A plot of the positions of the residuals in space may also reveal associations among the errors.) For instance, residuals that increase over time indicate a strong positive dependence. A statistical test of independence can be constructed from the first autocorrelation,
(719)
of residuals from adjacent periods. A popular test based on the statistic
j~ (Bj n
Bj_I)2
/
J~ BT == n
2(1  rd is called the DurbinWatson test. (See (14]
for a description of this test and tables of critical values.)
Example 7.7 (Residual plots) Three residual plots for the computer data discussed in Example 7.6 are shown in Figure 7.3. The sample size n == 7 is really too small to allow definitive judgments; however, it appears as if the regression assumptions are tenable. _
e
•
1.0
1.0
z, 1.0
••
•••
•
••
1.0
(a) (a)
•
• (b)
(b)
••
1.0
r~y
••
1.0
•
(c)
(c)
(d)
Figure 7.2 Residual plots.
Figure 7.3 Residual plots for the computer data of Example 7.6. I
Model Checking and Other Aspects of Regression
384 Chapter 7 Multivariate Linear Regression Models
If several observations of the response are available for the same values of the predictor variables, then a formal test for lack of fit can be carried out. (See [13] for a discussion of the pureerror lackoffit test.) .
Leverage and I!lfluence Although a residual analysis is useful in assessing the fit of a model, departures from the regression model are often hidden by the fitting process. For example, there may be "outliers" in either the response or explanatory variables that can have a considerable effect on the analysis yet are not easily detected from an examination of residual plots. In fact, these outIiers may determine the fit. The leverage h jj the (j, j) diagonal element of H = Z(Z' Zrl Z, can be interpret" ed in two related ways. First, the leverage is associated with the jth data point measures, in the space of the explanatory variables, how far the jth observation is from the other n  1 observations. For simple linear regression with one explanatory variable z, 1 n
(ZjZ)2
h·=+":'~~
JI
n
2: (z; 
z)2
;=1
The average leverage is (r + l)/n. (See Exercise 7.8.) Second, the leverage hjj' is a measure of pull that a single case exerts on the fit. The vector of predicted values is
385
Selecting predictor variables from a large set. In practice, it is often difficult to formulate an appropriate regression function immediately. Which predictor variables should be included? What form should the regression function take? When the list of possible predictor variables is very large, not all of the variables can be included in the regression function. Techniques and computer programs designed to select the "best" subset of predictors are now readily available. The good ones try all subsets: ZI alone, Z2 alone, ... , ZI and Z2, •.•. The best choice is decided by examining some criterion quantity like Rl. [See (79).] However, R2 always increases with the inclusion of additional predict~r variables. Although this problem can be circumvented by using the adjusted Rl, R2 = 1  (1  Rl) (n  l)/(n  r  1), a better statistic for selecting variables seems to be Mallow's C p statistic (see [12]), residual sum of squares for subset model) with p parameters, including an intercept Cl' = ( (residual variance forfull model)  (n  2p) A plot of the pairs (p, C p ), one for each subset of predictors, will indicate models that forecast the observed responses well. Good models typically have (p, C p) coordinates near the 45° line. In Figure 7.4, we have circled the point corresponding to the "best" subset of predictor variables. If the list of predictor variables is very Jong, cost considerations limit the number of models that can be examined. Another approach, called step wise regression (see [13]), attempts to select important predictors without considering all the possibilities.
y = ZjJ = Z(Z'Z)IZy = Hy where the jth row expresses the fitted value Yj in terms of the observations as Yj = hjjYj
+
2: h jkYk
k*j
Provided that all other Y values are held fixed ( change in Y;)
= hjj ( change in Yj)
If the leverage is large relative to the other hjk> then Yj will be a major contributor to the predicted value Yj· Observations that significantly affect inferences drawn from the data are said to be influential. Methods for assessing)nfluence are typically based on the change in the vector of parameter estimates, fJ, when observations are deleted. Plots based upon leverage and influence statistics and their use in diagnostic checking of regression models are described in [3], [5], and [10]. These references are recommended for anyone involved in an analysis of regression models. If, after the diagnostic checks, no serious violations of the assumptions are detected, we can make inferences about fJ and the future Y values with some assurance that we will not be misled.
1800 1600 1200 11
10
9
7 6
5 4
(1.2.3)
Additional Problems in Linear Regression We shall briefly discuss several important aspects of regression that deserve and receive extensive treatments in texts devoted to regression analysis. (See [10], [11], [13], and [23].)
1<..7==~~7= P = r
+1
Figure 7.4 C p plot for computer data from Example 7.6 with three predictor variables (z) = orders, Z2 = adddelete count, Z3 = number of items; see the example and original source).
386 Chapter 7 Multivariate Linear Regression Models
Multivariate Multiple Regression 387
The procedure can be described by listing the basic steps (algorithm) involved in the computations: Step 1. All possible simple linear regressions are considered. The predictor variable that explains the largest significant proportion of the variation in Y (the that has the largest correlation with the response) is the first variable to enter the regression function. Step 2. The next variable to enter is the one (out of those not yet included) makes the largest significant contribution to the regression sum of squares. The nificance of the contribution is determined by an Ftest. (See Result 7.6.) The of the Fstatistic that must be exceeded before the contribution of a variable is deemed significant is often called the F to enter. Step 3. Once an additional variable has been included in the equation, the indivi
Ale =
residual sum of squares for subset mOdel) with p parameters, including an intercept nln ( n
+ 2p
It is desirable that residual sum of squares be small, but the second term penalizes for too many parameters. Overall, we want to select models from those having the smaller values of Ale.
Colinearity. If Z is not of full rank, some linear combination, such as Za, must equal O. In this situation, the columns are said to be colinear. This implies that Z'Z does not have an inverse. For most regression analyses, it is unlikely that Za = 0 exactly. Yet, iflinear combinations of the columns of Z exist that are nearly 0, the calculation l of (Z'Zr l is numerically unstable. Typically, the diagoqal entries of (Z'Zr will be large. This yields large estimated variances fqr the f3/s and it is then difficult to detect the "significant" regression coefficients /3i. The problems caused by coIinearity can be overcome somewhat by (1) deleting one of a pair of predictor variables that are strongly correlated or (2) relating the response Y to the principal components of the predictor variablesthat is, the rows zj of Z are treated as a sample, and the first few principal components are calculated as is subsequently described in . Section 8.3. The response Y is then regressed on these new predictor variables.
Bias ca~sed by a misspecified model. Suppose some important predictor variables are omItted f~om the. proposed regression model. That is, suppose the true model has Z = [ZI i Z2] WIth rank r + 1 and
(720)
where E(E).= 0 and Var(E) = (1"21. However, the investigator unknowingly fits a model usmg only the fIrst q predictors by minimizing the error sum of squares_ (Y  ZI/3(I»'(Y  ZI/3(1). The least squares estimator of /3(1) is P(I) = (Z;Zd lZ;Y. Then, unlike the situation when the model is correct , 1 E(P(1» = (Z;Zlr Z;E(Y) = (Z;Zlr1Z;(ZI/3(I) + Z2P(2) + E(E» =
p(])
+ (Z;Zd1Z;Z2/3(2)
(721)
That is, P(1) is a biased. estimator of /3(1) unless the columns of ZI are perpendicular to those of Z2 (that IS, ZiZ2 = 0>. If important variables are missing from the model, the least squares estimates P(1) may be misleading.
1.1 Multivariate Multiple Regression In this section, we consider the problem of modeling the relationship between m respon~es Y1,Y2,· .. , Y,n and a single set of predictor variables ZI, Zz, ... , Zr. Each response IS assumed to follow its own regression model, so that
Yi =
Yz
f301
= f302
Ym =
f30m
+ +
f311Z1 f312Z1
+ ... + f3rlZr + el + ... + /3r2zr + e2
+ /31mZl + ... +
f3rmzr
(722)
+ em
The error term E' = [el' e2, ... , em] has E(E) = 0 and Var(E) = .I. Thus the error terms associated with different responses may be correlated. ' To establish notation conforming to the classical linear regression model, let [ZjO,~jI, ... ,Zjr] denote the values of the predictor variables for the jth trial, let Yj = [ljJ, ~2' ... , .ljm] be the responses, and let El = [ejl, ej2, ... , Ejm] be the errors. In matnx notatIOn, the design matrix
Z (nX(r+1)
=
Z10
Zll
Z20 :
Z21 :
ZnO
Znl
r
ZlrJ
Z2r
Znr
Multivariate Multiple Regression 389
388 Chapter 7 Multivariate Linear Regression Models
Collecting these univariate least squares estimates, we obtain
is the same as that for the singleresponse regression model. [See (73).] The matrix quantities have multivariate counterparts. Set Yl2
_ Y
=
(nXm)
fJ «r+l)Xm)
Yn1
Y n2
Ynm
For any choice of parameters B = [b(l) i b(2) i ... i b(m»), the matrix of errors is Y  ZB. The error sum of squares and cross products matrix is
[Po.
f302 f312
pom] f3~m ~ [P(J) i P(2) i ... i P(m)]
(Y  ZB)'(Y ; ZB)
:
f3!I'
" = [Y(!) . i Y(2) i
(726)
'" i Y(",)]
(Y(1)  Zb(l»)'(Y(1)  Zb(1»
(Y(1)  Zb(I»'(Y(m)  Zb(m» ]
f3rm
f3r2
=
(nXrn)
.00
or
:
=
!
122
[Y"Y~l
f3r1
e
[fl(1) i fl(2) i ... i fl(m)] = (Z'Zr IZ '[Y(1) i Y(2)
¥Om] 12",
:
=
jJ =
[
(Y(m)  Zb(m);'(Y(1)  Zb(l)
['"
E~l
E22
82m , "m] : = [E(1) " i E(2) i .. , i E(",»)
En 2
e nm
:
selection
b(i) = p(iJ
p.
Residuals:
The multivariate linear regression model is Z
the
ith
diagonal
sum
of
squares
/3.
Predicted values:
Y=
minimizes
Zb(i)'(Y(i)  Zb(i).Consequently,tr[(Y  ZB)'(Y  ZB») is minimized Also, the generalized variance I (Y  ZB)' (Y  ZB) I is minby the choice B = (See Exercise 7.11 for an additional generalimized by the least squares estimates ized sum of squares property.) , Using the least squares estimates fJ, we can form the matrices of (Y(i) 
~ [~;J (nxm)
Zb(m» (727)
The
Enl
Zb(m»~(Y("') 
(Y(nt) 
EI2
(728)
The orthogonality conditions among the residuals, predicted values, and columns of Z, which hold in classical linear regression, hold in multivariate multiple regression. They follow from Z'[I  Z(Z'ZrIZ') = Z'  Z' = O. Specifically,
p+e
(nX(r+I» «r+1)Xm)
Y = ZjJ = Z(Z'Zrlz,y i = Y  Y = [I  Z(Z'ZrIZ')Y
(/lXm)
with
z'i = Z'[I  Z(Z'Zr'Z']Y = 0 The m observations on the jth trial have covariance matrix I = {O"ik}, but ob. servations from different trials are uncorrelated. Here p and O"ik are unknown parameters; the design matrix Z has jth row [ZjO,Zjl,'''' Zjr)'
c '
(729)
so the residuals E(i) are perpendicular to the columns of Z. Also,
Y'e =
jJ'Z'[1 Z(Z'ZrIZ'jY = 0
(730)
confirming that the predicted values Y(iJ are perpendicular to all residual vectors' Simply stated, the ith response Y(il follows the linear regression model Y(iJ= ZPU)+E(i)'
L
Y + e, Y'Y = (Y + e)'(Y + e) = Y'Y + e'e + 0 + 0'
Because Y =
i=1,2, ... ,m
with Cov (£(i) = uijl. However, the errors for different responses on the same trial can be correlated. Given the outcomes Y and the values of the predic!or variables Z with column rank, we determine the least squares estimates P(n exclusively from observations Y(i) on the ith response. In conformity with the solution, we take lie;
E(k).
or
Y'Y
Y'Y
total sum of squares) = (predicted sum of squares) ( and cross products and cross products
+ +
e'e residual ( error) sum) of squares and ( cross products (731)
390
Multivariate Multiple Regression
Chapter 7 Multivariate Linear Regression Models
391
The residual sum of squares and cross products can also be written as
E'E
=
y'y = Y'Y  jJ'Z'ZjJ
Y'Y 
OF 1
,
\
Type 11/ SS 40.00000000
F Value 20.00
Mean Square 40.00000000'
Pr> F 0.0208
Example 1.8 {Fitting a multivariate straightline regression model) To illustrate the
.1
calculations of
jJ, t, and E, we fit a straightline reg;ession model (see Panel? Y;l
Y;z
= f101 + f1ll Zjl + Sjl = f10z + f112Zjl + Sj2, . .
j
Tfor HO: Parameter = 0 0.91 4.47
Std Error of Estimate 1.09544512 0.44721360
Pr> ITI 0.4286 0.02011
= 1,2, ... ,5
to two responses Y 1 and Yz using the data in Example? 3. These data, augmented by observations on an additional response, are as follows:
Y:t Y2
o
1
1
4 1
1
2 3 2
The design matrix Z remains unchanged from the singleresponse problem. We find that
,_[1 111IJ
(Z'Zr1 = [
Z01234
PANEL 7.2
.6 .2
OF 1 3 4
Sum of Squares 10.00000000 4.00000000 14.00000000
Mean Square 10.00000000 1.33333333
RSquare 0.714286
C.V. 115.4701
Root MSE 1.154701
OF
Type III SS 10.00000000
Mean Square 10.00000000
Source Model Error Corrected Total
4 9 2
3 8 3
.2J .1
Source Zl
Tfor HO: Parameter = 0 1.12 2.74
SAS ANALYSIS FOR EXAMPLE 7.8 USING PROe. GlM.
title 'Multivariate Regression Analysis'; data mra; infile 'Example 78 data; input y1 y2 zl; proc glm data = mra; model y1 y2 = zllss3; manova h = zl/printe;
PROGRAM COMMANDS
'IE= Error SS & CP Matrix Y1 Y1 Y2
General Linear Models Procedure loepelll:lenwariable: Source Model Error Corrected Total
Y~ I
RSquare 0.869565
Sum of Squares. 40.00000000 6.00000000 46.00000000 e.V. 28.28427
Mean Square 40.00000000 2.00000000
Root MSE 1.414214
F Value 20.00
Pr> F 0.0208
Y1 Mean 5.00000000
Pr> F 0.0714
Y2 Mean 1.00000000
FValue 7.50
Pr> F 0.0714 Std Error of Estimate 0.89442719 0.36514837
Pr> ITI 0.3450 0.0714
I
Y2
I~
Manova Test Criteria and Exact F Statistics for the Hypothesis of no Overall Zl Effect E = Error SS&CP Matrix H = Type 1/1 SS&CP Matrix for Zl S=l M=O N=O
OUTPUT
OF 1 3 4
F Value 7.50
Statistic Wilks' lambda Pillai's Trace HotellingLawley Trace Roy's Greatest Root
Value 0.06250000 0.93750000 15.00000000 15.00000000
F 15.0000 15.0000 15.0000 15.0000
Num OF 2 2 2 2
OenOF 2 2 2 2
Pr> F 0.0625 0.0625 0.0625 0.0625
394
MuItivariate Multiple Regression
Chapter 7 Multivariate Linear Regression Models
Dividing each entry E(i)E(k) of E' Eby n  r  1, we obtain the unbiased estimator of I. Finally, CoV(P(i),E(k» = E[(Z'ZrIZ'EUJE{k)(I  Z(Z'Zr IZ ')]
so each element of
=
(Z'ZrIZ'E(E(i)E(k»)(I  Z(Z'Zr1z'y
=
(Z'ZrIZ'O"ikI(I  Z(Z'Zr IZ ')
=
O"ik«Z'ZrIZ'  (Z'ZrIZ') = 0
E(/J) = fJ and Cov (p(i), P(k»
=
[ZOP(l)
is an unbiased estiffiator zoP since E(zoP(i» = zoE(/J(i» = zofJ(i) for each component. From the covariance matrix for P (i) and P (k) , the estimation errors zofJ (i)  zOP(i) have covariances E[zo(fJ(i)  P(i»)(fJ(k)  p(k»'zol = zo(E(fJ(i)  P(i))(fJ(k)  P(k»')ZO =
O"ikZO(Z'Zr1zo
(735)
Vo
The related problem is that of forecasting a new observation vector = [Y(ll, Yoz ,.··, Yoml at Zoo According to the regression model, YOi = zofJ(i) + eOi ,,:here the "new" error EO = [eOI, eoz, ... , eo m ] is independent of the errors E and satIsfies E( eo;) = 0 and E( eOieok) = O"ik. The forecast error for the ith component of Vo is 1'Oi  zo/J(i) = YOi  zofJ(i) + z'ofJU) 
= eOi  zo(/J(i) 
ZOP(i)
fJ(i)
so E(1'Oi  ZOP(i» = E(eo;)  zoE(PU)  fJ(i) = 0, indicating that ZOPU) is an unbiased predictor of YOi . The forecast errors have covariances E(YOi  ZOPU» (1'Ok  ZOP(k» =
E(eo;  zO(P(i)  fJ(i))) (eok  ZO(P(k)  fJ(k»)
=
E(eoieod + zoE(PU)  fJm)(P(k)  fJ(k»'ZO
l
= U'ik(Z'Zr . Also,
lAA
The maximized likelihood L (IL,
i) =
+ zo(Z'Zr1zo)
Note that E«PU)  fJ(i)eOk) = 0 since Pm = (Z'ZrIZ' E(i) + fJ(iJ is independelllt of EO. A similarresult holds for E(eoi(P(k)  fJ(k»)'). Maximum likelihood estimators and their distributions can be obtained when the errors e have a normal distribution.
A
(27Trmn/2/i/n/2emn/2.
•
Proof. (See website: www.prenhall.com/statistics) supp~rt
for using least squares estimates.
When the errors are normally distributed, fJ and nJE'E are the maximum likelihood estimators of fJ and ::t, respectively. Therefore, for large samples, they have nearly the smallest possible variances.
Comment. The multivariate mUltiple regression model poses no new computational problem~ ~~t squares (maximum likelihood) estimates,p(i) = (Z'Zr1Z'Y(i)' are computed mdlVldually for each response variable. Note, however, that the model requires that the same predictor variables be used for all responses. Once a multivariate multiple regression model has been fit to the data, it should be subjected to the diagnostic checks described in Section 7.6 for the singleresponse model. The residual vectors [EjJ, 8jZ, ... , 8jm] can be examined for normality or outliers using the techniques in Section 4.6. The remainder of this section is devoted to brief discussions of inference for the normal theory multivariate mUltiple regression model. Extended accounts of these procedures appear in [2] and [18].
likelihood Ratio Tests for Regression Parameters The multiresponse analog of (712), the hypothesis that the responses do not depend on Zq+l> Zq+z,·.·, Z,., becomes
Ho: fJ(Z)
=0
where
fJ =
[~~~)I~njJ fJ(Z)
 zoE«p(i)  fJ(i)eok)  E(eo;(p(k)  fJ(k»')ZO = O"ik(1
/J
A
Result 7.10 provides additional
1 ZOP(2) 1... 1ZoP(m)]
fJ and fJ ,has a normal distribution with
is independent of the maximum likelihood estimator of the positive definite I given by 1 I = E'E = (V  Z{J)'(Y  zfJ) n n and ni is distributed as Wp •n r  J (I) A
The mean vectors and covariance matrices determined in Result 7.9 enable us to obtain the sampling properties of the least squares predictors. We first consider the problem of estimating the mean vector when the predictor variables have the values Zo = [l,zOI, ... ,ZOr]. The mean of the ith response variable is zofJ(i)' and this is estimated by ZOP(I)' the ith component of the fitted regression relationship. Collectively,
zoP
Result 7.10. Let the multivariate multiple regression model in (723) hold with full rank (Z) = r + 1, n ~ (r + 1) + m, and let the errors E have a normal distribution. Then is the maximum likelihood estimator of
Pis uncorrelated with each ele~ent of e.
395
«rq)Xm)
Setting Z = [
Zl
(nX(q+ I»
E(Y)
! i
Zz
], we can write the general model as
(nX(rq»
= zfJ = [Zl i, Zz]
[!!~!~J = ZlfJ(l) + zzfJ(Z) fJ(2)
(737)
396
Multivariate Multiple Regression
Chapter 7 Multivariate Linear Regression Models
+ e and the likelihood ratio test of Ho is
Under Ho: /3(2) = 0, Y = Zt/J(1) on the quantities involved in the
extra sum ofsquares and cross products f
=: (Y 
ZJJ(1»)'(Y  ZJJ(I»  (Y  Zp), (Y  Zp)
= n(II  I)
where P(1) = (ZlZlrIZ1Y and II = nI(Y  ZIP(I»)' (Y  ZIP(I»' From Result 7 .10, the likelihood ratio, A, can be expressed in terms of generallizec variances:
Example 7.9 (Testing the importance of additional predictors with a multivariate response) The service in three locations of a large restaurant chain was rated according to two measures of quality by male and female patrons. The first servicequality index was introduced in Example 7.5. Suppose we consider a regression model that allows for the effects of location, gender, and the locationgender interaction on both servicequality indices. The design matrix (see Example 7.5) remains the same for the tworesponse situation. We shall illustrate the test of no locationgender interaction In either response using Result 7.11. A compl,1ter program provides
(
residual sum of squares) = nI = [2977.39 1021.72J and cross products 1021.72 2050.95 extra sum of squares) ( and cross products
Equivalently, Wilks'lambda statistic A2/n =
= n(I
I~I
=
lnil nIn ln:£ + n(:£1 :£)1
For n large,5 the modified statistic
 [n  r  1 
.!. (m 2
 r + q + 1) ] In (
has, to a close approximation, a chisquare distribution with
Proof. (See Supplement 7A.)
= [441.76
246.16
246.16J 366.12
~ In~1 ~)
Result 7.11. Let the multivariate multiple regression model of (723) hold with. of full rank r + 1 and (r + 1) + m:5 n. Let the errors e be normally Under Ho: /3(2) = 0, nI is distributed as Wp,norol(I) independently of n(II which, in turn, is distributed as Wp,rq(I). The likelihood ratio test of Ho is . to rejecting Ho for large values of
III)
i)
Let /3(2) be the matrix of interaction parameters for the two responses. Although the sample size n = 18 is not large, we shall illustrate the calculations involved in the test of Ho: /3(2) = 0 given in Result 7.11. Setting a = .05, we test Ho by referring
can be used.
lId
_ I
lId
2lnA = nln (
397
I~ I ) lId
mer  q) dJ.
P
If Z is not of full rank, but has rank rl + 1, then = (Z'Zrz'Y, (Z'Zr is the generalized inverse discussed in [22J. (See also Exerc!se 7.6.) distributional conclusions stated in Result 7.11 remain the same, proVIded that r replaced by rl and q + 1 by rank (ZI)' However, not all hypotheses concerning can be tested due to the lack of uniqueness in the identification of Pca.used. by linear dependencies among the columns of Z. Nevertheless, the gene:abzed allows all of the important MANOVA models to be analyzed as specIal cases of multivariate multiple regression model. STechnicaUy, both n  rand n  m should also be large to obtain a good chisquare applroxilnatlf
[nrll.!.(mrl+ql'+l)]ln( 2 InI + n(II  I)I = [18  5  1 
~(2 
5
+ 3 + 1)}n(.7605)
= 3.28
toa chisquare percentage point with m(rl  ql) = 2(2) = 4d.fSince3.28 < ~(.05) = 9.49, we do not reject Ho at the 5% level. The interaction terms are not needed. _ Information criterion are also available to aid in the selection of a simple but adequate multivariate mUltiple regresson model. For a model that includes d predictor variables counting the intercept, let
id = .!.n (residual sum of squares and cross products matrix) Then, the multivariate mUltiple regression version of the Akaike's information criterion is AIC = n In(1 I)  2p X d
id
This criterion attempts to balance the generalized variance with the number of paramete~s. Models with smaller AIC's are preferable. In the context of Example 7.9, under the null hypothesis of no interaction terms, we have n = 18, P = 2 response variables, and d = 4 terms, so AIC =
n
In (I I I)  2
p
X d =
181
n
(1~[3419.15 18 1267.88
1267.88]1)  2 X 2 X 4 2417.07
= 18 X In(20545.7)  16 = 162.75
More generally, we could consider a null hypothesis of the form Ho: c/3 = r o, where C is (r  q) X (r + 1) and is of full rank (r  q). For the choices
Multivariate Multiple Regression 399
398 Chapter 7 Multivariate Linear Regression Models C
= [0
ill
and fo = 0, this null hypothesis becomes H[): c/3
(rq)x(rq)
= /3(2)
== 0,
the case considered earlier. It can be shown that the extra sum of squares and cross products generated by the hypothesis Ho is ,n(II  I) = (CP  fo),(C(Z'ZrICT1(CjJ  fo)
.
.
Under the null hypothesis, the statistic n(II  I) is distributed as Wrq(I) independently of I. This distribution theory can be employed to develop a test of Ho: c/3 = fo similar to the test discussed in Result 7.11. (See, for example, [18].)
Predictions from Multivariate Multiple Regressions Suppose the model Y = z/3 + e, with normal errors e, has been fit and checked for any inadequacies. If the model is adequate, it can be employed for predictive purposes. One problem is to predict the mean responses corresponding to fixed values Zo of the predictor variables. Inferences about the mean responses can be made using the distribution theory in Result 7.10. From this result, we determine that
jJ'zo isdistributedas Nm(/3lzo,zo(Z'Z)lzoI) and nI
Other Multivariate Test Statistics Tests other than the likelihood ratio test have been proposed for testing Ho: /3(2) == 0 in the multivariate multiple regression model. Popular computerpackage programs routinely calculate four multivariate test statistics. To connect with their output, we introduce some alternative notation. Let. E be the p X P error, or residual, sum of squares and cross products matrix
Wn  r  1 (~)
is independently distributed as
The unknown value of the regression function at Zo is /3 ' ZOo So, from the discussion of the T 2 statistic in Section 5.2, we can write
T2 = (
~~:~;~~~:J' C;
1
Ir ~~:~z~~~~:J 1
(739)
(
and the 100( 1  a) % confidence ellipsoid for /3 ' Zo is provided by the inequality
E = nI that results from fitting the full model. The p X P hypothesis, or extra, sum of squares and crossproducts matrix .
(740)
H = n(II  I) The statistics can be defined in terms of E and H directly, or in terms of the nonzero eigenvalues 7JI ~ 1]2 ~ .. , ~ 1]s of HEI , where s = min (p, r  q). Equivalently, they are the roots of I (II  I)  7JI I = O. The definitions are •
n s
WIIks'lambda = PilIai's trace =
1=1
1 IEI 1 . = lE HI + 1], +
±~
i=1 1 + 1]i
= tr[H(H
+ Efl]
s
HotellingLawley trace
= 2: 7Ji
=
tr[HEI]
;=1
1]1 Roy's greatest root = 1+ 1]1 Roy's test selects the coefficient vector a so that the univariate Fstatistic based on a a ' Y. has its maximum possible value. When several of the eigenvalues 1]i are moderatel~ large, Roy's test will perform poorly relative to the other three. Simulation studies suggest that its power will be best when there is only one large eigenvalue. Charts and tables of critical values are available for Roy's test. (See [21] and [17].) Wilks' lambda, Roy's greatest root, and the HotellingLawley trace test are nearly equivalent for large sample sizes. If there is a large discrepancy in the reported Pvalues for the four tests, the eigenvalues and vectors may lead to an interpretation. In this text, we report Wilks' lambda, which is the likelihood ratio test.
where Fm,nrm( a) is the upper (100a)th percentile of an Fdistribution with m and . n  r  md.f. The 100(1  a)% simultaneous confidence intervals for E(Y;) = ZOP(!) are
~
ZOP(i) ±
I 1 (n \jl(m(nr1») n _ r  m Fm,nrm(a) \j zo(Z'Zf Zo n _ r
)
_ 1 Uii ,
i = 1,2, ... ,m
(741)
where p(;) is the ith column of jJ and Uji is the ith diagonal element of i. The second prediction problem is concerned with forecasting new responses Vo = /3 ' Zo + EO at Z00 Here EO is independent of e. Now, Vo  jJ'zo = (/3  jJ)'zo
+
EO
is distributed as
Nm(O, (1 + zb(Z'Z)lzo)I)
independently of ni, so the 100(1  a)% prediction ellipsoid for Yo becomes (Vo  jJ' zo)' (
n
nr:s;
(1
1 i)l (Yo  jJ' zo)
] + zo(Z'Z)lzO) [( m(nr1») Fm nrm( a) nrm '
(742)
The 100( 1  a) % simultaneous prediction intervals for the individual responses YOi are
~
z'oP(i) ±
I (n) \jl(m(nr1») n  r _ m Fm,nrm(a) \j (1 + zo(Z'Z)lZO) n _ r _ 1 Uii i=1,2 •... ,m
,
(743)
, 400 Chapter 7 Multivariate Linear Regression Models
The Concept of Linear Regression 40 I
where Pc;), aii, and Fm,nrm(a) are the same quantities appearing in (741). paring (741) and (743), we see that the prediction intervals for the actual values the response variables are wider than the corresponding intervals for the PYI"'~'~..l values. The extra width reflects the presence of the random error eo;·
Response 2 380
dPrediction ellipse Example 7.10 (Constructing a confidence ellipse and a prediction ellipse for responses) A second response variable was measured for the cOlmp,utt!rI'eQluirlemerit
problem discussed in Example 7.6. Measurements on the response Yz, input/output capacity, corresponding to the ZI and Z2 values in that example were
yz =
= 1.812. Thus, P(2) p(1)
zbP(l) = 151.97, and zb(Z'Zrlzo = .34725
We find that
zbP(2) = 14.14 + 2.25(130) + 5.67(7.5) = 349.17
Zo
=
[~l~~] Zo = [_zo~~~2] a' z' a 1"(2)
n
= 7,
ellipse
=
01"(2)
o
I'l.''''~_'_+
The classical linear regression model is concerned with the association between a single dependent variable Yand a collection of predictor variables ZI, Z2,"" Zr' The regression model that we have considered treats Y as a random variable whose mean depends uponjixed values of the z;'s. This mean is assumed to be a linear function of the regression coefficients f30, f3J, .. , f3r. The linear regression model also arises in a different setting. Suppose all the variables Y, ZI, Z2, ... , Zr are random and have a joint distribution, not necessarily I . Partitioning J.L normal, with mean vector J.L and covariance matrix (r+l)Xl (r+l)X(r+l) and ~ in an obvious fashion, we write
[151.97J 349.l7
zofJ(2)
(740), the set
J.L =
G::~
5.30Jl [zofJ(1)  151.97J 13.13 zbfJ(2)  349.17 $
(.34725)
Response I
1 + zb(Z'Z)I Z0 = 1.34725. Thus, the 95% prediction ellipse for Yb = [YOb YozJ is also centered at (151.97,349.17), but is larger than the confidence ellipse. Both ellipses are sketched in Figure 7.5. It is the prediction ellipse that is relevant to the determination of computer • requirements for a particular site with the given Zo.
. . for pa' Zo = [zbfJ(1)J' r = 2, and m = 2, a 95% confIdence ellIpse , IS, f rom
[zofJ(1)  151.97,zbfJ(2)  349.17](4)
confidence and prediction ellipses for the computer data with two responses.
7.8 The Concept of Linear Regression
and
P'
~onfidence
Figure 7.5 95%
h = 14.14 + 2.25z1 + 5.67zz = [14.14,2.25, 5.67J. From Example 7.6,
= [8.42,1.08, 42J,
Since
340
[301.8,396.1,328.2,307.4,362.4,369.5,229.1]
Obtain the 95% confidence ellipse for 13' Zo and the 95% prediction ellipse 'for Yb = [YOl , Yoz ] for a site with the configuration Zo = [1,130,7.5]. Computer calculations provide the fitted equation
with s
360
[C~4»)F2'3(.05)]
with F2,3(.05) = 9.55. This ellipse is centered at (151.97,349.17). Its orientation and the lengths of the m~jor and minor axes can be determined from the eigenvalues and eigenvectors of n~. Comparing (740) and (742), we see that the only change required for the calculation of the 95% prediction ellipse is to replace zb(Z'Zrlzo = .34725 with
[~r:~J
:'] [t~~l~~~' Uyy : UZy
(IXl) : (1Xr)
and
(rXl)
I
=
with UZy = [uYZ"uYZz,···,uyzJ
(744)
6
Izz can be taken to have full rank. Consider the problem of predicting Yusing the linear predictor
= bo + bt Z l + ... + brZr = bo + b'Z
(745)
6If l:zz is not of full rank, one variablefor example, Zkean be written lis a linear combination of the other Z,s and thus is redundant in forming the linear regression function Z' p_ That is, Z may be replaced by any subset of components whose n~>nsingular covariance matrix has the same rank as l:zz·
402
The Concept of Linear Regression 403
Chapter 7 Multivariate Linear Regression Models
For a given predictor of the form of (745), the error in the prediction of Y is prediction error
=Y
 bo  blZI  ...  brZr
=Y
or
 ho  b'Z
[Corr(bo
Because this error is random, it is customary to select bo and b to minimize the mean square error = E(Y  bo  b'Z)2
Now the mean square error depends on the joint distribution of Y and Z only through the parameters p. and I. It is possible to express the "optimal" linear predictor in terms of these latter quantities. Result 1.12. The linear predictor /30
/3 = Iz~uzy,
 p.z) is the linear predictor having maxi
mum correlation with Y; that is, Corr(Y,/3o + /3'Z) = ~~Corr(y,bo /3'I zz /3 /Tyy Proof. Writing bo + b'Z
E(Y  bo  b'Z)2
= =
with equality for b = l;z~uzy = p. The alternative expression for the maximum correlation follows from the equation UZyl;ZIZUZy = UZyp = uzyl:z~l;zzP = p'l;zzp· • The correlation between Yand its best linear predictor is called the population mUltiple correlation coefficient
py(Z) = +
/30 = /Ly  P'p.z
E(Y  /30  p'Z)2 = E(Y  /Ly  uZrIz~(Z  p.Z»2 = Uyy  uzyIz~uzy
= /Ly + uzyIz~(Z
Uyy
+ /3' Z with ~efficients
has minimum mean square among all linear predictors of the response Y. Its mean square error is Also, f30 + P'Z
+ b'Z,Y)f:s; uhl;z~uzy
= bo + b'Z + (/LY 
+ b'Z) uzyl;z~uzy Uyy
=
b' p.z)  (p.y  b' p.z), we get
+ (p.y  bo  b'p.z)f E(Y  /Ld + E(b' (Z  p.z) i + (p.y  bo  b' p.d
E[Y  /Ly  (b'Z  b'p.z)
(748)
The square of the population mUltiple correlation coefficient, phz), is called the population coefficient of determination. Note that, unlike other correlation coefficients, the multiple correlation coefficient is a positive square root, so 0 :s; PY(Z) :s; 1. . The population coefficient of determination has an important interpretation. From Result 7.12, the mean square error in using f30 + p'Z to forecast Yis , I Uyy  uzyl;zzuzy
= !Tyy  !Tyy (uzyl;z~uzy) = !Tyy(1  phz» !Tyy
(749)
If phz) = 0, there is no predictive power in Z. At the other extreme, phz) = 1 implies that Y can be predicted with no error. Example 7.11 (Determining the best linear predictor, its mean square error, and the multiple correlation coefficient) Given the mean vector and covariance matrix of Y, ZI,Z2,
 2E[b'(Z  p.z)(Y  p.y») = /Tyy
+ b'Izzb + (/Ly  bo 
b' p.zf  2b' UZy
Adding and subtracting uzyIz~uzy, we obtain
E(Y  bo .:.. b'zf
=
/Tyy  uzyIz~uzy + (/LY  bo  b' p.z? + (b  l;z~uzy )'l;zz(b  l;z~uzy)
The mean square error is minimized by taking b = l;z1zuzy = p, making the last term zero, and then choosing bo = /Ly  (IZ1Zuzy)' p'z = f30 to make the third term zero. The minimum mean square error is thus Uyy  Uz yl;z~uz y. Next, we note that Cov(bo + b'Z, Y) = Cov(b'Z, Y) = b'uzy so , 2_ [b'uZy)2 [Corr(bo+bZ,Y)]  /Tyy(b'Izzb)'
determine (a) the best linear predictor f30 + f3 1Z1 + f32Z2, (b) its mean square error, and (c) the multiple correlation coefficient. Also, verify that the mean square error equals !Tyy(1  phz». First,
p = f30
l;z~uzy =
= p.y
G~Jl~J
= [::
~
 p' P.z = 5  [1, 2{ ]
~:~J [~J = [~J
=3
forallbo,b so the best linear predictor is f30
Employing the extended CauchySchwartz inequality of (249) with B = l;zz, we obtain
!Tyy 
+ p'Z
uzyl;z~uzy = 10 
= 3
+ Zl  2Z2. The mean square error is
[1,1] [_::
~:~J [~J = 10 
3 = 7
404
The Concept of Linear Regression 405
Chapter 7 Multivariate Linear Regression Models
Consequently, the maximum likelihood estimator of the linear regression function is
and the multiple correlation coefficient is PY(Z)
Note that CTyy(1 
=
..?hz) =
(T' l;1 (T Zy zz Zy CTyy
10(1 
fo)
Po + P'z = y
=~  = .548 10
•
= 7 is the mean square error.
~ n  1 ,1 CTyy·Z = (Syy  SZySZZSZY)
n
1
2
1 PY(Z) = Pyy
(750)
where Pyy is the upperlefthand corner of the inverse of the correlation matrix determined from l;. The restriction to linear predictors is closely connected to the assumption of normality. Specifically, if we take
Proof. We use Result 4.11 and the invariance property of maximum likelihood estimators. [See (420).] Since, from Result 7.12, f30 = JLy  (l;Z~(TzY)'/LZ, f30
= JLy
+ (Thl;z~(z  /Lz)
= CTyy·Z = CTyy
 (Tzyl;z~(Tzy
the conclusions follow upon substitution of the maximum likelihood estimators to be d;",ibulod" N,., (p, X)
then the conditional distribution of Y with
+
+ /J'z
and mean square error
N(JLy
 Z)
and the maximum likelihood estimator of the mean square error E[ Y  f30  /J' Z f is
It is possible to show (see Exercise 7.5) that
[1:1
+ SZySz~(z
Z I, Zz, ... , Zr
fixed (see Result 4.6) is
for
(TZyl;ZIZ(Z  JLz), CTyy  (TZyl;Zlz(TZY)
The mean of this conditional distrioution is the linear predictor in Result 7.12. That is,
E(Y/z 1 , Z2,'''' Zr) = JLy + CTzyIz~(z  JLz)
(751)
= f30 + fJ'z and we conclude that E(Y /Z], Z2, ... , Zr) is the best linear predictor of Y when the population is N r + 1(/L,l;). The conditional expectation of Y in (751) is called the regression function. For normal populations, it is linear. When the population is not normal, the regression function E(Y / Zt, Zz,···, Zr) need not be of the form f30 + /J'z. Nevertheless, it can be shown (see [22]) that E(Y / Z], Z2,"" Zr), whatever its form, predicts Y with the smallest mean square error. Fortunately, this wider optimality among all estimators is possessed by the linear predictor when the population is normal. Result T.13. Suppose the joint distribution of Yand Z is Nr+1(/L, l;). Let
~ = [¥J
and
S
=
[~;Hi~~J
be the sample mean vector and sample covariance matrix, respectively, for a random sample of size n from this population. Then the maximum likelihood estimators of the coefficients in the linear predictor are
P= SZ~SZy,
Po = y
 sZrSz~Z = y 
P'Z
• It is customary to change the divisor from n to n  (r + 1) in the estimator of the mean square error, CTyy.Z = E(Y  f30  /J,zf, in order to obtain the unbiased estimator n
) (Syy ( _n__1_ nr 1 
SZySZ~SZY)
2: (If =
j=t
A.... 2 f30  /J'Zj) 1
nr
(752)
Example T.12 (Maximum likelihood estimate of the regression functionsingle response) For the computer data of Example 7.6, the n = 7 observations on Y (CPU time), ZI (orders), and Z2 (adddelete items) give the sampJe mean vector and sample covariance matrix:
#
~ [i] ~ [:~~;J
s
~ [~~I~:]~ [~!:j~:~!~~!]
406 Chapter 7 Multivariate Linear Regression Models
The Concept of Linear Regression
Assuming that Y, Zl> and Z2 are jointly normal, obtain the estimated regression function and the estimated mean square error. Result 7.13 gives the maximum likelihood estimates
zz~ZY
= [
.003128 _ .006422
Po = y  plZ = 150.44 
.006422J [41B.763J = [1.079J .086404 35.983 .420 [1.079, .420J
[13~:~:7 ]
= 150.44  142.019
. .
fio +
fi'Z =
8.42  1.0Bz1 + .42Z2
The maximum likelihood estimate of the mean square error arising from the prediction of Y with this regression function is
=
I
Sl
Szy ZZSZy
Result 7.14. Suppose Yand Z are jointly distributed as Nm+r(p,I). Then the regression of the vector Y on Z is
Po + fJz = py 
)
E(Y 
.006422J [418.763J) .086404 35.983
= .894
•
Prediction of Several Variables
IyzIz~(z  Pz)
= Iyy.z = I yy  IyzIzIZIzy
Based on a random sample of size n, the maximum likelihood estimator of the regression function is
Po + pz = Y + SyzSz~(z 
Z)
and the maximum likelihood estimator of I yy·z is
I yy.z
=
(n : 1) (Syy  SyzSZ~Szy)
Proof. The regression function and the covariance matrix for the prediction errors follow from Result 4.6. Using the relationships
(mXI)
is distributed as Nm+r(p,'l:,)
Po
(rXI)
with
+ 'l:,yzIz~z = py +
Po  fJZ) (Y  Po  fJZ)'
The extension of the previous results to the prediction of several responses Yh Y2 , ... , Ym is almost immediate. We present this extension for normal populations. Suppose
l l
'l:,yzIz~Pz
The expected squares and crossproducts matrix for the errors is
(%) (467.913  [418.763, 35.983J [ _::!~~
.~.Y
(754)
Because P and 'l:, are typically unknown, they must be estimated from a random sample in order to construct the multivariate linear predictor and determine expected prediction errors.
and the estimated regression function
Syy 
Y  py  'l:,yz'l:,z~(Z  Pz) 'l:,yy·z = E[Y  Py 'l:,yz'l:,z~(Z  pz)J [Y  /Ly 'l:,yz'l:,z~(Z  PZ)J' = 'l:,yy 'l:,yz'l:,zIz('l:,yz)' 'l:,yz'l:,z~'l:,zy + 'l:,yz'l:,z~'l:,zz'l:,z~('l:,yZ)' = 'l:,yy  'l:,yz'l:,z~'l:,zy
= 8.421
1) (
The error of prediction vector has the expected squares and crossproducts matrix
P= Sl
n ( n
407
= py  Iyz'l:,z~Pz,
fJ
=
'l:,yzIz~
Po + fJ z = py + Iyz'l:,zlz(z  Pz) I yy·z
=
I yy  IyzIz~Izy
=
'l:,yy  fJIzzfJ'
we deduce the maximum likelihood statements from the invariance property (see (420)J of maximum likelihood estimators upon substitution of By Result 4.6, the conditional expectation of [Yl> Y2, •• . , YmJ', given the fixed values Zl> Z2, ... , Zr of the predictor variables, is E(Y IZl> Zz,···, zrJ = py
+ 'l:,yzIz~(z  Pz)
(753)
'This conditional expected value, considered as a function of Zl, Zz, ... , z" is called the multivariate regression of the vector Y on Z. It is composed of m univariate regressions. For instance, the first component of the conditional mean vector is /LYl + 'l:,Y1Z'l:,Z~(Z  Pz) = E(Y11 Zl, Zz,···, Zr), which minimizes the mean square error for the prediction of Yi. The m X r matrix = 'l:,yz'l:,zlz is called the matrix of regression coefficients.
p
It can be shown that an unbiased estimator of I yy.z is n  1 ) ( n  r  1 (Syy _·SYZSZlZSZY) =
1
n
2: (Y 
n  r  1 j=l
J
. '
Po  fJz J) (YJ

. '
Po  fJz J)
I
(755)
The Concept of Linear Regression 409
408 Chapter 7 Multivariate Linear Regression Models
Example 1.13 (M aximum likelihood estimates of the regression functionstwo responses) We return to the computer data given in Examples 7.6 and 7.10. For Y1 = CPU time, Y2 = disk 110 capacity, ZI = orders, and Z2 = adddelete items, we have
'"t
+
1
and S =
'~~YL~x~J lSzy 1 Szz
467.913 1148.556/ 418.763 35. 983 = ~8.556 3072.4911 ~008.97~_~~0.~?~ 418.763 1008.9761 377.200 28.034 35.983 140.5581 28.034 13.657
The first estimated regression function, 8.42 + 1.08z1 + .42z2 , and the associated mean square error, .894, are the same as those in Example 7.12 for the singlerespons.e case. Similarly, the second estimated regression function, 14.14 + 2.25z1 + 5.67z2, IS the same as that given in Example 7.10. We see that the data enable us to predict the first response, ll, with smaller error than the second response, 1'2. The positive covariance .893 indicates that overprediction (underprediction) of CPU time tends to be accompanied by overprediction (underprediction) of disk capacity. Comment. Result 7.14 states that the assumption of a joint normal distribution for the whole collection ll, Y2, ... , Y"" ZI, Z2,"" Zr leads to the prediction equations
r
y + SyzSz~(z 
=
~Ol +
f3llZ1
+ ... +
f3rl zr
~
=
~02 +
f312Z1
+ ... +
f3r2 zr
Ym =
Assuming normality, we find that the estimated regression function is
Po + /Jz =
YI
~Om + ~lmZl + ... + ~rmZr
We note the following:
z)
1. The same values, ZI, Z2,'''' Zr are used to predict each Yj. 2. The ~ik are estimates of the (i, k )th entry of the regression coefficient matrix p = :Iyz:Iz~ for i, k ;:, 1.
150.44J [418.763 35.983J = [ 327.79 + 1008.976 140.558 X [
.003128  .006422J .006422 .086404
[ZI Z2 
130.24J 3.547
We conclude this discussion of the regression problem by introducing one further correlation coefficient.
[1.079(ZI  13014) + .420(Z2  3.547)J
150.44J
= [ 327.79 + 2.254 (ZI  13014) + 5.665 (Z2  3.547) Thus, the minimum mean square error predictor of l'! is. 150.44 + 1.079( Zl  130.24) + .420( Z2  3.547)
Partial Correlation Coefficient Consider the pair of errors
= 8.42 + 1.08z 1 + .42Z2
Y1
Similarly, the best predictor of Y2 is

1'2 
14.14 + 2.25z 1 + 5.67z2 The maximum likelihood estimate of the expected squared errors and crossproducts matrix :Iyy·z is 'given by
(n : 1) (Syy  SyzSZ~SZy)
/LY l  :IYlZ:IZ~(Z  /Lz) /LY2 
:IY2Z:IZ~(Z  /Lz)
obtained from using the best linear predictors to predict Y1 and 1'2. Their correlation, determined from the error covariance matrix :Iyy·z = :Iyy  :Iyz:Iz~:IZy, measures the association between Y1 and Y2 after eliminating the effects of ZI, Z2"",Zr'
We define the partial correlation coefficient between by
II and Y2 , eliminating ZI>
= • r. r
(756)
Z2""'Z"
= (
6) ([ 467.913 1148.536} 1148.536 3072.491
'7.
35.983J [ .003128 .006422J [418.763 _ [418.763 .086404 35.983 1008.976 140.558 .006422 = (
6) [1.043 1.042J [.894 .893J 1.042 2.572 = .893 2.205
7
PY l Y 2' Z
l008.976J) 140.558
vayly!'z
vaY2Y f Z
where aYiYk'Z is the (i, k)th entry in the matrix :Iyy·z = :Iyy  :Iyz:Izlz:IZY' The corresponding sample partial cor.relation coefficient is (757)
410 Chapter 7 Multivariate Linear Regression Models
Comparing the Tho Formulations of the Regression Model 41 I
with Sy;y.·z the (i,k)th element ofS yy  SYZSZ'zSzy.Assuming that Y and Z have a joint multivariate normal distribution, we find that the sample partial correlation coefficient in (757) is the maximum likelihood estimator of the partial correlation coefficient in (756).
with f3. = f30 + f311.1 + ... + f3rzr. The mean corrected design matrix corresponding to the reparameterization in (759) is
z<{
Example 7.14 (Calculating a partial correlation) From the computer data Example 7.13, 1
_
Syy  SyzSzzSZy 
Zll Z21 
Zl ZI ... ZZr  Zr '"
"'"J
Znl  Zl
Znr  zr
where the last r columns are each perpendicular to the first column, since
[1.043 1.042J 1.042 2.572
n
2: 1(Zji j=l
Therefore,
z;) = 0,
i = 1,2, ... ,r
Further, setting Zc = [1/ Zd with Z~21 = 0, we obtain
Calculating the ordinary correlation coefficient, we obtain rYl Y 2 = .96. Comparing the two correlation coefficients, we see that the association between Y1 and Y2 has been sharply reduced after eliminating the effects of the variables Z on both responses.
•
z'z c
= [ 1'1 c
Z~zl
l'ZczJ =
Z~ZZc2
[n0
0'
Z~zZcz
]
so
7.9 Comparing the Two Formulations of the Regression Model In Sections 7.2 and 7.7, we presented the multiple regression models for one and several response variables, respectively. In these treatments, the predictor variables had fixed values Zj at the jth trial. Alternatively, we can startas in Section 7.8with a set of variables that have a joint normal distribution. The process of conditioning on one subset of variables in order to predict values of the other set leads to a conditional expectation that is a multiple regression model. The two approaches to multiple regression are related. To show this relationship explicitly, we introduce two minor variants of the regression model formulation.
Mean Corrected Form of the Regression Model For any response variable Y, the multiple regression model asserts that
(760)
That is, t.!I e regression coefficients [f3h f3z, ... , f3r J' are unbiasedly estimated by (Z~zZcz) ;.1Z~zY and f3. is estimated by y. Because the definitions f31> f3z, ..• , f3r remain unchanged by the reparameterization in (759), their best estimates computed from the design matrix Zc are exactly the same as the best estimates computed from the design matrix Z. Thus, setting p~ = [Ph PZ, ... , Pr J, the linear predictor of Y can be written as (761)
The predictor variables can be "centered" by subtracting their means. For instance, f31Z1j = f31(Z'j  1.,) + f3,1.1 and we can write
lj
=
(f3o + f3,1., + .. , + f3r1. r) + f3'(Z'j ., 1.,) + ... + f3r(Zrj  1.r) + Sj
= f3.
+ f3,(z'j
 1.,)
+ ... + f3r(Zrj
 1.r)
+ Sj
with(z  z) = [Zl  1.bZZ  zz"",Zr  zr]'.Finally, Var(P.) [ Cov(Pc,
P.)
(762)
Multiple Regression Models with Time Dependent Errors 413
412 Chapter 7 Multivariate Linear Regression Models the same mean Commen.t The multivariate multiple regression model yields . f h corrected design matrix for each response. The least squares estImates 0 t e coeffi· cient vectors for the ith response are given by
P A
(i)
Y{i) ] = (Z~2ZC2rlZ~2 Y{iJ
i
[
= 1,2, ... ,m
'
Sometimes, for even further numerical stability, "standardized" input variables
~ (ZI'.' _ Z.)2 (Zji _ )/ Zi VI £.i ,
= (z.· .
I"
z·)/'V(n  J)sz.z·
are used. In this case, the
I I
slope coefficie~~~ f3i in the regression model are ~placed by ~i =
~i Y(n 
Although the two formulations of the linear prediction problem yield the same predictor equations, conceptually they are quite different. For the model in (73) or (723), the values of the input variables are assumed to be set by the experimenter. In the conditional mean model of (751) or (753), the values of the predictor variables are random variables that are observed along with the values of the response variable(s). The assumptions underlying the second approach are more stringent, but they yield an optimal predictor among all choices, rather than merely among linear predictors. We close by noting that the multivariate regression calculations in either case can be couched in terms of the sample mean vectors y and z and the sample sums of squares and crossproducts:
1) SZiZ;,
The least squares estimates ofthe beta coefficients' f3; beco~e 11; = /3.; Y~ n  1) ~Z;Zi' i = 1,2, ... , r. These relationships hold for each response In the multIvanate mUltIple regression situation as well.
Relating the Formulations . bl es Y ,), Z Z2,"" Zr areJ'ointlynormal, the estimated predictor of Y Wh en th evana (see Result 7.13) is ~o + jrz = y + SZySz~(z  z) = [Ly + uh:Iz~(z  p;z) (764) A
z/s.
where the estimation procedure leads naturally to the i~troduction of centered Recall from the mean corrected form of the regreSSIOn model that the best lm· ear predictor of Y [see (761)] is y = ~. + ~~(z  z) ·th {3A • = y and Pc a'  Y'z c2 (Z'c2 Z c2 )1 . Comparing (761) and (764), we see that 7 {3. = y = {3o and Pc = P smce sZrSz~ = Y'ZdZ~2Zdl (765)
WI A
_
A
,
' .
Therefore, both the normal theory conditional me~n and the classical regression model approaches yield exactly the same linear predIctors. . A similar argument indicates that the best linear predictors of the responses m the two multivariate multiple regression setups are also exactly the same. Example 7./5 (Two approaches yield the same Iin~r predictor) The ~mputer d~ta ~th . I e V  CPU tinIe were analyzed m ExanIple 7.6 USIng the classlcallin
the smg e respons 'I . . 12' ear regression model. The same data were analyzed agam In Example 7.. ' assuIIUD? . bl es Y1> Z I, and Z2 were J' oindy normal so that the best predIctor edict of Y1 IS tha t the vana . the conditional mean of Yi given ZI and Z2' Both approaches YIelded the same pr or,
y=
8.42
+ l.08z1 + .42Z2
•
+ jil so that + jil'Zc2 = (y  jil)'Zc2 + 0' = (y  jil)'Zc2
7The identify in (7·65) is established by writing y = (y  jil) y'Zc2
= (y 
jil)'Zc2
Consequently,
yZc2(Z~2Zd' = (y 
jil)'ZdZ;2Zd'
= (n 
zzr' = SZySZ'Z
l)s'zy[(n  l) S
This is the only information necessary to compute the estimated regression coefficients and their estimated covariances. Of course, an important part of regression analysis is model checking. This requires the residuals (errors), which must be calculated using all the original data.
7.10 Multiple Regression Models with Time Dependent Errors For data collected over time, observations in different time periods are often related, or autocorrelated. Consequently, in a regression context, the observations on the dependent variable or, equivalently, the errors, cannot be independent. As indicated in our discussion of dependence in Section 5.8, time dependence in the observations can invalidate inferences made using the usual independence assumption. Similarly, inferences in regression can be misleading when regression models are fit to time ordered data and the standard regression assumptions are used. This issue is important so, in the example that follows, we not only show how to detect the presence of time dependence, but also how to incorporate this dependence into the multiple regression model. Example 7.16 (Incorporating time dependent errors in a regression model) power companies must have enough natural gas to heat all of their customers' homes and businesses, particularly during the cold est days of the year. A major component of the planning process is a forecasting exercise based on a model relating the sendouts of natural gas to factors, like temperature, that clearly have some relationship to the amount of gas consumed. More gas is required on cold days. Rather than use the daily average temperature, it is customary to nse degree heating days
Multiple Regression Models with Time Dependent Errors 417
416 Chapter 7 Multivariate Linear Regression Models
When modeling relationships using time ordered data, regression models with noise structures that allow for the time dependence are often useful. Modern software packages, like SAS, allow the analyst to easily fit these expanded models.
PANEL 7.3
Lag 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
SAS ANALYSIS FOR EXAMPLE 7.16 USING PROC ARIMA
data a; infile 'T7 4.d at'; time =_n...; input obsend dhd dhdlag wind xweekend; proc arima data = a; identify var = obsend crosscor dhd dhdlag wind xweekend ); estimate p = (1 7) method = ml input = ( dhd dhdlag wind xweekend ) plot; estimate p = (1 7) noconstant method = ml input = ( dhd dhdlag wind xweekend ) plot;
PROGRAM COMMANDS
=(
OUTPUT
Maximum Likelihood Estimation
EstimatEl! 2.12957 . 0.4700/,1 0.23986 5.80976 1.42632 1.20740 10.10890
Constant Estimate
0.61770069
I
228.89402.8\
Std Error Estimate AIC SBC Number of Residuals
15.1292441 528.490321 543.492264 63
Variance Estimate
=
Approx. Std Error 13.12340 0.11779 0.11528 0.24047 0.24932 0.44681 6.03445
Lag 0
T Ratio 0.16 3.99 2.08 24.16 5.72 2.70 1.68
7 0 0 0 0
Variable OBSENO OBSENO OBSEND DHO OHDLAG WIND XWEEKEND
Shift 0 0 0 0 0 0 0
0.127 0.056 0.079 0.069
0.161 0.108 0.018 0.051
Autocorrelation Check of Residuals To Lag 6 12 18 24
Chi Square 6.04 10.27 15.92 23.44
Autocorrelations OF 4 10 16 22
Probe
0:1:961 0;4#"
~~1t1~,
0.079 0.144 0.013 0.018
Covariance 228.894 18.194945 2.763255 5.038727 44.059835 29.118892 36.904291 33.008858 15.424015 25.379057 12.890888 12.777280 24.825623 2.970197 24.150168 31.407314
Correlation 1.00000 0.07949 0.01207 0.02201 0.19249 0.12722 0.16123 0.14421 0.06738 0.11088 0.05632 0.05582 0.10846 0.01298 0.10551 0.13721
1
I I
I I I I I I
I I I I I I I I
9 8 7 6 543 2
o1
0.012 0.067 0.106 0.004
0.022 0.111 0.137 0.250
0.192 0.056 0.170 0.080
234 5 6 7 891
1*******************1
1** I I 1**** .
*** I 1*** 1*** *1 **1 *1 *1 **1 I 1** . *** I
" ." marks two standard errors
ARIMA Procedure
Parameter MU AR1,l AR1,2 NUMl NUM2 NUM3 NUM4
Autocorrelation Plot of Residuals
I I I I I I I I I I I I I I I
The Distribution of the Likelihood Ratio for the Multivariate Multiple Regression Model 419
Supplement
and the eigenvalues of Zl(ZlZd1Z; are 0 or 1. Moreover, tr(Zl(Z;Zlr1Z l) 1 ) == q + 1 = Al + A2 + ... + A +1> where (q+I)X(q+l) q Al :2! A2 :2! '" :2! Aq+1 > 0 are the eigenvalues of Zj (ZiZlr1Zi. This shows that Zj(ZlZjrlZl has q + 1 eigenvalues equal to 1. Now, (Zj(ZiZlrIZi)ZI == Zt> so any linear combination Zlb c of unit length is an eigenvector corresponding to the eigenvalue 1. The orthonormal vectors gc, e = 1,2, ... , q + 1, are therefore eigenvectors of ZI(ZiZlrIZl, since they are formed by taking particular linear combinations of the c~~lmns of Zl' By the spectral decomposition (216), we have
= tr«ZiZlrIZiZI) =
Zl(ZiZlflZi =
THE DISTRIBUTION OF THE LIKELIHOOD RATIO FOR THE MULTIVARIATE MULTIPLE REGRESSION MODEL
tr(
2: gcge. Similarly, by writing (Z (Z' ZrIZ') Z =
Z, we readily see
C=l
that the linear combination Zb c == gc, for example, is an eigenvector of Z (Z'Z flZ' r+l
2: gcge.
with eigenvalue A = 1, so that Z (Z'Zr1Z' ==
C=1
Continuing; we have PZ == [I  Z(Z'ZrIZ')Z = Z  Z == 0 so gc = Zb c, r + 1, are eigenvectors of P with eigenvalues A = O. Also, from the way the ge, r + 1, were constructed, Z'gc = 0, so that Pg e = gc. Consequently, these gc's are eigenvectors of P corresponding to the n  r  1 unit eigenvalues. By the spec
es e>
n
The development in this supplement establishes Result 7.1l. We know that nI == Y'(I  Z(Z'ZfIZ')Y and under Ho, nil == Y'[I  Zl(ZiZlr1zUY with Y == zd3(1) + e. Set P == [I  Z(Z'Zf1Z'). Since 0 = [I  Z(Z'ZfIZ')Z = [I  Z(Z'ZrIZ'j[ZI i Zz) = [PZ I i PZ 2) the columns of Z are perpendicular to P. Thus, we can write nI
= (z/3 + e),P(Z/3 + e) = e'pe
nil = (ZI/3(i)
+ e)'PI(Zd3(J) + e)
=
gl,gZ, ... ,gq+l> gq+Z,gq+3,···,gr+I' gr+Z,gr+3,···,gn
~'
r
Let (A, e) be an eigenvalueeigenvector pair of Zl(ZiZd1Zl' Then, since [Zl(ZlZdlZ1J[Zl(ZlZdlZll == ZI(Z;ZdIZl, it follows that 2 Ae = Zl(Zi Z lf1Z;e = (ZI(ZlZlrIZl/e == A(ZI(ZlZdIZDe == A e 418
:±
(E'gc)(E'gc)' =
l=r+2
.
:±
VcVe
C=r+2
where, because Cov(Vei , ljk) = E(geE(i)l'(k)gj) = O"ikgegj = 0, e oF j, the e'ge = Vc = [VC1,"" VCi ,";" VcmJ' are independently distributed as Nm(O, I). Consequently, by (422), nI is distributed as Wp,nrl(I). In the same manner,
P
_ 19C 
{gC e> q + 1 0
es
q + 1
n
so PI =
2:
ge gc· We can write the extra sum of squares and cross products as
(;q+2
"
n(I 1
,... 
I) = E'(P1
r+1 
P)E =
2:
r+l
(E'ge) (E'ge)' ==
f=q+2
J~
from columns from columns of Zz arbitrary set of of ZI but perpendicular orthonormal to columns of Z I vectors orthogonal to columns of Z
L
nI = E'PE =
E'PIE
where PI = 1  ZI(ZiZlfIZj. We then use the GramSchmidt process (see Result 2A.3) to construct the orthonormal vectors (gl' gz,···, gq+l) == G from the columns of ZI' Then we continue, obtaining the orthonormal set·from [G, Z2l, and finally complete the set to n dimensions by constructing an arbitrary orthonormal set of n  r  1 vectors orthogonal to the previous vectors. Consequently, we have
2: gegc and (=r+2
tral decomposition (216),P =
2:
VeVc
e=q+2
where the Ve are independently distributed as Nm(O, I). By (422), n(I 1  i) is since n(I 1  i) involves a different distributed as Wp,r_q(I) independently of set of independent Vc's.
ni,
The large sample distribution for [ n  r  1  ~ (m  r
+ q + 1) ]In (/i II/I 1 /) follows from Result 5.2, with P  Po = m(m + 1)/2 + mer + 1)  m(m + 1)/2 m(q + 1) = mer  q) dJ. The use of (n  r  1  ~(m  r + q + 1) instead of n in the statistic is due to Bartlett [4J following Box [7J, and it improves the chisquare approximation.
420
Chapter 7 Multivariate Linear Regression Models
Exercises 421
Exercises 7.1.
1.6.
Given the data ZI
I
5
10
19
7
11
8
9325713
;=1
is a generalized inverse of Z'Z.
fit the linear regression model lj =)3 0 + f3IZjl + Bj, j = 1,2, ... ,6. Specifically, calculate the least squares estimates /3, the fitted values y, the residuals E, and the . residual sum of squares, E' E. .
7.2.
P
(b) The coefficients that minimize the sum of squared errors (y  ZP)'(y  ZP) satisfy ~e normal equ~tions (Z'Z)P = Z'y. Show that these equations are satisfied for any P such that ZP is the projection of y on the columns of Z. (c) Show that ZP = Z(Z'Z)Z'y is the projection ofy on the columns of Z. (See Footnote 2 in this chapter.)
Given the data 10 2
5 3
7 3
19 6
11 7
18
Z2
y
15
9
3
25
7
13
ZI
9
P
(d) Show directly that = (Z'ZrZ'y is a solution to the normal equations (Z'Z)[(Z'Z)Z'y) = Z'y.
ZP
= {3IZjl
+ {32Zj2 + ej'
j = 1,2, ... ,6.
to the standardized form (see page 412) of the variables y, ZI, and Z2' From this fit,deduce the corresponding fitted regression equation for the original (not standardized) variables. 7.3.
ZP
Hint: (b) If is the projection, then y is perpendicular to the columns of Z. (d) The eigenvalueeigenvector requirement implies that (Z'Z)(Ai1ej) = e;for i ~ rl + 1 and 0 = ei(Z'Z)ej for i > rl + 1. Therefore, (Z'Z) (Ai1ej)eiZ'= ejeiZ'. Summing over i gives
fit the regression model Yj
(Generalized inverse of Z'Z) A matrix (Z'Zr is caJled a generalized inverse of Z'Z if ':' z'z. Let rl + 1 = rank(Z) and suppose Al ;:" A2 ;:" ... ;:" Aq + 1 > 0 are the nonzero elgenvalues of Z'Z with corresponding eigenvectors el, e2,"" e'I+I' (a) Show that ',+1 = ~ "I:' A:Ie.e~ ( Z'Z)./ I I I
z'z (Z'Z)Z'Z
(Z'Z)(Z'Z)Z'
',+1
)
~ Aileiei Z'
(Weighted least squares estimators.) Let
y
(nXI)
=
Z
/3
(/lX('+I)) ((,+1)XI)
+
=
E
(nXI)
7.7.
y (nXI)
If (T2 is unknown, it may be estimated, unbiasedly, by
Ilzzl (O'yy  uzylz~uzy) III Ilzz I Uyy IIzzluyy yy From Result 2A.8(c),u YY = IIzz IIII I, where O' is theentry.ofl I in the first row and first column. Since (see Exercise 2.23) p = V I/2l VI/2 and pI = (V I/ 2I V I/ 2fl = VI/2IIVI/2, the entry in the (1,1) position of pI is Pyy = O' yy (Tyy. =
(Tyy  Uzylz~uzy O'yy
=
ZI P(1) (/lX(q+I)) ((q+I)XI)
+
Z'
~
(nX(,q))
=r+
1, written as
P(2)
+ e
((rq)xJl
1
(P(2)  P(2))' [ZZZ2  ZzZI(Zj Z lr Zj Z 2] (P(2)  P(2)
(nXI)
~ ~2(r 
q)F,q,/lrl(a)
Hint: By ExerCise 4.12, with 1 's and 2's interchanged,
C
22
= [ZZZ2 
l Z zZI(ZjZIlI Z ;Z2r ,
where (Z'Z)I
=
[~~: ~:~J
Multiply by the squareroot matrix (C 22 rI/2, and conclude that (C 22 )If2(P(2)  P(2)1(T2 is N(O, I), so that l (P(2)  p(2)),( C22 (p(2)  P(2) iS~~_q.
Establish (750): phz) = 1  I/pYY. Hint: From (749) and Exercise 4.11 2
=
=
:=
Use the weighted least squares estimator in Exercise 7.3 to derive an expression for the estimate of the slope f3 in the model lj = f3Zj + ej' j = 1,2, ... ,n, when (a) Var (Ej) = (T2, (b) Var(e) = O' 2Zj, and (c) Var(ej) = O'2z;' Comment on tQe manner in which the unequal variances for the errors influence the optimal choice of f3 w·
PY(Z)
) eie; Z' = IZ'
1=1
where rank(ZI) q + 1. and r~nk(Z2) = r  q. If the parameters P(2) are identified beforehand as bemg ofpnmary mterest,show that a 100(1  a)% confidence region for P(2) is given by
ZPw).
Hint: V I/ 2y = (V I/ 2Z)/3 + V I/2e is of the classical linear regression form y* = " I Z*p + e*,withE(e*) = OandE(e*E*') =O' 2I.Thus,/3w = /3* = (Z*Z*) Z*'Y*.
. 1 
(r+1
= ~
since e;Z' = 0 for i > rl + 1. Suppose the classical regression model is, with rank (Z)
Pw = (Z'VIZrIZ'VIy (n  r  lr l x (y  ZPw),VI(y 
rl+l) ( ~ eiej Z' l=l
where E ( e) = 0 but E ( EE') = 0'2 V, with V (n X n) known and positive definite. For V of full rank, show that the weighted least squares estimator is
7.S.
= Z'Z (
r
7.S.
Recall that the hat matrix is defined by H = Z (Z'Z)_I Z ' with diagonal elements h jj • (a) Show that H is an idempotent matrix. [See Result 7.1 and (76).) (b) Show that 0 < h jj < 1, j
=
n
1,2, ... , n, and that
2: h jj = j=1
r + 1, where r is the
number of independent variables in the regression model. (In fact, (lln) ~ h jj < 1.)
Exercises 423
422 Chapter 7 Multivariate Linear Regression Models
(c) Verify, for the simple linear regression model with one independent variable the leverage, hji' is given by
z, that
7.13. The test scores for college students described in Example 5.5 have
Z
7.9.
Consider the following data on one predictor variable ZI and two responses Y1 and Y2:
"12 YI
Y2
5 3
1 3 1
0 4 1
2
·1 2 2
1 3
Determine the least squares estimates of the parameters in the bivariate straightline regression model ljl = {301
+ {3llZjl + Bjl
lj2 = {302 + {312Zjl +
Bj2'
j
Y
=
[
~2
Z3
=
[527.74] 54.69, 25.13
i
with
Y
=
[YI
i Y2)'
Y'Y + i'i
7.11. (Generalized least squares for multivariate multiple regression.) Let A be a positive defmite matrix, so that d7(B) = (Yj  B'zj)'A(Yj  B'zj) is a squared statistical choice distance from the jth observation Yj to its regression B'zj' Show that the n
jJ = (Z'Zr1z'Y minimizes the sum of squared statistical distances, ~ d7(B), , )=1
for any choice of positive definite A. Choices for A i.nc1u~~ II and I. Jl,int: Repeat the steps in the proof of Result 7.10 With I replaced by A. 7.12. Given the mean vector and covariance matrix of Y, ZI, and Z2,
determine each of the following. (a) The best linear predictor Po + {3I Z 1 + {32Zz of Y (b) The mean square error of the best linear predictor (c) The population multiple correlation coefficient (d) The partial correlation coefficient PYZ(Z,
S ;,
569134 600.51 [ 217.25
] 126.05 2337 23.11
Assume joint normality. (a) Obtain the maximum likelihood estimates of the parameters for predicting ZI from Z2 andZ3 • (b) Evaluate the estimated multiple correlation coefficient RZ,(Z2,Z,), (c) Determine the estimated partial correlation coefficient R Z "Z2' Z" 7.14. 1Wentyfive portfolio managers were evaluated in terms of their performance. Suppose Y represents the rate of return achieved over a period of time, ZI is the manager's attitude toward risk measured on a fivepoint scale from "very conservative" to "very risky," and Z2 is years of experience in the investment business. The observed correlation coefficients between pairs of variables are
Y
7.10. Using the results from Exercise 7.9, calculate each of the following. (a) A 95% confidence interval for the mean response E(Yo1 ) = {301 + {311Z01 corresponding to ZOI = 0.5 (b) A 95 % prediction interval for the response Yo 1 corresponding to Zo 1 = 0.5 Cc) A 95% prediction region for the responses Y01 and Y02 corresponding to ZOI = 0.5
B =
ZI]
= 1,2,3,4,5
Also calculate the matrices of fitted values and residuals Verify the sum of squares and crossproducts decomposition
Y'y
=
R =
['0 .35 .82
ZI 35 1.0.60
Z2
B2]
.60 1.0
(a) Interpret the sample correlation coefficients ryZ,
= .35 and rYZ2 = .82.
(b) Calculate the partial correlation coefficient rYZ!'Z2 and interpret this quantity with respect to the interpretation provided for ryZ, in Part a. The following exercises may require the use of a computer. 7.1 S. Use the realestate data in Table 7.1 and the linear regression model in Example 7 A. (a) Verify the results in Example 704. (b) AnaJyze the residuals to check the adequacy of the model. (See Section 7.6.) (c) Generate a 95% prediction interval for the selling price (Yo) corresponding to total dwelling size ZI = 17 and assessed value Z2 = 46. (d) Carry out a likelihood ratio test of Ho: {32 = 0 with a significance level of a = .05. Should the original model be modified? Discuss. 7.16. Calculate a Cp plot corresponding to the possible linear regressions involving the realestate data in Table 7.1. 7.17. Consider the Forbes data in Exercise 1.4. (a) Fit·a linear regression model to these data using profits as the dependent variable and sales and assets as the independent variables. (b) Analyze the residuals to check the adequacy of the model. Compute the leverages associated with the data points. Does one (or more) of these companies stand out as an outlier in the set of independent variable data points? (c) Generate a 95 % prediction interval for profits corresponding to sales of 100 (billions of dollars) and assets of 500 (billions of dollars). (d) Carry out a likelihood ratio test of Ho: {32 = 0 with a significance level of a = .05. Should the original model be modified? Discuss. .
Exercises 425
424 Chapter 7 Multivariate Linear Regression Models 7.18. Calculate (a) a C plot corresponding to the possible regressions involving the Forbes data p Exercise 1.4. (b) the AIC for each possible regression. 7.19. Satellite applications motivated the development of a silverzinc battery. Tab~e ~.5 contains failure data collected to characterize the performance of the battery dunng Its , life cycle. Use these d a t a . ' (a) Find the estimated linear regression of In (Y) on an appropriate ("best") subset of predictor variables. ' (b) Plot the residuals from the fitted model chosen in Part a to check the assumption.
Data Zt
Charge rate (amps) .375 1.000 1.000 1.000 1.625 1.625 1.625 .375 1.000 1.000 1.000 1.625 .375 1.000 1.000 1.000 1.625 1.625 .375 .375
Discharge rate (amps) 3.13 3.13 3.13 3.13 3.13 3.13 3.13 5.00 5.00 5.00 5.00 5.00 1.25 1.25 1.25 1.25 1.25 1.25 3.13 3.13
Z3
Z4
Depth of discharge (% ofrated amperehours)
Temperature
60.0 76.8 60.0 60.0 43.2 60.0 60.0 76.8 43.2 43.2 100.0 76.8 76.8 43.2 76.8 60.0 43.2 60.0 76.8 60.0
(QC)
40 30 20 20 10 20 20 10 10
30 20 10 10 10 30 0 30 20 30 20
Y
Zs End of charge voltage (volts)
Cycles to failure
2.00 1.99 2.00 1.98 2.01 2.00 2.02 2.01 1.99 2.01 2.00 1.99 2.01 1.99 2.00 2.00 1.99 2.00 1.99 2.00
101 141 96 125 43 16 188 10 3 386 45 2 76 78 160 3 216 73 314 170
S Sidik ,H. Leibecki "and J Bozek , Failure of Si/verZinc Cells with Competing Source' Se Iecte d from, Le . R Failure ModesPreliminary Dala Analysis, NASA Technical Memorandum 81556 (Cleveland:
WIS
h
esearc
Center, 1980),
7.20. Using the batteryfailure data in Table 7.5, regress In~Y) on the first princi~~s~ftm~ nent of the predictor variables Zb Z2,"" Zs· (See SectIOn 8.3.) Compare the the fitted model obtained in Exercise 7.19(a).
7.21. Consider the airpollution data in Table 1.5. Let Yi = N02 and Y2 = 03 be the two responses (pollutants) corresponding to the predictor variables Zt = wind and Z2 = solar radiation. (a) Perform a regression analysis using only the first response Yi, (i) Suggest and fit appropriate linear regression models. (ii) Analyze the residuals. (iii) Construct a 95% prediction interval for N02 corresponding to Zj = 10 and Z2 = 80. (b) Perform a muItivariate mUltiple regression analysis using both responses Yj and 12· (i) Suggest and fit appropriate linear regression models. (H) Analyze the residuals. (Hi) Construct a 95% prediction ellipse for both N02 and 0 3 for Zt = 10 and Z2 = 80. Compare this ellipse with the prediction interval in Part a (iii). Comment. 7.22. Using the data on bone mineral content in Table 1.8: (a) Perform a regression analysis by fitting the response for the dominant radius bone to the measurements on the last four bones. (i) Suggest and fit appropriate linear regression models. (ii) Analyze the residuals. (b) Perform a multivariate multiple regression analysis by fitting the responses from both radius bones. (c) Calculate the AIC for the model you chose in (b) and for the full model. 7.23. Using the data on the characteristics of bulls sold at auction in Table 1.10: (a) Perform a regression analysis using the response Yi = SalePr and the predictor variables Breed, YrHgt, FtFrBody, PrctFFB, Frame, BkFat, SaleHt, and SaleWt. (i) Determine the "best" regression equation by retaining only those predictor variables that are individually significant. (ii) Using the best fitting model, construct a 95% prediction interval for selling price for the set of predictor variable values (in the order listed above) 5,48.7, 990,74.0,7, .18,54.2 and 1450. (Hi) Examine the residuals from the best fitting model. (b) Repeat the analysis in Part a, using the natural logarithm of the sales price as the response. That is, set Yj = Ln (SalePr). Which analysis do you prefer? Why? 7.24. Using the data on the characteristics of bulls sold at auction in Table 1.10: (a) Perform a regression analysis, using only the response Yi = SaleHt and the predictor variables Zt = YrHgt and Zz = FtFrBody. (i) Fit an appropriate model and analyze the residuals. (ii) Construct a 95% prediction interval for SaleHt corresponding to Zj = 50.5 and Z2 = 970. (b) Perform a multivariate regression analysis with the responses Y j = SaleHt and Y2 = SaleWt and the predictors Zj = YrHgt and Z2 = FtFrBody. (i) Fit an appropriate multivariate model and analyze the residuals. (ii) Construct a 95% prediction ellipse for both SaleHt and SaleWt for Zl = 50.5 and Z2 = 970. Compare this eilipse with the prediction interval in Part a (H). Comment.
Exercises 42,7 426
Chapter 7 Multivariate Linear Regression Models (c) Perform a multivariate multiple regression analysis using both responses Yi and yz. (i) Suggest and fit appropriate linear regression models. (ii) Analyze the residuals. (iii) Construct a 95% prediction ellipse for both Total TCAD and Amount of amitriptyline for Zl = 1, Z2 = 1200, Z3 = 140, Z4 = 70, and Z5 = 85. Compare this ellipse with the prediction intervals in Parts a and b. Comment.
.. .' 'bed b some physicians as an antidepressant. However, there 7.25. Amltnptyh~e IS prdes~rdl ff Yts that seem to be related to ttie use of the drug: irregular hit d' are also conjecture SI e e ec I bl d ssures, and irregular waves on tee ec rocar wgram, heartbeat, abno~ma D ~o P~ered on 17 patients who were admitted to the hospital among other t~mg~. a a ga . ' Table 7.6. The two response variables after an amitrIptyhne overdose are given ID are Y I = Total TCAD plasma le~el (TOT) yz = Amount of amitriptyline present in TCAD plasma level (AMI)
7.26. Measurements of properties of pulp fibers and the paper made from them are contained in Table 7.7 (see also [19] and website: www.prenhall.com/statistics). There are n = 62 observations of the pulp fiber characteristics, Zl = arithmetic fiber length, Z2 = long fiber fraction, Z3 = fine fiber fraction, Z4 = zero span tensile, and the paper properties, Yl = breaking length, Y2 = elastic modulus, Y3 = stress at failure, Y4 = burst strength.
The five predictor variables are ZI = Gender: liffemale,Oifmale (GEN)
Z2 = Amount of antidepressants taken at time of overdose (AMT)
Table 7.7 Pulp and Paper Properites Data
Z3 = PR wave measurement (PR)
Y1
Z4 = Diastolic blood pressure (DIAP) Z5 = QRS wave measurement (QRS)
Table 7.6 Amitriptyline Data Yl TOT
Y2 AMI
3389 1101 1131 596 896 1767 807 1111 645 628 1360 652 860 500 781 1070 1754
3149 653 810 448 844 1450 493 941 547 392 1283 458 722 384 501 405 1520
Zl
Z2
Z3
GEN
AMT
PR
1 1 0 1 1 1 1 0 1 1 1 1 1 0 0 0 1
7500 1975 3600 675 750 2500 350 1500 375 1050 3000 450 1750 2000 4500 1500 3000
220 200 205 160 185 180 154 200 137 167 180 160 135 160 180 170 180
Z4
Z5
DIAP
QRS 140 100 111 120 83 80 98 93 105 74 80 60 79 80 100 120 129
0 0 60 60 70 60 80 70 60 60 60 64 90 60 0 90 0
Y3 SF
Y4 BS
Zl
Z2
Z3
Z4
AFL
LFF
FFF
ZST
21.312 21.206 20.709 19.542 20.449
7.039 6.979 6.779 6.601 6.795
5.326 5.237 5.060 4.479 4.912
.932 .871 .742 .513 577
.030 .015 .025 .030 .Q70
35.239 35.713 39.220 39.756 32.991
36.991 36.851 30.586 21.072 36570
1.057 1.064 1.053 1.050 1.049
16.441 16.294 20.289 17.163 20.289
6.315 6.572 7.719 7.086 7.437
2.997 3.017 4.866 3.396 4.859
.400 .478 .239 .236 .470
.605 .694 .559 .415 .324
84554 81.988 8.786 5.855 28.934
1.008 .998 1.081 1.033 1.070
:
2.845 1.515 2.054 3.018 17.639
:
(a) Perform a regression analysis using each of the response variables Y1, yz, 13 and Y4 • (i) Suggest and fit appropriate linear regression models. (ii) Analyze the residuals. Check for outliers or observations with high leverage. (iii) Construct a 95% prediction interval for SF (13) for Zl = .330, Z2 = 45.500, . Zl = 20.375, Z4 = 1.010. (b) Perform a muItivariate multiple regression analysis using all four response variables, Y1 , Yz, 13 and Y4 ,and the four independent variables, Zl, ZZ,Z3 and Z4' (i) Suggest and fit an appropriate linear regression model. Specify the matrix of estimated coefficients /J and estimated error covariance matrix (ii) Analyze the residuals. Check for outliers. (iii) Construct simultaneous 95% prediction intervals for the individual responses Yoi,i = 1,2, 3,4,for the same settings of the independent variables given in part a (iii) above. Compare the simultaneous prediction interval for Y03 with the prediction interval in part a (iii). Comment.
i.
(a) Perform a regression analysis using only the fi~st response Y1 • (i) Suggest and fit appropriate linear regressIOn models. (ii) Analyze the residuals. (iii) Construct a 95% prediction interval for Total TCAD for Z3 = 140, Z4 = 70, and Z5 = 85. • (b) Repeat Part a using the second response Yz.
:
Source: See Lee [19].
Source: See [24].
'
Y2 EM
BL
_ Zl 
=
1,
Z2
1200 '
7.27. Refer to the data on fixing breakdowns in cell phone relay towers in Table 6.20. In the initial design, experience level was coded as Novice or Guru. Now consider three levels of experience: Novice, Guru and Experienced. Some additional runs for an experienced engineer are given below. Also, in the original data set, reclassify Guru in run 3 as
428 Chapter 7 Multivariate Linear Regression Models
References 429
Experienced and Novice in run 14 as Experienced. Keep all the other numbers for these two engineers the same. With these changes and the new data below, perform a multivariate multiple regression analysis with assessment and implementation times as the responses, and problem severity, problem complexity and experience level as the predictor variables. Consider regression models with the predictor variables and two factor interaction terms as inputs. (Note: The two changes in the original data set and the additional. data below unbalances the design, so the analysis is best handled with regression· methods.) Problem severity level
Problem complexity level
Engineer experience level
Problem· assessment time
Problem implementation time
Total resolution time
Low Low High High High
Complex Complex Simple Simple Complex
Experienced Experienced Experienced Experienced Experienced
5.3 5.0 4.0 4:5 6.9
9.2 10.9 8.6 8.7 14.9
14.5 15.9 12.6 13.2 21.8
13. D~aper, N. R., and H. Smith. Applied Regression Analysis (3rd ed.). New York' John WIley, 1998. . 14. Durbi.n, 1., a~d G. S. Watson. "Testing for Serial Correlation in Least Squares Regression H." BLOmetnka, 38 (1951), 159178. ' 15. Galto~, F. "R~gression Toward Mediocrity in Heredity Stature." Journal of the AnthropologlcalInstltute, 15 (1885),246263: 16. Goldberger,A. S. Econometric Theory. New York: John Wiley, 1964. 17. Heck, D. ~. ':Charts ?,f Some Upper Percentage Points of the Distribution of the Largest Charactenstlc Root. Annals of Mathematical Statistics, 31 (1960), 625":'642. 18. Khattree, R. and D. .N. Naik. Applied Multivariate Statistics with SAS® Software (2nd ed.) Cary, Ne: SAS Institute Inc., 1999. 19. Lee, 1. "R.elati0!lshil?s Between Properties of PulpFibre and Paper." Unpublished doctoral theSIS, Umverslty of Toronto, Faculty of Forestry, 1992. 20. Neter, 1., W. Was~erman,.M. Kutner, and C. Nachtsheim. Applied Linear Regression Models (3rd ed.). ChIcago: RIchard D. Irwin, 1996. 21. Pillai,~. C. ~. "Upper Percentage Points of the Largest Root of a Matrix in Multivariate AnalYSIS." BLOmetrika, 54 (1967), 189193. 22. Rao, C. ~. Linear ~tatistical Inference and Its Applications (2nd ed.) (paperback). New York: WIIeyIntersclence, 2002.
References 1. Abraham, B. and 1. Ledolter. Introduction to Regression Modeling, Belmont, CA: Thompson Brooks/Cole, 2006. 2. Anderson, T. W. An Introduction to Multivariate Statistical Analysis (3rd ed.). New York: John Wiley, 2003. 3. Atkinson, A. C. Plots, Transformations and Regression: An Introduction to Graphical Methods of Diagnostic Regression Analysis. Oxford, England: Oxford University Press, 1986. 4. Bartlett, M. S. "A Note on Multiplying Factors for Various ChiSquared Approximations." Journal of the Royal Statistical Society (B), 16 (1954),296298. 5. Bels!ey, 0. A., E. Kuh, and R. E. Welsh. Regression Diagnostics: Identifying Influential Data and Sources of Collinearity (Paperback). New York: WileyInterscience, 2004. 6. Bowerman, B. L., and R. T. O'Connell. Linear Statistical Models: An Applied Approach (2nd ed.). Belmont, CA: Thompson Brooks/Cole, 2000. 7. Box, G. E. P. "A General Distribution Theory for a Class of Likelihood Criteria." Biometrika,36 (1949),317346. 8. Box, G. E. P., G. M. Jenkins, and G. C. Reinsel. Time Series Analysis: Forecasting and Control (3rd ed.). Englewood Cliffs, NJ: Prentice Hall, 1994. 9. Chatterjee, S., A. S. Hadi, and B. Price. RegreSSion Analysis by Example (4th ed.). New York: WiJeyInterscience, 2006. 10. Cook, R. D., and S. Weisberg. Applied Regression Including Computing and Graphics. New York: John Wiley, 1999. 11. Cook, R. D., and S. Weisberg. Residuals and Influence in Regression. London: Chapman and Hall, 1982. 12. Daniel, C. and F. S. Wood. Fitting Equations to Data (2nd ed.) (paperback). New York: WileYInterscience,1999.
23. Seber, G. A. F. Linear Regression Analy;is. New York: John Wiley, 1977. 24. Rudorfer, ,~. V. "Cardiov~scular Ch~nges and Plasma Drug Levels after Amitriptyline Overdose. Journal o/ToxlcologyClznical Toxicology, 19 (1982), 6771.
Population Principal Components 431
Cha pter
with X b X z, .. , , X p as th e coord'mate axes. The new axes represen t the d'Irect'lOns , bT' with max' ion Im~m vana Ilty and proVIde a simpler and more parsimo nious descript th f e covanan ce structur e. o ;ependXsoThlely . 0dn the covarian ce Y. elf evelopm ent does p' 2, ... , I, " ' t al " hand no require a ~ultJvanate normal assumpt ion, On the other a ini:;~~ useful have ions populat normal ~omp?nents derred for multiva riate es can be made Ions III terms 0 the constan t density ellipsoids. Further , inferenc normal. (See riate ~~~o~h~.;.)mple compon ents when the popUlation is· multiva
mat~S (~r S::ellc~~~l!t~~~c::!ri~opm)p~~~ts
PRINCIPAL COMPONENTS
. Jh vector X' = [X1, X 2, ' . , , X pave Let the Irandom the covanan ce matrix Y. \ ues ",I ~ A2 ~ '" ~ Ap ~ 0, Conslde r the hnear combina tions
WI'th' elge~va
8.1 Introduction
(81)
covariance A principal compon ent analysis is concerned with explaining the variance . Its variables these of tions combina linear few a structlir e of a set of variables through ation. interpret (2) and reduction data (1) are es general objectiv y, Althoug h p components are required to reproduce the total system variabilit printhe of k number small a by for d accounte be can ty often much of this variabili components cipal compon ents. If so, there is (almost) as much information in the k replace then can ents compon principal k The . as there is in the original p variables ments on .' the initial p variables, and the original data set, consisting of n measure k principal p variable s, is reduced to a data set consisting of n measurements on compon ents. were not An analysis of principal components often reveals relationships that y ordinaril not would that ations interpret allows thereby and d previou sly suspecte discussed in result. A good example of this is provided by the stock market data Exampl e 8.5. than an Analyse s of principal components are more of a means to an end rather much in steps iate intermed as serve ly end in themselves, because they frequent multiple a to inputs be may nts compone l principa , example For ations. larger investig r, (scaled) regressi on (see Chapter 7) or cluster analysis (see Chapter 12), Moreove the fact9r for matrix ce covarian the of g" "factorin one are principa l compon ents analysis model considered in Chapter 9.
Yp = a~X =
J J
430
a p2 X 2
+ '" +
appXp
Var(Y;) = aiY.ai
i = 1,2,.,., p
(82)
Cov(Y;, Yk ) = aiY.ak
i, k = 1,2, ... , p
(83)
Y,
y yThhe principa l co~ponents are those un correlat ed linear combina tions I, 2"·,, p w ose var~ances m (82) are as large as possible. maximu . The first. p?ncip~l ,compon ent is the linear combina tion with a'Y.a ca~ = (Yd Var that clear is It a1Y.al. var!ance. That IS, It m:U:Ir~llZeS Var(l}) = mi~acy indete~ this e eliminat To . constant some by al any Ing ~e. mcrease~ by multiplY We there~ length. unit of vectors nt coefficie to n attentio restrict to nt ~o~: ~~~:eme First principa l compon ent
. . es th a t maXlmIZ = linearcombinatl'on a'X 1
Var(a1X ) subject to alal = 1 " es Second principa l compon ent = linear combina tion a'2X th a t maxImiz Var (a2X) subject to a2a2 = 1 and Cov(a1 X,a2X) = 0
8.2 Population Principal Components of the p ranAlgebraically, principal components are particular linear combinations represen t tions combina linear these ically, Geometr Xp. , .• . , X Xl' s dom variable 2 original the rotating by the selectio n of a new coordinate system obtained
+
Then, using (245), we obtain
)
)
ap1X1
At the ith step, ith principa l compon ent = linear combina tion at X that maximiz es Var(aiX ) subject to aia; = 1 and Cov(a;X , a"X) = 0 for
k < i
432
Population Principal Components 433
Chapter 8 Principal Components
Proof. From Definition 2A.28, CTU +
Result 8.1. Let :t be the covariance matrix associated with the random vector X' = [XI, X 2, ... , Xp]. Let :t have the eigenvaIueeigenvector pairs (AI, el), . \ e) (A e) where Al ~ A2 ~ ... ~ Ap ~ O. Then the ith principal ( 1l2' 2,"·' P' P ponent is given by Y; =..eiX = enXI + ej2X2 + ... + ejpXp, i = 1,2, ... ,p
CT22
+ ... +
tr(:t) = tr(PAP') = tr(AP'P) = tr(A) = Al + A2 + ... + Ap p
L Var(X;}
i = 1,2, ... ,p
max For the choice a
e; : t e l , = Al =  ,  = el:tel elel
= Var(YI
)
a':ta  , = Ak+1 k = 1,2, ... ,p  1 aa
= ek+l, with ek+1ej = 0, for i
=
1,2, ... , k and k
= 1,2, ... , p
 1,
e"+1:tek+Iiele+lek+1 = ek+l:tek+1 = Var(Yk+d But ele+I(:tek+d = Ak+lek+lek+1 = Ak+1 so Var(Yk:l) = Ak+l· It remains to show that ej perpendicular to ek (that is, eiek = 0, i k) gives COy (Y;, Yk ) = O. ~~w, the eigenvectors of:t are orthogonal if all the :igenvalues AI, A2,···, A{' are dIstmct. If the eigenvalues are not all distinct, the eIgenvectors correspondm~ to common eigenvalues may be chosen to be orthogonal. There~o~e, f~r any t';o .eIgenvectors ej and ek' ejek = 0, i k. Since :tek = Akek, premultlplicatlOn by ej gIves
'*
'*
Cov(Y;, Yk ) = eiIek
= eiAkek
= Akeiek
=0
•
for any i * k, and the proof is complete.
From Result 8.1, the principal components are uncorrelated and have variances equal to the eigenvalues of :to
Result 8.2. Let X' =
[XI' X 2, .. . , Xp] have covariance matrix:t, with eigenvalue
eigenvector pairs (AJ,el)' (A2,e2), .. ·, (Ap,ep) where Al ~ A2 ~ ... ~ Ap Let Y = ejX, Y2 = e2X, ... , Yp = e;,x be the principal components. Then
~ O.
I
p
CTu +
CTn
+ ... + er pp
= 2: Var(Xj) i=1
Total population variance = CTII + CT22 = Al + A2 +
Proportion of total ) population variance _ Ak due to kth principal  Al + A2 + ... + Ap ( component
Similarly, using (252), we get • J. "l>e2, .. .,ek
•
+ ... + CT pp ... + Ap
(86)
( attained when a = el)
But el el = 1 since the eigenvectors are no~malized. Thus, a':ta max,.*0 a a
~I
and consequently, the proportion of total variance due to (explained by) the kth principal component is
Proof. We know from (251), with B = :t, that a':ta = Al .*0 a a
L Var(Y;)
Result 8.2 says that
If some Aj are equal, the choices of the corresponding coefficient vectors, ej, and. hence Y;, are not unique.
max,
p
= tr(:t) = tr(A) =
~I
=0
Cov (Y;, Yk ) ~ ei:tek
= tr(:t). From (220) with
Thus,
With these choices, Var(Y;) = ei:tej = Aj
CT pp
A = :t, we can write:t = PAP' where A is the diagonal matrix of eigenvalues and P = [el, e2,· .. ,ep ] so that PP' = P'P = I. Using ResuIt 2A.11(c),we have
p
=
Al + A2 + ... + Ap
=
2: Var(Y;) /=1
k = 1,2, ... ,p
(87)
If most (for instance, 80 to 90%) of the total population variance, for large p, can be· attributed to the first one, two, or three components, then these components can "replace" the original p variables without much loss of information. Each component of the coefficient vector ei = [ejJ, ... , ejk, ... , eip] also merits inspection. The magnitude of ejk measures the importance of the kth variable to the ith principal component, irrespective of the other variables. In particular, ejk is proportional to the correlation coefficient between Y; and X k •
Result 8.3. If 1] = e;X, 12 = ezX, ... , ~) = obtained from the covariance matrix :t, then PY;,X k =
ejkv% .~ VCTkk
e~X
are the principal components
i,k = 1,2, ... ,p
(88)
are the correlation coefficients between the components Y; and the variables X k · Here (A1> el)' (A2, e2),· .. , (Ap, e p ) are the eigenvalueeigenvector pairs for:t. Proof. Set ale = [0, ... ,0, 1, 0, ... , 0] so that X k = a"X and COy (Xk , Y;) = Cov(aleX, eiX) = alc:tej, according to (245). Since :tej = Ajej, COV(Xk, Y;) = a"Ajej= Aieik. Then Var(Y;) = Aj (see (85)J and Var(Xk ) = CTkkyield Cov(Y;, X k ) Aiejk e·k VX; PYiX.= _~./ = . r .  . r  = :,: , vVar(Y;) vVar(Xk ) vA; VCTkk VCTkk
i,k=1,2, ... , p .
Although the correlations of the variables with the principal components often help to interpret the components, they measure only the univariate contribution of an individual X to a component Y. That is, they do not indicate the importance of an X to a component Y in the presence of the other X's. For this reason, some
Population Principal Components 435
434 Chapter 8 Principal Components
statisticians (see, for example, Rencher [16]) recommend that only the coefficients eib and not the correlations, be used to interpret the components. Although the coefficients and the correlations can lead to different rankings as measures of the importance of the variables to a given component, it is our experience that these rankings are often not appreciably different. In practice, variables with relatively large coefficients (in absolute value) tend to have relatively large correlations, so the two measures of importance, the first multivariate and the second univariate, frequently give similar results. We recommend that both the coefficients and the correlations be examined to help interpret the principal components. The following hypothetical example illustrates the contents of Results 8.1,8.2, and 8.3. Example S.I (Calculating the population principal components) random variables Xl' X 2 and X3 have the covariance matrix
It may be verified that the eigenvalueeigenvector pairs are
Al
= 5.83,
A2 = 2.00, A3 = 0.17,
ei = [.383, .924,0] e2 = [0,0,1] e3 = [.924, .383, 0]
Therefore, the principal components become
Yi. = eiX = .383X1  .924X2 12 = e2X = X3 }\ = e3X = .924X1 + .383X2 The variable X3 is one of the principal components, because it is uncorrelated with the other two variables. Equation (85) can be demonstrated from first principles. For example, Var(Yd = Var(.383Xl  .924X2) = (.383?Var(X1) + (.924?Var(X2)
+ 2( .383) (  .924) Cov (Xl> X 2) + .854(5)  .708( 2)
= .147(1)
Cov(Y1 , 12)
= 5.83 = Al = Cov(.383Xl
 .924X2, X 3)
= .383 Cov(Xl> X 3)  .924 COV(X2' X 3)
= .383(0)
 .924(0)
=0
It is also readily apparent that
0"11
+ 0"22 + 0"33 = 1 + 5 + 2
=
Al
+ A2 + A3 = 5.83 + 2.00 + .17
validating Equation (86) for this example. The proportion of total variance accounted for by the first principal component isAJ/(A l + A2 + A3 ) = 5.83/8 = .73.Further,the first two components account for a proportion (5.83 + 2)/8 = .98 of the population variance. In this case, the components Y1 and Y2 could replace the original three variables with little loss of information. Next, using (88), we obtain
.924v'5.83
VS
= .998
Notice here that the variable X 2 , with coefficient .924, receives the greatest weight in the component YI . It also has the largest correlation (in absolute value) with Yi.. The correlation of Xl, with YI , .925, is almost as large as that for X 2 , indicating that the variables are about equally important to the first principal component. The relative sizes of the coefficients of Xl and X 2 suggest, however, that X 2 contributes more to the determination of YI than does Xl' Since, in this case, both coefficients are reasonably large and they have opposite signs, we would argue that both variables aid in the interpretation of Yi., Finally, (as it should) The remaining correlations can be neglected, since the third component is unimportant. _ It is informative to consider principal components derived from multivariate normal random variables. Suppose X is distributed as Np(IA' l;). We know from (47) that the density of X is constant on the lA centered ellipsoids
which have axes ±cVA; ei' i = 1,2, ... , p, where the (Ai, e;) are the eigenvalueeigenvector pairs of l;. A point lying on the ith axis of the ellipsoid will have coordinates proportional to ej = [ei I, ei2, ... , ei p] in the coordinate system that has origin lA and axes that are parallel to the original axes XI, X2, •.. , X p' It will be convenient to set lA = 0 in the argument that follows. l From our discussion in Section 2.3 with A = l;l, we can write ,~1 x = 1 ( el,)2 2 = x...... x
C
Al
+ 1 ( e2, x)2 + ... + 1 (e' x) 2 A2
Ap
p
IThis can be done without loss of generality because the normal random vector X can always be translated to the normal random vector W = X  p. and E(W) =~. However, Cov(X) = Cov(W).
Population Principal Components 437
436 Chapter 8 Principal Components where et x, eZ x, ... , e~ x are recognized as the principal components of x. Setting YI = el x, Y2 = ezx, ... , Yp = e~x, we have C
11
\1 11 1I
i 11
I I'I
1
1 2 1 2 1 2 z = ;Yl + ; Y2 + ... + A' Yp "I
"2
P
and this equation defines an ellipsoid (since Aj, A2,' .. , Ap are positive) in a coordinate system with axes YI,)2, ... , Yp lying in the ?irect~o~s o~ ej, e2,:'" ~p, tively. If Al is the largest eigenvalue, then the major aXIs hes ill the dIrectIOn el· The remaining minor axes lie in the directions defined by ez,···, e p • To summarize, the principal components YI' = et x, )2 = x, ... , Yp = e~x lie in the directions of the axes of a constant density ellipsoid. Therefore, any point on the ith ellipsoid axis has x coordinates proportional to e; = [e;I' ei2,"" eip] and,· necessarily, principal component coordinates of the form [0, ... ,0, Yi' 0, ... ,0). When /L =P 0, itis the meancentered principal component Yi = ei(x  /L) that has mean and lies in the direction ei' A constant density ellipse and the principal components for a bivariate normal __ ..~15L random vector with /L = 0 and p = .75 are shown in Figure 8.1. We see that the principal components are obtained by rotating the original coo~dina~e axes ~hrough an angle () until they coincide with the axes of the constant denSIty ellIpse. This result holds for p > 2 dimensions as well.
ez
°
In matrix notation,
(810) where the diagonal standard deviation matrix VI/2 is defined in (235). Clearly, E(Z) = 0 and l l Cov (Z) = (V I/2r l:(V I/2r = p by (237). The principal components of Z may be obtained from the eigenvectors of the correlation matrix p of X. All our previous results apply, with some simplifications, since the variance of each Z; is unity. We shall continue to use the notation Y; to refer to the ith principal component and (A;, e;) for the eigenvalueeigenvector pair from either p or l:. However, the (A;, e;) derived from :t are, in general, not the same as the ones derived from p. Result 8.4. The ith principal component of the Z' = [ZI,Z2, ... ,Zp)withCov(Z) = p,is given by
standardized
variables
i = 1,2, ... , p Moreover, p
2: Var(Y;)
p
=
;=1
2: Var(Z;)
=p
(811)
i=I
and
y, = e;x
i,k = 1,2, ... ,p In this case, (AI, et>, (Az, e2)"'" p, with Al ~ Az ~ ... ~ Ap ~ 0.
CAp, e p) are
the eigenvalueeigenvector pairs for
Proof. Result 8.4 follows from Results 8.1,8.2, and 8.3, with ZI, Z2 • ... , Zp in place of XI. X 2 • .•.• Xp and p in place of l:. • Figure 8.1 The constant density ellipse x'Il x = c Z and the principal components YI , Y2 for a bivariate normal random vector X having meanO.
11=0 P = .75
We see from (811) that the total (standardized variables) population variance is simply p, the sum of the diagonal elements of the matrix p. Using (87) with Z in place of X, we find that the proportion of total variance explained by the kth principal component of Z is Proportion of (standardized») A population variance due = ~, ( to kth principal component p
Principal Components Obtained from Standardized Variables Principal components may also be obtained for the standardized variables Z _ (Xj ILIl 1
~
z _ (X2 2 
1L2)
va:;
where the
Ak'S
k=1,2, ... ,p
(812)
are the eigenvalues of p.
Example 8.2 (Principal components obtained from covariance and correlation matrices are different) Consider the covariance matrix
l:=[!
lO~J
438
Population Principal Components 439
Chapter 8 Principal Components
When the first principal component obtained from p is expressed in terms of Xl and X 2 , the relative magnitudes of the weights .707 and .0707 are in direct opposition to those of the weights .040 and .999 attached to these variables in the principal component obtained from l:. •
and the derived correlation matrix
p=
[.~ '~J
The eigenvalueei.,genvector pairs from I are Al
= 100.16,
e;
= [.040, .999]
.84,
e2
= [.999, .040]
A2 =
Similarly, the eigenvalueeigenvector pairs from pare Al
=1+P=
A2
= 1  p = .6,
e; =
1.4,
[.707, .707J
e2 = [.707, .707]
The respective principal components become Yj = .040XI + .999X2 I: Y = .999X  .040X 2 I 2
The preceding example demonstrates that the principal components derived from I are different from those derived from p. Furthermore, one set of principal components is not a simple function of the other. This suggests that the standardization is not inconsequential. Variables should probably be standardized if they are measured on scales with widely differing ranges or if the units of measurement are not commensurate. For example, if Xl represents annual sales in the $10,000 to $350,000 range and X 2 is the ratio (net annual income)/(total assets) that falls in the .01 to .60 range, then the total variation will be due almost exclusively to dollar sales. In this case, we would expect a single (important) principal component with a heavy weighting of Xl' Alternatively, if both variables are standardized, their subsequent magnitudes will be of the same order, and X 2 (or Z2) will play a larger role in the construction of the principal components. This behavior was observed in Example 8.2.
and
+ .707Z2 =
YI = .707Z1
=
p: Yz = .707Z1
XI  ILl) .707 (   1 .707(XI ·ILI)
 IL2) + .707 (X2 10
+
.0707(X2  IL2)
XI  ILl) (X2  IL2)  .707Z2 = .707 (  1   .707 10
Principal Components for Covariance Matdces with Special Structures There are certain patterned covariance and correlation matrices whose principal components can be expressed in simple forms. Suppose l: is the diagonal matrix
l:
Because of its large variance, X 2 completely dominates the first prin~ipal compon~nt determined from I. Moreover, this first principal component explams a proportion _A_I_ = 100.16 = .992 Al + A2 101
of the total population variance. . . When the variables XI and X 2 are standardized, however, the resultmg variables contribute equally to the principal components determined from p. Using Result 8.4, we obtain z = ell v'X"; = .707v1.4 = .837 py1·1
and PY1,Z2
= e21 VI;" =
.707v1.4
= .837
In this case, the first principal component explains a proportion Al P
= 1.4 = .7 2
of the total (standardized) population variance. . Most strikingly, we see that the relative importance of the vanables. to,.for instance, the first principal component is greatly affected by the standardIZatIOn.
all
= .707(XI  ILl)  .0707(X2  IL2) =
o.
.. .
an .. . ..
..
. 0
fo
..
(813)
.
Setting e; = [0, ... ,0,1,0, ... ,0], with 1 in the ith position, we observe that 0 0
fT
a22
n
0 1 0
1aii
or
Ie; = aije;
and we conclude that (aj;, e;) is the ith eigenvalueeigenvector pair. Since the linear combination et X = Xi, the set of principal components is just the original set of uncorrelated random variables. For a covariance matrix with the pattern of (813), nothing is gained by extracting the principal components. From another point of view, if X is distributed as Np(p, l:), the contours of constant density are ellipsoids whose axes already lie in the directions of maximum variation. Consequently, there is no need to rotate the coordinate system.
Summarizing Sample Variation by Principal Components 441
440 Chapter 8 Principal Components Standardization does not substantially alter the situation for the 1: in (813). In that case, p = I, the p X P identity matrix. Clearly, pe; = le;, so the eigenvalue 1 has multiplicity p and e; = [0, ... ,0, 1,0, ... ,0], i = 1,2, ... , p, are convenient choices for the eigenvectors. Consequently, the principal components determined from p are also the original variables Zlo"" Zp. Moreover, in this case of equal eigenvalues, the multivariate normal ellipsoids of constant density are spheroids. Another patterned covariance matrix, which often describes the correspondence among certain biological variables such as the sizes of living things, has the general form
The first principal component l] =
(815)
is also the covariance matrix of the standardized variables. The matrix in (815) implies that the variables Xl' X 2 , . •• , Xp are equally correlated. It is not difficult to show (see Exercise 8.5) that the p eigenvalues of the correlation matrix (815) can be divided into two groups. When p is positive, the largest is Al = 1 + (p  l)p
1
p
2: Z; Vp;=l
= 
is proportional to the sum of the p standarized variables. It might be regarded as an "index" with equal weights. This principal component explains a proportion Al
=
p
The resulting correlation matrix
el Z
1
+ (p  l)p p
1 p p
=p+
(818)
of the total population variation. We see that Adp == p for p close to 1 or p large. For example, if p = .80 and p = 5, the first component explains 84 % of the total variance. When p is near 1, the last p  1 components collectively contribute very little to the total variance and can often be neglected. In this special case, retaining only the first principal component Yj = (l/vP) [1,1, ... ,1] X, a measure of total size, still explains the same proportion (818) of total variance. If the standardized variables Zl, Z2,' .. , Zp have a multivariate normal distribution with a covariance matrix given by (815), then the ellipsoids of constant density are "cigar shaped," with the major axis proportional to the first principal component Y1 = (I/Vp) (1,1, ... ,1] Z. This principal component is the projection ofZ on the equiangular line I' = [1,1, ... ,1]. The minor axes (andremainingprincipal components) occur in spherically symmetric directions perpendicular to the major axis (and first principal component).
with associated eigenvector ej =
[~,~, ,~J
(817)
...
The remaining p  1 eigenvalues are A2 = A3 = .,. = Ap = 1  P
and one choice for their eigenvectors is
ez = [~. v;~ 2· 0, ... ,oJ e3
=
e~ = I
e~
[k'V21X3'V;~3,0, ... ,oJ 1 [
VU 
1 {i  1) ,0, ... ,0 1)(''''~' v'(il)i
(p  1) ] 1 1 = [ V(p _ l)p"'" V(p  1)/ V(p  l)p
J
8.3 Summarizing Sample Variation by Principal Components We now have the framework necessary to study the problem of summarizing the variation in n measurements on p variables with a few judiciously chosen linear combinations. Suppose the data Xl, X2,"" Xn represent n ipdependent drawings from sOme pdimensional popUlation with mean vector p. and covariance matrix 1:. These data yield the sample mean vector x, the sample covariance matrix S, and the sample correlation matrix R. Our objective in this section will be to construct uncorrelated linear combinations of the measured characteristics that account for much of the variation in the sample. The uncorrelated combinations with the largest variances will be called the sample principal components. Recall that the n values of any linear combination j = 1,2, ... ,n
have sample mean 8J.X and sample variance 81S81' Also, the pairs of values (8J.Xj,8ZXJ, for two linear combinations, have sample covariance 8jS8z [see (336)].
442
Summarizing Sample Variation by Principal Components 443
Chapter 8 Principal Components The sample principal components are defined as those linear ,",VJ,uumanr which have maximum sample variance. As with the population quantities, strict the coefficient vectors ai to satisfy aiai = 1. Specifically,
I .
ljli I1 11I.
First sample linear combination aixj that maximizes principal component = the sample variance of a;xj subject to a1al = 1 Second sample linear combination a2Xj that maximizes the sample principal component = variance of a2Xj subject to a2a2 = 1 and zero cOvariance for the pairs (a;xj, a2Xj)
We shall denote the sample principal components by )11,52, ... , )lp, irrespective of whether they are obtained from S or R.2 The components constructed from Sand R are not the same, in general, but it will be clear from the context which matrix is being used, and the single notation Yi is convenient. It is also convenient to label the component coefficient vectors ei and the component variances Ai for both situations. The observations Xj are often "centered" by subtracting x. This has nO effect on the sample covariance matrix S and gives the ith principal component
.vi
= ei(x  x),
(821)
i = 1,2, ... ,p
for any observation vector x. If we consider the values of the ith component (822)
j = 1,2, ... ,n
At the ith step, we have ith sample principal component
generated by substituting each observation Xj for the arbitrary x in (821), then Yi;; = l~A'( ~ ei Xj n j=l
linear combination aixj that maximizes the sample
= variance of aixj subject to aiai = 1 and zero sample covariance for all pairs (aixj, a"xj), k < i
The first principal component maximizes a\Sa J or, equivalently, a1 Sa l a1 a l
lA'(~( ~ Xj 
x_) =  ei
n
») x
= lA, ej 0 = 0
j=l
n
(823)
That is, the sample m!?an of each principal component is zero. The sample variances are still given by the A;'s, as in (820). Example 8.3 (Summarizing sample variability with two sample principal components)
By (251), the maximum is the largest eigenvalue Al attained for the al = eigenvectqr el of S. Successive choices of ai maximize (819) subject o = aiSek = aiAkek> or ai perpendicular Jo ek' Thus, as in the proofs of 8.18.3, we obtain the following results conceming sample principal cornDCln€:ni
A census provided information, by tract, on five socioeconomic variables for the Madison, Wisconsin, area. The data from 61 tracts are listed in Table 8.5 in the exercises at the end of this chapter. These data produced the following summary statistics: X'
If S = {sid is the p X P sample covariance matrix with ·P'",nIVl'IJue··ei!>emlectod"··
e
pairs (AI' ed, (,1.2, e2),"" (Ap, p), the ith sample principal component is by i = 1,2, ... ,p
where Al ~ ,1.2 ~ .' . ~ Ap ~ 0 and x is any observation on the )(1,)(2,···,)(p·A1so,
= Ab
Sample variance(Yk) Sample covariance()li, )lk)

=
k = 1,2, ... , P 0, i #' k
=
[4.47, total population (thousands)
3.96, professional degree (percent)
71.42, employed age over 16 (percent)
26.91, government employment (percent)
1.64] median home value ($100,000)
and
33~
[ 1.102 S = 4.306 2.078 0.027
1.102 9.673 1.5l3 10.953 1.203
4.306 1.5l3 55.626 28.937 0.044
2.078 10.953 28.937 89.067 0.957
Oill7]
1.203 0.044 0.957 0.319
Can the sample variation be summarized by one or two principal components?
In addition, Total sample variance =
Lp Sii = Al" + A2 + ... + Ap'
i=l and
i, k = 1, 2, ... , p
2Sample principal components also can be obtained from I = Sn, the maximum likelihood estimate of the covariance matrix I, if the Xj are nonnally distributed. (See Result 4.11.) In this case, provided that the eigenvalues of I are distinct, the sample principal components can be viewed as the maximu~ likelihood estimates of the corresponding population counterparts. (S!!e [1].) We shall not consider J. because the assumption of nonnality is not required in this section. Also, I has eigenvalues [( n  1)/n]A; and c,?rresponding eigenvectors e;, where (A;, ei) are the eigenvalueeigenvector pairs for S. Thus, both S and I give the same sample principal components eix [see (820)] and the same proportion of explained variance A;/(.~l + A2 + ... + Ap). Finally, both S a!.1d I give the same sample correlation matrix R, so if the variables are standardized, the choice of S or I is irrelevant.
444
Summarizing Sample Variation by Principal Components 445
Chapter 8 Principal Components We find the following:
I!
!\ I
I
I
Coefficients for the Principal Coefficients in
e2
e3
Variable
el (rh,xk)
Total population Profession Employment (%) Government employment (%) Medium home value
 0.039(  .22) 0.105(.35) 0.492(  .68)
0.071(.24) 0.130(.26) 0.864(.73)
0.188 0.961 0.046
0.977 0.171 0.091
0.863(.95)
0.480(.32)
0.153
0.030
0.009(.16)
0.015(.17)
0.125
0.082
Variance (Ai): Cumulative percentage of total variance
107.02
39.67
8.37
2.87
67.7
92.8
98.1
e4
e5
99.9
The first principal component explains 67.7% of the total sample variance. The first two principal components, collectively, explain 92.8% of the total sample ance. Consequently, sample variation is summarized very well by two principal ponents and a reduction in the data from 61 observations on 5 observations to observations on 2 principal components is reasonable. Given the foregoing component coefficients, the first principal cOlnp,one:nl appears to be essentially a weighted difference between the percent employed government and the percent total employment. The second principal cOIloponelllr' appears to be a weighted sum of the two. As we said in our discussion of the population components, the component coefficients eik and the correlations ryi,Xk should both be exami?ed to inte.rpret the principal components. The correlations allow for differences m. t~e vanan~s the original variables, but only measure the importance of an indJVldual X Without regard to the other X's making up the component. We notice in Example 8.3, however, that the correlation coefficients displayed in the table confirm the interpretation provided by the component coefficients.
The Number of Principal Components
~'.
~.
~
There is always the question of how many components to retain. There is no defin , itive answer to this question. Things to consider include the amount of total variance explained, the relative sizes of the eigenvalues (the variances of the pIe components), and the subjectmatter interpretations of the components. In dition, as we discuss later, a component associated with an eigenvalue near and, hence, deemed unimportant, may indicate an unsuspected linear in the data.
Figure 8.2 A scree plot.
A useful visual aid to determining an appropriate number of principal components is a scree plot. 3 With the eigenvalues ordered from largest to smallest, a scree plot is a plot of Ai versus ithe magnitude of an eigenvalue versus its number. To determine the appropriate number of components, we look for an elbow (bend) in the scree plot. The number of components is taken to be the point at which the remaining eigenvalues are relatively small and all about the same size. Figure 8.2 shows a scree plot for a situation with six principal components. An elbow occurs in the plot in Figure 8.2 at about i = 3. That is, the eigenvalues after A2 are all relatively small and about the same size. In this case, it appears, without any other evidence, that two (or perhaps three) sample principal components effectively summarize the total sample variance. Example 8.4 (Summarizing sample variability with one sample principal component) In a study of size and shape relationships for painted turtles, Jolicoeur and Mosimann [11] measured carapace length, width, and height. Their data, reproduced in Exercise 6.18, Table 6.9, suggest an analysis in terms of logarithms. (Jolicoeur [10] generally suggests a logarithmic transformation in studies of sizeandshape relationships.) Perform a principal component analysis. 3 Scree
is the rock debris at the bottom of a cliff.
446
Summarizing Sample Variation by Principal Components 447
Chapter 8 Principal Components The natural logarithms of the dimensions of 24 male turtles have sample mean vector i' = [4.725,4.478,3.703) and covariance matrix
I11r
S = 103
11
iI
11
11
illI! I
11.072 8.019 8.160] 8.019 6.417 6.005 [ 8.160 6.005 6.773
A principal component analysis (see Panel 8.1 on page 447 for the output from the SAS statistical software package) yields the following summary:
PANEL 8.1 SAS ANALYSIS FOR EXAMPLE 8.4 USING PROC PRINCOMP.
title 'Principal Component Analysis'; data turtle; infile 'E84.dat'; input length width height; xl = log(length); x2 =Iog(width); x3 =Iog(height); proc princomp coy data = turtle out = result; var xl x2 x3;
1
PROGRAM COMMANDS
Principal Components Analysis
Coefficients for the Principal Components (Correlation Coefficients in Parentheses) Variable In (length) In (width) In (height) Variance (A;): Cumulative percentage of total variance
el{ryj,Xk)
e2
e3
.683 (.99) .510 (.97) .523 (.97)
.159 .594 .788
.713 .622 .324
.60 x'1O 3
.36 X 103
23.30
X
96.1
103
98.5
Mean StD
=
Xl 4.725443647 0.105223590
100
A scree plot is shown ih Figure 8.3. The very distinct elbow in this plot occurs at i = 2. There is clearly one dominant principal component. The first principal component, which explains 96% of the total variance, has an interesting subjectmatter interpretation. Since
YI
OUTPUT
24 Observations 3 Variables Simple Statistics X2 4.477573765 0.080104466
X3 3.703185794 0.082296771
I
Covariance Matrix
Xl
X2
X3
0.0080191419
0.0081596480
1
Xl
0.0110720040
X2
0.0080191419
0.0064167255
X3
0.0081596480
0.0060052707
I
0.0060052707 0.00677275851
.683 In (iength) + .510 In (width) + .523 In (height) Total Variance = 0.024261488
= In [(iength)·683(width).51O(height).523)
Eigenvalues of the Covariance Matrix
~i X 10 3
PRINl PRIN2 PRIN3
20
Eigenvalue 0.023303 0.000598 0.000360
Difference 0.022705 0.000238
1
Proportion 0.960508 0.024661 0.014832
Eigenvectors 10 ."
Xl X2 X3 oL~~==~~
3
Figure 8.3 A scree plot for the
turtle data.
. PRINl '0.683102 0.510220. 0:572539. ,
PRIN.2 .159479 .,..594012
> ().7884~
PRIN3 .712697 0.62.1953 . 0.324401
Cumulative 0.96051 0.98517 1.00000
Summarizing Sample Variation by Principal Components 449
448 Chaptet 8 Principal Components the first principal component may be viewed as the In (volume) of a box with adjusted dimensions. For instance, the adjusted height is (height).5Z3, which ... in some sense, for the rounded shape of the carapace. •
"2' (x  xl'S' (x  x) = c2
!I It
Interpretation of the Sample Principal Components
11
The sample principal components have several interpretations. First, suppose the underlying distribution of X is nearly Ni 1', I). Then the sample principal components, Yj = e;(x  x) are realizations of population principal components Y; = e;(X  I' which have an Np(O, A) distribution. The diagonal matrix A has entries AI, Az,· " , Ap and (A j , e;) are the eigenvalueeigenvector pairs of I. . . Also, from the sample values Xj' we can approximate I' by xand I by S. If S positive definite, the contour consisting of all p X 1 vectors x satisfying
\i 11
(x  X)'SI(X  x)
=
(xx)'S'(xx)=c2 =x,~~x,
Figure 8.4 Sample principal components and ellipses of constant distance.
cZ 2
estimates the constant density contour (x  p.),II(X  1') = c of the underlying normal density. The approximate contours can be drawn on the scatter plot to indicate the normal distribution that generated the data. The normality assumption is useful for the inference procedures discussed in Section 8.5, but it is not required for the development of the properties of the sample principal components summarized in (820). Even when the normal assumption is suspect and the scatter plot may depart somewhat from an elliptical pattern, we can still extract the eigenvalues from S and obtain the sample principal components. Geometrically, the data may be plotted as n points in pspace. The data can then be expressed in the new coordinates, which coincide with the axes of the contour of (824). Now, (824) defines .a hyperellipsoid that is centered at x and whose axes are given by the eigenvectors of SI or, equivalently, of S. (See Section 2.3 and Result 4.1, with S in place of I.) The lengths of these hyperellipsoid axes are proportional to i = 1,2, ... , p, where Al ;:: Az ;:: ... ;:: Ap ;:: 0 are the eigenvalues of S. . Because ej has length 1, the absolute value of the ith principal component, 1yd = 1e;(x  x) I, gives the length of the projection of the vector (x  x) on the unit vector ej. [See (28) and (29).] Thus, the sample principal components Yj = e;(x  x), i = 1,2, ... , p, lie along the axes of the hyperellipsoid, and their absolute values are the lengths of the projections of x  x in the directions of the axes ej. Consequently, the sample principal components can be viewed as the result of translating the origin of the original coordinate system to x and then rotating the coordinate axes until they pass through the scatter in the directions of maximum variance. The geometrical interpretation of the sample principal components is illustrated in Figure 8.~ for E. = 2. Figure 8.4(a) shows an ellipse of constant distanc~, centered at x, with Al > Az . The sample principal components are well determmed. They lie along the axes of the ellipse in the perpendicular directions of ~ampl~ variaflce. Fjgure 8.4(b) shows a constant distance ellipse, cen~ered at x, Ai == Az . If AI = Az, the axes of the ellipse (circle) of constant distance are uniquely determined and can lie in any two perpendicular directions, including
directions of the original coordinate axes. Similarly, the sample principal components can lie in any two perpendicular directions, including those of the original coordinate axes. When the contours of constant distance are nearly circular or, equivalently, when the eigenvalues of S are nearly equal, the sample variation is homogeneous in all directions. It is then not possible to represent the data well in fewer than p dimensions. If the last few eigenvalues Aj are sufficiently small such that the variation in the corresponding ej directions is negligible, the last few sample principal components can often be ignored, and the data can be adequately approximated by their representations in the space of the retained components. (See Section 8.4.) Finally, Supplement 8A gives a further result concerning the role of the sample principal components when directly approximating the meancentered data Xj 
x.
0;,
Standardizing the Sample Principal Components Sample principal components are, in general, not invariant with respect to changes in scale. (See Exercises 8.6 and 8.7.) As we mentioned in the treatment of population components, variables measured on different scales or on a common scale with widely differing ranges are often standardized. For the sample, standardization is accomplished by constructing Xjl 
XI
~ XjZ 
I Zj = n /2(Xj 
x) =
Xz
VS;
j = 1,2, ... , n
450 Chapter 8 Principal Components
Summarizing Sample Variation by Principal Components 451
The n X p data matrix of standardized observations
ZI]
[ZlI
Z12
... ZIP] '.' . Z?
~ = Z:~ = Z~l Z~2
[zn
Znl Xl
Xli 
~ X21 
Xl
vs;;
Xnl 
Zn2 Xl2 
Xl
~
If Zl, Z2, ... , Zn are standard ized observations with covariance matrix R, the ith sample principal compon ent is
i = 1,2, ... , p
where (Ai, e;) is the ith eigenvalueeigenvector pair of R with Al ~ Az ~ ... ~ Ap ~ O. Also,
znp X2
Xl p  Xp
vS;;
.VS;;
X22  Xz
X2p  Xp
VS;
VS;;
Xn2  Xz
Xnp  Xp
VS;
VS;;
Sample variance (Yi) = Ai Sample covariance (Yi, Yk) ~ 0 (826)
(829)
Total (standar dized) sample variance
= tr(R) = p = Al + Az + ... + Ap
and i,k = 1,2, ... ,p
1 ' Z' 1 z=(I ) =1 Z' 1=
n
=0
n
(827)
Using (829), we see that the proport ion of the total sample variance explaine d by the ith sample principal compon ent is Proport ion of (standar diZed») sample variance due to ith ( sample principa l compon ent
and sample covariance matrix [see (327)]
S = _l_(Z z
n1
i = l.
i = 1,2, ... ,p
!n'z) '(z  !n'z) n n
n 1
=_l_ Z 'Z n 1
Example
(n  l)SI1
(n  l)S12
(n  l)Slp
Sl1 (n  l)S12
~VS; (n  l)s22
(n  l)szp
~VS;
sZ2
Vs;~
(n  l)Slp
(n  l)s2p
(n  1)spp
~vs;;, VS; vs;;,
(830)
p
A rule of thumb suggests retaining only those compon ents whose variance s Ai are greater than unity or, equivalently, only those compon ents which, individu ally, explain at least a proport ion 1/p of the total variance. This rule does not have a great deal of theoreti cal support , however, and it should not be applied blindly. As we have mention ed, a scree plot is also useful for selecting the appropr iate number of components.
= _l_(Z  li')'(Z  lz')
n1
= 1,2, ... , p
In addition,
yields the sample mean vector [see (324)]
n
i
~~
=R
(828)
spp
The sample principal components of ~he standardized .observations ar:; given br, (820), with the matrix R in place of S. ~mce the observatlO?S are already centered by construction, there is no need to wnte the components In the form of (821).
8.S (Sample principal components from standardized data) The weekly rates of return for five stocks (JP Morgan , Citibank , Wells Fargo, Royal Dutch Shell, and ExxonMobil) listed on the New York Stock Exchang e were determi ned for the period January 2004 through Decemb er 2005. The weekly rates of return are defined as (current week closing pricep revious week closing price )/(previ ous week closing price), adjusted for stock splits and dividends. The data are listed in Table 8.4 in the Exercises. The observations in 103 successive weeks appear to be indepen dently distributed, but the rates of return across stocks are correlat ed, because as one 6xpects, stocks tend to move togethe r in respons e to general economic conditions. Let xl, Xz, ... , Xs denote observe d weekly rates of return for JP Morgan , Citibank, Wells Fargo, Royal Dutch Shell, and ExxonMobil, respectiv ely. Then
x'
= [.0011, .0007, .0016, .0040, .0040)
Summarizing Sample Variation by Principal Components 453
452 Chapter 8 Principal Components and
R
=
[L~
.632 .511 .115 .632 1.000 .574 .322 .574 1.000 .183 .511 .115 .322 .183 1.000 .155 .213 .146 .683
Example 8.6 (Components from a correlation matrix with a special structure) Geneticists are often concerned with the inheritance of characteristics that can be measured several times during an animal's lifetime. Body weight (in grams) for n = 150 female mice were obtained immediately after the birth of their first four litters. 4 The sample mean vector and sample correlation matrix were, respectively,
m]
.213 .146 .683
LOoo
x'
We note that R is the covariance matrix of the standardized observations Zl
=
Xl  XI ~ ,Zz
Xz  Xz
= VS; , ... ,Zs =
Xs  Xs
The eigenvalues and corresponding normalized eigenvectors of R, determined by a computer, are AI
= 2.437,
ej = [
Az
= 1.407,
e2 = [.368, .236, .315,
A3
= .501,
e) = [ .604,  .136,
A4
= .400,
e4 = [
.363,  .629, .289, .381,
As
= .255,
e5 = [
.384,  .496, .071, .595, .498)
.469, .532, .465, .387, .585,
1.000
R =
~.
.361) .606)
.772, .093, .109) .493)
= [39.88,45.08,48.11,49.95]
and .7501 [ .6329 .6363
.7501 1.000 .6925 .7386
.6329 .6925 1.000 .6625
.6363] .7386 .6625 1.000
The eigenvalues of this matrix are
Al = 3.085, A2 = .382,
A3
=
.342,
and
A4
= .217
We note that the first eigenvalue is nearly equal to 1 + (p  1)1' = 1 + (4  1) (.6854)
= 3.056, where I' is the arithmetic average of the offdiagonal elements of R. The remai~ing eig~nvalues are small and about equal, although A4 is somewhat smaller than Az and A3 . Thus, there is some evidence that the corresponding population correlation matrix p may be of the "equalcorrelation" form of (815). This notion is explored further in Example 8.9. The first principal component
Using the standardized variables, we obtain the first two sample principal components:
'vI = elz = .49z1 + .52zz + .49z3 + .50z4
.h = elz = .469z 1 + .532z2 + .465z3 + .387z4 + .361z s Yz = ezz =  .368z1  .236z2  .315z3 + .585z4 + .606zs
accounts for loo(AJ/p) % = 100(3.058/4)% = 76% of the total variance. Although the average postbirth weights increase over time, the variation in weights is fairly well explained by the first principal component with (nearly) equal coefficients. _
These components, which account for
Cl ; A2)
100%
=
C.437 ; 1.407) 100% = 77%
of the total (standardized) sample variance, have interesting interpretations. The first component is a roughly equally weighted sum, or "index," of the five stocks. This component might be called a general stockmarket component, or, simply, a market component. The second component represents a contrast between the banking stocks (JP Morgan, Citibank, Wells Fargo) and the oil stocks (Royal Dutch Shell, ExxonMobil). It might be called an industry component. Thus, we see that most of the variation in these stock returns is due to market activity and uncorrelated industry activity. This interpretation of stock price behavior also has been suggested by King [12). The remaining components are not easy to interpret and, collectively, represent variation that is probably specific to each stock. In any event, they do not explain • much of the total sample variance.
Comment. An unusually small value for the last eigenvalue from either the sample covariance or correlation matrix can indicate an unnoticed linear dependency in the data set. If this occurs, one (or more) of the variables is redundant and should be deleted. Consider a situation where Xl, xz, and X3 are subtest scores and the total score X4 is the sum Xl + Xz + X3' Then, although the linear combination e'x = [1,1,1, I)x = Xl + X2 + X3  X4 is always zero, rounding error in the computation of eigenvalues may lead to a small nonzero value. If the linear expression relating X4 to (Xl> XZ,X3) was initially overlooked, the smallest eigenvalueeigenvector pair should provide a clue to its existence. (See the discussion in Section 3.4, pages 131133.) Thus, although "large" eigenvalues and the corresponding eigenvectors are important in a principal component analysis, eigenvalues very close to zero should not be routinely ignored. The eigenvectors associated with these latter eigenvalues may point out linear dependencies in the data set that can cause interpretive and computational problems in a subsequent analysis. 4Data courtesy of 1. 1. Rutledge.
454 Chapter 8 Principal Components
Graphing the Principal Components 455
8.4 Graphing the Principal Components ,04
Plots of the principal components can reveal suspect observations, as well as provide checks on the assumption of normality. Since the principal components are combinations of the original variables, it is not unreasonable to expect them to nearly normal. it is often necessary to verify that the first few principal components are approximately normally distributed when they are to be used as the input for additional analyses. The last principal components can help pinpoint suspect observations. Each observation can be expressed as a linear combination Xj =
(xjedel + (xje2)e2
•••
,,,"
o.
./
•
.3
ez, ... ,
of the complete set of eigenvectors el , ep of S. Thus, the magnitudes of the principal components determine how well the firs~ fe,w fit the o~se~vations. That is, YiJeJ + Yj2 e2 + ... + Yj,qle ql differs from Xj by Yjqe q + '" + Yjpe p, the square of whose length is YJq + "; + YJp.,Suspect,obs~rvation~ will oftednlbe SUhCh t.hllabt atlleast one of the coordinates Yjq' ... , Yj p contnbutmg to this square engt Wl e arge. (See Supplement 8A for more general approximation results.) The following statements summarize these ideas.
1. To help check the normal assumption, construct scatter diagrams for pairs of the first few principal components. Also, make QQ plots from the sample values generated by each principal component. 2. Construct scatter diagrams and QQ plots for the last few principal components, These help identify suspect observations.
Example 8.7 (Plotting the principal components for the turtle data)
,I
:V,
•• ••
•
•• •
.3
•
•
• • • :. Figure 8.6 Scatter plot of the principal components ,h and Yz of the data on male turtles.
:V2
The diagnostics involving principal components apply equally well to the checking of assumptions for a multivariate multiple regression modeL In fact, having fit any model by any method of estimation, it is prudent to consider the
W~ illustra~e
 4.478)
+ .523(X3
 3,703)
52 =
.159(XI  4.725)  .594(X2  4.478)
+ .788(X3
 3.703)
5'3
,713(xI  4.725)
=
+ .51O(x2
••• ••
.1
the plotting of principal components for the data on male turtles discussed m Example 8.4. The three sample principal components are .683(XI  4,725)
Figure 8.S A
+ .,. + (xjep)e p
= Yjle, + Yj2 e2 + ... + Yipe p
Yl =
QQ plot for the second principal component Yz from the data on male turtles.
 ,04 L'_'_ _i  _  L _   . J 2 \ 0 2
+ ,622(X2  4.478) + .324(X3  3,703)
where Xl = In (length), X2 = In (width), and X3 = In (height), respectively. Figure 8.5 shows the QQ plot for Yz and Figure 8.6 sh~ws the scatte~ plot of (Yl, 52), The observation for the first turtle is circled and lies 10 the l0:ver nght corner of the scatter plot and in the upper right corner of the QQ plot; It may be suspect, This point should have been checked for recording errors, or the turtle have been examined for structural anomalies. Apart from the first turtle, the plot appears to be reasonably elliptical. The plots for the other sets of principal ponents do not indicate any substantial departures from normality.
Residual vector
=
(observation vector) _ (v(ect?r of pr)edicted) esttmated values
or
P\
Ej = Yj (pXI) (pXI) (pXI)
j = 1,2, .. " n
(831)
for the multivariate linear model. Principal components, derived from the covariance matrix of the residuals,
;;:)(Ae·  e·), ;;:  1. £ .~(A J e·  e· n  P
j=l
J
J
J
J
(832)
can be scrutinized in the same manner as those determined from a random sample. You should be aware that there are linear dependencies among the residuals from a linear regression analysis, so the last eigenvalues will be zero, within rounding error.
Large Sample Inferences 457
456 Chapter 8 Principal Components
8.S large Sample Inferences We have seen that the eigenvalues and eigenvectors of the covariance (correlation) matrix are the essence of a principal component analysis. The eigenvectors determine the directions of maximum variability, and the eigenvalues specify the variances. When the first few eigenvalues are much larger than the rest, most of the total variance can be "explained" in fewer than p dimensions. In practice, decisions regarding the quality of the principal component approximation must be made on the basis of the eigenvalueeigenvector pairs (Ai, Ci) extracted from S or R. Because of sainpling variation, these eigenvalues and eigenvectors will differ from their underlying population counterparts. The sampling distributions of Ai and Ci are difficult to derive and beyond the scope of this book. If you are interested, you can find some of these derivations for multivariate normal populations in [1], [2], and [5]. We shall simply summarize the pertinent large sample results.
Large Sample Properties of Ai and
ei
Currently available results concerning large sample confidence intervals for Ai and ei assume that the observations XI' X 2, ... , Xn are a random sample from a normal population. It must also be assumed that the (unknown) eigenvalues of :t are distinct and positive, so that Al > A2 > ... > Ap > o. The one exception is the case where the number of equal eigenvalues is known. Usually the conclusions for distinct eigenvalues are applied, unless there is a strong reason to believe that :t has a special structure that yields equal eigenvalues. Even when the normal assumption is violated the confidence intervals obtained in this manner still provide some indication of the uncertainty in Ai and Ci· Anderson [2] and Girshick [5] have established the following large sample distribution theory for the eigenvalues A' = [Ab.··' Ap] and eigenvectors Cl,···, p of S:
c
1. Let A be the diagonal matrix of eigenvalues Ab···' Ap of:t, then is approximately Np(O, 2A 2).
Vii (A 
A).
2. Let
then
Vii (ei
where z(a/2) is the upper 100(a/2)th percentile of a standard normal distribution. Bonferronitype simultaneous 100(1  a)% intervals for m A/s are obtained by replacing z(a/2) with z(a/2m). (See Section 5.4.) Result 2 implies that the e/s are normally distributed about the corresponding e/s for large samples. The elements of each ei are correlated, and the correlation ?epends to a large extent on the separation of the eigenvalues AI, A2, ... , Ap (which IS unknown) and the sample size n. Approximate standard errors for the coeffis.ients eik are given by the square rools of the diagonal elements of (l/n) Ei where Ei is derived from Ei by substituting A;'s for the A;'s and e;'s for the e;'s. Example 8.8 (Constructing a confidence interval for '\1) We shall obtain a 95% confidence interval for AI, the variance of the first population principal component, using the stock price data listed in Table 8.4 in the Exercises. Assume that the stock rates of return represent independent drawings from an N5(P,,:t) population, where :t is positive definite with distinct eigenvalues Al > A2 > ... > A5 > O. Since n = 103 is large, we can us~ (833) with i = 1 to construct a 95% confidence interval for Al. From Exercise 8.10, Al = .0014 and in addition, z(.025) = 1.96. Therefore, with 95% confidenc~,
.0014
(1
, (2)
+ 1.96 V 103
:5 Al
:5
.0014 ,!2 (1  1.96 V ~ )
.0011:5 Al
:5
.0019
•
Whenever an eigenvalue is large, such as 100 or even 1000, the intervals generated by (833) can be quite wide, for reasonable confidence levels, even though n is fairly large. In general, the confidence interval gets wider at the same rate that Ai gets larger. Consequently, some care must be exercised in dropping or retaining principal components based on an examination of the A/s.
Testing for the Equal Correlation Structure The special correlation structure Cov(Xj , X k ) = Yajjakk p, or Corr (Xi, X k ) = p, all i ~ k, is one important structure in which the eigenvalues of :t are not distinct and the previous results do not apply. To test for this structure, let
 ei) is approximately Np(O, E;).
Ho: P = po =
3. Each Ai is distributed independently of the elements of the associated ei· Result 1 implies that, for n large, the Ai are independently distributed. Moreover, Ai has an approximate N(Aj, 2Ar/n) distribution. Using this normal distribution, we obtainP[lAi  Ad:5 z(a/2)Ai V271i] = 1  a. A large sample 100(1  a)% confi
or
[~~ ~]
(pxp)·
p
.
p
1
and
dence interval for Ai is thus provided by A,·
_ _ _!....;=: <:
(1
+ z(a/2)V271i) 
A.
<:
1
A·I
(1  z(a/2)v2fn)
A test of Ho versus HI may be based on a likelihood ratio statistic, but Lawley [14] has demonstrated that an equivalent test procedure can be constructed from the offdiagonal elements of R.
458
Monitoring Quality with Principal Components 459
Chapter 8 Principal Components 4
Lawley's procedure requires the quantities 1
p
rk =  P  1 A
y =
2: 'ik
k = 1,2, ... ,p;
i=I i.,k
2: (rk  7')2 = (.6731  .6855f
r= p (P 
l)2:2:rik i
(p ~ 1f[1  (1  r)2] ~ 2 P  (p  2)(1  r)
=
:)2 [2:2: (rik 
(n (1  r)
7')2 
r
i
±
(rk  r)2] > XtP+I)(p2)/2(a)
k=l
where XtP+I)(p2)/2(a) is the upper (100a)th percentile of a chisquare distribution with (p + 1)(p  2)/2 d.f.
Example 8.9 (Testing for equicorrelation structure) From Example 8.6, the sample correlation matrix constructed from the n = 150 postbirth weights of female
l
mice is _ R 
.7501 .6329 .6925 1.0 .7501 .6329 .6925 1.0 .6363 .7386 .6625
['0
63 .7386 @ .6625 1.0
We shall use this correlation matrix to illustrate the large sample test in (835). Here p = 4, and we set
H,p
~ ~ r~ 7 : p,
p
p
HJ:p
'* Po
:l
1 p p 1
.00245
Y =
(4  1f[1  (1  .6855)2]
~'"'':~
4  (4  2)(1  .6855f
= 2.1329
'13 (.7501 + .6329 + .6363) = .6731, r3 = .6626, r4 = .6791
rI =
r2 = .7271,
r = _2_ (.7501 + .6329 + .6363 + .6925 + .7386 + .6625) = .6855 4(3)
2:2: (rik  r)2 = (.7501  .6855)2 i
.01277
T
(150  1) ~ .6855)2 [.01277  (2.1329)(.00245)]
= (1
= 11.4
Since (p + 1) (p  2)/2 = 5(2)/2 = 5, the 5% critical value for the test in (835) is X~(.05) = 11.07. The value of our test statistic is approximately equal to the large sample 5% critical point, so the evidence against Ho (equal correlations) is strong, but not overwhelming. As we saw in Example 8.6, the smallest eigenvalues A2 , A3 , and A4 are slightly different, with A4 being somewhat smaller than the other two. Consequently, with the large sample size in this problem, small differences from the equal correlation structure show up as statistically significant. _ Assuming a multivariate normal population, a large sample test that all variables are independent (all the offdiagonal elements of l: are zero) is contained in Exercise 8.9.
8.6 Monitoring Quality with Principal Components In Section 5.6, we introduced multivariate control charts, including the quality ellipse and the T2 chart. Today, witlI electronic and other automated methods of data collection, it is not uncommon for data to be collected on 10 or 20 process variables. Major chemical and drug companies report measuring over 100 process variables, including temperature, pressure, concentration, and weight, at various positions along the production process. Even witlI 10 variables to monitor, there are 45 pairs for which to create quality ellipses. Clearly, another approach is required to both visually display important quantities and still have the sensitivity to detect special causes of variation.
Checking a Given Set of Measurements for Stability
Using (834) and (835), we obtain
=
=
and
It is evident that rk is the average of the offdiagonal elements in the kth column (or row) of Rand r is the overall average of the offdiag<;mal elements. The large sample approximate 'alevel test is to reject Ho in favor of HI if
T
+ ... + (.6791  .6855)2
k=I
2
+ ... + (.6625  .6855)2
Let Xl, X 2, ... , Xn be a random sample from a multivariate normal distribution with mean p. and covariance matrix l:. We consider the first two sample principal components, YiI = el(xi  x) and Yi2 = eZ(xi  x). Additional principal components could be considered, but two are easier to inspect visually and, of any two components, the first two explain tlIe largest cumulative proportion of the total sample variance. If a process is stable over time, so that the measured characteristics are influenced only by variations in common causes, then the values of the first two principal components should be stable. Conversely, if tlIe principal components remain stable over time, tlIe common effects that influence tlIe process are likely to remain constant. To monitor quality using principal components, we consider a twopart procedure. The first part of the procedure is to construct an ellipse format chart for the pairs of values (Yjl, Yi2) for j = 1, 2, ... , n.
460 Chapter 8 Principal Components
Monitoring Quality with Principal Components 461
By (820), the sample variance of the first principal component YI is given by the largest eigenvalue AI, and the sample variance of the second principal component is the secondlargest eigenvalue '\2' The two sample components are uncorrelated, • so the quality ellipse for n large (see Section 5.6) reduces to the collection of pairs of.· possible values CYI, .rz) such that '2
•
'2
YI < X22( a ) , + Y2 ,Al A2
§ ,,..,
Example 8.10 (An ellipse format chart based on the first two principal components) Refer to the police department overtime data given in Table 5.8. Table 8.1 contains the five normalized eigenvectors and eigenvalues of the sample covariance matrix S.
S
Appearances overtime Extraordinary event Holdover hours COA hours Meeting hours
e2
e3
e4
es
(x) (xz) (X3) (X4) (xs)
.046 .039 .658 .734 .155
.048 .985 .107 .069 .107
.629 .077 .582 .503 .081
.643 .151 .250 .397 .586
.432 .007 .392 .213 .784
Aj
2,770,226
1,429,206 '628,129
221,138
99,824
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Yjl
2044.9 2143.7 177.8 2186.2 878.6 563.2 403.1 1988.9 132.8 2787.3 283.4 761.6 498.3 2366.2 1917.8 2187.7
..
• •
• • •
•
§
e)
'"I 2000
o
2000
F·igure 8.7 The 95% control ellipse based on the first two principal components of overtime hours.
4000
Let us construct a 95% ellipse format chart using the first two sample principal components and plot the 16 pairs of component values in Table 8.2. Although n = 16 is not large, we use x~(.05) = 5.99, and the ellipse becomes
:1 + :2z
Table 8.2 Values of the Principal Components for the Police Department Data
Period
•
• +. •
I
Table 8.1 Eigenvectors and Eigenvalues of Sample Covariance Matrix for Police Department Data
Variable
8
The first two sample components explain 82 % of the total variance. The sample values for all five components are displayed in Table 8.2.
•
•
M
Yj2
Yj3
Yj4
YjS
588.2 686.2 464.6 450.5 545.7 1045.4 66.8 801.8 563.7 213.4 3936.9 256.0 244.7 1193.7 782.0 373.8
425.8 883.6 707.5 184.0 115.7 281.2 340.6 1437.3 125.3 7.8 0.9 2153.6 966.5 165.5 82.9 170.1
189.1 565.9 736.3 443.7 296.4 620.5 135.5 148.8 68.2 169.4 276.2 418.8 1142.3 270.6 196.8 84.1
209.8 441.5 38.2 325.3 437.5 142.7 521.2 61.6 6115 202.3 159.6 28.2 182.6 344.9 89.9 250.2
'z
'z
Al
A
:5
5.99
This ellipse centered at (0,0), is shown in Figure 8.7, along with the data. One point is out of control, because the second principal component for this point has a large value. Scanning Table 8.2, we see that this is the value 3936.9 for period 11. According to the entries of e2 in Table 8.1, the second principal component is essentially extraordinary event overtime hours. The principal component approach has led us to the same conclusion we came to in Example 5.9. • In the event that special causes are likely to produce shocks to the system, the second part of our twopart procedurethat is, a second chartis required. This chart is created from the information in the principal components not involved in the ellipse format chart. Consider the deviation vector X  /L, and assume that X is distributed as Np(/L, I,). Even without the normal assumption, Xj  /L can be expressed as the sum of its projections on the eigenvectors of I, X  /L = (X  /L)'elel + (X  /L)'eZ e 2
+ (X  /L)'e3e3 + ... + (X 
/L)'epe p
462
Monitoring Quality with Principal Components 463
Chapter 8 Principal Components or
Example 8.11 (A T 2 chart for the unexplained [orthogonal] overtime hours)
x11
!I 11
,I
p
= Yjel + Y2e2 + Y3e3 + ... + Ypep
where Yi = (X  p) I ei is the population ith principal c~m~onent centered to have mean O. The approximation to X  p by the first two pnnclpal components has the form Y1el + Y2e2' This leaves an unexplained component of
11
!i I: 1\
Consider the quality control analysis of the police department overtime hours in Example 8.10. The first part of the quality monitoring procedure, the quality ellipse based on the first two principal components, was shown in Figure 8.7. To illustrate the second step of the twostep monitoring procedure, we create the chart for the other principal components. Since p = 5, this chart is based on 5  2 = 3 dimensions, and the upper control limit is X~(.05) = 7.81. Using the eigenvalues and the values of the principal componentl', given in Example 8.10, we plot the time sequence of values
X  p  Y1ej  Y2e2 Let E = [el, e2, ... , epJ be the orthogonal matrix whose columns are the eigenvectors of~. The orthogonal transformation of the unexplained part,
E'(X 
TJ~ =
~ y~,)" m [!] [1] ~ m~ UJ
'2 Yj3 '
+
A3
'2 Yj4
+
'2 YjS
A
A
A4
As
where the first value is T2 = .891 and so on. The T 2chart is shown in Figure 8.8.
Y,', 
F~~~~UCL
so the last p  2 principal components are obtained as 2an orthogonal transfo~mat~on of the approximation errors. Rather than base the.T . chart on the approxImatIOn errors, we can, equivalently, base it on these last prmclpal components. Recall that
6
f..
Var (Y;) = Ai for i = 1,2, ... , P
4
"*
and Cov(Yi, Yk ) = 0 for i k. Consequently, the statistic Y(2)~Y~2).Y(2)Y(2)' based on the last p  2 population principal components, becomes
Y~ + _Y~ + ... + _ Y~ _ A3
A4
(838)
o 
Ap
This is just the sum of the squares of p  2 independent standard normal variables, A1/2y; and so has a chisquare distribution with p  2 degrees of freedom. k Ink ~erms of the sample data, the principal components and eigenval ueS must be estimated. Because the coefficients of the linear combinations ej are also estimates, the principal components do not have a normal distrib~tion even when the pop~l~ tion is normal. However, it is customary to create a T chart based on the statistic '2
T}
Yj3
= ;;A3
'2
2
'2
Yj4
Yjp
A4
Ap
+ ;; + ... + ,
which involves the estimated eigenvalues and vectors. Further, it is usual to appeal to the large sample approximation described by (838) and set the upper control 2 limit of the T 2chart as UCL = c = ~2( a). This T 2statistic is based on highdimensional data. For example, when p = 20 variables are measured, it uses the information in the 18dimensional space per?endicular to the first two eigenvectors el and e2' Still, this T2 based on the unexplamed variation in the original observations is reported as highly effective in picking up special causes of variation.
o
5
10
15 Period
2
Figure 8.8 A T chart based on the last three principal components of overtime hours.
Since points 12 and 13 exceed or are near the upper control limit, something has happened during these periods. We note that they are just beyond the period in which the extraordinary event overtime hours peaked. From Table 8.2, Y3j is large in period 12, and from Table 8.1, the large coefficients in e3 belong to legal appearances, holdover, and COA hours. Was there some adjust_ ing of these other categories following the period extraordinary hours peaked?
Controlling Future Values Previously, we considered checking whether a given series of multivariate observations was stable by considering separately the first two principal components and then the last p  2. Because the chisquare distribution was used to approximate the UCL of the T 2chart and the critical distance for the ellipse format chart, no further modifications are necessary for monitoring future values.
464 Chapter 8 Principal Components
Monitoring Quality with Principal Components 465
In Example 8.10, determined that case 11 was out of control. We drop this point and recalculate eigenvalues and eigenvectors based on the covariance of the remaining 15 tions. The results are shown in Table 8.3. Example 8.12 (Control ellipse for future principal components)
Appearances overtime Extraordinary event Holdover hours COA hours Meeting hours
(Xl) (X2) (X3) (X4) (xs)
. In ~ome applications of multivariate control in the chemical and pharmaceutical mdustnes, more than 100 variables are monitored simultaneously. These include numerous process variables as well as quality variables. Typically, the space orthogonal to the first few principal components has a dimension greater than 100 and some of the eigenvalues are very small. An alternative approach (see [13]) to constructing a control chart, that avoids the difficulty caused by dividing a small squared principal co~ponent by a very small eigenvalue, has been successfully applied. To implement thIS approach, we proceed as follows. . For each stable observation, take the sum of squares of its unexplained component
db j
e
= (Xj  X  Yjlel  Yj2 e 2) , (Xj  X  Yjlel  Yj2 2)
Note that, by inserting EE! = I, we also have
The principal components have changed. The component consisting primarily extraordinary event overtime is now the third principal component and is not inclUded in the chart of the first two. Because our initial sample size is only 16, dropping a single case can make a substantial difference. Usually, at least 50 or more observations are needed, from stable operation of the process, in order to set future limits. Figure 8.9 gives the 99% prediction (836) ellipse for future pairs of values for the new first two principal components of overtime. The 15 stable pairs of principal components are also shown.
which is just the sum of squares of the neglected principal components. Using either form, the dbj are plotted versus j to create a control chart. The lower limit of the chart is 0 and the upper limit is set by approximating the distribution of db j as the distribution of a constant c times a chisquare random variable with IJ degrees of freedom. For the chisquare approximation, the constant c and degrees of freedom IJ are chosen to match the sample mean and variance of the db j, j = 1,2, ... , n. In particular, we set 2"
du
~ 2 = 1 "'' d u j = C IJ n j=l
80
'"
•
and detennine
•
0 0
S <;J::
••
§
•
I
5000
2000
•
.. • te
•• • •
The upper control limit is then cx;(a), where a = .05 or .01.
•
o
2000
4000
Figure 8.9 A 99% ellipse , format chart for the first two principal components of future values of overtime.
The Geometry of the Sample Principal Component Approximation 467
Supplement
where
[Yjl, Yj2,···, YjrJ' = [el(Xj  x), e2(Xj  x), ... , e~(xj  x) l' are the valu~s of the first r sample principal components for the jth unit. Moreover,
L (Xj  x 
,=1
where Ar+1 ~ ... ~
8j)' (Xj 
x
8j) = (n  1) (A +1 r
+ ... + A) p
Ap are the smallest eigenvalues of S.
proo::
C?nsider first any A whose transpose A' has columns a· that are a linear matlOn of a flXe.d .set of r perpendicular vectors UI, ~2' ... ' Un so that . t e db y .t  [u(, . .u2, ... , u r] satIsfies U'U = I. For fixed U ' x,  XI·S bes t approxlma I S projectIon on the space spanned by U(, u2, ... , Ur (see Result 2A.3), or
~~
THE GEOMETRY OF THE SAMPLE PRINCIPAL COMPONENT ApPROXIMATION
(Xj  X)'UIUI
+ (Xj  X)'U2U2 + ... + (x,  i)'u r Ur =
UHXj  X)l uz(Xj  x) : =UU'(xji)
[UbU2, ... ,Ur]
(SA2)
[ u;(Xj  x)
,
This follows because, for an arbitrary vector b '
Xj  i  Ubj = Xj  i  UU'(Xj  i) In this supplement, we shall present interpretations for approximations to the data based on the first r sample principal components. The interpretations of both the pdimensional scatter plot and the ndimensional representation rely on the algebraic result that follows. We consider approximations of the form A = [ab a2, ... , an]' (nXp) to the mean corrected data matrix [Xl  X, X2  X, ... , Xn  X]' The error of approximation is quantified as the sum of the np squared errors (SAI)
Result SA. I Let
A be any matrix with rank(A)
~r<
min (p, n). Let Er =
+ UU'(Xj  x)  Ubj
= (I  UU') (Xj  i) + U(U'(Xj  i)  b j ) so the error sum of squares is
+0 + (V' (Xj  x)  b j )' (U' (Xj  x)  b j)
(Xj  i  Ubj)'(xj  i  Ubj) = (Xj  i)'(1  UV')(Xj  x)
where the cross product vanishes because (I  UU') U = U  UU'U = U  U _== 0 .The . posl~lve.unless .. . chosen so that b = U'(x  i) , last t IS b j IS and UbI UU (Xj  x) IS the projectIOn of x  i on the plane' , Further, with the choice a= Ub= UV"( xx) (SA 1) b· n ' " ' ecomes
!r:n
L (Xj  x 
UU'(Xj  i»' (x  i  UU'(x  x»
,=1
"
n
(nXp)
[eb e2, ... , er], where ei is the ith eigenvector of S. The error of approximation sum of squares in (8Al) is minimized by the choice
=
2: (Xj 
i)' (I  UV') (x  i)
,=1
'
t
±
(Xj  i)' (Xj  x) (Xj  i)'UU'(x  x) (SA3) ,I j=1 ' We are now in a position to minimize the error over choices of U b . _. h last term· (SA 3) B h YmaxlmlZmg t e m  . y t e properties of trace (see Result 2A.12), =
n
so the jth column of its transpose A' is 8j = hlel +
2: (Xj ,=1 e + ... + }ljrer
Yj2 2
i)'UU'(xj  x) =
±
tr[(x  i)'UU'(x  i)]
j=I" n
=
L j=1
466
tr[UU'(xj  i)(x  i)'] '
= (n  1) tr[UU'S] = (n  1) tr[U'SU]
(SA4)
468
Chapter 8 Principal Components
The Geometry of the Sample Principal Component Approximation 469
That is, the best choice for U maximizes the sum of the diagonal elements of U'SU. From (S19), selecting DJ to maximize D1SD], the first diagonal element of U' SU, gives 01 = el' For ~2 perpendicular to ~2SD2 i~m~ed by e2. [See (252).] Continuing, we find that U = [e], e2,"" er] = Er and A' = ErE;[Xl  X, X2  x, ... , XIl  x],as asserted. With this choice the ith diagonal element of U'SU is e:Sei = e:p'iei) = Ai so
3
e],
A
tr [U'siJ] = AJ + A2 + ... + Ar . Also,
,
d"
"
Ln (Xj  i)' (Xj  x) = tr [ Ln (Xj  i) (Xj  i)' j=1
j=1
(n  1) tr(S) = (n  l)(Al + A2 + ... + error bound follows.
=
A/?).
Let U =
U in
(SA3), and the •
r   "       ____+_ 2 Figure 8_10 The r = 2dimensional plane that approximates the scatter n
The pDimensional Geometrical Interpretation
plot by minimizing
The geometrical interpretations involve the determination of best approximating planes to the pdimensional scatter plot. The plane through the origin, determined by uJ, U2,"" u" consists of all points x with for some b This plane, translated to pass through a, becomes a + Ub for some b. We want to select the rdimensional plane a + Ub that minimizes the sum of
" squared distances L dJ between the observations Xj and the plane. If Xj is approxij=1
" mated by a + Ubj with L b j =
0,5 then
L"
j=J
2,
ob~rvati,ons. Fr':.m (8A 2 the projection of the deviation x;  i on the plane Ub is Vj : l!U (Xj  x). Now, v = 0 and the sum a/the squared lengths a/the projection d ev la t/O I1S n
11
.? vjVj =
J1
is maximized by U
=
~ (Xj  i)'UU'(xj 
J=1
E. Also, since v =
x) = (n  1) tr[U'SU]
0, n
It
(n  l)S.
j=1
=
L (v; 
v)(Vj  v)'
L V·V~
=
j=1
(Xj 
a  Ubj)'(xj  a  Ubi)
2: dY.
j=1
J J
and this plane also maximizes the total variance
j=1
n
=
L
(Xj 
x
Ubj + ia)' (Xj 
(Xj 
x
Ubj)' (Xj 
x
"
x
Ubj +
x
a)
j=1
=
L"
x
Ubj )
+ n(x  a)' (x  a)
j=1
n
2:
L
(Xj 
A
ErE;(xj  i»'(xj 
x
"
A
ErE;(xj 
x»
j=1
by Result SA.1, since [Ublo'''' Ub,,] = A' has rank (A) :;; r. The lower bound is reached by taking a = x, so the plane passes through the sample mean. This plane is determined bye], e2"'" er' The coefficients of ek are ek(xj  x) = Jjb the kth sample principal component evaluated at the jth observation. . The approximating plane interpretation of sample principal components IS illustrated in Figure S.10. An alternative interpretation can be given. The investigator places a plane through xand moves it about to obtain the largest spread among the shadows of the 5 If
~
~
i=l
b·J =. nb
tr(S.) = ( 11
L
_ 1 1) tr [" Vjvj ] = j=1
1
(11 
1)
L
tr [" v'v· ] ;=1
)
J
The nDimensional Geometrical Interpretation Let ~s now consider, by columns, the approximation of the meancentered data matnx by A. For: = 1, t~e ith column [Xli  Xi' X2i  Xi,' .. , X"i  X;]' is approximated by a multIple cib of a fixed vector b' = [b], b2 , ..• , bIll. The square of the length of the error of approximation is n
LT = .L (Xji j=1
Xi 
ci bj )2
Considering ( An to be of rank one we conclude from Result SA . 1 that xp)'
Exercises 471
470 Chapter 8 Principal Components 3
(b) Compare the components calculated in Part a with those obtained in Exercise 8.1. Are they the same? Should they be? (c) Compute the correlations PYI>ZI' PYI>Z2' and PY2,z!, 8.3. Let
2 0 0] [0 0 4
1= 0 4 0 ~~~2
~... 2
Determine the principal components YI , Y2 , and Y3' What can you say about the eigenvectors (and principal components) associated with eigenvalues that are not distinct? 8.4. Find the principal components and the proportion of the total population variance explained by each when the covariance matrix is
(a) Principal component of S
(b) Principal component of R
Figure 8.11 The first sample principal component,'vI' minimizes the sum of the squares of the distances, L from the deviation vectors, d; = [Xli  Xi, X2i  Xi,"" Xni  Xi], to a line.
r;
1
1
v2
v2

p
minimizes the sum of squared lengths
2: LT. That is, the best direction is determined i=1
'
P=
by the vector of values of the first principal component. This is illustrated in Figure 8.11( a). Note that the longer deviation vectors (the larger s;;'s) have the most
Are your results consistent with (816) and (817)? (b) Verify the eigenvalueeigenvector pairs for the p X P matrix p given in (815).
p
influence on the minimization of
2: LT·
i=1
If the variables are first standardized, the resulting vector [(Xli  Xi)/YS;;, (XZ'  X)/vs:. (X .  x)/vs:.] has length n  1 for all variables, and each vec~or e~erts ~~~~i infl~~nce ~n th~'choice of direction. [See Figure 8.11(b).] In either case, the vector b is moved around in nspace to minimize the sum of P the squares of the distances L7. In the former case Liz·IS the squared d'Istance
2:
i=1
between [Xli  Xi> XZi  Xi,"" Xni  Xi)' and its projection on the line determined by b. The second principal component minimizes the same quantity among all vectors perpendicular to the first choice.
Exercises 8.1. Determine the population principal components YI and Yz for the covariance matrix
I =
1 P p] pIp [P P 1
[~ ~J
Also, calculate the proportion of the total population variance explained by the first principal component. 8.2. Convert the covariance matrix in Exercise 8.1 to a correlation matrix p. (a) Determine the principal components YI and Y2 from p and compute the proportion of total population variance explained by YI .
8.6. Data on XI = sales and X2 = profits for the 10 largest companies in the world were listed in Exercise 1.4 of Chapter 1. From Example 4.12 i = [155.60J
14.70 '
s=
[7476.45 303.62J 303.62 26.19
(a) Determine the sample principal components and their variances for these data. (You may need the quadratic formula to solve for the eigenvalues of S.) (b) Find the proportion of the total sample variance explained by 'vI' (c) Sketch the constant density ellipse (x  X)'SI(X  x) = 1.4, and indicate the principal components 511 and 512 on your graph. (d) Compute the correlation coefficients 'Yl>}(k' k = 1,2. What interpretation, if any, can you give to the first principal componeflt? 8.7. Convert the covariance matrix S in Exercise 8.6 to a sample correlation matrix R. (a) Find the sample principal components 511, Yz and their variances. (b) Compute the proportion of the total sample variance explained by 511' (c) Compute the correlation coefficients 'YI>Zk' k = 1,2. Interpret 'vI' (d) Compare the components obtained in Part a with those obtained in Exercise 8.6(a). Given the original data displayed in Exercise 1.4, do you feel that it is better to determine principal components from the sample covariance matrix or sample correlation matrix? Explain.
Exercises 473
472 Chapter 8 Principal Components Hint:
8.8. Use the results in Example 8.5.
(a) Compute the correlations r,;,Zk for i = 1,2 and k = 1,2, ... ,5. Do?these ~orrela • tions reinforce the interpretations given to the first two components. Explam.
(a) max L(JL,:t) is given by (510), and max L(JL, :to) is the product of the univariate p,};'
likelihoods, maX(27T)n/2O'i;n12eXP[±(XjjJLY/2O'il]. Hence ILi
(b) Test the hypothesis 1 p p p p 1 p P
Ho:
P
Po
=
p
p p
p
p
p
p' p
(b) Verify 0 2 =
Xl)2
+ ... +
±(Xjp  xp/J/n
p under Ho. Again,
/=1
the divisors n cancel in the statistic, so S may be used. Use Result 5.2 to calculate the chisquare degrees of freedom. The following exercises require the use of a computer.
at the 5% level of significance. List any assumptions required in carrying out this test. (A test that all variables are independent.)
(a) Consider that the normal theory likelihood ratio test of Ho: :t is the diagonal matrix
o 0'22
IT
o
A
s In/2
= I   = IR In/2 < p TI ;=1
n/2
For a large sample size, 2ln A is approximately X~(pl)/~' Bartlett [3] suggests th~t the test statistic 2[1  (2p + 1l)/6nJlnA be used m place of :~lnA ..Th~s results in an improved chisquare approximation. The larg~ sample a CrItical pomt IS 2 )1 (a) . Note that testing:t = :to is the same as testmg p = I. X p(pI 2
(1,(8)/ p
l
IT
A.
= 0'21
]n12
)"'~ ~ ~ i,), 
IS Inl2
.
,~I
(;,
A
I
in (820). (Note that the sample mean vector x is displayed in Example 8.5.) (b) Determine the proportion of the total sample variance explained by the first three principal components. Interpret these components. (c) Construct Bonferroni simultaneous 90% confidence intervals for the variances AI, A2 , and A3 of the first three population components YI , Y2 , and Y 3 • (d) Given the results in Parts ac, do you feel that the stock ratesofreturn data can be summarized in fewer than five dimensions? Explain.
rejects Ho if .
[Mithm'ti' moon
npl2
J A
geometrIC mean Aj
_
JP Morgan
Citibank
1 2 3 4 5 6 7 8 9 10
0.01303 0.00849 0.01792 0.02156 0.01082 0.01017 0.01113 0.04848 0.03449 0.00466
0.00784 0.01669 0.00864 0.00349 0.00372 0.01220 0.02800 0.00515 0.01380 0.02099
0.00319 0.00621 0.01004 0.01744 0.01013 0.00838 0.00807 0.01825 0.00805 0.00608
94 95 96 97 98 99 100 101 102 103
0.03732 0.02380 0.02568 0.00606 0.02174 0.00337 0.00336 0.01701 0.01039 0.01279
0.03593 0.00311 0.05253 0.00863 0.02296 0.01531 0.00290 0.00951 0.00266 0.01437
0.02528 0.00688 0.04070 0.00584 0.02920 0.02382 0.00305 0.01820 0.00443 0.01874
Week
c
Sji
(b) Show that the likelihood ratio test of Ho: :t
8.10. The weekly rates of return for five stocks listed on the New York Stock Exchange are given in Table 8.4. (See the stockprice data on the following website: www.prenhal1.comlstatistics.) (a) Construct the sample covariance matrix S, and find the sample principal components
Table 84 StockPrice Data (Weekly Rate Of Return)
Show that the test is as follows: Reject Ho if
~
(xj1 
/=1
versus
A
j=l
(Xjj  Xj)2. The divisor n cancels in A, so S may be used.
j=1
1 p 1
[ p
=
.
8.9.
± [±
and ojj = (1In)
= nI±xjj
j=l
J.LjUjj
<
:
C
for a large sample size, Bartlett [3] suggests that 2[1  (2p2 + P + 2)/6pn) In A .. al 'nt is is approximately Xtp+2){p1)/2' Thus, the large sample a CrItIc pO! . 2 (a) This test is called a sphericity test, because the constant denSIty . X(p+2){pl)/2 • 2 contours are spheres when:t = 0' I.
Wells Pargo
Royal Dutch Shell
Exxon Mobil
0.04477 0.01196 0 0.02859 0.02919 0.01371 0.03054 0.00633 0.02990 0.02039
0.00522 0.01349 0.00614 0.00695 0.04098 0.00299 0.00323 0.00768 0.01081 0.01267
0.05819 0.01225 0.03166 0.04456 0.00844 0.00167 0.00122 0.01618 0.00248 0.00498
0.01697 0.02817 0.01885 0.03059 0.03193 0.01723 0.00970 0.00756 0.01645 0.01637
:
Exercises 475 474 Chapter 8 Principal Components 'der the censustract dat~ listed in Table 8.5. Suppose the observations on d' lue home were recorded in ten thousands, rather than hundred thousands, Xs = me Jan va . h . h I fth table by 10 of dollars; that is, multiply all the numbers listed m t e SlXt co umn 0 e . C t the sample covariance matrix S for the censustract data when lue home is recorded in ten thousands of dollars. (Note that . (a) on~truc d' Xs  me lan va . ' . . E I . atrix can be obtained from the covanance matnx given m xamp e 8.3 covanance m . h f'f h i d ow by 10 by multiplying the offdiagonal elements m t e I t co umn an r an d th e diagonal element S55 by 100. Why?) . . (b) Obtain the eigenvalueeigen~e~tor pairs and the first two sample pnnclpal components for the covariance matnx m Part a. . , . c Corn ute the proportion of totar variance explained .by the f~r~t two pnnclpal ( ) p t obtained in Part b Calculate the correlatIOn coefficients, ry;.Xk' and ~omponetnths e components if p' ossible. Compare your results with the results in' es f h' h . I h mterpre 3 Wh at. can you say about the effects 0 t IS C ange m sca e on t e . Exampe I 8., principal components? 'd h . II tion data listed in Table 1.5. Your job is to summarize these data in Sl2ConslertealTpoU .' ' 0 fh • . _ 7 d' ensions if possible. Conduct a pnnclpal componen t ana IYSls t e·· fewer ~an bP th thel~ovariance matrix S and the correlation matrix R. What have you . ' . ' ? C an th e d at a be data usmg matnx IS chosen for anaI YSls. ? D 0 't make any difference which d oes I learne. . h' . I t ? . d' th e or fewer dimensions? Can you mterpret t e prmclpa componen s. summarIZe m re
S.II. Consl
8.13. In the radiotherapy data listed in Table 1.7 (see also the radiotherapy data on the website www.prenhall.com/statistics). the n = 98 observations on p = 6 variables represent patients' reactions to radiotherapy. (a) Obtain the covariance and correlation matrices Sand R for these data. (b) Pick one of the matrices S or R (justify your choice), and determine the eigenvalues and eigenvectors. Prepare a table showing, in decreasing order of size, the percent that each eigenvalue contributes to the total sample variance. (c) Given the results in Part b, decide on the number of important sample principal components. Is it possible to summarize the radiotherapy data with a single reactionindex component? Explain. (d) Prepare a table of the correlation coefficients between each principal component you decide to retain and the original variables. If possible, interpret the components. 8.14. Perform a principal component analysis using the sample covariance matrix of the sweat data given in Example 5.2. Construct a QQ plot for each of the important principal components. Are there any suspect observations? Explain. S.IS. The four sample standard deviations for the postbirth weights discussed in Example 8.6
are
v'5,';' = 32.9909,
VS22
= 33.5918,
Vs))
= 36.5534,
and
VS44
= 37.3517
Use these and the correlations given in Example 8.6 to construct the sample covariance matrix S.Perform a principal component analysis using S.
Tract
1 2 3 4 5 6 7 8 9 10
52 53 54 55 56 57 58 59 60 61
Median home value ($100,000)
Total population (thousands)
Professional degree (percent)
Employed age over 16 (percent)
Government employment (percent)
2.67 2.25 3.12 5.14 5.54 5.04 3.14 2.43 5.38 7.34
5.71 4.37 10.27 7.44 9.25 4.84 4.82 2.40 4.30 2.73
69.02 72.98 64.94 71.29 74.94 53.61 67.00 67.20 83.03 72.60
30.3 43.3 32.0 24.5 31.0 48.2 37.6 36.8 19.7 24.5
1.48 1.44 2.11 1.85 2.23 1.60 1.52 1.40 2.07 1.42
1.16 2.93 4.47 2.26 2.36 6.30 4.79 5.82 4.71 4.93
78.52 73.59 77.33 79.70 74.58 86.54 78.84 71.39 78.01 74.23
23.6 22.3 26.2 20.2 21.8 17.4 20.0 27.1 20.6 20.9
1.50 1.65 2.16 1.58 1.72 2.80 2.33 1.69 1.55 1.98
7.25 5.44 5.83 3.74 9.21 2.14 6.62 4.24 4.72 6.48
:
. f d' nt census tracts are likely to be correlated. That is, these 61 observations may not Note''. ObservatIOns rom aI Jace . . . C plete data set available at www.prenhall.com/statJstlcs. constitute a random samp e. om
S.16. Over a period of five years in the 1990s, yearly samples of fishermen on 28 lakes in Wisconsin were asked to report the time they spent fishing and how many of each type of game fish they caught. Their responses were then converted to a catch rate per hour for Xl
= Bluegill
X2
= Black crappie
X3
= Smallmouth bass
X4
= Largemouth bass
Xs
= Walleye
X6
= Northern pike
The estimated correlation matrix (courtesy of Jodi Barnet)
R=
1 .4919 .2635 .4653 .2277 .0652
.4919 .3127 .3506 .1917 .2045
.2636 .3127 .4108 .0647 .2493
.4653 .3506 .4108 .2249 .2293
.2277  .1917 .0647 .2249 .2144
.0652 .2045 .2493 .2293 .2144 1
is based on a sample of about 120. (There were a few missing values.) Fish caught by the same fisherman live alongside of each other, so the data should provide some evidence on how the fish group. The first four fish belong to the centrarchids, the most plentiful family. The walleye is the most popular fish to eat. (a) Comment on the pattern of correlation within the centrarchid family XI through X4' Does the walleye appear to group with the other fish? (b) Perform a principal component analysis using only Xl through X4' Interpret your results. (c) Perform a principal component analysis using all six variables. Interpret your results.
Exercises 477 476 Chapter 8 Principal Components 8.11. Using the data on bone mineral content in Table 1.8, perform a principal component analysis of S. 8.18. The data on national track records for women are'listed in Table 1.9. (a) Obtain the sample correlation matrix R for these data, and determine its ~·5""·'alU". and eigenvectors. (b) Determine the first two principal components for the standardized variables. Prepare a table showing the correlations of the standardized variables with the nents, and the cumulative percentage of the total (standardized) sample explained by the two components. (c) Interpret the two principal components obtained in Part b. (Note that the first component is essentially a normalized unit vector and might measure the athletic excellence of a given nation. The second component might measure the relative strength of a nation at the various running distances.) (d) Rank the nations based on their score on the first principal component. Does this ranking correspond with your inituitive notion of athletic excellence for the various
countries? 8.19. Refer to Exercise 8.18. Convert the national track records for women in Table 1.9 to speeds measured in meters per second. Notice that the records for 800 m, 1500 m, 3000 m, and the marathon are given in minutes. The marathon is 26.2 miles, or 42,195 meters, long. Perform a principal components analysis using the covariance matrix S of the speed data. Compare the results with the results in Exercise 8.18. Do your interpretations of the components differ? If the nations are ranked on the basis of their s~ore on the first principal component, does the subsequent ranking differ from that in Exercise 8.18? Which analysis do you prefer? Why? 8.20. The data on national track records for men are listed in Table 8.6. (See also the data on national track records for men on the website www.prenhall.comlstatistics) Repeat the principal component analysis outlined in Exercise 8.18 for the men. Are the results consistent with those obtained from the women's data? 8.21. Refer to Exercise 8.20. Convert the national track records for men in Table 8.6 to speeds measured in meters per second. Notice that the records for 800 m, 1500 m, 5000 m, 10,000 m and the marathon are given in minutes. The marathon is 26.2 miles, or 42,195 meters, long. Perform a principal component analysis using the covariance matrix S of the speed data. Compare the results with the results in Exercise 8.20. Which analysis do you prefer? Why? 8.22. Consider the data on bulls in Table 1.10. Utilizing the seven variables YrHgt, FtFrBody, PrctFFB, Frame, BkFat, SaleHt, and Sale Wt, perform a principal component analysis using the covariance matrix S and the correlation matrix R. Your analysis should include the following: (a) Determine the appropriate number of components to effectively summarize the sample variability. Construct a scree plot to aid your determination. (b) Interpret the sample principal components. (c) Do you think it is possible to develop a "body size" or "body configuration" index from the data on the seven variables above? Explain. (d) Using the values for the first two principal components, plot the data in a twodimensional space with YI along the vertical axis and Yz along the horizontal axis. Can you distinguish groups representing the three breeds of cattle? Are there any outliers? (e) Construct a QQ plot using the first principal component. Interpret the plot.
Table 8.6 National1rack Records for Men Country Argentina Australia Austria Belgium Bermuda Brazil Canada Chile China Columbia Cook Islands Costa Rica Czech Republic Denmark DominicanRepublic Finland France Germany Great Britain Greece Guatemala Hungary India Indonesia Ireland Israel Italy Japan Kenya Korea, South Korea, North Luxembourg Malaysia Mauritius Mexico Myanmar(Burma) Netherlands New Zealand Norway Papua New Guinea Philippines Poland Portugal Romania Russia Samoa Singapore Spain Sweden Switzerland Taiwan Thailand Thrkey USA
800 m 1500 m
5000 m
10,000 m Marathon
lOOm (s)
200 m
400 m
(s)
(s)
(min)
(min)
(min)
(min)
(min)
10.23 9.93 10.15 10.14 10.27 10.00 9.84 10.10 10.17 10.29 10.97 10.32 10.24 10.29 10.16 10.21 10.02 10.06 9.87 10.11 10.32 10.08 10.33 10.20 10.35 10.20 10.01 10.00 10.28 10.34 10.60 10.41 10.30 10.13 10.21 10.64 10.19 10.11 10.08 10.40 10.57 10.00 9.86 10.21 10;11 10.78 10.37 10.17 10.18 10.16 10.36 10.23 10.38 9.78
20.37 20.06 20.45 20.19 20.30 19.89 20.17 20.15 20.42 20.85 22.46 20.96 20.61 20.52 20.65 20.47 20.16 20.23 19.94 19.85 21.09 20.11 20.73 20.93 20.54 20.89 19.72 20.03 20.43 20.41 21.23 20.77 20.92 20.06 20.40 21.52 20.19 20.42 20.17 21.18 21.43 19.98 20.12 20.75 20.23 21.86 21.14 20.59 20.43 20.41 20.81 20.69 21.04 19.32
46.18 44.38 45.80 45.02 45.26 44.29 44.72 45.92 45.25 45.84 51.40 46.42 45.77 45.89 44.90 45.49 44.64 44.33 44.36 45.57 48.44 45.43 45.48 46.37 45.58 46.59 45.26 44.78 44.18 45.37 46.95 47.90 46.41 44.69 44.31 48.63 45.68 46.09 46.11 46.77 45.57 44.62 46.11 45.77 44.60 49.98 47.60 44.96 45.54 44.99 46.72 46.05 46.63 43.18
1.77 1.74 1.77 1.73 1.79 1.70 1.75 1.76 1.77 1.80 1.94 1.87 1.75 1.69 1.81 1.74 1.72 1.73 1.70 1.75 1.82 1.76 1.76 1.83 1.75 1.80 1.73 1.77 1.70 1.74 1.82 1.76 1.79 1.80 1.78 1.80 1.73 1.74 1.71 1.80 1.80 1.72 1.75 1.76 1.71 1.94 1.84 1.73 1.76 1.71 1.79 1.81 1.78 1.71
3.68 3.53 3.58 3.57 3.70 3.57 3.53 3.65 3.61 3.72 4.24 3.84 3.58 3.52 3.73 3.61 3.48 3.53 3.49 3.61 3.74 3.59 3.63 3.77 3.56 3.70 3.35 3.62 3.44 3.64 3.77 3.67 3.76 3.83 3.63 3.80 3.55 3.54 3.62 4.00 3.82 3.59 3.50 3.57 3.54 4.01 3.86 3.48 3.61 3.53 3.77 3.77 3.59 3.46
13.33 12.93 13.26 12.83 14.64 13.48 13.23 13.39 13.42 13.49 16.70 13.75 13.42 13.42 14.31 13.27 12.98 12.91 13.01 13.48 13.98 13.45 13.50 14.21 13.07 13.66 13.09 13.22 12.66 13.84 13.90 13.64 14.11 14.15 13.13 14.19 13.22 13.21 13.11 14.72 13.97 13.29 13.05 13.25. 13.20 16.28 14.96 13.04 13.29 13.13 13.91 14.25 13.45 12.97
27.65 27.53 27.72 26.87 30.49 28.13 27.60 28.09 28.17 27.88 35.38 28.81 27.80 27.91 30.43 27.52 27.38 27.36 27.30 28.12 . 29.34 28.03 28.81 29.65 27.78 28.72 27.28 27.58 26.46 28.51 28.45 28.77 29.50 29.84 27.14 29.62 27.44 27.70 27.54 31.36 29.04 27.89 27.21 27.67 27.90 34.71 31.32 27.24 27.93 27.90 29.20 29.67 28.33 27.23
129.57 127.51 132.22 127.20 146.37 126.05 130.09 132.19 129.18 131.17 171.26 133.23 131.57 129.43 146.00 131.15 126.36 128.47 127.13 132.04 132.53 132.10 132.00 139.18 129.15 134.21 127.29 126.16 124.55 127.20 129.26 134.03 149.27 143.07 127.19 139.57 128.31 128.59 130.17 148.13 138.44 129.23 126.36 132.30 129.16 161.50 144.22 127.23 130.38 129.56 134.35 139.33 130.25 125.38
Source: lAAFlATES Track and Field Statistics Handbook for the Helsinki 2005 Olympics. Courtesy of Ottavio Castellini.
478 Chapter 8 Principal Components Exercises 479 8.23. A naturalist for the Alaska Fish and Game Department studies grizzly bears with the goal of maintaining a healthy population. Measurements on n = 61 bears provided following summary statistics: .
Variable
Sample mean x
Weight (kg)
Body length (cm)
95.52
164.38
Neck (cm)
55.69
Girth (cm)
Head length (cm)
Head width (cm)
93.39
17.98
31.13
Covariance matrix
s=
3266.46 1343.97 731.54 1175.50 162.68 238.37
1343.97 721.91 324.25 537.35 80.17 117.73
(b) Interpret the sample principal components. (c) D? you t~ink it it i.s possible to develop a "paper strength" index that effectively contams the mformatlOn in the four paper variables? Explain. (d) Using the values for the first two principal components, plot the data in a twodimensional space with YI along the vertical axis and Y2 along the horizontal axis. Identify any outliers in this data set. 8.28. ~urvey data were coll.ected as part of a study to assess options for enhancing food secunty.through the sustaInable use of natural resources in the Sikasso region of Mali (West Afnca). A total of n = 76 farmers were surveyed and observations on the nine variables
XI = Family (total number of individuals in household)
731.54 1175.50 162.68 238.37 324.25 537.35 80.17 117.73 179.28 281.17 56.80 39.15 281.17 474.98 63.73 94.85 39.15 63.73 13.88 9.95 56.80 94.85 13.88 21.26
(a) Perform a principal component analysis using the covariance matrix. Can the data be effectively summarized in fewer than six dimensions? (b) Perform a principal component analysis using the correlation matrix. (c) Comment on the similarities and differences between the two analyses. 8.24. Refer to Example 8.10 and the data in Table 5.8, page 240. Add the variable X6 = regular overtime hours whose values are (read across) 6187 7679
7336 8259
6988 10954
6964 9353
8425 6291
6778 4969
5922 4825
7307 6019
and redo Example 8.10. 8.25. Refer to the police overtime hours data in Example 8.10. Cons~ruct an .al~ern~te cont~ol chart, based on the sum of squares db j, to monitor the unexplaIned vanatlon m the onginal observations summarized by the additional principal components. 8.26. Consider the psychological profile data in Table 4.6. Using the five var~abl~s, Indep, Sup~, Benev, Conform and Leader, performs a principal component analYSIS usmg the cov~n ance matrix S and the correlation matrix R Your analysis should include the followmg: (a) Determine the appropriate number .of. components t~ e~ectively summarize the variability. Construct a scree plot to aid m your determInation. (b) Interpret the sample principal components. . (c) Using the values for the: first two principal co~pone~ts, plot the dat~ m a tW?dimensional space with YI along the vertical aXIs and Y2 along the honzontal axiS. Can you distinguish groups representing the two socioeconomic levels and/or the two genders? Are there any outliers? . . (d) Construct a 95% confidence interval for Ab the variance of the first population principal component from the covariance matrix. 8.27. The pulp and paper properties data is given in Table 7.7. Using the four paper variables, BL (breaking length), EM (elastic modulus), .SF .(Stress at f~ilure) and. BS strength), perform a principal component analYSIS USIng the covanance matnx Sand correlation matrix R. Your analysis should include the following: (a) Determine the appropriate number of components to effectively summarize variability. Construct a scree plot to aid in your determination.
X2
=
X3
=
DistRd (distance in kilometers to nearest passable road) Cotton (hectares of cotton planted in year 2000)
X4
=
Maize (hectares of maize planted in year 2000)
Xs
= Sorg (hectares of sorghum planted in year 2000)
X6
=
Millet (hectares of miJIet planted in year 2000)
X7
= Bull (total number of bullocks or draft animals)
Xs
=
Cattle (total);
X9 =
Goats (total)
were recorded. The data are listed in Table 8.7 and on the website www.prenhall.com/statistics (a) Construct twodimensional scatterplots of Family versus DistRd, and DistRd versus Cattle. Remove any obvious autliers from the data set. Table 8.7 Mali Family Farm Data Family
DistRD
12 54 11 21 61 20 29 29 57 23
80 8 l3 13 30 70 35 35 9 33
20 27 18 30
0 41 500 19 18 500 100 100 90 90
Cotton
Maize
Sorg
Millet
Bull
1.5 6.0 .5 2.0 3.0 0 1.5 2.0 5.0 2.0
1.00 4.00 1.00 2.50 5.00 2.00 2.00 3.00 5.00 2.00
3.0 0 0 1.0 0 3.0 0 2.0 0 1.0
.25 1.00 0 0 0 0 0 0 0 0
2 6 0 1 4 2 0 0 4 2
1 5 0 5 0 3 0 0 2 7
1.5 1.1 2.0 2.0 8.0 5.0 .5 2.0 2.0 10.0
:
:
0 32 0 0 21 0 0 0 5 1
1.00 .25 1.00 2.00 4.00 1.00 .50 3.00 1.50 7.00
3.0 1.5 1.5 4.0 6.0 3.0 0 0 1.5 0
0 1.50 .50 1.00 4.00 4.00 1.00 .50 1.50 1.50
1 0 1 2 6 1 0 3 2 7
6 3 0 0 8 0 0 14 0 8
0 1 0 5 6 5 4 10 2 7
:
77 21 l3 24 29 57
Source: Data courtesy of Jay Angerer.
Cattle
Goats
:
480
Chapter 8 Principal Components (b) Perform a principal component analysis using the correlation matrix R. Determine the number of components to effectively summarize the variability. Use the propor" tion of variation explained and a scree plot to aid in your determination. (c) Interpret the first five principal components. Can you identify, for example, a size" component? A, perhaps, "goats and distance to road" component?
8.29. Refer to Exercise 5.28. Using the covariance matrix S for the first 30 cases of car assembly data, obtain the sample principal components. (a) Construct a 95% ellipse format chart using the first two principal components.vl Yz. Identify the car locations that appear to be out of control. (b) Construct an alternative control chart, based on the sum of squares db j, to the variation in the original observations summarized by the remaining four princi" pal components. Interpret this chart.
References 1. Anderson, T. W. An Introduction to Muftivariate Statistical Analysis (3rd ed.). New John Wiley, 2003. 2. Anderson, T. W. "Asymptotic Theory for Principal Components Analysis." Annals of Mathe/1zatical Statistics, 34 (1963), 122148. 3. Bartlett, M. S. "A Note on Multiplying Factors for Various ChiSquared Approximations." Journal of the Royal Statistical Society (B), 16 (1954), 296298. 4. Dawkins, B. "Multivariate Analysis of National Track Records." The American Statistician,43 (1989), 110115. 5. Girschick, M. A. "On the Sampling Theory of Roots of Determinantal Equations." Annals of Mathematical Statistics, 10 (1939),203224. 6. Hotelling, H. "Analysis of a Complex of Statistical Variables into Principal Components." Journal of Educational Psychology, 24 (1933),417441,498520. 7. Hotelling, H. "The Most Predictable Criterion." Journal of Educationaf Psychology, 26 (1935), 139142. 8. Hotelling, H. "Simplified Calculation of Principal Components." Psychometrika, 1 (1936),2735. 9. Hotelling, H. "Relations between Two Sets ofVariates." Biometrika, 28 (1936),321377. 10. Jolicoeur, P. "The Multivariate Generalization of the Allometry Equation." Biometrics, 19 (1963),497499. 11. Jolicoeur, P., and 1. E. Mosimann. "Size and Shape Variation in the Painted Turtle: A Principal Component Analysis." Growth, 24 (1960),339354. 12. King, B. "Market and Industry Factors in Stock Price Behavior." Journal of Business, 39 (1966), 139190. 13. Kourti, T., and 1. McGregor, "Multivariate SPC Methods for Process and Product Monitoring," Journal of Quality Technology, 28 (1996),409428. 14. Lawley, D. N. "On Testing a Set of Correlation Coefficients for Equality." Annals of Mathematical Statistics, 34 (1963), 149151. 15. Rao, C. R. Linear Statistical Inference and Its Applications (2nd ed.). New York: WileyInterscience,2oo2. 16. Rencher, A. C. "Interpretation of Canonical Discriminant Functions, Canonical Variates and Principal Components." The American Statistician, 46 (1992),217225.
FACTOR ANALYSIS AND INFERENCE FOR STRUCTURED COVARIANCE MATRICES 9.1 Introduction Factor analy~is ~as p~o~oked rather turbulent controversy throughout its history. Its modern begInnIngs he m the early20thcentury attempts of Karl Pearson, Charles Spea~m?n, a~d others to define and measure intelligence. Because of this early aSSOCIatIOn With constructs such as intelligence, factor analysis was nurtured and developed primarily by scientists interested in psychometrics. Arguments over the psychological interpretations of several early studies and the lack of powerful computing facilities impeded its initial development as a statistical method. The advent of highspeed computers has generated a renewed interest in the theoretical and computational aspects of factor analysis. Most of the original techniques have been ~ba?doned and early controversies resolved in the wake of recent developments. It IS std I true, however, that each application of the technique must be examined on its own merits to determine its success. . ~e e~sential purpose of factor analysis is to describe, if possible, the covariance relatIOnshIps a~ong many variables in terms of a few underlying, but un observable, rando~ quantities called factors. Basically, the factor model is motivated by the follOWIng argument: Suppose variables can be grouped by their correlations. That is, suppose all variables within a particular group are highly correlated among them~e~ves, bu~ have relatively small correlations with variables in a different group. Then It IS concelvabl.e that each group of variables represents a single underlying construct, or factor, that IS responsible for the observed correlations. For example, correlations from the group of test scores in classics, French, English, mathematics, and music colIect.ed by Spearman suggested an underlying "intelligence" factor. A second group of variables, repr~se~ting physicalfitness scores, if available, might correspond to another factor. It IS thiS type of structure that factor analysis seeks to confirm. 481
482
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices The Orthogonal Factor Model
FaCtor analysis can be considered an extension of principal component analysis, Both can be viewed as attempts to approximate the covariance matrix l:. However the approximation based on the factor analysis model is more elaborate. Th~ primary question in factor analysis is whether the data are consistent with a prescribed structure.
and that F and e are independent, so Cov(e,F)
£llFl
0 (pXm)
Orthogonal Factor Model with m Common Factors
The observable random vector X,.with p components, has mean p, and C01varian,.,..' matrix l:. The factor model postulates that X is linearly dependent upon a few unobservable random variables Fl , F2, ... , Fm, called common factors, and p additional sources of variation El, E2, ... , Ep' called errors or, sometimes, specific factors. 1 In particular, the factor an~lysis model is
£2l F l
= E(eF') =
These assumptions and the relation in (92) constitute the orthogonal factor model.2
9.2 The Orthogonal Factor Model
Xl  ILl = X 2  IL2 =
4¥3
X=p,+L F+e (pXl) (pXl) (pXm)(mXl) (pXl) ILi = mean of variable i Ei = ith specific factor
Fj
+ £12F2 + ... + flmFm + El + £22 F2 + ... + f2mFm + E2
(94)
= jth common factor
eij =
loading ofthe ith variable on the jth factor
The unobservable random vectors F and e satisfy the following conditions: F and e are independent E(F) = 0, Cov (F) = I
or, in matrix notation, XIL= L F (pXm)(mXl) (pXl)
+ E
E( e) = 0, Cov (e) = 'It, where 'I' is a diagonal matrix
(pXl)
The coefficient £ij is called the loading of the ith variable on the jth factor, so the matrix L is the matrix of factor loadings. Note that the ith specific factor Ei is associated only with the ith response Xi' The p deviations Xl  ILl, X 2  IL2,' .. , Xp  ILp are expressed in terms of p + m random variables Fj, F2, . .. , Fm, El, E2, ... , Ep which are unobservable. This distinguishes the factor model of (92) from the multivariate regression model in (7 23), in which the independent variables [whose position is occupied by Fin (92)] can be observed. With so many unobservable quantities, a direct verification of the factor model from observations on Xl, X 2, ... , Xp is hopeless. However, with some additional assumptions about the random vectors F and e, the model in (92) implies certain covariance relationships, which can be checked. We assume that E(F) =
E(e) =
0 ,
(mxI)
0 , (pXl)
Cov (F) = E[FF'] =
Cov(e) = E[ee'] = .
Th~ orthogonal factor model implies a covariance structure for X From the model In (94), . (X  p,) (X  p,)' = (LF + e) (LF + e), = (LF + e) «LF)' + e') =
so that
l:
I (mXm)
'It = (pXp)
LF(LF)' + e(LF)' + LFe' + ee'
= Cov(X) = E(X  p,) (X  p,)' = LE(FF')L' + E(eF')L' + LE(Fe') + E(ee') = LL'
["'?
0/2
:
jJ
(93)
1 As Maxwell [12] points out, in many investigations the E, tend to be combinations of measurement error and factors that are uniquely associated with the individual variables.
+ 'It
according to (93). Also by independence, Cov (e, F) = E( e F') = 0 Also, by the model in (94), (X  p,) F' = (LF + F' = LF F' Cov(X,F) = E(X  p,)F' = LE(FF') + E(eF') = L.
e)
2 AllOWing. the factors F to be correlated so that Cov (F) is not diagonal ~ m?deL The obhque model presents some additional estimation difficulties a .
thiS book. (See [10].)
l)Y
+ eF'.
484 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
The Orthogonal Factor Model 485 The equality
Covariance Structure for the Orthogonal Factor Model 1. Cov(X) = LL'
+
['9 3. 12J [4 1] 2 30 57 5 23 2 5 38 47 12 23 47 68
'If
or Var(Xi) =
e'rl + '" + Crm + I/Ii
=
7 2 1 6 1 8
4 [1
~J [~
1
7 2
6
0 4 0 0
+
0 0 1 0
or
COv(X;,Xk) = CilC kl + .,. + CimC km
~J
l: = LL' + 'If may be verified by matrix algebra. Therefore, l: has the structure produced by an m = 2 orthogonal factor model. Since
2. Cov(X,F) = L or
L = The model X  p. = LF + e is linear in the common factors. If the p responsesX are, in fact, related to underlying factors, but the relationship is nonlinear, such as in Xl  ILl = Cl1 F1F3 + Bl,X2  IL2 = C21 F2F3 + e2,andsoforth,th~nthecovari_ ance structure LV + 'If given by (95) may not be adequate. The very lmportant assumption of linearity is inherent in the formulation of the traditional factor model. That portion of the variance of the ith variable contributed by the m common factors is called the ith communality. That portion of Var (XJ = (J"ii due to the spe . cific factor is often called the uniqueness, or specific variance. Denoting the ith communality by hr, we see frOm (95) that
C22
C2l C3l
e32
£41
£42
'If =
0 0 0
1/12 0 0
_ 
7 2 1 6 '
1 8
0 0
I/Ii
0 0 0 0 1 0 0 4
1/13
the communality of Xl is, from (96),
hi = cL + e1 2 =
CrI + CT2 + '" + CYm +
42
+
12
= 17
and the variance of Xl can be decomposed as
~
communality
['" ' 'J [4 lJ r"' JJr~ ~J
+ specific variance
(J"ll= (erl+Cfz)
+ I/Il=hr+I/Il
or
or
(96) and i
19
+
~
'v'
variance
communality
2
variance A similar breakdown occurs for the other variables.
Example 9.1 (Verifying the relation
l: =
LL'
+
'I' for two factors) Consider the co
variance matrix 19 30
l:
=
25 23 12] 30 57 [ 2 5 38 47 12 23 47 68
+2
+ specific
= 1,2, ... , P
The ith communality is the sum of squares of the loadings of the ith variable on the m common factors.
17
~
•
Thefactor model assumes thatthe p + pep  1 )/2 = pep + 1 )/2 variances and covariances for X can be reproduced from the pm factor loadings Cij and the p specific variances I/Ii' When m = p, any covariance matrix l: can be reproduced exactly as LV [see (911)], so 'I' can be the zero matrix. However, it is when m is' small relativp to p that factor analysis is most useful. In this case, the factor model provides a"'" pIe" explanation of the covariation in X with fewer parameters than the pep parameters in l:. For example, if X contains p = 12 variables, and the factr (94) with m = 2 is appropriate, then the pep + 1)/2 = 12(13)/2 = '7~ l: are described in terms of the mp + p = 12(2) + 12 = 36 pararr the factor model.
/
/
The Orthogonal Factor Model 487
486 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices Unfortunately for the factor analyst, most covariance matrices cannot be factored as LL' + '11, where the number of factors m is much less than p. The follOWing example demonstrates one of the problems that can arise when attempting to determine the parameters Cij and o/i from the variances and covariances of the observable variables. Example 9.2 (Nonexistence of a proper solution) Let p = 3 and m = 1, and suppose the random variables Xl> Xz, and X3 have the positive definite covariance matrix
I
=
.4
which is unsatisfactory, since it gives a negative value for Var (e1) = 0/1' Thus, for this example with m = 1, it is possible to get a unique numerical solution to the equations I = LL' + '1'. However, the solution is not consistent with the statistical interpretation of the coefficients, so it is not a proper solution. _
1
x
Using the factor model in (94), we obtain
p = LF
+ E = LTT'F +
E
= L*F*
+
E
(97)
where
+ El C21 Fl + E2
Xl 
ILl
= C11 Fl
z
IL2
=
X3 
IL3
= C31 Fl
X
0/1 = 1  1.575 = .575
When m > 1, there is always some inherent ambiguity associated with the factor model. To see this, let T be any m X m orthogonal matrix 1 so that TT' = T'T = I. Then the expression in (92) can be written
. [1.9 .91 .7] .4 .7
gives
L* = LT
and
F* = T'F
Since
+ E3
E(F*) = T' E(F) = 0
and
The covariance structure in (95) implies that
Cov(F*) = T'Cov(F)T
I = LV + '11 or .90 = C11 C21
·70 = C11 C31
1 = C~l
AD = C21 C3l
+ o/z
1
=
C~1 + 0/3
The pair of equations
=
T'T =
I
(mXm)
it is impossible, on the basis of observations on X, to distinguish the loadings L from the loadings L*. That is, the factors F and F* = T'F have the same statistical properties, and even though the loadings L* are, in general, different from the loadings L, they both generate the same covariance matrix I. That is,
I
=
LV
+ '11 =
LTT'L'
+ 'I' = (L*) (L*), + 'I'
(98)
This ambiguity provides the rationale for "factor rotation," since orthogonal matrices correspond to rotations (and reflections) of the coordinate system for X .
.70 = C11 C31
.40 == C21 C31 Factor loadings L are determined only up to an orthogonal matrix T. Thus, the loadings
implies that
L*
=
LT
and
L
(99)
both give the same representation. The communalities, given by the diagonal elements of LL' = (L*) (L*), are also unaffected by the choice of T .
Substituting this result for C21 in the equation .90 = C11 C21
yieldS efl = 1.575, or Cl1 = ± 1.255. Since Var(Fd = 1 (by assumption) and Var(XI ) = 1, C11 = Cov(XI,Fd = Corr(X1 ,FI ). Now, a correlation coeffic~ent cannot be greater than unity (in absolute value), so, from this point of View, ICll l = 1.255 is too large. Also, the equation
1 =' Cl1 + o/l> or 0/1
=
1  Cl1
The analysis of the factor model proceeds by imposing conditions that allow one to uniquely estimate Land '11. The loading matrix is then rotated (multiplied by an orthogonal matrix), where the rotation is determined by some "easeofinterpretation" criterion. Once the loadings and specific variances are obtained, factors are identified, and estimated values for the factors themselves (called factor scores) are frequently constructed.
488
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Methods of Estimation 489
9.3 Methods of Estimation Given observations XI, x2,' .. , xn on p generally correlated variables, factor analysis. seeks to answer the question, Does the factor model of (94), with a small number of. factors, adequately represent the data? In essence, we tackle this statistical model_ building problem by trying to verify the covariance relationship in (95). The sample covariance matrix S is an estimator of the unknown population covariance matrix 1:. If the offdiagonal elements of S are small or those ofthe sample correlation matrix R essentially zero, the variables are not related, and a factor analysis will not prove useful. In .these circumstances, the specific factors play the . dominant role, whereas the major aim of factor analysis is to determine a few important common factors. . If 1: appears to deviate significantly from a diagonal matrix, then a factor model can be entertained, and the initial problem is one of estimating the factor loadings f.;. and specific variances !/Ii' We shall consider two of the most popular methods of para~ meter estimation, the principal component (and the related principal factor) method and the maximum likelihood method. The solution from either method can be in order to simplify the interpretation of factors, as described in Section 9.4. It is always prudent to try more than one method of solution; if the factor model is appropriate for the problem at hand, the solutions should be consistent with one another. Current estimation and rotation methods require iterative calculations that must be done on a computer. Several computer programs are now available for this purpose.
approach, when the last p  m eigenvalues are small, is to neglect the contribution of A,?,+lem+l e :r,+l .+ .. , + Apepe~ to 1: in (910). Neglecting this contribution, we obtam the apprOlumation
1: ==
[VAr" el ! ~ e2
~elJ
[
.~.~~..
! ... ! \lA,;; em]
=
:
L
L'
(pXm) (mXp)
(912)
\lA,;;e:r,
The appr.oxi~ate representation in (912) assumes that the specific factors e in (94) are of mm~r Import~nce and can also be ignored in the factoring of 1:. If specific factors are mcluded m the model, their variances may be taken to be the diagonal elements of 1:  LL', where LL' is as defined in (912). Allowing for specific factors, we find that the approximation becomes
I==LL'+'IJt
[~elj __ . __ •••.••••
_ 
[~el
:
'1'1
~ei
"
: \IX; e2 i ... i \lA,;; em]
::::::c:::;::
r'"~
+
o
~em
m
2: th for i
= 1,2, ... , p.
The Principal Component (and Principal Factor) Method
where!/li
The spectral decomposition of (216) provides us with one factoring of the covariance matrix 1:. Let 1: have eigenvalueeigenvector pairs (Ai. ei) with A1 ;:=: A2 ;:=: ••• ;:=: Ap;:=: O. Then
To apply this approach to a data set xl> X2,"" Xn , it is customary first to center the observations by subtracting the sample mean x. The centered observations
~
.~
ivA,.,
'.~
i
vA,.,
,
i··· ,
[
~e;l
VA;ei vA,.,] ::~~:
'.1>
(pXp)
L
L'
(pxp)(pXp)
+ 0
(pXp)
= LV
(Tu 
j=l
Xj
(910)
r:;~l x
:
Xjp

r;~J r:;~ =;~l =
:
xp
:
Xjp 
j = 1,2, .. . ,n
(914)
xp
have the same sample covariance matrix S as the original observations. . In cases in whi~h the units of the variables are not commensurate, it is usually deSirable to work WIth the standardized variables
This fits the prescribed covariance structure for the factor analysis model having as many factors as variables (m = p) and specific variances !/Ii = 0 for all i. The loading matrix has jth column given by VAj ej. That is, we can write
1:
=
(911)
Apart from the scale factor VAj, the factor loadings on the jth factor are the coefficients for the jth principal component of the population. Although the factor analysis representation of I in (911) is exact, it is not particularly useful: It employs as many common factors as there are variables and does not allow for any variation in the specific factors £ in (94). We prefer models that . explain the covaiiance structure in terms of just a few common factors. One
(Xjl 
Xl)
~ (Xj2 
X2)
VS; (Xjp 
j = 1,2, ... ,n
xp)
~ whose sample covariance matrix is the sample correlation matrix R of the observations xl, ~2' ... , Xn • St~ndardization avoids the problems of having one variable with large vanance unduly mfluencing the determination of factor loadings.
490 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Methods of Estimation 491
The representation in (913), when applied to the sample covariance matrix S Or the sample correlation matrix R, is known as the principal component solution. The name follows from the fact that the factor loadings are the scaled coefficients of the first few sample principal components. (See Chapter 8.)
Principal Component Solution of the Factor Model The principal component factor analysis of the sample covariance matrix S is specified in terms of its eigen.valueeigenvector pairs (AI, ed, (A2, ~), ... , (Ap, p ), where Al ~ A2 ~ ... ~ Ap. Let m < p be ,!.he number of common factors. Then the matrix of estimated factor loadings {f ij } is given by
e
I: = [~e1 ! Vfze2 ! ... ! VA:em ]
(915)
The estimat~
~ r~'~' ~] =
o
with
(916)
'iil p
Communalities are estimated as ~2
~2
hi = fi
~2
1
~2
+ fi 2 + ... + f im
(917)
The prirlcipal component factor anl!lysis of the sample correlation matrix is obtained by starting with R in place of S.
For the principal component solution, the estimated loadings for a given factor do not change as the number of factors is increased. For example, if m = 1,
I: = [~ed, and if m
2, I: = [~e1 ! ~ e2]' where (AI, e1) and (A2,C2) are the first two eigenvalueeigenvector pairs for S (or R). By the d~initio~ of 'iili, the diagonal elements of S are equal to the diagonal elements of LV + '1'. However, the offdiagonal elements of S are not usually reproduced by 1:1:' + 'if. How, then, do we select the number of factors m? If the number of common factors is not determined by a priori considerations, such as by theory or the work of other researchers, the choice of m can be based on the estimated eigenvalues in much the same manner as with principal components. Consider the residual matrix =
(918) resulting from the approxinlation of S by the principal component solution. The diagonal elements are zero, and if the other elements are also small, we may subjectively take the m factor model to be appropriate. AnalytiCally, we have (see Exercise 9.5) Sum of squared entries of (S 
(1:1:' + 'if»
s A~+l + ... + A~
Consequently, a small value for the sum of the squares of the neglected eigenvalues implies a small value for the sum of the squared errors of approximation. Ideally, the contributions of the first few factors to the sample variances of the variables should be large. The contribution to the sample variance s··It from the .........2 fIrst common factor is f il' The contribution to the total sample variance, s]] + S22 + ... + sPI' = tr(S), from the first common factor is then
e1\ + e~1 + '" + e:1= (~Cl)'(~el) = Al since the eigenvector el has unit length. In general,
~f tot~l)
Proportion sample vanance ( due to jth factor 
l
Sll
=
A.
+ S22 ~ .•. + s pp for a factor analysis of S (92D)
Aj
for a factor analysis of R
p
C~iterion (920) is frequently used as a heuristic device for determining the appropnate number of common factors. The number of common factors retained in the model is increased until a "suitable proportion" of the total sample variance has been explained. Another convention, frequently encountered in packaged computer programs, is to Set m equal to the number of eigenvalues of R greater than one if the sample correlation matrix is factored, or equal to the number of positive eigenvalues of S if the sample covariance matrix is factored. These rules of thumb should not be applied indiscriminately. For example, m = p if the rule for S is obeyed, since all the eigenvalues are expected to be positive for large sample sizes. The best approach is to retain few rather than many factors, assuming that they provide a satisfactory interpretation of the data and yield a satisfactory fit to S or R.
Example 9.3 (Factor analysis of consumerpreference data) In a consumerpreference study, a random sample of customers were asked to rate several attributes of a new product. The responses, on a 7point semantic differential scale, were tabulated and the attribute correlation matrix constructed. The correlation matrix is presented next: Attribute (Variable) Taste Good buy for money 2 Flavor 3 Suitable for snack 4 Provides lots of energy 5
1
TOO
.02 .96 .42 .01
2 .02 1.00 .13 .71 .85
3
4 .42 .13 .71 1.00 .50 .50 1.00 .11 .79
®
5
@ Oll .11
®
1.00
It is clear from the circled entries in the correlation matrix that variables 1 and 3 and variables 2 and 5 form groups. Variable 4 is "closer" to the (2,5) group than the (1,3) group. Given these results and the small number of variables, we might expect that the apparent linear relationships between the variables can be explained in terms of, at most, two or three common factors.
492
Methods of Estimation 493
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
nearly reproduces the correlation matrix R. Thus, on a purely descriptive basis, we would judge a twofactor model with the factor loadings displayed in Table 9.1 as providing a good fit to the data. The communalities (.98, .88, .98, .89, .93) indicate that the two factors account for a large percentage of the sample variance of each variable. We shall not interpret the factors at this point. As we noted in Section 9.2, the factors (and loadings) are unique up to an orthogonal rotation. A rotation of the factors often reveals a simple structure and aids interpretation. We shall consider this example again (see Example 9.9 and Panel 9.1) after factor rotation has been discussed. _
The first two eigenvalues, A1 = 2.85 and A2 = 1.81, of R are the only eigenval_ ues greater than unity. Moreover, m = 2 common factors will account for a cumula_ tive proportion A1
+
A2 = 2.85
P
+ 1.81
= .93
5
of the total (standardized) sample variance. The estimated factor loadings, communalities, and specific variances, obtained using (915), (916), and (917), are given in Table 9.1.
,I
Example 9.4 (Factor analysis of stockprice data) Stockprice data consisting of n = 103 weekly rates of return on p = 5 stocks were introduced in Example 8.5. In that example, the first two sample principal components were obtained from R. Taking m = 1 and m = 2, we can easily obtain principal component solutions to the orthogonal factor model. Specifically, the estimated factor loadings are the sample principal component coefficients (eigenvectors of R), scaled by the square root of the corresponding eigenvalues. The estimated factor loadings, communalities, specific variances, and proportion of total (standardized) sample variance explained by each factor for the m = 1 and m = 2 factor solutions are available in Table 9.2. The communalities are given by (917). So, for example, with
Table 9.1 Estimated factor loadings ~VI Aieij ij =
e
Communalities ~2
1. Taste 2. Good buy for money 3. Flavor 4. Suitable for snack 5. Provides lots of energy Eigenvalues Cumulative proportion of total (standardized) sample variance
';fri
=
1  h~
.56
F2 .82
.98
.02
.78 .65
.53 .75
.88 .98
.12 .02
.94
~.10
.89
.11
.80
.54
.93
.07
2.85
1.81
F1
Variable
Specific variances
hi
~2
.56 .78 ~ = .65 [ .94 .80
.82] [.56 .53 .75 .82 .10 .54
. [0.02
+ 0
o o
.78 .53
2
Onefactor solution Estimated factor loadings F1
Twofactor solution
Specific variances ifJi = 1  hi
~
.732 .831 .726 .605 .563
JPMorgan Citibank Wells Fargo Royal Dutch Shell ExxonMobil
Cumulative proportion of total (standardized) sample variance explained
.94 .80J .10 .54
2
(.732) + (.437) = .73.
~2
.46 .31 .47 .63 .68
.487
Estimated factor loadings F1 F2 .732 .831 .726 .605 .563
.437 .280 .374 .694 .719
.487
.769
The residual matrix corresponding to the solution for m
o
0.12 0 .02 0
0 0
.65 .75
~2
Table 9.2
1. 2. 3. 4. 5.
.932
.571
~2
= 2, h1 = ell + e12 =
Variable
Now,
Ll> +
m
o o
o o 0 .11
o
~ .07
.01 1.00 ]
=
[1.00
.97
.11 1.00
.44 .79 .53 1.00
0 R 
LL' 
.099 ~ = .185 [ .025 .056
.099 0 .134 .014 .054
.185 .134
Specific . variances ;Pi = 1 hJ
= 2 factors is
o
.025 .014 .003
.003 .006
.156
o
.056] .054 .006 .156 0
.27 .23 .33 ~15
.17
494
Methods of Estimation 495
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
The proportion of the total variance explained by the twofactor solution!.~appreciably larger than that for the onefactor solution. However. for. m = 2: ~L' p~oduces numbers that are, in general, larger than the sample correlatIons. ThIs IS partIcularly true for '13' . It seems fairly clear that the first factor, Flo represents general economIc conditions and might be called a market factor. All of the stocks load highly on this f~c tor, and the loadings are about equal. The second factor contr~sts the. banktng stocks with the oil stocks. (The banks have relatively large negattve loadtngs, and the oils have large positive loadings, on the factor.) Thus, F2 seems to differentiate stocks in different industries and might be called an industry factor. To rates of return appear to be determined by general market co?ditions a?d activi~ies that are unique to the different industries, as well as a res~du~ or ftrm speC:lfic · factor. This is essentially the conclusion reached by an exammatton of the sample principal components in Example 8.5. •
A Modified Approachthe Principal Factor Solution A modification of the principal component approach is sometimes considered. We describe the reasoning in terms of a factor analysis of R, although the procedure is also appropriate for S. If the factor model p = LV + 'I' is correctly specified, the m common factors should account for the offdiagonal elements of p, as well as the communality portions of the diagonal elements Pii
= 1 = IzT + "'i
"'i
If the specific factor contribution is removed from the diagonal or, equivalently, the 1 replaced by the resulting matrix i~ p  'I' = ~~'. . . Suppose, now, that initial estimates of t~~ speCIfIC ~anances .are "avallabl~; Then replacing the ith diagonal element of R by hi = 1 we obtam a reduced sample correlation matrix
hr,
"'i
"'i .
Now, apart from sampling variation, all of the elements of the reduced sampl.e correlation matrix Rr should be accounted for by the m common factors. In partIcular, Rr is factored as (921)
L;
where = {e;j} are the estimated loadings. . . The principal factor method of factor analysIs employs the esttmates
L; = [vAfe~ i vA;e;
i'"
i
~e~l
where (A;, e7), i = 1,2, ... , m are the (largest) eigenvalueeigenvector pairs determined from R r . In turn, the communalities would then be (re)estimated by
};*2 = £.J ~e*? l I)
The principal factor solution can be obtained iteratively, with the communality estimates of (923) becoming the initial estimates for the next stage. In the spirit of the principal component solution, consideration of the estimated eigenvalues Ai, A;, ... , A; helps determine the number of common factors to retain. An added complication is that now some of the eigenvalues may be negative, due to the use of initial communality estimates. Ideally, we should take the number of common factors equal to the rank of the reduced popUlation matrix. Unfortunately, this rank is not always well determined from R" and some judgment is necessary. Although there are many choices for initial estimates of specific variances, the most popular choice, when one is working with a correlation matrix, is "'; = 1/rU , where rii is the ith diagonal element of R I . The initial communality estimates then become *2
hi
=1
"" e*2 £.J ij j=1
=1
"'i = 1 •
1
(924)
,; r"
which is equal to the square of the multiple correlation coefficient between Xi and the other p  1 variables. The relation to the multiple correlation coefficient means that h? can be calculated even when R is not of full rank. For factoring S, the initial specific variance estimates use Sii, the diagonal elements of SI. Further discussion of these and other initial estimates is contained in [6]. Although the principal component method for R can be regarded as a principal factor method with initial communality estimates of unity, or specific variances equal to zero, the two are philosophically and geometrically different. (See [6].) In practice, however, the two frequently produce comparable factor loadings if the number of variables is large and the number of common factors is small. We do not pursue the principal factor solution, since, to our minds, the solution methods that have the most to recommend them are the principal component method and the maximum likelihood method, which we discuss next.
The Maximum likelihood Method If the common factors F and the specific factors
E can be assumed to be normally distributed, then maximum likelihood estimates of the factor loadings and specific variances may be obtained. When Fj and Ej are jointly normal, the observations Xj  /L = LFj + Ej are then normal, and from (416), the likelihood is
L(/L,~)
= (21T) 
i, ~ '~e m [r tr
1 (
~1 (Xji)(XjiY+n(iIL)(iILY)] [ (...
)]
(nl)p (nl) ~!) = (21T)2'~'2e\2 tr II {;1(xji)(Xji)'
m
"'i•
(923)
j=1
X
(21T) ~, ~ ,!e (~)(iIL)'Il (iIL)
(925)
496
Methods of Estimation
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
which depends on L and 'I' through l: = LV + qr. This mo~el is still not well defined, because of the multiplic;ity of choices for L made ?ossl~le by orthogonal transformations. It is desirable to make L well defined by lffiposlOg the computationally convenient uniqueness condition (926)
a diagonal matrix
The maximum likelihood estimates I, and q, must be obtained by numerical maximization of (925). Fortunately, efficient computer programs now exist that en: . able one to get these estimates rather easily. We summarize some facts about maximum likelihood estimators and, for now, rely on a computer to perform the numerical details.
Result 9.1. Let X l ,X 2, ... ,Xn be a random sample from Np(JL,l:), where l: = LL' + 'I' is the covariance matrix for the m common factor model of (94). The maximum likelihood estimators 1" q" and jL = x maximize (925) subject to l:.jrli being diagonal. The maximum likelihood estimates of the communalities are (927)
for i = 1,2, ... , P
so
'2
Proportion oftotal sample = ( variance due to jth factor )
'2
+ ... +
e'2im ·
•
I If, as in (810), the variables are standardized so that Z = V /2(X  JL), then the covariance matrix p of Z has the representation
p = Vl/2l:VI/2 = (VI/2L) (Vl/2L), + Vlf2 '1'Vl/2
(929) I L  V /2Land. . hio~d'109.mat' fiX • Thus, p has a factorization analogous to (95 ) Wit specific variance matrix '1'. = V l/2'1'V l/2. By the lO~anance pro.perty of maXimum likelihood estimators, the maximum likelihood estimator of p IS
jJ = CVl/21,) CVI/2I,)' + yI/2q,yl/2 =
"
L.L~
+
.
,(proportion of total (standardiZed») = sample variance due to jth factor
'2 f lj
'2
'2
+ e2j + ... +
fpj
(932)
p
To avoid more tedious notations, the preceding ei/s denote the elements of i •.
Comment. Ordinarily, the observations are standardized, and a sample correlation matrix is factor analyzed. The sample correlation matrix R is inserted for [en  l)/njS in the likelihood function of (925), and the maximum likelihood estimates i. and .jr. are obtained using a computer. Although the likelihood in (925) is appropriate for S, not R, surprisingly, this practice is equivalent to obtaining the maximum likelihood estimates i and .jr based on the sample covariance matrix S, setting i. = yl/2i and .jr. = V l /2.jrV I/2. Here V I/2 is the diagonal matrix with the reciprocal of the sample standard deviations (computed with the divisor vn) on the main diagonaL Going In the other direction, given the estimated loadings i. and specific variances '1'. obtained from R, we find that the resulting maximum likelihood estimates for a factor analysis of the co variance matrix [(n  1 )/n j S are i = yl/2i. and .jr = yl/2.jr. V 1/2, or
(928)
functions of L and 'I' are estimated by the same functIOns of L and '1'. In particular, the communalities hr = erl + ... + erm have maximum likelihood estimates
ea
the maximum likelihood estimates of the communalities, and we evaluate the importance of the factors on the basis of
'2
e·+e·+···+e I} 2} P} Sll + S22 + .. , + spp
Proof. By the invariance property of maximum likeliho~d estim~tes (se!? Section ~.3),
'2 '2 hi =
497
.T.
(930)
'r.
where Uii is the sample variance computed with divisor n. The distinction between divisors can be ignored with principal component solutions. _ The equivalency between factoring Sand R has apparently been confused in many published discussions of factor analysis. (See Supplement 9A.) Example 9.S (Factor analysis of stockprice data using the maximum likelihood method) The stockprice data of Examples 8.5 and 9.4 were reanalyzed assuming
an m = 2 factor model and using the maximum likelihood method. The estimated factor loadings, communalities, specific variances, and proportion of total (standardized) sample variance explained by each factor are in Table 9.3. 3 The corresponding figures for the m = 2 factor solution obtained by the principal component method (see Example 9.4) are also provided. The communalities corresponding to the maximum likelihood factoring of R are of the form [see (931)] h;2 = ei~ + ei~' So, for example,
hI = (.115)2 + (.765f = .58
l
where V l /2 and i are the maximum likelihood estimators of V /2 and L, respec. .' tively. (See Supplement 9A.) As a consequence of the factorization of (930), whenever the maximum hkelihood analysis pertains to the correlation matrix, we call i = 1,2, ... ,p
3 The maximum likelihood solution leads to a Heywood case. For this example, the solution of the likelihood equations give estimated loadings such that a specific variance is negative. The software program obtains a feasible solution by slightly adjusting the loadings so that all specific variance estimates are nonnegative. A Heywood case is suggested here by the .00 value for the specific variance of Royal Dutch Shell.
498
Methods of Estimation 499
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
The patterns of the initial factor loadings for the maximum likelihood solution are constrained by the uniqueness condition that L',plL be a diagonal matrix. Therefore, useful factor patterns are often not revealed until the factors are rotated (see Section 9.4). •
3
Principal components
Maximum likelihood
Variable 1. 2. 3. 4. 5.
.115 J PMorgan .322 Citibank .182 Wells Fargo Royal Dutch Shell 1.000 .683 Texaco
Cumulative proportion of total (standardized) sample variance explained
'
"'i =
F2
FI
.755 .788 .652 .000 .032
Estimated factor loadings
Specific variances
Estimated factor .loadings
'2
1  hi
.42 .27 .54 .00 .53
Fl .732 .831 .726 .605 .563
F2 .437 .280 374 .694 .719
Specific variances
';(,i =
1
h?
Example 9.6 (Factor analysis of Olympic decathlon data) Linden [11] originally con
I
ducted a factor analytic study of Olympic decathlon results for all 160 complete starts from the end of World War 11 until the midseventies. Following his approach we examine the n = 280 complete starts from 1960 through 2004. The recorded values for each event were standardized and the signs of the timed events changed so that large scores are good for all events. We, too, analyze the correlation matrix, which based on all 280 cases, is
.27 .23 .33 .15 .17
R= .323
.647
.487
.769
The residual matrix is
R 
fi'  ,p
.001 .002 0 .002 .001 [ 0 .002 0 = .002 .000 .000 .000 .001 .052 .033
.000 .000 .000 0 .000
ffi2]
.033 .001 .000 0
The elements of R  LL'  ,p are much smaller than those of the residual matrix corresponding to the principal component factoring of R presented in Example 9.4. On this basis, we prefer the maximum likelihood approach and typically feature it in subsequent examples. The cumulative proportion of the total sample variance explained by the factors is larger for principal component factoring than for maximum likelihood factoring. It is not surprising that this criterion typically favors principal component factori~g. Loadings obtained by a principal component factor analysis are related to the prmcipal components, which have, by design, a variance optimizing property. [See the discussion preceding (819).] . Focusing attention on the maximum likelihood solution, we see that all :a~ abIes have positive loadings on FI . We call this factor the market factor, as we dId m the principal component solution. The interpretation of the second factor is not as clear as it appeared to be in the principal component solution. The bank stocks have large positive loadings and the oil stocks have negligible loadings on the second factor F2 . From this perspective, the second factor differentiaties the bank stocks from the oil stocks and might be called an industry factor. Alternatively, the second factor might be simply called a banking factor.
1.000 .6386 .4752 .3227 .5520 .3262 .3509 .4008 .1821 .0352
.6386 1.0000 .4953 .5668 .4706 .3520 .3998 .5167 .3102 .1012
.4752 .4953 1.0000 .4357 .2539 .2812 .7926 .4728 .4682 .0120
.3227 .5668 .4357 1.0000 .3449 .3503 .3657 .6040 .2344 .2380
.5520 .3262 .4706 .3520 .2539 .2812 .3449 .3503 1.0000 .1546 .1546 1.0000 .2100 .2553 .4213 .4163 .2116 .1712 .4125 .0002
.3509 .3998 .7926 .3657 .2100 .2553 1.0000 .4036 .4179 .0109
.4008 .5167 .4728 .6040 .4213 .4163 .4036 1.0000 .3151 .2395
.1821 .3102 .4682 .2344 .2116 .1712 .4179 .3151 1.0000 .0983
.0352 .1012 .0120 .2380 .4125 .0002 .0109 .2395 .0983 1.0000
From a principal component factor analysis perspective, the first four eigenvalues, 4.21, 1.39, 1.06, .92, of R suggest a factor solution with m = 3 or m = 4. A subsequent interpretation, much like Linden's original analysis, reinforces the choice m = 4. In this case, the two solution methods produced very different results. For the principal component factorization, all events except the 1,500meter run have large positive loading on the first factor. This factor might be labeled general athletic ability. Factor 2, which loads heavily on the 400meter run and 1,500meter run might be called a running endurance factor. The remaining factors cannot be easily interpreted to our minds. For the maximum likelihood method, the first factor appears to be a general athletic ability factor but the loading pattern is not as strong as with principal component factor solution. The second factor is primarily a strength factor because shot put and discus load highly on this factor. The third factor is running endurance since the 400meter run and 1,500meter run have large loadings. Again, the fourth factor is not easily identified, althoug~ it may have something to do with jumping ability or leg strength. We shall return to an interpretation of the factors in Example 9.11 after a discussion of factor rotation. The fourfactor principal component solution accounts for much of the total (standardized) sample variance, although the estimated specific variances are large in some cases (for example, the javelin). This suggests that some events might require unique or specific attributes not required for the other events. The fourfactor maximum likelihood solution accounts for less of the total sample
Methods of Estimation 50 I
variance, bpt, as t}1e following residual matrices indicate, the maximum likelihood estimates ~ and 't do a better job of reproducing R than the principal component estimates L and "It. Principal component: R 
~8l~;;68
~~"1q\C1
I
LL'  'If = o .082
.006 .021 .068 .031 .016 .082 0 .046 .033 .107 .078 .048 .006 .046 0 .006 .010 .014 .003 .021 .033.006 0 .038 .204 .015 .068 .107 .010 .038 0 .096 .025 .031 .078 .014 .204 .096 0 .015 .016 .048 .003 .015 .025 .015 o .003 . ~.059 .013 .078 .006 .124 .029 .039 .042  .151  .064 .030 .119  .210 .062 .006 .055 .086 .074 .085 .064
.003 .059 .013 .078 .006 .124
.039 .042 .151 .064 .030 .119
.062 .006 .055 .086 .074 .085
.029
.210  .026 0 .078
.064  .084  .078 0
o  .026 .084
Maximum likelihood: R 
2.::o ~+~r~ S 8
C;;
....
0..
B
~
4<
"g
J:
~
('10\000\", rOOM,....IO\
tIl
"1C"!q""l~
bJJ
I
S '2. ..... '0 t
.~ ..9
if;  ~ = o .000
.000.  .000 .000 0 .002 .023 .000 .002 0 .004 .000 .023 .004 o .000 .005 .001  .002 .000  .017  .009  .030  .000  .003 .000  .004 .000  .030  .001  .006 .001 .047 .001  .042 .000  .024 .000 .010
 .000 .000 .005 .017  .000  .009 .002 .030 0  .002  .002 0 .001 .022 .001 .069 .001 .029 .001 .019
'.000  .003 .000 .004 .001 .022
o  .000 .000 .000
.000  .001 000  .030 .047  .024 .001 .001 .000 .006 .042 .010 .001 .000  .001 .069 .029  .019 .000 .000 .000 o .021 .011 .021 0 .003 0 .011 .003
'"
•
~
A Large Sample Test for the Number of Common Factors The assumption of a normal population leads directly to a test of the adequacy of the model. Suppose the m common factor model holds. In this case l: = LV + "It, and testing the adequacy of the m common factor model is equivalent to testing Ho:
l:
(pXp)
=
L
L'
(pXm) (mxp)
+ "It
(pXp)
(933)
versus Hi: l: any other positive definite matrix. When l: does not have any special form, the maximum of the likelihood function [see (418) and Result 4.11 with i = ((n l)/n)S = SnJisproportionalto 500
(934)
504
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices Factor Rotation
Using Bartlett's correction, we evaluate the test statistic in (939): [n  1  (2p
v~ct0.fS al~ng perpe?dicular co~rdinate axes. A plot of the pairs of factor loadings ( Cil , Cd YIel~s p pomts, each pomt corresponding to a variable. The coordinate axes
ILl: + q, I + 4m + 5)/6) In I Sn I = [ 103  1 
~an tp*en be vIsually rotated through an anglecall it and the new rotated load(10
+ 8 + 5)] 6
mgs Cij are determined from the relationships In (1.0216) = 2.10
i.
r" [~'~ (pX2)
Since ~[(p  m)2  p  m) = ![(5  2)2  5  2) = 1, the 5% critical value Xy( .05) = 3.84 is not exceeded, and we fail to reject Ho. We conclude that the data do
sm
not contradict a twofactor model.. In fact, the observed significance level, or Pvalue, P[Xy > 2.10) == .15 implies that Ho would not be rejected at any reasonable level.. •
where
T=[COS sin
~
!arge sample variances and covariances for the maximum likelihood estimates £;., !J!i have been derived when these estimates have been determined from the sample U:variance matrix S. (See [10).) The expressions are, in general, quite complicated.
9.4 Factor Rotation As we indicated in Section 9.2, all factor loadings obtained from the initialloadings by an orthogonal transformation have the same ability to reproduce the covariance (or correlation) matrix. [See (98).) From matrix algebra, we know that an orthogonal transformation corresponds to a rigid rotation (or reflection) of the coordinate axes. For this reason, an orthogonal transformation of the factor loadings, as well as the implied orthogonal transformation of the factors, is called factor rotation. If L is the p X m matrix of estimated factor loadings obtained by any method (principal component, maximum likelihood, and so forth) then L* = LT,
where TT'
= T'T =
I
=
i
T
sin ] cos sin ] cos
hr,
clockwise rotation counterclockwise rotation
Example 9.8 (A ~irst look ~t factor rotation) Lawley and Maxwell [10] present the sa~ple correlatIOn matrIX of examination scores in p = 6 subject areas for
n  220 male students. The correlation matrix is Gaelic
English
History
Arithmetic
Algebra
Geometry
1.0
.439 1.0
.410 .351 1.0
.288 .354 .164 1.0
.329 .320 .190 .595 1.0
.248 .329 .181 .470 .464 1.0
R=
(943) Equation (943) indicates that the residual matrix, Sn~ LL'  q, = Sn  L*L*'  q" unchanged. Moreover, the specific variances !J!i, and hence the communalitie,!' ~are unaltered. Thus, from a mathematical viewpoint, it is immaterial whether L or L * is obtained. Since the originalloadings may not be readily interpretable, it is usual practice to rotate them until a "simpler structure" is achieved. The rationale is very much akin to sharpening the focus of a microscope in order to see the detail more clearly. Ideally, we should like to see a pattern of loadings such that each variable loads highly on a single factor and has small to moderate loadings on the remaining factors. However, it is not always possible to get this simple structure, although the rotated loadings for the decathlon data discussed in Example 9.11 provide a nearly ideal pattern. We shall concentrate on graphical and analytical methods for determining an orthogonal rotation to a simple structure. When m = 2, or the common factors are considered two at a time, the transformation to a simple structure can frequently be determined graphically. The uncorrelated common factors are regarded as unit
(944)
(pX2)(2X2)
~e relati?n~hip ~n (944) is rarely implemented in a twodimensional graphical analysIs. In thIS sItuat~on, c!usters of variables are often apparent by eye, and these c~usters enable one to Ident~ the common factors without having to inspect the magmt~des. of ~e rotated loadmgs. On the other hand, for m > 2; orientations are not easIly v~suahz.ed, and the. magnitudes of the rotated loadings must be inspected to find a mean~n~ful mterpretatIOn of the original data. The choice of an orthogonal matrix T that satisfies an analytical measure of simple structure will be considered shortly.
(942)
is a p X m matrix of "rotated" loadings. Moreover, the estimated covariance (or correlation) matrix remains unchanged, since
~emains
505
~nd a maximum likelihood solution for m = 2 common factors yields the estimates m Table 9.5. Table 9.S
Variable 1. 2. 3. 4. 5. 6.
Gaelic English History Arithmetic Algebra Geometry
Estimated factor loadings FI F2 .553 .568 .392 .740 .724 .595
:429 .288 .450 .273 .211 .132
Communalities ~2 hi .490 .406 .356 .623 .569 .372
Factor Rotation 507
506 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices All the variables have positive loadings on the first factor. Lawley Maxwell suggest that this factor reflects the overall response of the students to instruction and might be labeled a general intelligence factor. Half the loadings ' : positive and half are negative on the second factor. A fact~r with this ?~ttern loadings is called a bipolar factor. (The assignment of negatIve and posltlve. '. is arbitrary, because the signs of the loadings on a factor can be reversed wIthout " affecting the analysis.) This factor is not easily identified,but is such that individuals who get aboveaverage scores on the verbal tests get aboveaver~ge Scores the factor. Individuals with aboveaverage scores on the mathematIcal tests belowaverage scores on the factor. Perhaps this factor can be classified as "math,nonmath" factor. The factor loading pairs (fil' f i2 ) are plotted as points in Figure 9.1. The poi.nt& are labeled with the numbers of the corresponding variables. Also shown is a clockwise orthogonal rotation of the coordinate axes through an a~gle of c/J == 20°. This angle was chosen so that one of the new axes passes throug~ (C41 • ( 42 )· W~~n this is done. all the points fall in the first quadrant (the factor loadmgs are all pOSltlve), and the two distinct clusters of variables are more clearly revealed. The mathematical test variables load highly on and have negligible loadings on F;. The first factor might be called a l/l~lhelllalica!abiliIY factor. Similarly, the three verbal test variables have high loadmgs on F and moderate to small The second factor might be l~beled a ver~alability factor; loadings on The generalintelligence factor identified initially IS submerged m the factors F I A
Fr
2
Fr.
and
F;.
.
.
matrices. We point out that Figure 9.1 suggests an oblique rotation of the coordinates. One new axis would pass through the cluster {1,2,3} and the othe~ through the {4, 5, 6} group. Oblique rotations are so named because they correspon~ to a non rigid rotation of coordinate axes leading to new axes that are not perpendIcular.
F1
I
.5
I
I
I
I
I
I
3 _I 2
Figure 9.1 Factor rotation for test
scores.
Estimated rotated factor loadings F; F~
Variable 1. 2. 3. 4. 5. 6.
Gaelic English History Arithmetic Algebra Geometry
.369 .433
l!J .789 .752 .604
Communali ties
aID .467 .558 .001 .054 .083
j,~ =
•
j,2
•
.490 .406 .356 .623 .568 .372
It is apparent, however, that the interpretation of the oblique factors for this
example would be much the same as that given previously for an orthogonal rotation.
•
Kaiser [9] has suggested an analytical measure of simple structure known as the 'l7j = f7/hi to be the rotated coefficients scaled by the square root of the communalities. Then the (normal) varimax procedure selects the orthogonal transformation T that makes varimax (or normal varimax) criterion. Define
(p2: ~*2)2/ ] 2: 2:p ~*4 £ij £ij P
1 m V = P J=I
°
The rotated factor loadings obtained from (944) wIth c/J = 20 and the corresponding communality estimates are shown in. Table 9.6. The magnitudes of the rotated factor loadings reinforce the interpretatIOn of the factors suggested by Figure 9.1. . . The communality. estimates are unchanged by the orthogonal rotatIOn, smce ii: = iTT'i' = i*i*', and the communalities are the diagonal elements of these
F2
Table 9.6
[
.=1
(945)
.=1
as large as possible. Scaling the rotated coefficients C;j has the effect of giving variables with small communalities relatively more weight in the determination of simple structure. After the transformation T is determined, the loadings 'l7j are multiplied by hi so that the original communalities are preserved. Although (945) looks rather forbidding, it has a simple interpretation. In words, V
<X
~ j=I
(variance of squares of (scaled) loadings for) jth factor
(946)
Effectively, maximizing V corresponds to "spreading out" the squares of the loadings on each factor as much as possible. Therefore, we hope to find groups of large and negligible coefficients in any column of the rotated loadings matrix L*. Computing algorithms exist for maximizing V, and most popular factor analysis computer programs (for example, the statistical software packages SAS, SPSS, BMDP, and MINITAB) provide varimax rotations. As might be expected, varimax rotations of factor loadings obtained by different solution methods (principal components, maximum likelihood, and so forth) will not, in general, coincide. Also, the pattern of rotated loadings may change considerably if additional common factors are included in the rotation. If a dominant single factor exists, it will generally be obscured by any orthogonal rotation. By contrast, it can always be held fixed and the remaining factors rotated.
508
Factor Rotation
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
509
PANEL 9.1 SAS ANALYSIS FOR EXAMPLE 9.9 USING PROC FACTOR.
Example 9.9 (Rotated loadings for the consumerpreference data) Let us return to . the marketing data discussed in Example 9.3. The original factor loadings lODltal1~pil by the principal component method), the communalities; and the (varimax) factor loadings are shown in Table 9.7. (See the SAS statistical software output Panel 9.1.)
Estimated factor loadings Fl F2
Variable 1. 2. 3. 4. 5.
.56 .78 .65 .94 .80
Taste Good buy for money Flavor Suitable for snack Provides lots of energy
Rqtated estimated factor loadings F~ F;
.82 .52 .75 .10 .54
Communalities
title 'Factor Analysis'; data consumer(type = corr); _type_='CORR'; input _name_$ taste money cards; taste 1.00 money 1.00 .02 flavor .96 .13 snack .42 .71 energy .01 .85
flavor snack energy;
1.00 .50 .11
PROGRAM COMMANDS 1.00 .79
1.00
hr
proc factor res data=consumer method=prin nfact=2rotate=varimax preplot plot; var taste money flavor snack energy;
.98 .88 .98 .89
!Initial Factor Method: Principal Components
.93
I
OUTPUT
Prior Communality Estimates: ONE
Cumulative proportion of total (standardized) sample variance explained
.571
.507
.932
Eigenvalues of the Correlation Matrix: Total = 5 Average = 1
.932
It is clear that variables 2, 4, and 5 define factor 1 (high loadings on factor 1, small or negligible loadings on factor 2), while variables 1 and 3 define factor 2 (high loadings on factor 2, small or negligible loadings on factor 1). Variable 4 is most closely aligned with factor 1, although it has aspects of the trait represented by factor 2. We might call factor 1 a nutritional factor and factor 2 a taste factor. The factor loadings for the variables are pictured with respect to the original and (varimax) rotated factor· axes in Figure 9.2. •
Eigenvalue Difference
1 2.853090 1.046758
Proportion Cumulative
0.5706 0.5706
2 1.806332 1.601842
0.j61~ 1 ...
0.931~
4 0.102409 0.068732
0.033677
0.0409 0.9728
0.0205 0.9933
0.0067 1.0000
2 factors will be retained by the NFACTOR criterion.
! Factor Pattern.
F2
I
/
F*2
/
/
I
FAcrORi .. ;FAdb~2; TASTE MONEY FLAVOR SNACK ENERGY
/1 • 3
/ .5
5
3 0.204490 0.102081
/
0.55986 0.77726
0.81610 .0.52420 0.64534' 074795 0.·93911:.o:1o/m 0.79821 :0.54323
/
/ 0
.5
/
.... .... .5 .... ....
• 1.0 4
F,
TASTE
.... 2· ...... 5
....
MONEY
FLAVOR
SNACK
ENERGY
0.878920
....
Figure 9.2 Factor rotation for hypothetical marketing data.
(continues on next page)
510
Factor Rotation
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices PANEL 9.1
I
specific variances and cumulative proportions of the total (standardized) sample variance explained by each factor are also given. An interpretation of the factors suggested by the unrotated loadings was presented in Example 9.5. We identified market and industry factors. The rotated loadings indicate that the bank stocks (JP Morgan, Citibank, and Wells Fargo) load highly on the first factor, while the oil stocks (Royal Dutch Shell and ExxonMobil) load highly on the second factor. (Although the rotated loadings obtained from the principal component solution are not displayed, the same phenomenon is observed for them.) The two rotated factors, together, differentiate the industries. It is difficult for us to label these factors intelligently. Factor 1 represents those unique economic forces that cause bank stocks to move together. Factor 2 appears to represent economic conditions affecting oil stocks. As we have noted, a general factor (that is, one on which all the variables load highly) tends to be "destroyed after rotation." For this reason, in cases where a general factor is evident, an orthogonal rotation is sometimes performed with the general factor loadings fixed. 5 _
(continued)
Rotation Method: Varimax
I Rotated Factor Pattern
TASTE
MONEY FlAVOR SNACK
ENERGY
FACTOR 1 0.01970 0.93744 0.12856 0.84244 0.96539
FACTOR2 0.98948 0.01123 0.97947 0.42805 0.01563
Variance explained by each factor
FACTOR 1 2.537396
FACTOR2 2.122027
Rotation of factor loadings is recommended particularly . for loadi~gs obtained by maximum likelihooq, sjpce, the initi~1 values are c.onstr~med to. s.atls~ the uniqueness condition that L''I'IL be a diagonal matnx. This condition. IS convenient for computational purposes, but may not lead to factors that can easily be interpreted. Example 9.10 (Rotated loadings for the stockprice data) Ta?le 9.8 shows the init.ial and rotated maximum likelihood estimates of the factor loadmgs for the stoc~pnce data of Examples 8.5 and 9.5. An m = 2 factor model is assumed. The estimated Table 9.8 Maximum likelihood estimates of facfOfloadings Variable JPMorgan Citibank Wells Fargo Royal Dutch Shell ExxonMobil Cumulative proportion of total sample variance explained
.....

5II
FI
F2
Fj
Fi
Specific variances ~r = 1  hf
.115 .322 .182 1.000 .683
.755 .788 .652 .000 .032
.821 .669 .118 .113
~
.024 .227 .104 (.993J .675
.42 .27 .54 .00 .53
.323
.647
.346
.647
Rotated estimated factor loadings
Example 9.11 (Rotated loadings for the Olympic decathlon data) The estimated factor loadings and specific variances for the Olympic decathlon data were presented in Example 9.6. These quantities were derived for an m = 4 factor model, using both principal component and maximum likelihood solution methods. The interpretation of all the underlying factors was not immediately evident. A varimax rotation [see (945)] was performed to see whether the rotated factor loadings would provide additional insights. The varimax rotated loadings for the m = 4 factor solutions are displayed in Table 9.9, along with the specific variances. Apart from the estimated loadings, rotation will affect only the distribution of the proportions of the total sample variance explained by each factor. The cumulative proportion of the total sample variance explained for all factors does not change. The rotated factor loadings for both methods of solution point to the same underlying attributes, although factors 1 and 2 are not in the same order. We see that shot put, discus, and javelin load highly on a factor, and, following Linden [11], this factor might be caUed explosive arm strength. Similarly, high jump, llDmeter hurdles, pole vault, andto some extentlong jump load highly on another factor. Linden labeled this factor explosive leg strength. The lOOmeter run, 400meter run, andagain to some extentthe long jump load highly on a third factor. This factor could be called running speed. Finally, the I5DOmeter run loads heavily and the 400meter run loads heavily on the fourth factor. Linden called this factor running endurance. As he notes, "The basic functions indicated in this study are mainly consistent with the traditional classification of track and field athletics."
5Some generalpurpose factor analysis programs allow one to fix loadings associated with certain factors and to rotate the remaining factors.
'"
5 12
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Factor Scores 513
9.9
1.0 I
Estimated rotated factor loadings, e7j
Variable lOOm run Long jump
F;
Estimated rotated factor loadings, f7j
Specific variances
F; F:
~
rpi = 1 
~2
hi
0.8 I
A
Fi
F; F;
F:
rpi = 1
N
i
Il<
.12
.204 .296
.055
.29
.280 1.5541 1;~~L
.302 .252 .097
.17
.182 1.8851 .205 .139 .291
Shot put High jump
F~
.267 .221
.293
400m run
1.0
Maximum likelihood
Principal component
1.8831 .278
.33
.254 1.7391
.17
.142 .151
.005
.01
.155
.39
.228 .045
.09
0.6 I
0.4 r
0.2
.242
.33
........
.23
•
0.0 
Cumulative proportion of total sample variance explained
.15
.22
.43
.62
.76
.001 .110 .070
.20
.37
2
J
f
f
0.2
0.4
0.6
J 0.8
Factor f
•
0.4
0.2
0.0
6
•
()
8
4
•
9
•
• 0.4
0.6
0.8
Factor I
often suggested after one views the estimated factor loadings and do not follow from our postulated model. Nevertheless, an oblique rotation is frequently a useful aid in factor analysis. . If we regard the m common factors as coordinate axes, the point with the m coordinates i 1, j2 , •.. , jl1l ) represents the position of the ith variable in the factor space. Assuming that the variables are grouped into nonoverlapping clusters, an orthogonal rotation to a simple structure corresponds to a rigid rotation of the coordi; nate axes such that the axes, after rotation, pass as closely to the clusters as possible. An oblique rotation to a simple structure corresponds to a nonrigid rotation of the coordinate system such that the rotated axes (no longer perpendicular) pass (nearly) through the clusters. An oblique rotation seeks to express each variable in terms of a minimum number of factorspreferably, a single factor. Oblique rotations are discussed in several sources (see, for example, [6] or [10]) and will not be pursued in this book .
(e
.002 .019 .075
()
9
~
Figure 9.3 Rotated maximum likelihood loadings for factor pairs (1, 2) and (1, 3)decathlon data. (The numbers in the figures correspond to variables.)
.28
run
0.6
B
CV 0 L 0.0
.057
0.8
.51
.62
Plots of rotated maximum likelihood loadings for factors pairs (1,2) and (1,3) are displayed in Figure 9.3 on page 513. The points are generally grouped along the factor axes. Plots of rotated principal component loadings are very similar. . •
Oblique Rotations Orthogonal rotations are appropriate for a factor model in which the common ~ac" tors are assumed to be independent. Many investigators in social sciences conSIder oblique (nonorthogonal) rotations, as well as orthogonal rotations. The former are
e
e
9.S Factor Scores In factor analysis, interest is usually centered on the parameters in the factor model. However, the estimated values of the common factors, called factor scores, may also be required. These quantities are often used for diagnostic purposes, as well as inputs to a subsequent analysis. . Factor scores are not estimates of unknown parameters in the usual sense. Rather, they are estimates of values for the unobserved random factor vectors Fj , j = 1,2, ... , n. That is, factor scores fj = estimate of the values fj attained by Fj (jth case)
514
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Factor Scores 5 15
The estimation situation is complicated by the fact that the unobserved quantities f. and Ej outnumber the observed Xj. To overcome this difficulty, some rather heUris~ tic, but reasoned, approaches to the problem of estimating factor values have been advanced. We describe two of these approaches. Both of the factor score approaches have two elements in common: 1. They treat the estimated factor loadings were the true values.
Factor Scores Obtained by Weighted Least Squares from the Maximum Likelihood Estimates
c=
e and specific variances ~i as if they
j
ij
=
(L'~JL)JL'~l(Xj _ jL) ..1JL'~l(x,.  i),
]. = 12 , , ... ,n
or, if the correlation matrix is factored
2. They involve linear transformations of the original data, perhaps centered or standardized. "TYpically, th.e estimated rotated loadings, rather than the . original estimated loadings, are used to compute factor scores. The computational formulas, as given in this section, do not change when rotated loadings are substituted for unrotated loadings, so we will not differentiate between them.
C·) =
(950)
(L'~JL z Z z )lL,·r.J Z T Z Zj j = 1,2, ... ,n
where Zj
= nlj2 (Xj
 i), as in (825), and
jJ = LzL~ +
~z.
The Weighted Least Squares Method
The factor scores generated by (950) have sample mean vector 0 and zero sample covariances. (See Exercise 9.16.)
Suppose first that the mean vector p" the factor loadings L, and the specific variance 'Ware known for the factor model
If rotated loadings L* = ~T are used in place of the originalloadings in (950), the subsequent factor scores, ' are related to C·} by} C* = T'C.J' ,. = 1" 2 ... , n .
Xp,
(pXl)
(pXJ)
L
F+E
(pXm)(mXJ)
(pxJ)
Further, regard the specific factors E' = [Bb B2' .•• , Bp] as errors. Since Var( Si) = I/Ii, i = 1, 2, ... , p, need not be equal, Bartlett [2] has suggested that weighted least squares be used to estimate the common factor values. The sum of the squares of the errors, weighted by the reciprocal of their variances, is (947)
f,·
Com~e.nt. If the factor loadings are estimated by the principal component method, It IS customary to generate factor scores using an unweighted (ordinary) least squares procedure. Implicitly, this amounts to assuming that the I/Ii are equal or nearly equal. The factor scores are.then 
or
Cj = (L~LzrrL~zj for standardized data. Since we have
L = [~el i ~ e2
Bartlett proposed choosing the estimates Cof f to minimize (947). The solution (see Exercise 7.3) is (951)
(948)
Motivated by (948), we take the estimates L, ~, and jL = i as the true values and obtain the factor scores for the jth case as
For these factor scores,
(949)
When L and ~ are determined by the maximum likelihood method, these estimates must satisfy the uniqueness condition, L'~JL = ..1, a diagonal matrix. We then have the following:
(sample mean) and 1
n ~ ~
~f.f'. = I
n  1 j=J
'
,
(sample covariance)
5 16
Factor Scores 5 17
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
In an attempt to reduce the effects of a (possibly) incorrect determination of the number of factors, practitioners tend to calculate the factor scores in (955) by using S (the original sample covariance matrix) instead of I = LL' + ,J,. We then have the following:
Comparing (951) with (821), we see that the fj are nothing more than the first m (scaled) principal components, evaluated at Xj'
The Regression Method Starting again with the original factor model X  11 = LF + E, we initially treat the loadings matrix L and specific variance matrix 'I' as known. When the common ~ factors F and the specific factors (or errors) E are jointly normally distributed with . meanS and covariances given by (93), the linear combination X  JL = LF + E has an Np(O, LV + '1') distribution.·(See Result 4.3.) Moreover, the joint distribution of (X  JL) and F is Nm+p(O, I*), where
(m+p~~m+p)
=
..........
f·}
= L'SI(X'J  x) '
j = 1,2, ... ,n
(958)
or, if a correlation matrix is factored, j = 1,2, ... ,n
II = (pxp) LV + 'I' i Ll i (pXm)
,
Factor Scores Obtained by Regression
;~~:;········T;:!:~
(952)
~_".".;,..;."
where, see (825),
and 0 is an (m + p) X 1. vector of zeros. Using Result 4.6, we find that the conditional distribution of Fix is multivariate normal with mean = E(Flx) = L'II(x  11) = L'(LL' + 'l'fl(X  11)
(953)
Again, if rotated loadings L* = LT are used in place of the original loadings in (958), the subsequent factor scores fj are related to fj by
and covariance = Cov(Flx) = I  L'I1L = I  L'(LL'
+ 'l'r1L
The quantities L'(LL' + 'l'rl in (953) are the coefficients in a (multivariate) regression of the factors On the variables. Estimates of these coefficients produce factor scores that are analogous to the estimates of the conditional mean values in multivariate regression analysis. (See Chapter 7.) Consequen!ly, giv~n any vector of observations xi' and taking the maximum likelihood estimates L and 'I' as the true values, we see that the jth factor score vector is given by
fj = i:II(xj  x) = L' (Li; + ,J,fl(Xj  x),
j
= 1,2, ... , n
(955)
The calculation of f) in (955) can be simplified by using the matrix identity (see Exercise 9.6) (956) L' (LL' + ,J,r1 = (I + L' ,plifl L' ,pl
.
(mXp)
(pXp)
(mxm)
= (L',J,ILrl(1
A numerical measure of agreement between the factor scores generated from two different calculation methods is provided by the sample correlation coefficient between scores On the same factor. Of the methods presented, none is recommended as uniformly superior.
Example 9.12 (Computing factor scores) We shall illustrate the computation of factor scores by the least squares and regression methods using the stockprice data discussed in Example 9.10. A maximum likelihood solution from R gave the estimated rotated loadings and specific variances
(mXp) (pxp)
This identity allows us to compare the factor scores in (955), generated by the regression argument, with those generated by the weighted least squares procedure , 'LS [see (950)]. Temporarily, we denote the former by ff and the latter by fj • Then, using (956), we obtain
fjS
j = 1,2, ... , n
(954)
+ L',J,IL)ff = (I + (L',plLrl)ff
For maximum likelihood estimates (L',J,li)'l = AI and if the elements of this diagonal matrix are close to zero, the regression and generalized least squares methods will give nearly the same factor scores.
Li
=
.763 .821 .669 [ .118 .113
.024]
.227 .104 .993 .675
[.42 and,J,z
=
0 0 0 0
o .27
o .54 o o o o
The vector of standardized observations, Z' =
[.50, 1.40, .20, .70; 1.40]
yields the following scores On factors 1 and 2:
o o
o o o .00
o
1]
Perspectives and a Strategy for Factor Analysis 519
518 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
on factor 2, and so forth. Data reduction is accomplished by replacing the standardized data by these simple factor scores. The simple factor scores are frequently highly correlated with the factor scores obtained by the more complex least squares and regression methods.
Weighted least squares (950):6
f=
Ci:'wli*)li*',j,l = z z z z z z
[.61J .61
Example 9.13 (Creating simple summary scores from factor analysis groupings) The principal component factor analysis of the stock price data in Example 9.4 produced the estimated loadings
Regression (958):
.526 .063
.221 .026
.137· .011J 1.023 .001
.50] ~.40
_ .20[ .70 1.40
In this case, the two methods produce very similar results. All of the factor scores, obtained using (958), are plotted in Figure 9.4.
Comment. Factor scores with a rather pleasing intuitive property can structed very simply. Group the variables with high (say, greater than absolute value) loadings on a factor. The scores for factor 1 are then summing the (standardized) observed values of the variables in the bined according to the sign of the loadings. The factor scores for sums of the standardized observations corresponding to variables with
•
'"B g
• •
•
•
••
•
•
• • •• •
•• • • • • •
tI.
•• •
)
•
2 2
• ••
•
• \
.437] .280 .374 .694
••
L*
and
=
LT
.719
.852 .851 = .813 [ .133 .084
.030] .214 .079 .911
.909
For each factor, take the loadings with largest absolute value in L as equal in magnitude, and neglect the smaller loadings. Thus, we create the linear combinations
!I
=
fz =
Xl + X4
X2
+ Xs
+
X3

Xl
+
X4
+
Xs
as a summary. In practice, we would standardize these new variables. If, instead of L, we start with the varimax rotated loadings L*, the simple factor scores would be
il =
Xl +
!2 = X4
o 2
L
.732 .831 = .726 [ .605 .563
X2
+
X3
+ Xs
The identification of high loadings and negligible loadings is really quite subjective. _ Linear compounds that make subjectmatter sense are preferable. Although multivariate normality is often assumed for the variables in a factor analysis, it is very difficult to justify the assumption for a large number of variables . As we pointed out in Chapter 4, marginal transformations may help. Similarly, the factor scores mayor may not be normally distributed. Bivariate scatter plots of factor scores can produce all sorts of nonelliptical shapes. Plots of factor scores should be examined prior to using these scores in other analyses. They can reveal outlying values and the extent of the (possible) nonnormality.
•
o Factor)
Figure 9.4 Factor scores using (958) for factors 1 and 2 of the stockprice
(maximum likelihood estimates of the factor loadings). 6 In order to calculate the weighted least squares factor scores, .00 in the fourth "'. was set to .01 so that this matrix could be inverted.
Perspectives and a Strategy for Factor Analysis There are many decisions that must be made in any factor analytic study. Probably the most important decision is the choice of m, the number of common factors. Although a large sample test of the adequacy of a model is available for a given rn, it is suitable only for data that are approximately normally distributed. Moreover, the test will most assuredly reject the model for small rn if the number of variables and observations is large. Yet this is the situation when factor analysis provides a useful approximation. Most often, the final choice of m is based on some combination of
Perspectives and a Strategy for Factor Analysis 521
520 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices The sample correlation matrix
(1) the proportion of the sample variance explained, (2) subjectmatter knowledge, and (3) the "reasonableness" of the results. The choice of the solution method and type of rotation is a less crucial decision. In fact, the most satisfactory factor analyses are those in which rotations are tried with more than one method and all the results substantially confirm the same
factor structure. At the present time, factor analysis still maintains the flavor of an art, and no single strategy should yet be "chiseled into stone." We suggest and illustrate one reasonable option: 1. Perform a principal component factor analysis. This method is particularly appropriate for a first pass through the data. (It is not required that R or S be nonsingular. ) (a) Look for suspicious observations by plotting the factor scores. Also, calculate standardized scores for each observation and squared distances as described in Section 4.6. (b) Try a varimax rotation.
2. Perform a maximum likelihood factor analysis, including a varimax rotation. 3. Compare the solutions obtained from the two factor analyses. (8) Do the loadings group in the same manner? (b) Plot factor scores obtained for principal components against scores from
the maximum likelihood analysis. 4. Repeat the first three steps for other numbers of common factors m. Do extra factors necessarily contribute to the understanding and interpretation of the data? 5. For large data sets, split them in half and perform a factor analysis on each part. Compare the two results with each other and with that obtained from the complete data set to check the stability of the solution. (The data might be divided by placing the first half of the cases in one group and the second half of the cases in the other group. This would reveal changes over time.)
R=
Head:
Xl = skull length { X 2 = skull breadth
Leg:
X3 = femurlength { X 4 = tibia length
Wing:
X5 = humerus length { X6 = ulna length
.505 .569 1.000 .422 .422 1.000 .467 .926 .482 .877 .450 .878
.602 .467 .926 1.000 .874 .894
.621 .603 .482 .450 .877 . .878 .874 .894 1.000 .937 .937 1.000
was factor analyzed by the principal component and maximum likelihood methods for an m = 3 factor model. The results are given in Table 9.10. 7 Table 9.10 Factor Analysis of ChickenBone Data Principal Component Variable 1. 2. 3. 4. 5. 6.
Skull length Skull breadth Femur length Tibia length Humeruslength Ulna length
Cumulative proportion of total (standardized) sample variance explained
Estimated factor loadings F2 F3 Fl .741 .604 .929 .943 .948 .945
.350 .720 .233 .175 .143 .189
.573 .340 .075 .067 .045 .047
.743
.873
.950
Rotated estimated loadings F; F~ Fi
.921 .904 .888 ~
.244 (.949) .164 .212 .228 .192
(.902) .211 .218 .252 .283 .264
.576
.763
.950
.355
~
~i .00 .00 .08 .08 .08 .07
Maximum Likelihood Variable
Example 9.14 (Factor analysis of chickenbone data) We present the results of several factor analyses on bone and skull measurements of white leghorn fowl. The original data were taken from Dunn [5]. Factor analysis of Dunn's data was originally considered by Wright [15], who started his analysis from a different correlation matrix than the one we use. The full data set consists of n = 276 measurements on bone dimensions:
1.000 .505 .569 .602 .621 .603
1. 2. 3. 4. 5. 6.
Skull length Skull breadth Femur length Tibia length Humerus length Ulna length
Cumulative proportion of total (standardized) sample variance explained
Estimated factor loadings F2 F3 Fl .602 .467 .926 1.000 .874 .894
.214 .177 .145 .000 .463 .336
.286 .652 .057 .000 .012 .039
.667
.738
.823
Rotated estimated loadings F; F~ Fi .467
~ .890 .936 .831
~
.559
(506 ) .792 .289 .345 .362 .325
.779
.128 .050 .084 .073 .396 .272
If, .51 .33 .12 .00 .02 .09
.823
. 7 Notice the estimated specific variance of .00 for tibia length in the maximum likelihood solution. TIlls su.ggests that maximizing the likelihood function may produce a Heywood case. Readers attempting ~ to replicate our results should try the Hey(wood) option if SAS or similar software is used.
522 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Perspectives and a Strategy for Factor Analysis 523
After rotation, the two methods of solution appear to give somewhat different results. Focusing our attention on the principal component method and the cumula_ tive proportion of the total sample variance explained, we see that a threefactor solution appears to be warranted. The third factor explains a "significant" amount of additional sample variation. The first factor appears to be a bodysize factor dominated by wing and leg dimensions. The second and third factors, collectively, represent skull dimensions and might be given the same names as the variables, skull breadth and skull length, respectively. The rotated maximum likelihood factor loadings are consistent with those generated by the principal component method for the first factor, but not for factors 2 . and 3. For the maximum likelihood method, the second factor appears to represent head size. The meaning of the third factor is unclear, and it is probably not needed. Further support for retaining three or fewer factors is provided by the resid~al matrix obtained from the maximum likelihood estimates:
Rii:~ z z z
=
.000 .000 .003 .000 .001 .004
3 I
.001 .000
•
11
••
.•
•
•
• • .... • • • •• • •• .. ••• • .$
• •• $$
$ • •• $ ••• $
$ •• $. $ ••• .$ $ $ . . $ $ . .$
•
•••
.000
$
$
$ •• $
$•
$$
I
3

• • .$•
$
. • ......... .. • • • • • • .... • . .. l$. • • · • • ·• • • •• • . • •• . · $ • $ $ •
•• $$ ••
•
•• • ••
$ (
•
$$
$
.
• $
•• ••
I
o
2
$•

•
• ••
••• •
••• ••••• •
2 l
All of the entries in this matrix are very small. We shall pursue the m = 3 factor model in this example. An m = 2 factor model is considered in Exercise 9.10. Factor scores for factors 1 and 2 produced from (958) with the rotated maximum likelihood estimates are plotted in Figure 9.5. Plots of this kind allow us to identify observations that, for one reason or another, are not consistent with the remaining observations. Potential outliers are circled in the figure. It is also of interest to plot pairs of factor scores obtained using the principal component and maximum likelihood estimates of factor loadings. For the chickenbone data, plots of pairs of factor scores are given in Figure 9.6 on pages 524526. If the loadings on a particular factor agree, the pairs of scores should cluster tightly about the 45° line through the origin. Sets of loadings that do not agree will produce factor scores that deviate from this pattern. If .the latter occurs, it is usually associated with the last factors and may suggest that the number of factors is too large. That is, the last factors are not meaningful. This seems to be the case with the third factor in the chickenbone data, as indicated by Plot (c) in Figure 9.6. Plots of pairs of factor scores using estimated loadings from two solution methods are also good tools for detecting outliers. If the sets of loadings for a factor tend to agree, outliers will appear as points in the neighborhood of the 45° line, but far from the origin and the cluster of the remaining points. It is clear from Plot (b) in Figure 9.6 that one of the 276 observations is not consistent with the others. It has an unusually large Fzscore. When this point, [39.1,39.3,75.7,115,73.4,69.1], was removed and the analysis repeate{l, the loadings were not altered appreciably. When the data set is large, it should be divided into two (roughly) equal sets, and a factor analysis should be performed on each half. The results of these analyses can be compared with each other and with the analysis for the full data set to
. •
•
• .000 .000
I
•
$.
.000 .000 .000 .000 .000
I
2
I
.000 .001 .000 .000 .001
I
I


I
2
3
Figure 9.S Factor scores for the first two factors of chickenbone data. test .the sta.bility of the solution. If the results are consistent with one another confIdence In the solution is increased. ' The .chickenbone data were divided into two sets of nr = 137 and n2 = 139 observatIOns, respectively. The resulting sample correlation matrices were .
Rr=
1.000 .696 .588 .639 .694 .660
1.000 .540 1.000 .575 .901 .606 .844 .584 .866
1.000 .835 1.000 .863 .931 1.000
1.000 .366 1.000 .572 .352 1.000 .587 .406 .950 .587 .420 .909 .598 .386 .894
1.000 .911 1.000 .927 .940 1.000
and
R2 =
Perspectives and a Strategy for Factor Analysis 525
524 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
11 2.7
9. o

7.5
~
I
I
I Principal component
I
I
I
T
I
T
I
10
I

I
I I 1 I 11
I
1.8
I
11 I I I 32 1121 12 I I 21 1I 1 rI2321 1 I 4 26311 I 21 24 3 1 I 33112 2 11 21 17 2 1
.9
o~
3
______________________~~~~~+r~ 11 43 1224331 13 223152 11 1411221 I 12121 115251 11133 1 1 2
.9
3 2 1
11 1
6.0


4.5


3.0 ,..

Maximum likelihood
III I 1 1
1.8
1.5
1 IJ 1 1 I 2 12111 1 23221 12346231 224331 21 4 46C61 1

1
o
2.7
1.5 I
3.6
3.5 3.0
2.5 2.0 1.5
1.0
.5
o
.5

.
1.0
1.5
2.0
2.5
3.0
(a) First factor
Figure 9.6 Pairs of factor scores for the chickenbone data. (Loadings are estimated by principal component and maximum likelihood methods.)
The rotated estimated loadings, specific variances, and proportion of the total (standardized) sample variance explained for a principal component solution of an m = 3 factor model are given in Table 9.11 on page 525. The results for the two halves of the chickenbone measurements are very similar. Factors F; and F; interchange with respect to their labels, skull length and skull breadth, but they collectively see<m to represent head size. The first factor, F~, again appears to be a bodysize factor dominated by leg and wing dimensions. These are the same interpretations we gave to the results from a principal component factor analysis of the entire set of data. The solution is remarkably stable, and we can be fairly confident that the large loadings are "real." As we have pointed out however, three factors are probably too many. A one or twofactor model is surely sufficient for the chickenbone data, and you are encouraged to repeat the analyses here with fewer factors and alternative solution methods. (See Exercise 9.10.) •
Figure 9.6
I 3.00
213625 572 121A3837 31 11525111 223 31 I 21 1 III 1 I II 1 I1 1 I I 2.25 1.50 .75 0 ~

Maximum likelihood
I
I
I
I
I
I
I
I
I
I~
~~
300
3~
UO
5~
~OO
6~
7~
( continued)
(b) Second factor
Table 9.11
First set (n} = 137 observations) Rotated estimated factor loadings Variable
Fi
F;
F;
1. Skull length
Skull breadth Femur length Tibia length Humerus length Ulna length
.360 .303 .914 .877 .830 .871
.361 (.899) .238 .270 .247 .231
(.853 ) .312 .175 .242 .395 .332 .
Cumulative proportion of total (standardized) sample variance' explained
.546
.743
.940
2. 3. 4. 5. 6.
Second set (n2 = 139 observations) Rotated estimated factor loadings
if,i
Fi
F;
F;
t/!i
.01 .00 .08
.352 .203 .930 .925 .912 .914
(.921 ) .145 .239 .248 .252 .272
.167 (.968) .130 .187 .208 .168
.00 .00 .06 .05 .06 .06
.593
.780
.962
.10 .11 .08
526
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Principal component
3.00
2.25 2
2 1.50
.75
o
1 111 1 1 1 1 1 2 1 11 I III . I I I 1 11 1111 21 32 1 I 21 22 121 I I 1 2 1 3141 I I I 2 I 111 11 I 11 11 I 1 1 2 II 2 I 11 3 I I 111 I 111
I 1 1
II1 22 1 I I I I I 1 2 1 2 I I 21 2 1 211 11 11 21 1 1 I 1 2 1 11 III 1 2 I I1 3 11 I 112 I II 1 2 I 1 I 1 I I· 1 I I 1 11 I 2 1
I
2
.75
1.50
2.25 .
I
Maximum likelihood 1
SOME COMPUTATIONAL DETAILS FOR MAXIMUM LIKELIHOOD ESTIMATION Althoug h a simple aqalyticaJ expressi on cannot be obtaine d for the maximu m likeliho od estimato rs L and 'It, they can be shown to satisfy certain equation s. Not surprisingly, the conditio ns are stated in terms of the maximu m likeliho od estimato r n S" = (l/n) (Xi  X) (Xi  X)' of an unstruc tured covarian ce matrix. Some i=1 factor analysts employ the usual sample covarian ce S, but still use the title maximu m likelihoo d to refer to resulting estimates. This modific ation, referenc ed in Footnot e 4 of this chapter, amounts to employi ng the likeliho od obtained from the Wishart It distribu tion of (Xi  X) (Xi  X)' and ignoring the minor contribu tion due to i=1 the normal density for X. The factor analysis of R is, of course, unaffec ted by the choice of Sn or S, since they both produce the same correlat ion matrix.
2:
3.00 3.0 2.4 1.8 1.2
.6
.6
1.2
(c) Third factor
Figure 9.6 (continued)
. ., If ral and social Factor analysis has a tremendous mtUltive appea or the behavio . . al sciences In these areas, it is natural to regard multivariate observatlOns. o~, apmmt . . b ac or and human processe s as manifestations of underlym g uno ser~ able "traits .'. t ., analysis provide s a way of explamm g t he 0 bserved variability ID behavlOr ID erms of these traits. Still when all is said and done, factor analysis remains very su b'Jec fIV.e· Our exam h f t pies, in dommon with most published sources, consist of situations i~ whlch ~ ~I a~ ~ analysis model provides reasonable explanations in terms of a few mt~rr::e ah e a tors. In practice the vast majority of attempted factor analyses do not Ylel l suc ~eru; cut results. unfortun ately, the criterion for judging the quality of any factor an YSIS has not been well quantified. Rather, that quality seems to depend on a WOW criterion . .' If, while scrutinizing the factor analYSIs, the mvestIga tor can s hout "Wow, I understand these factors," the application is deemed successful.
2:
Result 9A.I. Let x I, Xz, •.. , Xn be a random sample from a normal populat ion. The maximu m likeliho od estimate s i and .q, are obtaine d by maximiz ing (925) subject to the uniquen ess conditio n in (926). They satisfy
(9A1) so the jth column of .q,I/2i is thAe (nonnor malized ) eigenve ctor of .q,I/2Sn .q,1/2 correspo nding to eigenva lue 1 + ~i' Here n
Sn
= n 1 2: (Xj  i)(xj  i)' = nt(n j=l
521'
l)S
and
'&1 ~ '&2 ~ .,. ~ .&m
528
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Some Computational Details for Maximum Likelihood Estimation 529
Also, at convergence,
~i = ithdiagonalelementofSn  LL'
(9A2)
2. Given.f, compute the first m distinct eigenvalues, Al > A2 > ... > A > 1, and correspon?ing eigenvectors, el, e2, ... ,em, of the "uniquenessrescal;d" covariance matnx
and
(9A5)
We avoid the details of the proof. However, it is evident that jL = xand a consideration of the loglikelihood leads to the maximization of (nj2) [1nl ~ I + tr(~ISn)] over L and '1'. Equivalently, since Sn and p are constant with. respect to the maximization, We minimize (9A3) subject to L'qt1L
=
a, a diagonal matrix.
•
Comment. Lawley and Maxwell [10], along with many others who do factor analysis, use the unbiased estimate S of the covariance matrix instead of the maxi _ mum likelihood estimate Sn. Now, (n  1) S has, for normal data, a Wishart distribution. [See (421) and (423).] If we ignore the contribution to the likelihood in (925) from the second term involving (IL  x), then maximizing the reduced likelihood over L and 'I' is equivalent to maximizing the Wishart likelihood Likelihood ex I ~
1(n1)/2 e[(n1)/2]lr[:£'S]
over L and '1'. Equivalently, we can minimize 1nl ~ I + tr(rIS)
Under these conditions, Result (9A 1) holds with S in place of S". Also, for large n, S;.. and S11",are almost identical, and the corresponding maximum likelihood "estimates, • L and '1', would be similar. For testing the factor model [see (939)], ILL' + '1'1 should be compared with ISn I if the actual likelihood of (925) is employed, and I ii' + .fl should be compared with IS I if the foregoing Wishart likelihood is used to derive i and .f. A
,..
 1)j2constraints on the elements of Land '1', and the likelihood equations are solved, subject to these contraints, in an iterative fashion. One procedure is the following:
1. Compute initial estimates of the specific variances 1/11,1/12,"" I/Ip. J6reskog [8] suggests setting 2 P
where Sii is the ith diagonal element of Sl.
sI!
. .Comment. It ofte~ happens that the objective funct~on in'(9A3) has a relative
~Il1mm~~ correspondmg to negative values for some I/Ii' This solution is clearly
madm1sslble and is said to b~ improper, or a Heywood case. For most packaged computer p~o.grams, negative I/Ii, if they occur on a particular iteration, are changed to small pOSltlve numbers before proceeding with the next step.
+
'\{I z
matrix for the standardized variables is L. = V1/ 2L, and the corresponding specific variance matrix is '1'. = V1/2 qtV1/2, where V1/2 is the diagonal matrix with ith diagonal element O'i/f2. If R is substituted for S" in the objective function of (9A3), the investigator minimizes In (
IL.L~ + '1'. I) IRI
+ tr[(L.L~ + qt.flR) 
p
(9A7)
· , 1/2 I ntrod~cm~ the diagonal matrix V ,whose ith diagonal element is the square root of the lth dIagonal element of Sn, we can write the objective function in (9A7) as
Recommended Computational Scheme For m > 1, the condition L'qt 1L = a effectively imposes m(m
(1 _1.. m) (1,)
e
When ~ has the factor analysis structure ~ = LL' + '1', p can be factored as p = VI/2~V1/2 = (V1/2L) (V1/2L), + V1/2qtV I/2 = LzL~ + '1' •. The loading
1nl ~ I + tr(~lS)  InlSIp
.1•. =
(9A6) 3. Substitute i obtained in (9A6) into the likelihood function (9A3), and minimize the result with re~pe~t to ,'/11:. ,'/12, ... ,,'/1p' A numerical search routine must be used. The values 1/11,1/12, •.. ,1/1 p obtained from this minimization are employed at Step (2) to create a new L Steps (2) and (3) are repeated until convergencethat is, until the differences between successive values of ij and ~i are negligible.
Maximum likelihood Estimators of p = l z l'z
or, as in (9A3),
'1'1
Let ~ = [e1 i ~2 l~'" i e!?'] be the p X m matrix of normalized eigenvectors and A = diaglA lo A2'''~' Am] ~e th~ m ::< ~m diagonal matrix of eigenvalues. From (9A1), A = I + a and E = qt1/2LA1/2. Thus, we obtain the estimates
(9A4)
In (
IVI/211L L' + 'I' IIV1/21) ~ • z ~• + tr [(L L' + 'I' )lVI/2V1/2RVI/2V1/2) _ p IVl/211RIIV1/21 '" I (V 1/2L.) (V1/2L )' + VI /2qt V1/21) = In ( • z I Sn I
+ tr[ «VI/2Lz)(Vl/2L.)' + V 112 '1',V1/2) IS n) _ ~ln
'(I ii' + i I) ISnl
~~
~
1
+tr[(LL'+qtfSn)p
p
(9A8)
530
Exercises 531
Chapter 9 Factor Analysis and Inference for Structured C'avariance Matrices The last inequality follows because the maximum likelihood estimates I. and ~ minimize the objective function (9A3). [Equality holds in (9A8) for L. = yI/lL and = yl/2iYI/l.JTherefore,minimizing (9A7) over L. and '1'. is equivalent I to obtaining Land i from Sn and estimating L. = V /2L by L. = yI/lL and '1'. = VI/l'l'VI/l by = yI/2~yI/l. The rationale for the latter procedure comes from the invariance property of maximum likelihood estimators. [See (
Now, S i:i:' = Am+lem+te:"+l
i.
i.
9.6. Verify the following matrix identities. (a) (I + L''I' I Lr l L''Ir l L = I  (I + L''I'lLr l Hint: Premultiply both sides by (I + L''I'tL). (b) (LL' + 'l'r l = '1'1·_ 'I'IL(I + L''I'lL)lL''I't
Exercises 9.1.
Hint: Postmultiply both sides by (LL' + '1') and use (a). (c) L'(LL' + 'l'r t = (I + L''I' l Lr 1L''I'l
Show that the covariance matrix
P=
Hint: Postm.!lltiply the result in (b) by L use (a), and take the transpose, noting that (LL' + '1') 1, '1'1, and (I + L''I'tLr l are symmetric matrices.
1.0 .63 .45] .63 1.0 .35 [ .35 1.0
.45
9.7.
for the p = 3 standardized random variables 2 1 ,22 , and 23 can be generated by the ::::~.: m = 1 factor model 21 = .9FI + 61 22 =
23 =
.7FI + .5FI +
62
(The factor model parameterization need not be unique.) Let the factor model with p = 2 and m = 1 prevail. Show that O"ll
=
0"22
.19 0
A3
e;
0"21
=
Cll C2l
l: 
.9
.7
1
Show that there is a unique choice of L and 'I' with l: = LL' + '1', but that 0/3 < 0, so the choice is not admissible. 9.9. In a stU?y of liquor preference in France, Stoetzel [14] collected preference rankings of p = 9 lIquor types from n = 1442 individuals. A factor analysis of the 9 x 9 sample correlation matrix of rank orderings gave the following estimated loadings:
= [.625, .593, .507]
ez = [.219,.491,.843]
e3 =
Estimated factor loadings
[.749, .638, .177]
(a) Assuming an m = 1 factor model, calculate the loading matrix L and matrix of specific variances 'I' using the principal component solution method. Compare the results with those in Exercise 9.!. (b) What proportion of the total population variance is explained by the first common factor? 9.4. Given p and 'I' in Exercise 9.1 and an m = 1 factor model, calculate the reduced correlation matrix = p  'I' and the principal factor solution for the loading matrix L. Is the result consistent with the information in Exercise 9.1? Should it be? . 9.S. Establish the inequality (919). Hint: Since S  i:i>  ~ has zeros on the diagonal,
p
(sum of squared entries ofS 
=
_[1.4 .41 .9] .7
gl
That is, write p in the form p = LL' + '1'. 9.2. Use the information in Exercise 9.1. (a) Calculate communalities hT, i = 1,2,3, and interpret these quantities. (b) Calculate Corr(2j ,Ft ) for i = 1,2,3. Which variable might carry the greatest weight in "naming" the common factor? Why? 9.3. The eigenvalues and eigenvectors of the correlation matrix p in Exercise 9.1 are
= 1.96, = .68, = .36,
0"12
cL + 0/2
=
9.8.· (Unique but improper solution: Heywood case.) Consider an m = 1 factor model for the population with covariance matrix
'I' = Cov(e) = [ ~
A2
Ctl + 0/1,
and, for given O"ll, 0"22, and 0"12, there is an infinity of choices for L and '1'.
63
where Var (Ft) = 1, Cov (e, Ft) = 0, and
Al
+ ... +Apepe~ = P(2)A(2)P(2), where P(2) = [e m +li···i ep ]
and A(2) is the diagonal matrix with elements Am+l>"" Ap. Use (sum of squared entries of A) = tr AA' and Ir [P(2)A(2)A(2i(2)] =tr [A (2l A (2)).
i:i>  ~)
:s;
(sum of squared entries ofS 
i:l:')
Variable (Xl)
FI
F2
F3
Liquors Kirsch Mirabelle Rum Marc Whiskey Calvados Cognac Armagnac
.64
.02 .06 .24 .74 .66 .08 .20 .03 .17
.16 .10 .19 .97* .39 .09 .04 .42 .14
.50 .46 .17 .29 .29 .49 .52 .60
*This figure is too high. It exceeds the maximum value of .64, as a result of an approximation method for obtaining the estimated factor loadings used by Stoetzel.
Exercises 533 532 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices Given these results, Stoetzel concluded the following: The major principle of liquor preference in France is the distinction between sweet and strong liquors. The second motivating element is price, which can be understood by remembering that liquor is both an expensive commodity and an item of conspicuous consumption. Except in the case of the two most popular and least expensive items (rum and marc), this second factor plays. a much smaller role in producing preference judgments. The third factor concerns the sociological and primarily the regional, variability of the judgments. (See [14], p.ll.) (a) Given what you know about the various liquors involved, does Stoetzel's interpretation seem reasonable? (b) Plot the loading pairs for the first two factors. Conduct a graphical orthogonal rotation of the factor axes. Generate approximate rotated loadings. Interpret the rotated loadings for the first two factors. Does your interpretation agree with Stoetzel's interpretation of these factors from the unrotated loadings? Explain. . 9.10. The correlation matrix for chickenbone measurements (see Example 9.14) is 1.000 .505 1.000 .422 1.000 .569 .926 1.000 .467 .602 .874 1.000 .877 .482 .621 .894 .937 1.000 .878 .450 .603
The follo~ing maximum likelihood estimates of the factor loadings for an m = 1 model were obtamed: Estimated factor loadings Variable
FI
1. In(length) 2. In(width) 3. In(height)
.1022 .0752 .0765
Using the ~stimated factor loadings, obtain the maximum likelihood estimates of each of the followmg. (a) Specific variances. (b) Communalities. (c) Proportion of variance explained by the factor. (d) The residual matrix Sn  ii:  ,j. Hint: Convert S to Sn.
The following estimated factor loadings were extracted by the maximum likelihood
9.13. ~e~er ~ EX,ercise 9.1~. Compute the test statistic in (939). Indicate why a test of ?l:  LL + 'I' (WIth m = 1) versus HI: l: unrestricted cannot be carried out for thIS example. [See (940).]
procedure:
9.14. The maximum likelihood factor loading estimates are given in (9A6) by Estimated factor loadings Variable
1. 2. 3. 4. 5. 6.
Skull length Skull breadth Femur length Tibia length Humerus length Ulna length
FI
.602 .467 .926 1.000 .874 .894
Varimax rotated estimated factor loadings
F2
F;
F;
.200
.484 .375 .603 519 .861 .744
.411 .319 .717 .855 .499 .594
.154 .143 .000 .476 .327
Using the unrotated estimated factor loadings, obtain the maximum likelihood estimates of the following. (a) The specific variances. (b) The communalities. (c) The proportion of variance explained by each factor. (d) The residual matrix R  izi~  ~ z· 9.11. Refer to Exercise 9.10. COlllpute the value of the varimax criterion using both unrotated and rotated estimated factor loadings. Comment on the results. 9.12. The covariance matrix for the logarithms of turtle measurements (see Example 8.4) is . 11.072 ] S = 103 8.019 6.417 [ 8.160 6.005 6.773
i
=
,j1/2i'& 1/2
Verify, for this choice, that
where'& = A  I is a diagonal matrix . 9.IS. Hirsche! and Wichern [7] investigate the consistency, determinants, and uses of accou~tmg and ma~ketval~e measures of profitability. As part of their study, a factor analYSIS of accountmg p~ofIt me~sures and market estiJ?1ates of economic profits was conducted. The correlatIOn matnx. of ~~counting historical, accounting replacement, and marketvalue measures of profItabIlIty for a sample of firms operating in 1977 is as follows:
Variable Historical return on assets, HRA Historical return on equity, HRE Historical return on sales, HRS Replacement return on assets, RRA Replacement return on equity, RRE Replacement return on sales, RRS Market Q ratio, Q Market relative excess value, REV
HRA
HRE
HRS RRA RRE
RRS
Q
REV
1.000
.738 .731 .828 .681 .712 .625 .604
1.000 .520 1.000 .652 1.000 .688 .831 513 B87 1.000 .543 .826 .867 .692 .322 .579 .639 .419 .563 .352 .303 .617
1.000 .608 1.000 .610 .937 1.000
Exercises 535
534 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices The following rotated principal component estimates of factor loadings for an m :, factor model were obtained: Estimated factor loadings Variable
FI
F2
F3
Historical return on assets Historical return on equity Historical return on sales Replacement return on assets Replacement return on equity Replacement return on sales Market Q ratio Market relative excess value
.433 .125 .296 .406 .198 .331 .928 .910
.612 .892 .238 .708 .895 .414 .160 .079
.499 .234 .887 .483 .283 .789 .294.355
Cumulative proportion of total variance explained
.287
.628
.908
(a) Using the estimated factor loadings, determine the specific variances and communalities. (b) Determine the residual matrix, R  LzL~  ir z' Given this information and the cumulative proportion of total variance explained in the preceding table, does an m = 3 factor model appear appropriate for these data? (c) Assuming that estimated loadings less than.4 are small, interpret the three factors. Does it appear, for example, that marketvalue measures provide evidence of profitability distinct from that provided by accounting measures? Can you separate accounting historical measures of profitability from accounting replacement measures?
9.16. Verify that factor scores constructed according to (950) have sample mean vector 0 zero sample covariances. l
9.17. Refer to Example 9.12. Using the information in this example, evaluate (i;ir;IL.r . Note: Set the fourth diagonal element of ir z to .01 so that ir;1 can be determined. Will the regression and generalized least squares methods for constructing factors scores for standardized stock price observations give nearly the same results? Hint: See equation (957) and the discussion following it. The following exercises require the use of a computer.
9.18. Refer to Exercise 8.16 concerning the numbers of fish caught. (a) Using only the measurements XI  X4, obtain the principal component solution for factor models with m = 1 and m = 2. (b) Using only the measurements XI  X4, obtain the maximum likelihood solution for .. factor models with m = 1 and m = 2. (c) Rotate your solutions in Parts (a) and (b). Compare the solutions and comment on them. Interpret each factor. (d) Perform a factor analysis using the measurements XI  X6' Determine ~ relisonall>lc: number of factors m, and compare the principal component and maximum hood solutions aft~r rotation. Interpret the factors. 9.19. A firm is attempting to evaluate the quality of its sales staff and is trying to fin~ an amination or series of tests that may reveal the potential for good performance In
The firm has selected a random sample of 50 sales people and has evaluated each on 3 measures of performance: growth of sales, profitability of sales, and newaccount sales. These measures have been converted to a scale, on which 100 indicates "average" performance. Each of the 50 individuals took each of 4 tests, which purported to measure creativity, mechanical reasoning, abstract reasoning, and mathematical ability, respectively. The n = 50 observations on p = 7 variables are listed in Table 9.12 on page 536. (a) Assume an orthQgonal factor model for the standardized variables Zi = (Xi  }Li)/VU:;;, i = 1,2, ... ,7. Obtain either the principal component solution or the maximum likelihood solution for m = 2 and m = 3 common factors . (b) Given your solution in (a), obtain the rotated loadings for m = 2 and m = 3. Compare the two sets of rotated loadings. Interpret the m = 2 and m = 3 factor solutions. (c) List the estimated communalities, specific variances, and LL' + ir for the m = 2 and m = 3 solutions. Compare the results. Which choice of m do you prefer at this point? Why? (d) Conduct a test of Ho: I = LV + 'I' versus HI: I ;t. LV + 'I' for both m = 2 and m = 3 at the Cl' = .01 level. With these results and those in Parts band c, which choice of m appears to be the best? (e) Suppose a new salesperson, selected at random, obtains the test scores x' = [Xi> X2, ... ,X7] = [110,98,105,15,18,12,35]. Calculate the salesperson's factor score using the weighted least squares method and the regression method. Note: The components of x must be standardized using the sample means and variances calculated from the original data.
9.20. Using the airpollution variables Xl> X 2 , X 5 , and X6 given in Table 1.5, generate the sample covariance matrix. (a) Obtain the principal component solution to a factor model with m = 1 and m = 2. (b) Find the maximum likelihood estimates of L and 'I' for m = 1 and m = 2. (c) Compare the factorization obtained by the principal component and maximum likelihood methods. 9.21. Perform a varimax rotation of both m = 2 solutions in Exercise 9.20. Interpret the results. Are the principal component and maximum likelihood solutions consistent with each other? 9.22. Refer to Exercise 9.20. (a) Calculate the factor scores from the m = 2 maximum likelihood estimates by (i) weighted least squares in (950) and (ii) the regression approach of (958). (b) Find the factor scores from the principal component solution, using (951). (c) Compare the three sets of factor scores. 9.23. Repeat Exercise 9.20, starting from the sample correlation matrix. Interpret the factors for the m = 1 and m = 2 solutions. Does it make a difference if R, rather than S, is factored? Explain.
9.24. Perform a factor analysis of the censustract data in Table 8.5. Start with R and obtain both the· maximum likelihood and principal component solutions. Comment on your choice of m. Your analysis should include factor rotation and the computation of factor scores.
9.25. Perform a factor analysis of the "stiffness" measurements given in Table 4.3 and discussed in Example 4.14. Compute factor scores, and check for outliers in the data. Use the sample covariance matrix S.
536
Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
Exercises 537
9.26. Consider the miceweight data in Example 8.6. Start with the sample co variance matrix. . (See Exercise 8.15 for VS;;.)
Table 9.12 Salespeople Data Score on:
Index of:
Salesperson
1 2 3 4 5 6 7 8 9 10 11 12 13 '14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Sales growth
Newaccount sales
(xl)
(X2)
(X3)
(X4)
(X5)
(X6)
(x7)
96.0 91.8 100.3 103.8 107.8 97.5 99.5 122.0 108.3 120.5 109.8 111.8 112.5 105.5 107.0 93.5 105.3 110.8 104.3 105.3 95.3 115.0 92.5 114.0 121.0 102.0 118.0 120.0 90.8 121.0 119.5 92.8 103.3 94.5 121.5 115.5 99.5 99.8 122.3 119.0 109.3 102.5 113.8 87.3 101.8 112.0 96.0 89.8 109.5 118.5
97.8 96.8 99.0 106.8 103.0 99.3 99.0 115.3 103.8 102.0 104.0 100.3 107.0 102.3 102.8 95.0 102.8 103.5 103.0 106.3 95.8 104.3 95.8 105.3 109.0 97.8 107.3 104.8 99.8 104.5 110.5 96.8 100.5 99.0 110.5 107.0 103.5 103.3 108.5 106.8 103.8 99.3 106.8 96.3 99.8 110.8 97.3 94.3 106.5 105.0
09 07 08 13 10 10
12 10 . 12 14 15 14 12 20 17 18 17 18 17 10 10 09 12 14 14 17 12
09
20 15 26 29 32 21 25 51 31 39 32 31 34 34 34 16 32 35 30 27 15 42 16 37 39 23 39 49 17 44 43 10 27 19 42 47 18 28 41 37 32 23 32 15 24 37 14
93.0 88.8 95.0 101.3 102.0 95.8 95.5 110.8 102.8 106.8 103.3 99.5 103.5 99.5 100.0 81.5 101.3 103.3 95.3 99.5 88.5 99.3 87.5 105.3 107.0 93.3 106.8 106.8 92.3 106.3 106.0 88.3 96.0 94.3 106.5 106.5 92.0 102.0 108.3 106.8 102.5 92.5 102.8 83.3 94.8 103.5 89.5 84.3 104.3 106.0
(a) Obtain the principal component solution to the factor model with m = 1 and m = 2.
Mechanical Abstract MatheCreativity reasoning reasoning matics test test test test
Sales profitability
09 18 10 14 12 10 16 08 13 07 11 11 05 17 10 05 09 12 16 10 14 10 08 09 18 13 07 10 18 08 18 13 15 14 09 13 17 01 07 18 07 08 14 12
11
09 15 19 15 16 16 10 17 15 11
15 12 17 13 ' 16 12 19 20 17 15 20 05 16 13 15 08 12 16
10 09 12 12 11 09 15 13 11 12 08 11 11 08 05 11 11 13 11 07 11 07 12 12 07 12 11 13
11 10 08 11
11 10 14 08 14 12 12 13
06 10 09 11 12 11 08 12 11
09 36 39
(b) Find the maximum likelihood estimates of the loadings and specific variances for m = 1 and m = 2. (c) Perform a varimax rotation of the solutions in Parts a and b.
9.27. Repeat Exercise 9.26 by factoring R instead of the sample covariance matrix S. Also, for the mouse with standardized weights [.8, .2, .6, 1.5], obtain the factor scores using the maximum likelihood estimates of the loadings and Equation (958). 9.28. Perform a factor analysis of the national track records for women given in Table 1.9. Use the sample covariance matrix S and interpret the factors. Compute factor scores, and check for out/iers in the data. Repeat the analysis with the sample correlation matrix R. Does it make a difference if R, rather than S, is factored? Explain. .
9.29. Refer to Exercise 9.28. Convert the national track records for women to speeds measured in meters per second. (See Exercise 8.19.) Perform a factor analysis of the speed data. Use the sample covariance matrix S and interpret the factors. Compute factor scores, and check for outliers in the data. Repeat the analysis with the sample correlation matrix R. Does it make a difference if R, rather than S, is fadored? Explain. Compare your results with the results in Exercise 9.28. Which analysis do you prefer? Why?
9.30. Perform a factor analysis of the national track records for men given in Table 8.6. Repeat the steps given in Exercise 9.28. Is the appropriate factor model for the men's data different from the one for the women's data? If not, are the interpretations of the factors roughly the same? If the models are different, explain the differences. 9.31. Refer to Exercise 9.30. Convert the national track records for men to speeds measured in meters per second. (See Exercise 8.21.) Perform a factor analysis of the speed data. Use the sample covariance matrix S and interpret the factors. Compute factor scores, and check for outIiers in the data. Repeat the analYSis with the sample correlation matrix R. Does it make a difference if R, rather than S, is fadored? Explain. Compare your results with the results in Exercise 9.30. Which analysis do you prefer? Why?
9.32. Perform a factor analysis of the data on bulIs given in Table 1.10. Use the seven variables YrHgt, FtFrBody, PrctFFB, Frame, BkFat, SaleHt, and Sale Wt. Factor the sample covariance matrix S and interpret the factors. Compute factor scores, and check for outliers. Repeat the analysis with the sample correlation matrix R. Compare the results obtained from S with the results from R. Does it make a difference if R, rather than S, is factored? Explain. . 9.33. Perform a factor analysis of the psychological profile data in Table 4.6. Use the sample correlation matrix R constructed from measurements on the five variables, Indep, Supp, Benev, Conform and Leader. Obtain both the principal component and maximum likelihood solutions for m = 2 and m = 3 factors. Can you interpret the factors? Your analysis should include factor rotation and the computation of factor scores. Note: Be aware that a maximum likelihood solution may result in a Heywood case. 9.34. The pulp and paper properties data are given in Table 7.7. Perform a factor analysis using observations on the four paper property variables, BL, EM, SF, and BS and the sample correlation matrix R. Can the information in these data be summarized by a single factor? If so, can you interpret the factor? Try both the principal component and maximum likelihood solution methods. Repeat this analysis with the sample covariance matrix S. Does your interpretation of the factor(s) change if S rather than R is factored?
538 Chapter 9 Factor Analysis and Inference for Structured Covariance Matrices
9.3S. Repeat Exercise 9.34 using observations on the pulp fiber characteristic var!ables AFL, LFF, FFF, and ZST. Can these data be summarized by a single factor? Explam.
9.36. Factor analyze the Mali family farm data in Tabl~ 8.7. U~e t~e sample c~ITelation matrix R. Try both the principal component and maximum hkeh~ood solutlO~ methods for m = 3 4 and 5 factors. Can you interpret the factors? Justify your chOice of m. Your
analysi~ ;hould include factor rotation and the computation of factor scores. Can you identify any outliers in these data?
References 1. Anderson, T. W. An Introduction to Multivariate Statistical Analysis (3rd ed.). New York:
John Wiley, 2003. 2. Bartlett, M. S. "The Statistical Conception of Mental Factors." British Journal of Psychology, 28 (1937), 97104. 3. Bartlett, M. S. "A Note on Multiplying Factors for Various ChiSquared Approxima tions." Journal of the Royal Statistical Society (B) 16 (1954),296298. 4. Dixon, W. S. Statistical Software Manual to Accompany BMDP Release 71version 7.0 (paperback). Berkeley, CA: University of California Press, 1992. 5. Dunn, L. C. "The Effect of Inbreeding on the Bones of the Fowl." Storrs AgriculturalExperimental Station Bulletin, 52 (1928),1112. 6. Harmon, H. H. Modern Factor Analysis (3rd ed.). Chicago: The University of Chicago Press, 1976. 7. Hirschey,M., and D. W. Wichern. "Accounting and M~ketValue Measu~es of ~r?fitability: Consistency, Determinants and Uses." Journal of Busmess and Economic Statlstlcs, 2, no. 4 (1984),375383. 8. Joreskog, K. G. "Factor Analysis by Least Squares and .Maximum Likelihood." I~ Statistical Methods for Digital Computers, edited by K. Enslem, A. Ralston, and H. S. WIlf. New York: John Wiley, 1975. 9. Kaiser, H.F. "The Varimax Criterion for Analytic Rotation in Factor Analysis." Psychome
trika,23 (1958), 187200. 10. Lawley, D. N., and A. E. Maxwell. Factor Analysis as a Statistical Method (2nd ed.). New York: American Elsevier Publishing Co., 1971. 11. Linden, M. "A Factor Analytic Study of Olympic Decathlon Data." Research Quarterly, 48,no.3 (1977),562568. 12. Maxwell, A. E. Multivariate Analysis in Behavioral Research. London: Chapman and Hall, 1977. 13. Morrison, D. F. Multivariate Statistical Methods (4th ed.). Belmont, CA: Brooks/Cole Thompson Learning,2005. 14. Stoetzel, 1. "A Factor Analysis of Liquor Preference." Journal of Advertising Research,l (1960),711. . . 15. Wright, S. "The Interpretation of Multivariate Systems." In Statistics and ~athe~natlcs m Biology, edited by O. Kempthorne and others. Ames, lA: Iowa State UmvefSlty Press, 1954,1133.
CANONICAL CORRELATION ANALYSIS 10.1 Introduction Canonical correlation analysis seeks to identify and quantify the associations between two sets of variables. H. HoteIling ([5], [6]), who initially developed the technique, provided the example of relating arithmetic speed and arithmetic power to reading speed and reading power. (See Exercise 10.9.) Other examples include relating governmental policy variables with economic goal variables and relating college "performance" variables with precollege "achievement" variables. Canonical correlation analysis focuses on the correlation between a linear combination of the variables in one set and a linear combination of the variables in another set. The idea is first to determine the pair of linear combinations having the largest correlation. Next, we determine the pair of linear combinations having the largest correlation among all pairs uncorrelated with the initially selected pair, and so on. The pairs of linear combinations are called the canonical variables, and their correlations are called canonical correlations. The canonical correlations measure the strength of association between the two sets of variables. The maximization aspect of the technique represents an attempt to concentrate a highdimensional relationship between two sets of variables into a few pairs of canonical variables.
10.2 Canonical Variates and Canonical Correlations We shall be interested in measures of association between two groups of variables. The first group,ofp variables, is represented by the (p X 1) random vector X(l). The second group, of q variables, is represented by the (q X 1) random vector X(2). We assume, in the theoretical development, that X(l) represents the smaller set, so that p :5 q. 539
540
Chapter 10 Canonical Correlation Analysis
Canonical Variates and Canonical Correlations 541
For the random vectors X(J) and X(2), let
Linear combinations provide simple summary measures of a set of variables. Set
E(X(1» = p,(J);
Cov (X(1» = 1:11
V = a'X(l)
E(X(2» = p,(2);
Cov(X(2»
V = b'X(2)
=
1:22
for some pair of coefficient vectors a and b. Then, using (105) and (245), we obtain
Cov (X(1), X(2» = I12 = Ih
Var(V) = a' Cov(X(1»a
It will be convenient to consider X(J) and X(2) jointly, so, using results (238) through (240) and (101), we find that the random vector
Var(V)
x(1)
xi
We shall seek coefficient vectors a and b such that
)
((p+q)X1)
X(2)
=
(102)
p,
=
E(X) =
[§'(~~;2J = [.~~~;J E(X) P,
(103)
and covariance matrix
=
r
a'1: 12 b Ya ' 1:11 a Yb'I 22 b
(107)
is as large as possible. We define the following:
At the kth step, The kth pair of canonical variables, or kth canonical variate pair, is the pair of linear combinations Vb Vk having unit variances, which maximize the correlation (107) among all choices uncorrelated with the previous k  1 canonical variable pairs.
Ill ij (pxq) I12] (pXp) .......... j..........
The correlation between the kth pair of canonical variables is called the kth canonical correlation. The following result gives the necessary details for obtaining the canonical variables and their correlations. .
(104)
I21 i I22
(qXp)
(106)
The first pair of canonical variables, or first canonical variate pair, is the pair of linear combinations Vb V1 having unit variances, which maximize the correlation (107); The second pair of canonical variables, or second canonical variate pair, is the pair of linear combinations V 2 , V2 having unit variances, which maximize the correlation (107) among all choices that are uncorrelated with the first pair of canonical variables.
has mean vector
((p+q)X1)
a'1: 11 a
= b' Cov(X(2»b = b'I22 b = a' Cov(X(1),X(2»b = a'1: 12 b
Corr(V, V) = X(1)] = [ .........
=
Cov(V, V)
1
X
(105)
i
(qXq)
The covariances between pairs of variables from different se~sone v~riable from X(l) one variable from X(2)are contained in 1:12 or, equ1valently, m I 21 .. That is, th~ pq elements of I12 measure the association between th~ two.sets. ~he~ p and q are relatively large, interpreting the e~em~nts of 1: 12 .collectlvely 1S ~rdman~ ly hopeless. Moreover, it is often linear combmabons of vanabl~s that are mter~st ing and useful for predictive or comparative purposes. The mam task of can(o)mcal . the assoc1atlOns .. bet ween the X(1) and X 2 sets correlation analysis is to summanze in terms of a few carefully chosen covariances (or correlations) rather than the pq covariances in 1:12 .
Result 10.1. Suppose ps q and let the random vectors X(l) and X(2) have (pXl)
(qX1)
Cov (X(1» = 1:11 , Cov (X(2) = 1:22 and Cov (X(l), X(2» = 1:12 , where 1: has full (pXp)
(qXq)
(pXq)
rank. For coefficient vectors a and b , form the linear combinations U = a'X(l) (pX1) (qx1) and V = b'X(2). Then max Corr (V, V) a,b
=
p;:
attained by the linear combinations (first canonical variate pair) V1
=
eiI1i12 X(1) ai
and
Vi
=
fiIZY2x(2)
542
Canonical Variates and Canonical Correlations 543
Chapter 10 Canonical Correlation Analysis The kth pair of canonical variates, k
=
2,3, ... , p,
Uk = eic:t1flZX(l)
Vk = fic:tZ"1/2x(Z)
maximizes Corr(Ub Vk ) = P:
among those linear combinations uncorrelated with the preceding 1,2, ... , le . canonical variables. Here p? ~ pz*2 ~ ... ~ p;2 are the eigenvalues of :tlV2I12IZ"!I2III1/2 . e e2,' .. , e are the associated (p xl) eigenvect<;>rs. [The quantities p?, P2*2, •• a~~ also the; largest eigenvalues of the matrix :tZ"1/2I21 III :t12IZ"1/2 with ing (q xl) eigenvectors f l , f2, ... , f p • Each f; is proportional to IZ"1/2:t2III1/2e; The canonical variates have the properties Var (Uk )
=
Var (Vk )
=
1
'* e k '* e k '* e
where Var(X)1) = au, i = 1,2, ... , p. Therefore, the canonical coefficients for the standardized variables, z)1) = (x)1)  ILP)/v'U;;, are simply related to the canonical coefficients attached to the original variables x)1) . Specifically, if a" is the coefficient vector for the kth canonicalvariate Uk , then ale vlf is the coefficient vector for the kth canonical variate constructed from the standardized variables Z(l). Here vl{2 is the diagonal matrix with ith diagonal element v'U;;. Similarly, ble V!q is the coefficient vector for the canonical variate constructed from the set of standardized variables Z(2). In this case vg2 is the diagonal matrix with ith diagonal element v'U;; = VVar(Xf). The canonical correlations are unchanged by the standardization. However, the choice of the coefficient vectors ak, b k will not be unique if p",( = p~+ I, The relationship between the canonical coefficients of the standardized variables and the canonical coefficients of the original variables follows from the special structure of the matrix [see also (1011)]
for k,
Cov (Vb Ve)
=
Corr (Vk , Ve)
= 0
Cov (Ub Vf)
=
Corr (Uk , Ye)
= 0
and, in this book, is unique to canonical correlation analysis. For example, in principal component analysis, if ale is the coefficient vector for the kth principal component obtained from :t, then a,,(X  ,..,) = a" VI/2Z, but we cannot infer that a" VI/2 is the coefficient vector for the kthprincipal component derived from p.
e. = 1, 2, ... , p.
Proof. (See website: www.prenhall.com/statistics)
Z(2)
If the original variables are standardized with Z(I) = [Z\I), Z~I), .. . , Z~I)]' = [Z(2), Z~2), ... , Z~Z))', from first principleS, the canonical variates are
Example 10.1 (Calculating canonical variates and canonical correlations for standardized variables) Suppose Z(1) = [ZP), Z~l))' are standardized variables and Z(2) = [ZIZ), Z))' are also standardized variables. Let Z = [Z(1), Z(2)], and
Z1
l
1.0
Uk = aleZ(I) = eicPII/2Z(1)
Cnv(Z)
Vk = b"Z(Z) = f"PZ"!/2Z(2)
Here, Cov(Z(I) = PlI, COV(Z(2) = P2Z, COV(Z(I),Z(2) = P12 = P2b and fk are the eigenvectors of Pljl2 P12 PZ"! P21PII/2 and PZ"1f2 PZIPI! respectively. The canonical correlations, p~, satisfy Corr(Ub Vk ) = p~,
PIV2 p1zP2'iPzI PIVz
:tlfl2:t12:t2'!:t21:tlV2 or
Cov(UbUf,) = Corr(UbUc) = 0 k
1.5
.6l
~ [~;;i~J ~+'j!di .6
Then
k = 1,2, ... ,p
p? ~ p;Z ~ ... ~ p;z are the nonzero ei~envalues of Pljl2p12P2iP21PlJl2 (or, equivalently, the largest elgenvalues of P12PZ"Y2).
.4
where
.4:.2 1.0
1/2 _ [1.0681 PI I  .2229
.2229J 1.0681
[1.0417 .2083
.2083J 1.0417
I _
P22 and
Comment. Notice that
a/,(X(I)  ,..,(1) = akl(Xp) 
,..,P)
+ adX~I)
+ ... + akp(X~I) ~'. \
Ilii
l
= ak! ~
(XP) 
1/2 _ [.4371 .2178J
1
P11 P12P22PZIP11 
.2178
.1096
 fL~I) fLP)
~ all
+ ... + ak P va:: pp
1/2
 fL~I)
+ akZ vo:;;
(X~I)  JL~I)
v;;:::, O'pp
(X~I)

fL~I)
.V~Z r=
The eigenvalues, p?, p;Z, of Pll/2 P12PZ"!P21PIF2 are obtained from
I
0= /.4371  A .2178 = (.4371  A) (.1096  A)  (2.178f .2178 .1096  A = AZ

.5467A + .0005
544
Interpreting the Population Canonical Variables 545
Chapter 10 Canonical Correlation Analysis
yielding p? = .5458 and equation
p';!
= .0009. The eigenvector el follows from the vector
For these variates, Var(UI ) Var(Vd
.4371 .2178Je = (.5458)e [ .2178 .1096 I I
COV(UI'~)
Thus,ej = [.8947, .4466) and
bl
<X
<X
=
1/2 PII el
=
I _ [.3959 P22P21 a l  .5209
~
[.8561J .2776
P"2P P2IPJ.l/2el and b l
p~'ffl. Consequently,
=
.2292J [.8561J _ [.4026J .3542 .2776  .5443
The vector [.4026, .5443)' gives 1.0 [.4026, .5443) [ .2
.2J [.4026J 1.0 .5443
=
.5460
4.0 v 12.4 v2.4
= • r;;:;:;. MA = .73
The correlatio!!, b~ween the rather simple and, perhaps, easily interpretable linear • combinations Uj, l'I is almost the maximum value pi = .74. The procedure for obtaining the canonical variates presented in Result 10.1 has certain advantages. The symmetric matrices, whose eigenvectors determine the canonical coefficients, are readily handled by computer routines. Moreover, writing the coefficient vectors as ak = ~11/2ek and bk = ~"2y2fk facilitates analytic descriptions and their geometric interpretations. To ease the computational burden, many people prefer to get the canonical correlations from the eigenvalue equation 1~11~12~2~~21  p*2II = 0
(1010)
The coefficient vectors a and b follow directly from the eigenvector equations ~11~12~2i~2Ia = p*2a
Using V.5460 = .7389, we take
=
~
Corr(UI, VI)
We must scale b l so that Var(VI ) = Var(bjZ(2) = bjP22bl = 1
bl
= a'PI2b = 4.0
and al
From Result 10.1, fl
= a'P11 a = 12.4 = b' P22b = 2.4
1 [.4026J .7389 .5443
=
[.5448J .7366
The first pair of canonical variates is
+ .28Z~I) 2 .54Zi ) + .74Z~2)
~"2i~21~11~12b = p*2b
(1011)
The matrices ~11~12~2i~21 and ~2iI2IIll~12 are, in general, not symmetric. (See Exercise 10.4 for more details.)
UI = a;Z(1) = .86Z\1)
VI = b\Z(2)
=
and their canonical correlation is
p~ =
\I'P? =
V.5458
=
.74
This is the largest correlation possible between linear combinations of v:ariables from the Z(l) and Z(2) sets. The second canonical correlation, p; = V.0009 = .03, is very small, and consequently, the second pair of canonical variates, although uncorrelated with members of the first pair, conveys very little information about the association between sets. (The calculation of the second pair of canonical variates is considered in Exercise 10.5.) We note that UI and VI, apart from a scale change, are not much different from the pair ~ , I ZI(I)J (I) (I) UI = a Z( ) = [3, 1) [ Z~I) = 3Z1 + Z2
~
= b'Z(2)
Z(2)J
= [1, 1) [ Z~2) = zi2) + Z~2)
10.3 Interpreting the Population Canonical Variables Canonical variables are, in general, artificial. That is, they have no physical meaning. If the original variables X(I) and X(2) are used, the canonical coefficients a and b have units proportional to those of the X(l) and X(2) sets. If the original variables are standardized to have zero means and unit variances, the canonical coefficients have no units of measurement, and they must be interpreted in terms of the standardized variables. Result 10.1 gives the technical definitions of the canonical variables and canonical correlations. In this sectiop., we concentrate on interpreting these quantities.
Identifying the Canonical Variables Even though the canonical variables are artificial, they can often be "identified" in terms of the subjectmatter variables. Many times this identification is aided by computing the correlations between the canonical variates and the original variables. These correlations, however, must be interpreted with caution. They provide only univariate information, in the sense that they do not indicate how the original variables contribute jointly to the canonical analyses. (See, for example, [11}.)
546
Chapter 10 Canonical Correlation Analysis
Interpreting the Population Canonical Variables 547
For this reason, many investigators prefer to assess the contributions of the original variables directly from the standardized coefficients (108). Let A = [ab a2,"" ap ]' and B = [bb bz,···, b q ]" so that the vectors of
and
canonical variables are
With p
~x~
PI2
~x~
U
=
AX(1)
=
1,
V =BX(2)
A z = [.86, .28)
(qXI)
(pXI)
=
where we are primarily interested in the first p canonical variables in V. Then
[.5 .6J .3
.4
Bz
= [.54, .74)
so
Cov(U,X{l» = COV(AX(I),X{l» = Al:11 I
Because Var(Vi ) = 1, Corr(U;,Xi » is obtained by dividing Cov(O;,xiI » by VVar (XiI» = u}(1. Equivalently, Corr (Vb xiI» = Cov (0;, uk}/2 xiI». Int~o ducing the (p X p) diagonal matrix Vjlf2 with kth diagonal element uklf, we have, in matrix terms, PU,x(l) = Corr (U, X(l» = Cov (U, Vjfl2x(1» = Cov (AX(I), Vjjl2X(1» (pXp)
Similar calculations for the pairs (U, X(2», (V, X(2» and (V, X{l» yield PU,X(l) = Al:ll Vjlf2 (pxp)
PV,X(2) = Bl:22 V 2Y2
(qXq)
PU,x(2) = Al: 12V 2Y2
~x~
PV,x(l) ~x~
= Bl:2I Vjl/2 .
(1014)
where V 2Y2 is the (q X q) diagonal matrix with ith diagonal element [Var(Xi 2»). Canonical variables derived from standardized variables are sometimes interpreted by computing the correlations. Thus, Pu.z(l)
= A zPI I
PV,Z(2)
= B z P22 (1015)
where A z and Bz are the matrices whose rows contain the canonical coefficients (pxp)
(qxq)
for the Z(I) and Z(2) sets, respectively. The correlations in the matrices displayed in (1015) have the same numerical values as those appearing in (1014); that is, PU,X(l) = PU,z(l), and so forth. This follows because, for example, PU,X(l) = Al:11 Vjlf2 = AVlfVjfl2l: 1l VjV2 = A z PI I = PU,z(1l. The correlations are unaffected by the standardization. Example 10.2 (Computing correlations between canonical variates and their component variables) Compute the correlations between the first pair of canonical variates and their component variables for the situation considered in Example 10.1. The variables in Example 10.1 are already standardized, so equation (1015) is applicable. For the standardized variables, Pll = [
1.0
.4
.4J
1.0
P22
=[
1.0 .2
.2J
1.0
and
We conclude that, of the two variables in the set Z{l), the first is most closely associated with the canonical variate VI' Of the two variables in the set Z(2), the second is most closely associated with VI . In this case, the correlations reinforce the information supplied by the standardized coefficients A z and Bz. However, the correlations elevate the relative importance of Z~l) in the first set and Z~2) in the second set because they ignore the contribution of the remaining variable in each set. From (1015), we also obtain the correlations PU}oZ(2)
= A z PI2 =
[.86, .28)
[:~ :~ ]
= [.51, .63)
and PVloz(l)
= BzP21 = BzPiz = [.54,.74)
[:~
:!J =
[.71,.46)
Later, in our discussion of the sample canonical variates, we shall comment on the interpretation of these last correlations. _ The correlations PU,x(1) and PV,X(2) can help supply meanings for the canonical variates. The spirit is the same as in principal component analysis when the correlations between the principal components and their associated variables may provide subjectmatter interpretations for the components.
Canonical Correlations as Generalizations of Other Correlation Coefficients First, the canonical correlation generalizes the correlation between two variables. When X(I) and X(2) each consist of a single variable, so that p = q = 1, for all a, b
~
548
Chapter 10 Canonical Correlation Analysis
Interpreting the Population Canonical Variables
Therefore, the "canonical variates" UJ = x)1) and VI = X(2) have correlation pi = ICorr (XP), X(2»I. When X(I) and X(2) have more components, setting a' = [0, ... ,0,1, 0, ... ,0] with 1 in the ith position and b' = [0, ... ,0,1, 0, ... ,0] with 1 in the kth position yields ICorr (x)I), 2»1 = ICorr(a'X(l),b'X(Z»1
549
Example 10.3 (Canonical correlation as a poor summary of variability) Consider the
covariance matrix
xi
s max Corr(a'X(I), b'X(2» = a,b
pi
That is, the first canonical correlation is larger than the absolute value of any entry . in PIZ = vlllz~12v2~/2.· Second, the multiple correlation coefficient PI(X(2) [see (748)] is a special case of a canonical correlation whenX(I) has the single element XP)(p = 1). Recall that for
p = 1
When p > 1, P;: is larger than each of the multiple correlations of x)I) with X(2) or the multiple correlations of x)2) with X(I). Finally, we note that PUk(X(2)
= max Corr (Ub b'X(2» b
=
The reader may verify (see Exercise 10.1) that the first pair of canonical variates UI = X~I) and VI = xf) has correlation .
Corr (Ub Vk ) = p~,
(1018)
Yet UI = X~I) provides a very poor summary of the variability in the first set. Most of the variability in this set is in xjl), which is uncorrelated with UI . The same situ• ation is true for VI = X\Z) in the second set.
k = 1,2, ... , P
from the proof of Result 10.1 (see website: www.prenhall.comlstatistics). Similarly, PVk(x(l)
= m:xCorr(a'X(I), Vk ) = Corr(Ub Vk ) =
P:,
(1019)
k = 1,2, ... ,p
That is, the canonical correlations are also the multiple correlation coefficients of Uk with X(2) or the multiple correlation coefficients of Vk with X(1). Because of its multiple correlation coefficient interpretation, the kth squared canonical correlation p~2 is the proportion of the variance of canonical variate Uk "explained" by the set X(2). It is also the proportion of the variance of canonical variate Vk "explained" by the set X(!). Therefore, p? is often called the shared variance between the two sets X(!) and X(2). The largest value, p?, is sometimes regarded as a measure of set "overlap."
The First r Canonical Variables as a Summary of Variability The change of coordinates from X(I) to U = AX(I) and from X(2) to V = ·BX(Z) is chosen to maximize Corr (UI , VI) and, successively, Corr (Ui , Vi), where (Ui , Vi) have zero correlation with the previous pairs (UI , Yt), (Uz , Vz),···, (0;1> Vid. Correlation between the sets X(!) and X(2) has been isolated in the pairs of canonical variables By design, the coefficient vectors ai, bi are selected to maximize correlations, not necessarily to provide variables that (approximately) account for the subset covariances ~ll and ~22' When the first few pairs of canonical variables provide poor summaries of the variability in ~II and ~Z2' it is not clear how a high canonical correlation should be interpreted.
A Geometrical Interpretation of the Population Canonical Correlation Analysis A geometrical interpretation of the procedure for selecting canonical variables provides some valuable insights into the nature of a canonical correlation analysis. The transformation U = AX(1) from X(I) to U gives Cov(U)
= A~llA' =
I
From Result 10.1 and (222), A = E'~lfl2 = E'PIAlII2Pl where E' is an orthogonal matrix with rowel, and ~ll = PIAIPj. Now, P1X(I) is the set of principal components derived from X(I) alone. The matrix A11/ 2P 1X(l) has ith row (1jW;) piX(l), which is the ith principal component scaled to have unit variance. That is, Cov(A 11/2P 1X(1» = All/2Pl~llPIAll/2 = All/zP1PIAIP1PIAll/2
= All/2 AIA 1l/2 = I Consequently, U = AX(I) = E'P1 A1I/2P 1X(l) can be interpreted as (1) a transformation of X(I) to uncorrelated standardized principal components, followed by (2) a rigid (orthogonal) rotation PI determined by ~ll and then (3) another rotation E' determined from the full covariance matrix ~. A similar interpretation applies to V = BX(2).
550
The Sample Canonical Variates and Sample Canonical Correlations 551
Chapter 10 Canonical Correlation Analysis
10.4 The Sample Canonical Variates and Sample Canonical Correlations A random sample of n observations on each of the (p be assembled into the n x (p + q) data matrix
X
=
=
Result 10.2. Let Pfz:2: ~z :2: ... :2:?; be the p ordered eigenvalues of siF2 S12 Sl"iSzI SiI/2 with corresponding eigenvectors el, ez, ... , p , where the Ski are defined in (lO22) and p:5 q. Let £1' £z, ... , £p be the eigenvectors of SZi/2S21 Sil S12Sl"i/2, where the first p fS may be obtained from £k= (l/Pf)Sl"Y2SzISiFzeb
e
+ q) variables X(I), X(2) can
k = 1,2, ... , p. Then the kth sample canonical variate pairl is
Uk = e"sil/2x(l)
[X(I) i X(2)]
l
X~v
xW .,.
xW xW ." (I)
x 1l 1
(I) X 1I 2
(I) i (2) XIP i X11 (I): (2) X2p
i·X:I (I) 1 ~2) X"P i Xnl
(2)]
(2) XI2 (2)
Xn
'v'
[xli), i\ x(2)'] I I
Xlq (2) X2q
_
:

:
(i),
xn
(2)
(2) Xn2
i
i
:
where x(l) and x(Z) are the values of the variables X(1) and X(2) for a particular experimental unit. Also, the first sample canonical variate pair has the maximum sample correlation
(i),
i xn
Xnq
rv"v,
The vector of sample means can be organized as i
(p+q)xl
['~~~J x( )
=
where i(l)
=
i(2) =
± 1. ±
X)2)
(lO21)
j=1
Similarly, the sample covariance matrix can be arranged analogous to the representation (104). Thus,
SI1 ii S12] (p+q)~(p+q) = "·s~·~··t···s;;· (pXp)
(qXp)
where Ski
= _1_ n  1
±
is the largest possible correlation among linear combinations uncorrelated with the preceding k  1 sample canonical variates. The quantities Pr,~, are the sample canonical correlations. z
... ,;;;'
Proof. The proof of this result follows the proof of Result 10.1, with Ski substituted for I kl , k, I = 1,2. •
(pxq)
i
The sample canonical variates have unit sample variances
sv"v. = SVk, Vk = 1
(qxq)
(1025)
and their sample correlations are
(x(k) 
j=1
= pi
and for the kth pair,
1.n j=1 xjI) n
r
Vk = hSl"i/2x(2)
'v'
X(k)) (xY)  i(l))',
k, 1= 1,2
(1022)
rUk,V,
J
The linear combinations {; =
a'x(l);
(1023)
have sample correlation [see (336)]
ru ,v =
. ;;:;;:;;;. ~
The first pair of sample canonical variates is the pair of linear combinations ., !n general, the kth pair ofsample canonic~1 v~riates IS ~e pair of hnear combma?ons Ub V.k having unit sample variances that maxtmlZe the ratio (1024) among those lmear • . combinations uncorrelated with the prevjous k : 1 sample canorucal vanates. . The sample correlation between Uk and Vk is called the kth sample canonical correlation. The sample canonical variates and the sample canonical correlations can be obtained from the sample covariance matrices S11, SI2 = Sh, and S22 in a manner consistent with the population case described in Result 10.1.
k*C k*C
(1026)
The interpretation of Uk , Vk is often aided by computing the sample correlations between the canonical variates and the variables in the sets X(I) and X(2). We define the matrices A
(1024)
Va'S11 a Vb'Sn b
= rVk, V, = 0, rUk,V, = 0,
(pXp)
=
[31,32, ... ,3 p )'
B
(qXq)
=
[bt>bz, ... ,bq]'
(1027)
whose rows are the coefficient vectors for the sample canonical variates. 3 Analogous to (1012), we have
UI , VI having unit sample variances that maximize th.e ratio ~102~).
iJ
= Ax(1)
(pXI)
A
v
= Bx(Z)
(1028)
(qXI)
I When the distribution is normal, the maximum likelihood method can be employed using:I = S. in place of S. The sample canonical correlations are, therefore, the maximum likelihood estimates of and Yn/(n  1) ak> Yn/(n  1) bkare the maximum likelihood estimates of 8k and bb respectively. 2 If P > rank(S12) = PI, the nonzero sample canonical correlations are Pf, ... , pr, . 3 The vectors bp,+1 = Si"~/2fp'+1,bp,+2 = Si"~/2fp'+2, ... ,bq = Si"~/2rqaredetermin~fromachoiceof
p:
P:
the last q  PI mutually orthogonal eigenvectors f associated with the zero eigenvalue of Si"PS21 si1 S12S2~/2 .
552
The Sample Canonical Variates and Sample Canonical Correlations
Chapter 10 Canonical Correlation Analysis
553
have the sample correlation matrix
and we can define Rv,x(l)
= matrix of sample correlations orv with x(1)
Rv,x(l) = matrix of sample correlations of Vwith x(2) RV,x(l)
= matrix of sample correlations ofU with X(2)
RV,.(I)
=
matrix of sample correlations of Vwith x(l) A canonical correlation analysis of the head and leg sets of variables using R produces the two canonical correlations and corresponding pairs of variables
Corresponding to (1019), we have
RiJ;x(l) = AS llD 1lf2 Rv,x(l) = BS22 D i f2
.
z AS D zif2
RiJ
X(l)
=
Rv
X(I)
= BS 2I D 1
Pr =
(1029)
12
lf2
UI = .781zl1) + .345z~1) Vi = .060z12 ) + .944z~2)
.631
and
where Dl}f2 is the (p X p) diagonal matrix with ith diagonal element (sample var(xF»r l / 2 and D Y2 is the (q X q) diagonal matrix with ith diagonal element (sample var(xf»)1/2..
~ = .057
z
Comment, If the observations are standardized [see (825)], the data matrix becomes
U2 = .856zP) + 1.106Z~I) V2 =  2.648zi2 ) + 2.475zi2 )
Here zF) , i = 1,2 and z)2) , i = 1,2 are the standardized data values for sets 1 and 2, respectively. The preceding results were taken from the SAS statistical software output shown in Panel 10.1. In addition, the correlations of the original variables • with the canonical variables are highlighted in that panel.
ZI(I)' :i ZI(2)1]
Z=[Z(J) i Z(2)]=
[
:
i :
(1)': (2)1 Zn :, Zn
and the sample canonical variates become
U (pXI)
=
Az z(1)
v
(qXI)
=
Bz Z(2)
(1030)
where A z = ADlI? and Bz = BD!q, The sample canonical correlations are unaffected by the standardization, The correlations displayed in (1029) remain unchAanged and may be calculated, for standardized observations, by substituting A z for A B for Band R for S. Note that Dlfl2 = I and D Y2 = I for standardized , z , ( x ) ( xq) observations. p P q
z
Example 10.4 (Canonical correlation analysis of the chickenbone data) In Example 9.14, data consisting of bone and skull measurements of white leghom fowl were described. From this example, the chickenbone measurements for Head (X(1»:
Xli) = skull length { X~I) = skull breadth
Leg (X(2»:
X12) = femur length { X~2) = tibia length
Example I O.S(Canonical correlation analysis of job satisfaction) As part of a larger study of the effects of organizational structure on "job satisfaction," Dunham [4] investigated the extent to which measures of job satisfaction are related to job characteristics. Using a survey instrument, Dunham obtained measurements of p = 5 job characteristics and q = 7 job satisfaction variables for n = 784 executives from the corporate branch of a large retail merchandising corporation. Are measures of job satisfaction associated with job characteristics? The answer may have implications for job design. PANEL 10.1
SAS ANALYSIS FOR EXAMPLE 10.4 USING PROC CANCORR.
title 'Canonical Correlation Analysis'; data skull (type corr); _type_ = 'CORR'; input _name_S x1 x2 x3 x4; cards; x1 1.0 x2 .505 1.0 x3 .569 .422 1.0 x4 .602 .467 .926 1.0
=
PROGRAM COMMANDS
proc cancorr data" skull vprefix = head wprefix = leg; var x1 x2; with x3 x4;
(continues on next page)
554
Chapter 10 Canonical Correlation Analysis PANEL 10.1
'I1!e Sample Canonical Variates and Sample Canonical Correlations
(continued) Canonical Correlation Analysis Adjusted Approx Canonical Standard Correlation Error
0.628291
2
0.036286 0.060108
The original job characteristic variables, X(1l, and job satisfaction variables, X (2) , were respectively defined as Squared Canonical Correlation
X(1) =
0.398268 0.003226
Raw CanoniCal Coefficient for the ~VAR' Variables
~ ~
HEAD2 .:0.855973184 1.1061835145
HEAQl 0.7807924389 0.3445068301
[
X~I)l xii) X~l) X~I) X~I)
~
LEGl 0.0602508775 0.943948961
1
[ =
taskfeedback significance task variety task identity autonomy
supervisor satisfaction careerfuture satisfaction financial satisfaction workload satisfaction company identification kindofworksatisfaction general satisfaction
OUTPUT
Raw Canonical Coefficient forthe 'WITH' Variables
Q
555
LEG2 2.648156338 2.4749388913
Responses for variables X(1) and X(2) were recorded on a scale and then standardized. The sample correlation matrix based on 784 responses is
Canonical Structure Correlations Between the 'VAR' Variables and Their Canonical Variables
Xl X2
HEADl 0.9548 0.7388
HEAD2 .:0.2974 0.6739
(see 1029)
Correlations Between the 'WITH' Variables and Their Canonical Variables
X3 X4
LEGl 0.9343 0.9997
LEG2 .:0.3564 0.0227
(see 1029)
X2
LEGl 0.6025 0.4663
lEG2 .:0.0169 0.0383
(see 1029)
Correlations Between the WITH' Variables and the Canonical Variables of the 'VAR' Variables
X3 X4
HEADl 0.5897 0.6309
HEAD2 .:0.0202 0.0013
21
22
1.0 .49 1.0 .53 .57 .49 .46
1.0 .48
.33 .32 .30 .21 .31 .23 .24.22
1.0
...
.20 .16 .14 .12
.19 .08 .07 .19
.30 .27 .24 .21
.37 .35 .37 .29
.21 .20 .18 .16
:?~._._ ...:?~. ___ ._:?Z .. __.. :?Z___ }:Q.. __ L.}.~ .:}?.___ ._}_!__ .. __:~~ ......}~._._ .. }~......:~?
Correlations Between the 'VAR' Variables and the Canonical Variables of the 'WITH' Variables
Xl
R12] [Rll R !R
! R = .. .r..
(see 1029)
.33 .32 .20 .19 .30
.30 .21 .16 .08 .27
.31 .23 .14 .07 .24
.24 .22 .12 .19 .21
.38 .32 .17 .23 .32
i 1.0 i .43 i.27 i.24 i .34
~
~
~
~
~i~
~
~
~
~
.21
.20
.18
.16
.27 i.40
.58
.45
.27
.59
1.0 .33 1.0 .26 .25 .54 .46
1.0 .28
1.0
In .31
1.0
The min(p, q) = min(5,7) = 5 sample canonical correlations and the sample canonical variate coefficient vectors (from Dunham [4]) are displayed in the following table:
The Sample Canonical Variates and Sample Canonical Correlations 557 N .....
.~
N
~'" N
'"
:0 .~ .... >
'"
"0
N
~'"
N
'"
'"""!
N
V]
'"
"0
0\
"!
C')
"!
N
'"""!
00
q
I
0")
t
~
I C')
.:t:
0\
~
0\ ~
'1" ~
q
t
t
I
I
M
N
.:t: I
"2
o
~
N
q
.:t:
N
M
!"'!
M
.:t:
\0
0
<'l ~
.:t:
I
u
·S 0 !:l
N
N
~N
'"
C')
.:t:
q
<;&;
<"Q
00
V]
C')
!"'!
U
+ .21Z~I) + .17z~l)  .02zil ) + .44z~I)
A (2) VI = .42z1
+
(2)
.22z2
(2)
 .03z3
+
(2)
.01z4
+
(2)
.29zs
+
(2)
.52z6
(2)
 .12z7
with sample canonical correlation = .55. According Ato the coefficients, VI is primarily a feedback and autonomy variable, while VI represents supervisor, careerfuture, and kindofwork satisfaction, along with company identification. To provide interpretations for [;1 and}'i, the sample correlations between VI and its component variables and between Vi and its component variables were computed. Also, the following table shows the sample correlations bet.ween variables in one set and the first sample canonical variate of the other set. These correlations can be calculated using (1029).
t:
V]
VI = .42z~I)
Pr
\0
.:t:
t; I
For example, the first sample canonical variate pair is
I
I
N
~N
'1"
I
N
V]
~
q
U
0\
M
q
!:l N ~'"
'".... '0 ....
q
N
I
N
.g C;;'" '"
N
I
M
~.,.
~.... !:l
N
Sample Correlations Between Original Variables and Canonical Variables Sample canonical variates
N
"1 I
Sample canonical variates
"0
!:l
'"
"E'"
:'N
..:~
.:~
,:v.
<"Q
<"Q
<"Q
N
00
V)
.~
E
(*~
V)
C')
V]
"!
'"""!
~
M
~'"
q
q
V)
V)
0")
!"'!
N ~
U
~
·c
~
N
oq I
I
C;; u
·S 0 !:l
N
2'"
~.,.
q
'" .g '" '" "0
'"""!
U
oD
>
.~ "0
t
~'"
0\
0\
I
I
"!
I
V)
oq
.:t:
0\
'"""!
'1"
~ .<
N
~
'" '"
"0
!:l
C;;
~N
V)
\C!
.:t:
q
\0
N
.:t:
"1 I
(. = <e'
:N
. 556
'"
t
M
"!
'" ~N
'1"
q
.<
I
....
"'1 I
.:~
<"i
.....
\0
q
I
.<
\0
t
t:
Feedback Task significance Task variety Task identity Autonomy
VI
VI
.83 .74 .75 .62 .85
.46 .41 .42 .34 .48
variables
VI
VI
Supervisor satisfaction Careerfuture satisfaction Financial satisfaction Workload satisfaction Company identification Kindofwork satisfaction General satisfaction
.42 .35 .21 .21 .36 .44 .28
.75 .65 .39 .37 .65 .80 .50
X(2)
1. 2. 3. 4. 5. 6. 7.
All five job charact~ristic variables have roughly the same correlations with the first canonical variate VI. From this standpoint, VI might be interpreted as a job characteristic "index." This differs from the preferred interpretation, based on coefficients, where the task variables are not important. The other member of the first canonical variate pair, ~, seems to be representing, primarily, supervisor satisfaction, careerfuture satisfacti~n, company identification, and kindofwork satisfaction. As the variables suggest, VI might be regarded as a job satisfactioncompany identification index. This agrees with the preceding interpretation based on the c~nonical coe!ficients of the 2 ),s. The sample correlation between the two indices VI and Vi is pi = .55. There appears to be some overlap between job characteristics and job satisfaction. We explore this issue further in Example 10.7.
zi
!"'!
,..,. '" <=
<=
1. 2. 3. 4. 5.
variables
\0 ~
I
N
X(1)
•
Scatter plots of the first (VI, ~) pair may r~e~ atypical observations Xj requiring further study. If the canonical correlations pi, pj, ... are also moderately large,
558
Additional Sample Descriptive Measures 559
Chapter 10 Canonical Correlation Analysis
scatter plots of the pairs ([;2, V2 ), ([;3, V3)'" . may also be helpful in this Many analysts suggest plotting "significant" canonical variates against . nent variables as an aid in subjectmatter interpretation. These plots correlation coefficients in (1029). If the sample size is large, it is often desirable to split the sample in first half of the sample can be used to construct and evaluate the sampl cal variates and canonical correlations. The results can then be " the remaining observations. The change (if any) in the nature of the analysis will provide an indication of the sampling variability and the the conclusions. .
I O. S Additional Sample Descriptive Measures If the canonical variates are "good" summaries of their respective sets of then the associations between variables can be described in terms of variates and their correlations. It is useful to have summary measures of to which the canonical variates account for the variation in their respective also useful, on occasion, to calculate the proportion of variance in one set abies explained by the canonical variates of the other set.
If only the first r canonical pairs are used, so that for instance,
(1033)
and
then S12 is approximated by sample Cov(x(l), x(2». Continuing, we see that the matrices of errors of approximation are Sl1 
(8(1)8(1)1
S22  (b(1)b(1)I S12 
+
a(2)8(2)I
+ '" +
+ b(2)b(2)' + '" +
(pta(1)b(1)I
+
~a(2)b(2).
8(r+1)8(r+I)I
+ .. , +
a(p)a(p)I
b(r)b(r)/) = b(r+l)b(r+l).
+ .. , +
b(q)b(q)·
=
8(r)a(r)/)
+ '" + ,£;;a(r)b(r).) =
'£;;+la(r+1)b(r+l)I
+ '" +
p;a(p)b(p)I
(1034)
Matrices of Errors of Approximations
Gi~en the Il}atrices A a~d B d~fine'!.. in (lP27), let a;i) an9 b~) denote th: of AI and B 1 , respectively. Smce U = Ax(1) and V = Bx( ) we can wnte, x(1)
V
AI
=
(pXI)
(pXp) (pXl)
x(2J (qXl)
=
B1 V
(qXq) (qXI)
' , V')  A'S 12, B'I sample Cov(V) = ASI"lA' Because sampIe C ov (U sample COV (V) = BS 22B' = I ,
=
(qXq)
Pf 0
'1
S12
=
A
f
S1l = (AI)
S22
=
.. .
o
Pr .. . o
~ ~
.
8(I)a(1)I
+
8(2)a(2)I
= b(1)b(I)I
+
b(2)b(2) 1
(AI)' =
(B 1) (B 1),
+ .,. + a(P)8(P)I + .,. + b(q)b(q)I
Since x(1) = AIV and V has sample covariance I, the first/ ' contain the sample covariances of the first r canonical variate~ VI, V2 their component variables XP) , X~I) , ... , X~) . Similarly, the fIrst r contain the sample covariances of ~, V2 , ... , V,. with their component
The approximation error matrices (1034) may be interpreted as descriptive summaries of how well the first r sample canonical variates reproduce the sample covariance matrices. Patterns of large entries in the rows and/or columns of the approximation error matrices indicate a poor "fit" to the corresponding variable(s). Ordinarily, the first r variates do a better job of reproducing the elements of S12 = S21 than the elements of SI1 or S22' Mathematically, this occurs because the residual matrix in the former case is directly related to the smallest p  r sample canonical correlations. These correlations are usually all close to zero. On the other hand, the residual matrices associated with the approximations. to the matrices S11 and S22 depend only on the last p  rand q  r coefficient vectors. The elements in these vectors may be relatively large, and hence, the residual matrices can have "large" entries. For standardized observations, Rkl replaces Ski and a~k), b~/) replace 8(k) , b(1) in (1034). Example 10.6 (Calculating matrices of errors of approximation) In Example 10.4, we obtained the canonical correlations between the two head and the two leg variables for white leghorn fowl. Starting with the sample correlation matrix
R "
[~;;i~;;J "[1~1i_li:fiJ~:m]
l
.602
.467:
.926
1.0
Additional Sample Descriptive Measures
560 Chapter 10 Canonical Correlation Analysis
We see that the first pair of canonical variables effectively summarizes (reproduces) the intraset correlations in R!2' However, the individual variates are not particularly effective summaries of the sampli~g variability in the original z(1) and Z(2) sets, respectively. This is especially true for U1 • •
we obtained the two sets of canonical correlations and variables (I)
Pt =
.631
and
Pf. =
.057
(I)
+ .345z2 (2) (2) .060z1 + .944z2
A
UI = .781zl
VI A
=
Proportions of Explained Sample Variance
.856zil ) + 1.l06z~l) V2 = 2.648zi2) + 2.475z~2)
U2 =
where Z(I) i = 1, 2 and Z(2) i =·1 , 2 are the stand~dized data values for sets 1 andI' " 2, respectively. We fIrst calculate (see Panel 10.1) A I = [.781 .345J1 z .856 1.106
sample Cov(z(1), iJ)
.2974J .7388 .6739
so
Consequently, the matrices of errors of approximation created by using only the first canonical pair are
= (.057)
[:~~~:J [.3564
A(p)] = A1 = [A(!) A(2) Az az , 8 z , ... , a z
8! = [bAr!) bA(2) Z
Rll  sample Cov('Z(1» = [.2974J [.2974 .6739
" r
UI.Z (I) 1
TU 2 ,i:1
ru,.t(~) .
rU2'Z(~)
Z
,
Z
, ••• ,
bA(q)] = t
.6739]
l
rup,z(:)l
ru p,i~l
rup:,Z(~)
rU t ,1.{1)
.0227]
.006 .OOOJ [ .014 .001
= [
= sample Cov(A;IU, U) = A;1
and
A_I _ [.9343 .3564J .9997 .0227
=
When the observations are standardized, the sample covariance matrices Ski are correlation matrices R kf • The canonical coefficient vectors are the rows of the matrices A z and 8 z and the columns of A;1 and 8;1 are the sample correlations between the canonical variates and their component variables. Specifically,
= [.9548
Bz 
RJ2  sampleCov('Z(I),'Z(2»
561
rv,.z'~) rV2.z';) rv"z(;) rv;,z(;) . ... ..
rvq,z{;)
rVl>z(~l
rvq:,z(i)
rV2'Z(~)
rVq,z'i)l (1035)
where 'ui,il) and rVi,t(!) are the sample correlation coefficients between the quantities with subscripts. Using (1032) with standardized observations, we obtain
.088 .200J .200 .454
Total (standardized) sample variance in first set R22  sampleCov('Z(2» =
[:~~~~J [.3564
= [ .127 .008 ~(1) ~(2)'
where z , z respectively.
. are gIven by (1033) WIth r
.0227]
= tr(R22)
.001
A(I) bA(I)
8z
= tr(a~l)a~l),
+ a~2)a~2), + ... + a~p)a~p),) = p
(lO36a)
Total (standardized) sample variance in second set
'<~08J
= 1 and
= tr(Rl1)
,
z
A I A(I) b(I) rep ace 8 , ,
= tr(b~l)b~l),
+ b~2)b~2), + ... + b~q)b~q),) = q
(lO36b)
Since the correlations in the first r < p columns of A;1 and 8;1 involve only the sample canonical variates UI , U2 , ••. , Ur and VI, V2 , • .• , V" respectively, we define
562
Large Sample Inferences 563
Chapter 10 Canonical Correlation Analysis
the contributions of the first r canonical variates to the total (standardized) variances as
and
Example 10.T (Calculating proportions of sample variance explained by canonical variates) Consider the job characteristicjob satisfaction data discussed in Example 10.5. Using the table of sample correlation coefficients presented in that example, we find that R~(J)lu)
The proportions of total (standardized) sample variances "explained by" the canonical variates then become proportion of total standardized) R;
+ ... + a~r)a~r),)
1
1
5
= 5 k=l :L r1 l,oll.: = 5 [(.83f + (.74)2 + ... + (.85)2J = .58
1 R~(2)lvl = 7
(I)
1
7
:L r~
k=l
(2)
= 7 [(.75f
+ (,65)2 + .. , + (.50fJ
= .37
ItZI;
The first sample canonical variate UI of the job characteristics set accounts for 58% of th~ set's total sample variance. The first sample canonical variate of the job satisf~ction set explains 37% of the set's total saIllple variance. We might thus infer that UI is a "better" representative of its set than VI is of its set. The interested reader may wish to see how well U1 and Vi reproduce the correlation matrices RJ1and R 22 , respectively. [See (1O29).J •
Vi
tr (Rll)
10.6 Large Sample Inferences p
When :I12 = 0, a'X(I) and b'X(2) have covariance a':IJ2b = 0 for all vectors a and b. Consequently, all the canonical correlations must be zero, and there is no point in pursuing a canonical correlation analysis. The. next result provides a way of testing :IJ2 = 0, for large samples.
and
,
.
R;(2)l1i Jo v 2,""v, =
proportion of total standardized) sample variance)n ~econd,set ( explained by~, V2 , •.. , V; tr(b~l)b~I),
Result 10.3. Let
+ ... + b~r)b~r),)
j=1,2, ... ,n
tr(Rzz) r
q
:L :L?y i=1 k=l
be a random sample from an
Np+q(p..,:I)
q :I =
Descriptive measures (1037) provide some indication of how well the cal variates represent their respective sets. They provide singlenumber rip
! tr [R 22 q

a'(2)a,(2),  .. ,  a'(r)a,(r)'J = 1  R2z(!)lu 1> u 2,····Ur z z z z
l
i
:Ill : (pXq) :IJ2 (pxp)
,,+,,,,
J
:I21 j :I22
(qXp)
Then the likelihood ratio test of Ho: :I12 = large values of
b'(I)b'(I),  b'(2)b'(2),  ...  b'(r)b(r)'J = 1  R2(2)IV" v v z z Z z z z z 1> 2."·' r
according to (1036) and (1037).
population with
(2)
"ZI.:
21 A = I n n n
i
(qXq)
versus HI: :I12 #
0 (pXq)
0 rejects Ho for (pxq)
S (I SI1IIS 22I)i n In n (1~) ;=1 P, P
1
= 
(1038)
564
Large Sample Inferences 565
Chapter 10 Canonical Correlation Analysis
Bartlett [2] has argued that the kth hypothesis in (1040) ca'u be tested by the likelihood ratio criterion. Specifically,
where
Reject H~k)at significance level a if 
is the unbiased estimator of l:. For large n, the test statistic (1038) is distributed as a chisquare random variable with pq dJ.
(
n 
1
21 (p + q + 1) )
P ~ In J~t (1  pT2) >
xtPk)(qk)(a)
(1041)
where XfPk)(qk)(a) is the upper (100a)th percentile of a chisquare distribution with (p  k)(q  k) d.f. We point out that the test statistic in (1041) involves
Proof. See Kshirsagar [8].
P
The likelihood ratio statistic (1038) compares the sample generalized under Ho, namely,
II
~
(1  pj2), the "residual" after the first k sample canonical correlations have
i=k+1
been removed from the total criterion A2/n ==
P
II (1
.~
 pj2).
i=1
with the unrestricted generalized variance r S I· Bartlett [3] suggests replacing the mUltiplicative factor n in the ratio statistic with the factor n  1  (p + q + 1) to improve the X2 mation to the sampling distribution of 2 In A. Thus, for nand n large, we
!
If the members of the sequence Ho, H&I), H&2), and so forth, are tested one at . · a t lme untl·1 H(k). 0 IS not rejected for some k, the overall significance level is not a and, in fact, would be difficult to determine. Another defect of this procedure is the tendency it induces to conclude that a null hypothesis is correct simply because it is not rejected. To summarize, the overall test of significance in Result 10.3 is useful for multivariate normal data. The sequential tests implied by (1041) should be interpreted with caution and are, perhaps, best regarded as rough guides for selecting the number of important canonical variates.
Reject Ho: l:12 == 0 (p~ = P; == •.. = P~ == 0) at significance level a if
x;,q(
a) is the upper (100a )th percentile of a chisquare where pq dJ. If the null hYpothesis Ho: IJ2 = 0 (p~ = P; = ... = = 0) is rejected, ural to examine the "significance" of the individual canonical correlations. canonical correlations are ordered from the largest to the smallest, we can .. assuming that the first canonical correlation is nonzero and the relmaini!1! canonical correlations are zero. If this hypothesis is rejected, we assume two canonical correlations are nonzero, but the remaining p  2 Cal[IUJ.U';o'" tions are zero, and so forth. Let the implied sequence of hypotheses be
P;
H1: P;
~ 0, for some i 2:: k
+1
Example 10.8 (Testing the significance of the canonical correlations for the job satisfaction data) Test the significance of the canonical correlations exhibited by the job characteristicsjob satisfaction data introduced in Example 10.5. All the test statistics of immediate interest are summarized in the table on ~ge 566. Fro~Example 10.5, n = 784, p = 5, q == 7, = .55, = .23, = .12, P4 = .08, and Ps = .05. Assuming multivariate normal data, we find that the first two canonical correlations, p; and p;, appear to be nonzero, although with the very large sample size, small deviations from zero will show up as statistically significant. From a practical point of view, the second (and subsequent) sample canonical correlations can probably be ignored, since (1) they are reasonably small in magnitude and (2) the corresponding canonical variates explain very little of the sample variation in the variable setsX(I) andX(2).
Pr
P1
Pf
•
The distribution theory associated with the sample canonical correlations and the sample canonical variate coefficients is extremely complex (apart from the p = 1 and q == 1 situations), even in the null case, l:12 = O. The reader interested in the distribution theory is referred to Kshirsagar [8].
~
t::
:e
.§ t::
U
U
'ii)'
0..
It)
~
11
;:..eN
o ~.o .... 0 ....
.... ,.....
:.a
N..,
t;
0.. 0.. ~
"'11"
<'l
.... ,.....
... ...... q
q ....
~
''
on
"'N"
V")
<:>11
N
m
~ .... ,.....
2: N
E It)
1
t:l"<
Cl
I:l.. ....
.... N
r<) It)
(*0:
"''C!"
1
u
....
.~
t; t::
..=
ctI",c
....
U
V11=:! .E
m ........ m ....
~
"0 .... (I)
t:l"<
(I)
;>
.... Q)
~~ m t1:I
.o~
0''
+ ~
......
... N (*Ci:
1
~ ,..... V)
+
t:::~
~
~
+
+
......
1
......
t:l"<
+
+
"""IN
...... IN
.:::
00
"'"
......
1
c<)
~ ci
~
......
t:l"<
"'" 11
1
1
~ 0
......
11
~
\0
~
E([¥J)
~
1
......
1
N oci ......
1
0 0
"ff. ON
11 .on Cl.
i\.
11
m
'r;; Q)
.e ~
'3m
(I)
t:t:
....m
~
Cl
=
z3
W~
~ .e
11
N
.... 0
11
*~

~
tJ::
~
N
566
.
"ff.
11
.v> Cl.
11
Cl.
~O
''
~
Cl.
* Cl. '
Cl.
ON
Cl.
Eo ~
t'"i
3
2
7
(a) Calculate the canonical correlati ons p;, pi. (b) Determi ne the canonical variate pairs (U , Vd and (U , V ). I 2 2 (c) Let U = [UI , U2J' and V = [VI, V2J'. From first principles, evaluate
.E
~
~ [i;;ji;;] ~ F=~i:i jJ i 1
V)t:::~
+ ,.....IN
I
1
......
.E
I"
,.....IN 1
.... N (*Ci:
,.....
.E
,.....
+
~
~
,..... ~
.:::
X~l), VI = X~2) with canonical
~ ~ t~;~J~HJ
1
(I)
=
10.2. The (2 X 1) random vectors X(I) and X(2) have the joint mean vector and joint covariance matrix
......
... ,.....
It) c<)
:
!
Verify that the first pair of canonica l variates are U I correlati on p; = .95.
11
Q)
Co,
'><
~
Wi"J [100 0 0 0J fl:: ) ~ [~~;ti;;] ~!! 5y!,~~ (I):
11
NN
S
00
V")
~
V>
Conside r the covarian ce matrix given in Example 10.3:
Cl
N
~
10.1.
00 ~
~
0 "0
Clt::
t:t:
.~
T""""'I (,j....( .....
...
U Q)
l"
Exercises
.V'
'ii)'
~
t::
Q)
~
0
.S0
Exercises 567
t)
.8 m
0'" Cl.
and
Cov
(t¥J) = [~~~rI~~J
Compar e your results with the properti es in Result 10.1. 10.3. LetZ(!) = VjV2(X(!)  1'(1) andZ(2) = ViY2(X( 2)  1'(2) be two sets of standard ized variables. If p~, p;, ... , p; are the canonical correlati ons for the X (I) , X (2) sets and (U;, Vi) = (aiX(I), biX(2), i = 1,2, ... , p, are the associated canonica l variates, determine the canonical correlati ons and canonical variates for the Z(1), Z(2) sets. That is, eX8ress the canonical correlati ons and canonical variate coefficient vectors for the Z(I), Z ) sets in terms ofthose for the X(I), X (2) sets. 10.4. (Alternative calculation of canonical correlat ions and variates.) Show that, if Ai is an eigenvalue of Ijlf2I12Ii~I21Ijfl2 with associated eigenvec tor ei, then Ai is also an eigenvalue of IjII12Ii~I21 with eigenvec tor Ij!i2 . ei Hint: 1Ijlf2I12Ii~I2IIj!i2  Ail 1 = 0 implies that
o = 1Ijl f2 11 Ij!i2I12I2~I21Ij!i2 = 1IjII12I2~I21 
Ail 1
 Ail 11 IW 1
568
Chapter 10 Canonical Correlation Analysis
Exercises 569
10.5. Use the information in Example 10.1.
(a) Find the eigenvalues of II1I12I2t.~;21 and verify that these eigenvalues are same as the eigenvalues of IIV2I 12 IZ!I 21 IiJl2. (b) Determine the second pair of canonical variates (U2 , V2 ) and verify, from first pies, that their correlation is the second canonical correlation p; = .03.
(e) Twentyone observations on the 6:00 A.M. and noon wind directions give the correlationmatrix cos(8d
R = [ ..
10.6. Show that the canonical correlations are invariant under nonsingular linear tr.'n.f,,~,. tions of the X(1), X(2) variables ofthe form C X(l) and D X(2). (pXp) (pXl)
Hint: Consider Cov
(['~~'~~!'J) = [.~~J.!.~.~j ... ~~n~: ] Consider any linear DX(2) DI 21 C'i DInD' .
= [:
:J
andPII
.372
.243 i
Find the sample canonical correlation
structure where X(1) and X(2) each have two components. (a) Determine the canonical variates corresponding to the nonzero canonical correlation. (b) Generalize the results in Part a to the case where X(1) has p components and X(2) has q 2! P components. Hint: P12 = pll',wherelisa(p X 1)columnvectorof1'sandl'isa(q X 1) row 1 vector of l's. Note that PIll = [1 + (p  l)p]l so PI]l21 = (1 + (p 1)pr /21. 10.8. (Correlation for angular measurement.) Some observations, such as wind direction, are in the form of angles. An angle 82 can be represented as the pair x (2) = [cos( 82), sin( 82) Y.
(a) Show that b'X(2) = Vby + b~cos(82  f3) where bIiYby + b~ = cos(f3) b2lVbi + b~ = sin(,8). Hint: cos(8 2  ,8) = cos(8 2) cos(f3) + sin(8 2) sin(f3). (b) Let X(I) have a single component XP) . Show that the single canonical correlation is p~ = max Corr (x)1), cost 82  ,8». Selecting the canonical variable VI amounts to /3
selecting a new origin ,8 for the angle 82, (See Iohnson and Wehrly (7].) (c) Let x)1) be ozone (in parts per million) and 82 = wind direction measured from the north. Nineteen observations made in downtown Milwaukee, Wisconsin, give the sample correlation matrix
.181
1.0
Pt and VI, VI .
The following exercises may require a comput~r.
10.9. H. Hotelling [5] reports that n = 140 seventhgrade children received four tests on x(1) = reading speed, X~I) = reading power, X\2) = arithmetic speed, and X~2) = arithmetic power. The correlations for performance are
~JcorresPOndingtotheeqUalCOrrelation
= P22 = [:
sin(82)
~l~5X~::lii;··li:l
(qXq) (qXl)
nation ai(CX(1» = a'X(I) with a' = a;C. Similarly, consider bi(DX(2» = with b' = biD. Th