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Exact testing procedures for unbalanced random and mixed linear models

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Exact testing procedures for unbalanced random and mixed linear models
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Thesis (Ph. D.)--University of Florida, 1991.
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Includes bibliographical references (leaves 294-297).
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EXACT TESTING PROCEDURES FOR UNBALANCED
RANDOM AND MIXED LINEAR MODELS

















By

ROBERT CALVIN CAPEN


A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA


1991


































To my mother and the memory of my father













ACKNOWLEDGMENTS


I would like to express my most sincere appreciation to Dr.

Andre' I. Khuri for his patience, understanding, and guidance during

the writing of this dissertation. Also, I would like to thank Dr.

Malay Ghosh, Dr. Ramon Littell, Dr. P.V. Rao, and Dr. Timothy White

for serving on my committee.

I am grateful to my fiance, Debra Thurman, for all the love

and support she has provided me during these past two years. She

gave me the strength to continue.

I would also like to thank Jeff, Joe, Amber, and Julie for

their friendship and moral support.

Finally, I wish to acknowledge Leslie Easom for her superb

typing job -- thank you very much.


-iii-















LIST OF TABLES


Table Page

3.1 An example of an unbalanced random two-way model
with missing cells ...................................... 44

3.2 An example of an unbalanced random 3-fold nested
model ............................................... 91

3.3 Information needed to determine the vector v in
(3.257) for a three-way cross-classification model
in which Factors A and B are fixed while Factor C
is random ............................................. 129

3.4 Information needed to determine the vector x in
(3.288) for a three-way cross-classification model
in which Factor A is fixed while Factors B and C
are random ............................................ 139

3.5 A comparison of the powers of the exact tests for o
in an unbalanced mixed two-way cross-classification
model ................................................. 162

3.6 Numerical scores resulting from a comparison between
dyed cotton-synthetic cloth and a standard .......... 186

3.7 Some useful quantities used in the construction of the
exact tests for the variance components utilizing
the methodology developed in Section 3.3.3 .......... 188

3.8 Expected mean square values for the cloth example ... 189

3.9 Results from the analysis of the variance components
for the cloth example .............................. 190

3.10 Some useful quantities used in the construction of the
exact tests for the fixed-effects utilizing the
methodology developed in Section 3.3.4 .............. 191

3.11 Summary of results for the cloth example ............ 192

3.12 Performance values for three machines ............... 194


-iv-











Table
Page

3.13 Results from the analysis of the variance components
for the machine example ............................. 195

3.14 Some useful quantities needed in the development of
the exact confidence intervals for the machine
example ............................................... 196

3.15 Exact and approximate 95% confidence intervals for
the machine example ..................... .............. 198

3.16 Some useful quantities used in the construction of
the exact tests for the four variance components
in the example .......................................... 262

3.17 Results from our testing procedures ................. 264

















TABLE OF CONTENTS


Page


ACKNOWLEDGMENTS ............................................... iii

LIST OF TABLES ................................................ iv

ABSTRACT ...................................................... viii

CHAPTERS

ONE INTRODUCTION .......................................... 1

TWO LITERATURE REVIEW ..................................... 5

THREE EXACT TESTING PROCEDURES .............................. 13

3.1 The Unbalanced Random Two-Way model
with Some Cells Missing ........................... 13
3.1.1 Introduction .................................. 13
3.1.2 Preliminary Development ................... 19
3.1.3 An Exact Test Concerning .............. 26

3.1.4 An Exact Test Concerning a2 ............... 36
3.1.5 An Exact Test Concerning ap .............. 41
3.1.6 A Numerical Example ....................... 42
3.1.7 The Power of Exact Tests .................. 45
3.1.8 Concluding Remarks ........................ 47

3.2 The Unbalanced Three-Fold Nested Random Model .... 49
3.2.1 Introduction .................................. 49
3.2.2 Preliminary Development ................... 54
3.2.3 An Exact Test Concerning 2 ............... 58

3.2.4 An Exact Test Concerning a. ............... 71
3.2.5 Reduction of the Exact Tests when the
Design is Balanced ......................... 82
3.2.6 A Numerical Example ....................... 88
3.2.7 The Power of Exact Tests .................. 92
3.2.8 Concluding Remarks ........................ 98


-vi-










3.3 The General Unbalanced Mixed Model with Imbalance
Affecting the Last Stage Only .................... 99
3.3.1 Introduction .................................. 99
3.3.2 Preliminary Development ................... 100
3.3.3 Analyzing the Variance Component .......... 110
3.3.4 Analyzing the Fixed Effects Parameter ..... 119
3.3.5 Reduction of the Tests When the
Design is Balanced ........................ 141
3.3.6 The Power of the Exact Tests .............. 143
3.3.7 An Alternative Approach to the Analysis of
the Unbalanced Mixed Model with Imbalance
Affecting the Last Stage Only ............. 151
3.3.8 Numerical Examples ........................ 184
3.3.9 Concluding Remarks ........................ 199


3.4 The Random Unbalanced Nested Model with Imbalance
Affecting the Last Two Stages Only ............... 201
3.4.1 Introduction ............................. 201
3.4.2 Notations and Preliminary Development ..... 201
3.4.3 An Exact Test for as .......... ........... 204
3.4.4 An Exact Test for *-1 .................. 211
3.4.5 An Exact Test for a2, 92,**, sr2 ......... 227
3.4.6 Reduction of the Exact Tests When the
Design is Balanced ....................... 247
3.4.7 The Power of the Exact Tests .............. 254
3.4.8 A Numerical Example ....................... 261
3.4.9 Concluding Remarks ........................ 261

FOUR CONCLUSIONS AND DIRECTIONS FOR FUTURE RESEARCH ........ 265

APPENDICES

A MATHEMATICAL RESULTS .................................. 269

B STATISTICAL RESULTS ................................... 280

C AN ILLUSTRATION OF THE UTILITY OF RESAMPLING .......... 286

D HIROTSU'S APPROXIMATION ............................... 292

REFERENCES ...................................................... 294

BIOGRAPHICAL SKETCH ............................................. 298


-vii-

















Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy


EXACT TESTING PROCEDURES FOR UNBALANCED
RANDOM AND MIXED LINEAR MODELS

By

Robert Calvin Capen

August 1991

Chairman: Dr. Andr6 I. Khuri
Major Department: Statistics



Although much attention has been given to the problem of

estimating variance components in unbalanced random and mixed linear

models, the same cannot be said about the development of exact testing

procedures for the aforementioned quantities. One reason for this is

the mathematical complexity of the problem. Even in some balanced

random or mixed models, no exact analysis of variance tests exist for

testing certain variance components. Consequently, approximate

methods, such as Satterthwaite's procedure, are typically relied upon

to furnish the required tests. Unfortunately, the exact distributions

of these approximate procedures are generally unknown, even under the

null hypothesis.


-viii-















Recently, some progress has been made in the construction of

exact tests for the variance components and for estimable linear

functions of the fixed-effects parameters in unbalanced random and

mixed models. In this dissertation we extend some of these known

results and generate some new ones. Specifically, exact procedures are

developed to test hypotheses concerning the variance components in an

unbalanced random two-way model with some cells missing, an unbalanced

random 3-fold nested model, and an unbalanced random s-fold nested

model where the imbalance affects the last two stages only. Also,

methods are derived to test hypotheses about the variance components

and about estimable linear functions of the fixed-effects parameters in

a general unbalanced mixed model with imbalance affecting the last

stage only.


-ix-
















CHAPTER ONE
INTRODUCTION


Although much attention has been given to the problem of

estimating variance components (see, e.g., Rao and Kleffe (1988)),

the same cannot be said about the development of exact testing

procedures concerning the variance components in unbalanced random

and mixed linear models. Perhaps one reason for this is due to the

mathematical complexity of the problem. Even for some balanced

models no exact analysis of variance (ANOVA) tests exist for testing

certain variance components. For example, in a three-way cross-

classification random-effects model, no exact procedures exist for

testing hypotheses concerning the variance components associated with

the main effects of the model. Also, when exact tests do exist for

balanced models, they generally cannot be used to test the

corresponding hypotheses in unbalanced models. The reason for this

is that for unbalanced models, the traditional test statistics used

with balanced data are, in general, no longer based upon independent,

chi-squared-type sums of squares.

Approximate methods, such as Satterthwaite's (1946) procedure,

are typically relied upon to furnish tests for the variance

components when no such exact procedure exists. However, approximate

methods are just that -- approximate. Some may be relatively easy to

implement, and perform reasonably well in certain situations, but







-2-


they cannot be expected to always produce adequate solutions (see,

e.g., Cummings and Gaylor (1974), Tietjen (1974), and more recently,

Ames and Webster (1991)). Thus there is a clear need for the

development of exact testing procedures.

In this dissertation we have concentrated on developing exact

methodology for some unbalanced random and mixed linear models. Much

of this work has been inspired by the derivations given in Khuri

(1984, 1987, 1990), Khuri and Littell (1987), Gallo (1987), and Gallo

and Khuri (1990). Their results and those obtained in this

dissertation are based on certain transformations which reduce the

analysis of the unbalanced model to that of a balanced one. A

crucial theme in the construction of these transformations is a

scheme for resampling from the error vector. In this scheme, only a

portion of the error vector is utilized in making the transformation.

Consequently, the values of the resulting test statistics will not be

unique. However, the distributional properties of these statistics

will remain unchanged no matter what portion of the error vector is

used in the transformation. It is in this sense that our methodology

is similar to procedures involving resampling. Appendix C provides

an illustration of the utility of resampling as a means of


"counteracting" the imbalance in a design.

This thesis is divided into four chapters. Chapters one and

give a brief introduction to the area of exact testing procedures

unbalanced random and mixed linear models and a review of the

pertinent literature. The main contributions of this study are fo

in Chapter three. This chapter is divided into four sections. Ir


)und

i











the first section we develop exact tests for the variance components

in a random two-way model with some cells missing. Section two deals

with the unbalanced random 3-fold nested model. Here the imbalance

is assumed to affect all stages of the design. In the last two

sections we concentrate on more general models. The general mixed

model with imbalance affecting the last stage only is analyzed in

Section three. While in Section four we obtain exact procedures for

testing the variance components in a random s-fold nested model with

imbalance affecting the last two stages only. Finally, Chapter four

summarizes our findings and offers directions for future research.

Throughout this work we will try to motivate, as much as

possible, the reason behind each transformation we use. At times,

the mathematical derivations may become "intense" in the sense that

there may be pages containing nothing but equations. Unfortunately,

there seems to be no way to avoid this. A numerical example is given

at the end of each section in Chapter three to illustrate the

proposed methodology. These examples also serve as useful (and

probably needed) "rest stops" between pages full of formulas.

The following notation will be used throughout this thesis:

rank(A) will denote the rank of the matrix A

range(A) will denote the range space of the matrix A, that

is, the vector space spanned by the columns of A.

tr(A) will denote the trace of the matrix A.

A- will denote a generalized inverse of A, that is, any

matrix which satisfies AA-A=A.






-4-



* diag(A, B) will denote the matrix 0 B This can also be

represented by AEB where "D" is the direct sum operator.

* AOB will be used to represent the (right) direct (Kronecker)

product of two matrices. Briefly, if A is nxm and B is pxq

then AB is the npxmq partitioned matrix whose (i,j)th

matrix is aijB where aij is the (i,j)th element of A.
















CHAPTER TWO
LITERATURE REVIEW


The first paper that addresses the issue of exact tests for

variance components in unbalanced models is due to Wald (1940). Wald

constructed an exact test for a/.a in the random one-way model:



yij =p+ai+ij (2.1)



(i=l,2,. .,p; j=l,2,.--,mi), where p is an unknown constant

parameter; ai and eij are independent, normally distributed random

variables with zero means and variances a. and a respectively. He

later extended this result to the general random-effects model

without interactions. Wald (1947) also considered the mixed model



y=Xa+Zb+e (2.2)



where a is a vector of fixed effects; b-N(Q, oaI) independently of

f~-N(Q, aoI). Using a regression analysis approach, Wald was able to

construct an exact test for ab 2

As pointed out in Spj0tvoll (1968), Wald's idea cannot be used to

test variance components in all linear models. To illustrate this

phenomenon, Spj0tvoll considered the random two-way model. Based on











a modification of Wald's method he was able to derive an exact test

for the variance component associated with the interaction effect.

However, only under the assumption of additivity could he construct

exact tests for the main effects variance components.

By modifying an approach suggested by Graybill and Hultquist

(1961), Thomsen (1975) was able to construct exact tests for the

variance components in an unbalanced two-way random model. Thomsen

did this by first making an orthogonal transformation as outlined

below. Let y be the vector of cell means, then in matrix notation we

have (see Thomsen (1975, p. 258)):



y= Plrs+Biq+B2+ Irs(aP) +? (2.3)



where 1rs is the vector of ones of dimension rs; Irs is the rsxrs

identity matrix; B1 and B2 are defined as



B1 =Ir is

(2.4)

B2 = rIs



and r is the number of rows and s is the number of columns in the

design. Since A = B1B1' and A2 = B2B2' commute, there exists an

orthogonal matrix P which simultaneously diagonalizes A1 and A2.

Further, the first row of P may be chosen to be (1/,r-s)l's. Making

the transformation











z =P (2.5)



results in Thomsen's "semi-canonical" form. Thomsen used the last

rs-1 elements of z to derive exact tests for ag (the variance

component associated with the interaction effect) and for a2 and o

(the main effects variance components). However, as in Spj0tvoll

(1968), the development of the latter two tests rested on the

assumption that a2p =0. Also, Thomsen assumed the layout contained

no missing cells. (In a later section he dropped this assumption,

but still assumed the design was connected.)

In 1983, Seely and El-Bassiouni returned to Wald's variance

component test. In their paper, necessary and sufficient conditions

were given under which the Wald test can be used in mixed models.

They also developed a uniqueness property of Wald's test. Thus based

solely on matrix ranks, one can determine if a particular variance

component test is a Wald's test. They applied their notion to the

tests developed by Spjetvoll (1968) and Thomsen (1975). They

concluded that the tests derived by Thomsen are indeed Vald's tests,

while only the test for interaction developed by Spjetvoll is a

Wald's test.

Although both Spjotvoll and Thomsen needed to assume additivity in

their development of exact tests for the main effects variance

components in an unbalanced random two-way model, Khuri and Littell

(1987) demonstrated that this assumption was unnecessary. Initially,

their work proceeded similarly to Thomsen (1975). However, they also

made a particular transformation that reduced the analysis of the










unbalanced model to that of a balanced one. Their idea was based on

a scheme for resampling from the error vector. After making this

transformation they were left with a model of the form (see Khuri and

Littell (1987, p. 548)),



w= PBla +P1B2P +E* (2.6)



where a~N(O, oaIr); ~-N(O, aIs); *~N(Q, 6Irs-1) where 6= a2

+Amax Oa and Amax is the largest eigenvalue of the matrix P1KP1' where

K==diag(n ,n ,.. **,nn) and P1 consists of the last rs-1 rows of P

(see (2.5)). Furthermore, a, 8 and c* are mutually independent. B1

and B2 are defined in (2.4).

After partitioning w into w=(_a, W', wap)', Khuri and Littell

developed exact tests for a2 and a based on (Y, "al)' and (W

W1 )', respectively. In particular, their test statistic for testing

Ho:a =0 is given by



F= w/(r-1) (2.7)
W apw/[(r-1)(s-1)]


It should be noted that



rank(P1B1:P1B2) = r + s-2



rank(P1B1) = r-


rank(P1B2) = s-1.











Consequently, according to Seely and El-Bassiouni (1983), the tests

derived by Khuri and Littell (1987) for the main effects variance

components are Wald's tests based on the w model in (2.6).

One attractive feature of a Vald's test is that when applied in an

unbalanced one-way-random effects model, it may be interpreted as

being "almost equal to the most powerful invariant tests against

large alternatives" (Spjotvoll (1967, p.422)). This result is also

hinted at in Mostafa (1967). Also, under certain conditions, Mathew

and Sinha (1988) show that the Wald test for testing the variance

component associated with the unbalanced random two-way mixed model

without interaction is the uniformly most powerful invariant test

under a certain group of transformations. This is more fully

discussed in chapter four.

Besides the above mentioned results, other exact procedures do

exist in the literature. Khuri (1987) developed an exact test for

the nesting effect's variance component in an unbalanced random 2-

fold nested model. The idea of resampling from the error vector was

also utilized in this paper. Verdooren (1988) also analyzed the

unbalanced random 2-fold nested model. He was able to derive an

exact test for p, for each given value of P2. Here p1 is the ratio of

the nesting effect's variance component to the error variance

component and P2 is the ratio of the nested effect's variance

component to the error variance component.

Exact tests for the random and fixed effects in some unbalanced

mixed models were derived in Gallo (1987) and Gallo and Khuri (1990).

Hocking (1988) also considered the mixed linear model. Although he






-10-


was more interested in estimation than in testing, he did note that

his estimators could be used to develop tests of hypotheses for

designs with moderate imbalance. These tests, however, are not

exact. We only mention them since they may represent reasonable

alternatives to the methodology presented in Section three of Chapter

three.

Turning to more general models, Khuri (1990) developed exact tests

for random models with unequal cell frequencies in the last stage.

These designs include all cross-classification random effects models

with no missing cells. Utilizing resampling, Khuri was able to

construct a set of mutually independent sums of squares that are

distributed as scaled chi-squared variates. These sums of squares

were then used to test hypotheses concerning the variance components.

It should be noted that not all variance components can be tested

exactly using Khuri's procedure. This should not be too surprising,

however, since this phenomenon occurs even for some balanced random

(or mixed) models (cf. Kendall and Stuart (1968), p. 83).

Using a different approach, Ofversten (1989) was also able to

construct exact tests for the variance components in general mixed

linear models. His idea was to use a result found in Allen (1974) to

orthogonally transform the model matrices into so-called layer

triangular form. The nice thing about this type of transformation is

that when combined with orthogonal row permutations it can be used to

reduce matrices to ones with full row rank. This technique is

repeatedly utilized in Ofversten's work to produce exact tests for

the variance components in certain mixed models. However, when






-11-


certain rank conditions are not satisfied this approach cannot be

used to produce the exact procedures. When this occurs, Ofversten

relies on resampling to develop his exact tests. Ofversten's

methodology can be used in any unbalanced mixed model to construct

tests for the variance components. However, it is not altogether

clear as to how to adapt his methods to designs not examined in his

dissertation. More is said about Ofversten's results in Chapter four.

Partial motivation behind the need to develop exact testing

procedures can be found in Tietjen (1974) and Cummings and Gaylor

(1974). Both of these papers considered the random 2-fold nested

model. In his simulation study, Tietjen showed that the approximate

test based on Satterthwaite's (1946) procedure for testing the

nesting effect's variance component can be markedly off in

approximating the true p-value. This led him to recommend the use of

the conventional F-test (that is, the test used when the data are

balanced) as an approximate test for this hypothesis.

Cummings and Gaylor (1974) considered various 2-fold nested

designs in their simulations. These designs were specifically chosen

so that the mean squares involved in the approximate (Satterthwaite)

F-test were either dependent, failed to be distributed as scaled chi-

squared variates, or both. They demonstrated that when the mean

squares were dependent, the estimated test size was less than the

stated test size and thus, the approximate test was conservative.

When the mean squares were independent but did not have chi-squared

type distributions, the opposite occurred. The estimated test size

was greater than the stated test size. When the mean squares were







-12-


both dependent and not distributed as scaled chi-squared variates, a

"counter-balancing" effect was observed. The approximate F-test in

this case seemed to perform well. However, this counteracting effect

may not occur in higher level designs. As Cummings and Gaylor (1974,

p. 771) state "there is a possibility the test size disturbances due

to mean square dependence and nonchisquaredness will provide some

reinforcement."

In the next chapter we extend the work of Khuri (1987, 1990),

Khuri and Littell (1987), and Gallo and Khuri (1990). The idea of

resampling from the error vector is vital in our development and is

used repeatedly throughout this dissertation. The reader is referred

to Appendix C where the utility of this approach is illustrated.















CHAPTER THREE
EXACT TESTING PROCEDURES


3.1. The Unbalanced Random Two-Way Model with Some Cells Missing



3.1.1. Introduction

In this section we will consider the unbalanced random two-way

model with some cells missing. Specifically, we will concentrate on

the following model:



yijk=P++ai +j+(aP)ij +ijk (3.1)



((i,j)EY; k=l,2,...,nij); where p is an unknown constant parameter;

ai, Pj, (af)ij, and eijk are independently distributed normal random

variables with zero means and variances a,, a, #, and aa,

respectively; I is the set of subscripts, (i,j), which label the

nonempty cells of (3.1). That is,



Y={(i,j): 1O} (3.2)



where r is the number of rows and s the number of columns of the

design. Note that I is a proper subset of IXJ, where I=

{i:i=1,2,* -,r}; J={j:j=1,2,.. ,s}; and X denotes the Cartesian

product.


-13-






-14-


In matrix form, (3.1) becomes


y = 1n +Xlg+X2 +X3(%#) +f


(3.3)


where y is the vector of observations of dimension n= C nij; in
(i,j)E
is a vector of ones of dimension n; X1, X2, and X3 are known matrices

of zeros and ones of orders nxr, nxs, and nx(rs-p), respectively,

where p is the number of empty cells; a, /, and (apf) are column

vectors whose elements are the ai, )j, and (ap)ij, respectively

((i,j) )); e is the nxl vector of errors. Define


nij
Yij n- k=i jk
ii k=l


(3.4)


for all (i,j) E. From (3.1) we obtain


(3.5)


((i,j) E), where


(3.6)


nij
~1i j = eijk
ij i i k=1


for all (i,j) E?.


In matrix form, (3.5) becomes


(3.7)


Y=Prs-p +A +A2l+ Irs-p(al) +


where y is the (rs-p)xl vector of (filled) cell means; Irs-p is the






-15-


identity matrix of order rs-p; A1 and A2 are known matrices of zeros

and ones of orders (rs-p)xr and (rs-p)xs, respectively. A1 and A2

can be expressed as


r
Ai= El s-pi (3.8)
i=l 1i




(Is-pl: 0(s-pl)xpl)E1


A,= (Is-2 0(s-P2) 2)E2 (3.9)


(Is-r: (s-r) xpr)Er




where pi is the number of empty cells in row i (i=1,2,.* ,r);

O(s-p)xPi is a matrix of zeros of order (s-pi)xpi (i=1,2,. -,r);

and Ei is a product of interchange matrices (i=1,2,. -,r). We take

(Is-i : 0(s-pi)xp)=Is when pi=0 (i=1,2,...,r).



Example 3.1



Consider the following two-way table, an "X" represents a filled

cell.


X X X X
X X X
X X
x\ ~r







-16-


Then,


r=3

s=4

Pi =0

P2 = 1

p3 = 2
P3=2
p=3


furthermore,


El = 14 = E3



1 0 0 0 1 0 0 0
0 0 1 0 0 0 0 1
E2 0 1 0 0 0 0 1 0

0 0 0 0 0 0
0001 0100


1

0
0
0


consequently,


A1 =diag(14,13,12)






-17-


14


10 00
A= 0 0 1 0
O 001


(12:02 X2)



Besides the usual assumptions concerning the random effects, we will

also assume:



(i) The model used is model (3.1) (or model (3.3))

(ii) Both r and s are at least two

(iii) rank(Al:A2)=r+s-1 (3.10)

(iv) p<(r-1)(s-1)

(v) n >2(rs-p)-min(r,s)



Assumption (iii) implies that the design is connected (Gately (1962)

p. 46), while the last assumption (along with the others) is needed to

insure that the exact tests for the main effects' variance components

are valid.

The inclusion of missing cells into any design adds a degree of

difficulty to the analysis. The two-way random model is no

exception. One problem is that the matrices BI=A1A and B2 =A2A do

not, in general, commute. B1 and B2 will commute if all cells are

filled (see Thomsen (1975) or Khuri and Littell (1987)).






-18-


Reconsidering Example 3.1, we have


Bi=diag(J4, J3, J2)


where Ja = la While


1E (o3)

13


hence


BB,2= J3(I3:)E2


I
J4E' (o)

J3


Since, for example, J4( o= 2) while J2(:0) = (J2:0), we see that BIB2

is not symmetric. Thus B1 and B2 do not commute. Consequently no

orthogonal matrix exists that will simultaneously diagonalize B1 and

B2.


(2?)


(I1:0) E2


I )


I
J4( 02)

J3( 13:Q)E2 0


B2 = (I3:-)E2


2:0) (2:0)E I
(I,:0) ~o (I,0)


J2(12:0) J2(l2:0)E2 of






-19-


The next section lays the groundwork necessary for the

construction of the exact tests.



3.1.2 Preliminary Development



Recall (3.7). From the distributional properties of the random

effects, we can conclude that



~N(p4rs-p, E)


where


E= B+B2 +Irs-p+K
E = Bla2+B201+ Irs- po + +


(3.11)


(3.12)


K = E nA1
(i,Bj) E


B, = A,A;


(3.13)


B2 = A'A


The following lemma provides the ranks of B1 and B2.







-20-


Lemma 3.1



(a) rank(B1)=r



(b) rank(B2)= s



proof:


(a) rank(B1) =rank(A1A') =rank(A1).


r
rank(A1) = ( 1s-p)
i=1- 1


Since


(see (3.8))


r
=E (1)=r,
i=l


the result follows.


(b) Without loss of generality, we can assume that each column (and

row) in the layout includes at least one cell containing

observations. Thus, each column in A2 contains at least one "1".

Hence there exists a permutation matrix II such that



A2 = [**]



for some (rs-p-s)xs matrix 4. Consequently,



rank(B2) = rank(A2A )


=rank(A2)






-21-


=rank(IIA2)


ran. =S.
= ran = s.


Now, let


u = Hy (3.14)



where H1 consists of the last rs-p-1 rows of the (rs-p) x (rs -p)

Helmert matrix (see Searle (1982), p. 71). We note that

Hlrs-p =rs-p-1 Then, it is easily verified that


E(u)= Qrs-p-1


(3.15)


Var(u) = H1B1H'u + H1B2H + I-p-7'/ + HKH'
HIB a + Irs_p-l aP + H1cKHl 'T


The purpose of this transformation is to construct a vector which has

a zero mean vector. The ranks of H1B1H' and H1B2H' are examined in the

next lemma.



Lemma 3.2


(a) rank(HB1H'))=r-l






-22-


(b) rank(HB2H) = s-1



(c) rank(HI1B1H'+H1B2H) =r+s-2



Proof:


(a) rank(H1BHII)


=r(HIAAIAH1)


=r(H1A1)





(from Lemma 3.1(a)).


However,


Allr = I1is-Pi lr = rs-p.


Thus,


HiAlir = Hilrs-p = 9rs-p-1


Therefore,


rank(H1B1H') < r- 1.


Conversely, by the Frobenius Inequality (see Lemma A.1), we have






-23-


rank(H1BiH') = rank(H1A1)


> rank(H1) +rank(Al) -rank(Irs-p)


=rs-p-1+r-(rs-p) =r-1.



Consequently, rank(Hl1BlH) = r- 1.



(b) Similar to (a).


(c) rank(H1B1H' + H1B2H') < rank(H1B1H') + rank(H1B2H')


=r+s-2 (from (a) and (b)).



Conversely, the Frobenius Inequality gives us


rank(H1B1H' +HIB2HI) = rank(Hi(Bi + B2)H)


= rank(H1(A1A + A2A )H2)


=rank(H,[A,:A2] H1
A1


= rank(H [Ai:A2])


> rank(H1) + rank([Ai:A2])-rank(Irs-p)






-24-


=rs-p-1+r+s-l-(rs -p)



=r+s-2



(by assumption, rank ([Ai:A2])=r+s-1 (see (3.10))). 0



The next step in our analysis is to simultaneously diagonalize

HIB1HB and H1B2H1. Because B1 and B2 do not, in general, commute,

neither will H1B1H' and H1BZHi. So we cannot achieve simultaneous

diagonalization by way of an orthogonal transformation. However,

both H1BIH' and H1B2H' are n.n.d. Consequently a nonsingular matrix,

M, of order (rs-p-1)x(rs-p-1), will exist that will diagonalize

both H1B1H' and H1B2H' (see Lemma A.4). Define



Al = MH1BIHM' (3.16)



A2 = MH1B2H'M' (3.17)



Both Ai and A2 are diagonal matrices of order rs-p-1. By Corollary

A.4.1 of Lemma A, we can express A1 and A2 as



Al=diag(Ir1, Os-,_ 0(r-l)(s-l)-p) (3.18)



A2=diag(Or-_ I,_-1 0(r-1)(s-1)-p). (3.19)


Consider, then, the (rs-p-1)xl vector v defined as






-25-


S= Mu.


(3.20)


It is easily shown that


E() =


(3.21)


Var(y) = 2Al + A +F2 + G2


where


F = MM'


(3.22)


G = MHIKH'M'.



We can partition y into


(3.23)


y=(y'1, y, y-3


where vy consists of the r-1 elements of y whose covariance matrix is

free from ao; y2 consists of the s-1 elements of y whose covariance

matrix is free from a.; and v3 consists of the (r-l)(s-l)-p

elements of y whose covariance matrix is free from both ao and a .

Hence,






-26-


E(y) =
(3.24)

Var(y) =diag(Ir-10a' IS-1fl, O) +F ap+Ga.



This result follows from (3.18), (3.19), and (3.21).

We are now ready to develop our exact tests. To begin with, we

develop an exact test concerning a'.



3.1.3 An Exact Test Concerning aq



Initial development. In this section an exact test for testing

H0:o =0 vs. Ha:ua#O is derived by utilizing the subvectors y1 and Y3

of y in (3.23). To begin with, partition F and G (see (3.22)) as


and


G=


(3.25)


(3.26)


where F11(G11), F22(G22), and F33(G33) are, respectively, (r-1)x(r-1),

(s-1)x(s-1), and ((r-1)(s-1)-p)x((r-1)(s-1)-p) matrices. Also,

let V1:3 be the (s(r-1)-p)xl vector defined as






-27-


(3.27)


Yl
1 :3 =
Y3


where y1 is (r-1)xl while Y3 is ((r-l(s-l)-p)xl. Then, from

(3.21) we have


E(y1:3) = 9


(3.28)


Var(y1:3)= AlOa+FI:3Tap+GI:30f


where


r-1
A1 =
0


0


0(r-1)(s-1)-p


FIl

F113


F1:3=


G1l
Gl:3=
G13


(3.29)







(3.30)







(3.31)


G13

G33


From (3.22) notice that since both F and G are positive definite

(p.d.), F1:3 and GI:3 are also both p.d.






-28-


Now, there exists a nonsingular matrix N or order s(r-l)-p such

that (see Lemma A.3)



NFI:3N' =Is(r-l)-p (3.32)

and



NAlN' = A, (3.33)



where A1 is a diagonal matrix whose diagonal entries are solutions 0

to

IA1-0FI:3| = 0. (3.34)



We note that if F13 represents the first r-1 rows and columns of F;3

then the nonzero solutions to (3.34) are exactly the eigenvalues of

F1:3.

Without loss of generality we may express A, in (3.33) as



A= ] (3.35)
0 0


where A* is a diagonal matrix of order r-1 whose diagonal entries are

the eigenvalues of F:l3. (If A1 is not in this form then we can find

an orthogonal matrix P such that


A0 0
PAP'= 1
0 0






-29-


Thus, letting


N* = PN


we see that


N*AIN*' = PNA1N'P'



=PA1P' (see (3.33))




0 0


The purpose of introducing the matrix N was to simultaneously

diagonalize A1 and F1:3 in the expression for Var(yV:3) given in

(3.28).

Define the s(r-1)-p random vector x1:3 as


(3.36)


where Xi is (r-l)xl and x3 is ((r-1)(s-1)-p)xl. Then



E(x1:3) =0


(3.37)


Var(x:32) = A +2 2+LIr.
var(l:3) =Alaa+Is(rl)_p a# + -


X:3= = N = Ny1:3
13 Y3






-30-


This follows from (3.28), (3.32), and (3.33). L1 is the (s(r-l)-p)

x(s(r-l)-p) full rank matrix


L1 = NG: 3N'.


(3.38)


Ve can equivalently express Var(51:3) in (3.37) as



Var(xl:3)=diag(A* +Ir-1p I(r-1)(s-1)-p a0)+1E. (3.39)



Because L1 is not diagonal, x, and x3 are not independent. To achieve

independence between x1 and x3, we resample from the error vector.

The results of Lemma B.6 are utilized heavily in the forthcoming

development.

Utilizing resampling. Because model (3.1) or (3.3) is a special

case of model (B.1) the error sum of squares, SSE, associated with

this model can be expressed as


SSE = y'Ry


(3.40)


where R is the nxn symmetric, idempotent matrix


(3.41)


R = In W(W'VW)W'


where


W= [In:Xi:X2:X3]


(3.42)






-31-


and X1, X2, and X3 are described in (3.3). Since



range(In)C range(X3)


range(Xi)C range(X3)


(i = 1,2),


(3.41) can also be written as



R= In X3(X )-3


=In- (i (Jn./nij).
(i,j) EY ij


(3.43)


This last equality follows since


X = e lnij
(i,j)e3 TJ


(3.44)


Following Lemma B.6 (iv), there exists an orthogonal matrix C and a

diagonal matrix F such that


R= CrC'.


(3.45)


Partition C and r as


C= [C1C2:C:C3


(3.46)






-32-


r = diag(I,1,,12,0)



where



T1= s(r-1)-p
(3.47)

12 = n-2(rs- p) +s.



We note that + 72= n (rs-p) = rank(R) and 172>0 by (3.10)(v).

C1,C2, and C3 are matrices or orders nxYr, nx72 and nx(rs-p),
respectively. From (3.46), R can be written as


R = C1C+C +CA. (3.48)



Now, consider the (s(r-1)-p)xl random vector W1:3 defined to be



Y :3 = :3+ ( l,maxI i L1)2Cy (3.49)

1
where Al,max is the largest eigenvalue of L1 and (Al,maxI7q-L1)2 is a

symmetric matrix with eigenvalues equal to the square roots of the

eigenvalues of (A, maxl 1-L1), which are nonnegative.

Let y1:3 be partitioned just like x1:3 in (3.36), that is



13= (1) (3.50)



where wy and w3 are of orders r-1 and (r-1)(s-l)-p, respectively.






-33-


The major result of this section is given in the following

theorem.



Theorem 3.1



(a) E(w1)=O; E(Y3) =



(b) W, and yw are independent, normally distributed random vectors

and have the following variance-covariance matrices:



Var(wl) = Aa+ 61Ir_1


(3.51)


Var(w3) = 61i(r-1)(s-1)-p


where l = ap +Ai,max i.



Proof:



(a) From (3.49) we have


1
E(wl:3) = E(xl:3) + (Al,maxIz1 L1)2C'E(y)


From (3.37) we know that E(x1:3)=0. Since Rln= (see Lemma

B.6(ii)) and range(C1) Crange(R), we have that C1n=O. Thus the

result follows after noting that E(y) =Pin (see (3.3)).






-34-


(b) Clearly y1 and w3 are normally distributed. From Lemma B.6(v) we

have that y is independent of SSE. Consequently



DOR =0



where D is such that Dy=y, and


n = Var(y) = XXl, a + X2X'2 + X3XaZ0 + 2 EIn


(3.52)


(see (3.3)). However, since range(C1)Crange(R) it follows that



DC1 = 0.



Thus y is independent of C'y. Since x1:3 is a function of y, it

must also be independent of C'y. Hence


1 1
Var(-W:3) = Var(XZ1:3) + (A1,maxIgl L1)'C'1OC1(A1,maxIq7 -Lj)!.


(3.53)


From Lemma B.6(ii), ROR=R-2a2. Thus since range(C1)Crange(R)

and R is idempotent we have



C f2C1 = CCa = or 21



The last equality follows from the fact that C = [C:C2:C3] is

orthogonal. Therefore, (3.53) can be expressed as





-35-


Var(xl:3) = Var(x1:3) + (Al,maxI1 L)


= +2 'a + l 1 2 L1)a 2
=Ai +Is(r-1)_pp a + ,(Al,maxI, -l )



= A1o% + Is(r-l1)-p ap '1,max0)


Consequently, from (3.35),



Var(yw) = A Ia +61Ir_1


Var(w3) = 61I(r-1)(s-1)-p



where 61 = p + Amax' 0



As a result of Theorem 3.1 we obtain



(i) SS1 = '(A*+r 2 -1 (3.54)
(ii) SS3= yW3/61~ xr-1l)(s-l)-p (3.55)

(iii) SSI and SS3 are independent.


A test statistic for testing H0:Oi=0


vs. Ha:a:0O is then


F YW1/(r-1)
33/((r-1)(s-1)-p)


(3.56)






-36-


Under Ho, F1-Fr-l,(r-l)(s-l)-p* Note that



<^)J= 12a + (3.57)



where W1 is the average of the eigenvalues of F\:3 (see (3.35)). Since

F1:3 (see (3.30)) is p.d. it follows that F-13 is p.d. Thus Fl\1 is

p.d., implying that 01>0. Consequently, we would reject Ho at the 7-

level of significance if



F1> F,r-1,(r-1)(s-1)-p



where F,r-1,(r-1)(s-)-p is the upper 1007% point of the

corresponding F-distribution.

We now turn our attention to the construction of an exact test

concerning aa.



3.1.4 An Exact Test Concerning a2



Initial development. This section is concerned with the

construction of an exact test for a~. Much of the development in

this section parallels that of Section 3.1.4 except that we now focus

on the subvectors y2 and v3 of v in (3.23).

Let v2:3 be the (r(s-1)-p)xl random vector


S(3.58)
Y2:3 = (3.58)
Y3





-37-


where v2 is (s-l)xl and vy (recall) is ((r-1)(s-1)-p)x1. Similarly to

Section 3.1.3 we can make a nonsingular transformation, T, to obtain

a random vector x2:3 = [x:x']'=Tv2:3 with



E(x2:3) =
(3.59)

Var(x?2:3) = A2U + I(s-1)-ppr+L 2



where A2=-diag(Ah, 0) and Ah is a diagonal matrix of order s-1 whose

diagonal entries are the eigenvalues of F3. F2:3 represents the

first s-1 rows and columns of F-3 where

F22 F23
F2:3 = (3.60)
F'23 F33



(see (3.25)). Also, L2 is the (r(s-1)-p)x(r(s-1)-p) full rank matrix


L2 = TG2:3T'. (3.61)


T is the nonsingular matrix that simultaneously diagonalizes F2:3 and

A2 where


A2 1= (3.62)
0 0(r-1)(s-1)-p




(see(3.24) ; and G2:3 is defined to be






-38-


G22 G22
G2:3 = (3.63)
G23 G23




(see (3.26)).

Ve note that x3, defined in x2:3 above, is not the same as x3 in

(3.36) since a different nonsingular transformation was necessary to

simultaneously diagonalize F2:3 and A2.

Because x2 and x* are not independent (L2 is not diagonal (see

(3.39))), we will again need to employ resampling to aid in the

construction of the exact test.

Utilizing resampling. Consider again the matrix R defined in

(3.41) or (3.43). Unlike the partitioning of C and r described in

(3.46), we now partition C and r as



C= [C*:C*:C*]

(3.64)

rF=diag(I ,1I *,0)



where



l= r(s-1)-p

(3.65)

72 = n-2(rs-p) +r.



Note that q1 +1 =n-(rs-p)=rank(R) and >0 by (3.10)(v). Cl, C(,






-39-


and C* are matrices of orders nx i, nx 7, and nx(rs-p),

respectively. From (3.64), we can express R as



R= C' + C2' (3.66)



(see also (3.45)).

Now define the (r(s-l)-p)x1 random vector 2.:3 as



W2:3= 2:3+(A2,maxI L2)C'y (3.67)

1
where A2,max is the largest eigenvalue of L2 and (A2,maxI *-L2)2 is a

symmetric matrix with eigenvalues equal to the square roots of the

eigenvalues of (A2,maxI *-L2), which are nonnegative.

Let -2:3 be partitioned just like x2:3, that is


W2
"2:3= (3.68)
2:3


where y2 and w* are of orders s-1 and (r-1)(s-l)-p, respectively.

Theorem 3.2 represents the major result of this section.



Theorem 3.2



(a) E(w,)=O; E(wD =9



(b) y2 and w3 are independent, normally distributed random vectors

and have the following variance-covariance matrices:






-40-


Var(_2) = A + 62Is-1



Var(w) =2l(r-1)(s-1)-p


(3.69)


where 62= ao + A2,maxO*'



Proof:


The proof of Theorem 3.2 is very similar to the proof of


Theorem 3.1 and is thus omitted.



As a result of Theorem 3.2 we obtain



(i) SS = y(A2 + 62Is_1)-12 ~Xs-1

(ii) SS *= = ~x2r-2)-
3 -3 -3/2 (r-1)(s-1)-p

(iii) SSS2 and SS* are independent.



A test statistic for testing Ho:0 =0 vs. Ha:a 0 is then



FW2 y2/(s-1)
F2
2 W /((r-1)(s-1)-p)


Under Ho, F2 -Fs-l,(r-l)(s-l)-p. Note that


p(2-= Y 22 + 62
ks_-l) 20fl


where 02 is the average of the eigenvalues of F113.


(3.70)

(3.71)


(3.72)


(3.73)


Since F2:3 (see






-41-


(3.60)) is p.d. it follows that F'3 is p.d. Thus F113 is p.d.,

implying that 02>0. Consequently we would reject Ho at the y-level

of significance if F2>Fy,s-l,(r-l)(s-l)-p*

A test for the variance component associated with the

interaction effect is developed in the next section.



3.1.5 An Exact Test Concerning a2p



In this section we develop an exact test for au by using y3 (see

(3.23)). We note at the onset that this test is equivalent to

Thomsen's (1975) test.

From (3.24)-(3.26) we have



E(y3) =
(3.74)

Var(y3) = F33ap4 + G330a.



Since v is a function of y (see (3.14) and (3.20)) and y is

independent of SSE (Lemma B.6(v)), y is also independent of SSE. Thus

Y3 is independent of SSE as it is a subvector of v. Also, it clearly

follows from (3.74) that


Y-(F33 +G33 ) Y-l3~")(s l)- p (3.75)
3(V33 0 3G33a -3-Xr_1)(s-1)-p"



Because SSE/a2 X_(rs-p) a test statistic for testing Ho:ao f=

H:a) ^ 0 is





-42-


Y3IYV3/((r-1)(s-l)-p)
F3= (3.76)
SSE/(n-(rs-p))


Under Ho, F3 ~F(r-l)(s-l)-p,n-(rs-p). Note that


SyG334Y3 =3o 2 ap 2 (3.77)
(r-1)(s-1)-p 3 7


where 93 is the average of the eigenvalues of G3-F33. Since both G-1

and F33 are p.d. 93>0 (see Graybill(1983, p. 397)). Hence we would

reject Ho for large values of F3. In this case, we did not need to

resample to construct the exact test.



3.1.6 A Numerical Example



An automobile manufacturer wished to study the effects of

differences between drivers and differences between cars on gasoline

consumption. Four drivers were selected at random; also five cars of

the same model with manual transmission were randomly selected from

the assembly line. Each driver drove each car a maximum of three

times over a 40-mile test course and the miles per gallon were

recorded. For technical reasons the drivers did not drive the cars

the same number of times. Unfortunately, unforeseen mechanical

problems forced the cancellation of some of the test drives. The

data is recorded in Table 3.1

We first test for a significant interaction effect using (3.76).

We find that






-43-


F3 = 0.7167

numerator d.f. =9

denominator d.f. =22

p-value = 0.6885.



Thus there does not appear to be a significant interaction effect.

The results concerning the main effects variance components are now

presented.

For testing Ho0:O =0, regarding the driver variance component, we

use (3.56). For this data we find that



F1 = 261.1971

numerator d.f. =3

denominator d.f. =9

p-value <0.0001



The test for the car variance component resulted in (see (3.72))



F2 = 71.2372

numerator d.f. =4

denominator d.f. =9

p-value < 0.0001


All calculations were performed using S-plus.






-44-


Table 3.1


A(driver)


An example of an unbalanced random two-way model with
missing cells


1

25.3

25.2





33.6

32.9

31.8



27.7

28.5


4 29.2

29.3

(source: Dr. Andre I.


2

28.1

28.9

30.0



36.7

36.5


B(car)

3

24.8

25.1





31.7

31.9


30.7

30.4


32.4

32.4

Khuri)


4

28.4

27.9

26.7



35.6

35.0





29.7

30.2

31.5


31.8

30.7


33.7

33.9

32.9



29.2

28.9


30.3

29.9






-45-


3.1.7 The Power of the Exact Tests



The approximate power of the exact test concerning a is derived

in this section. Similar expressions can also be developed for the

test for a2 and the test for a2. Our development is based on the

results in Hirotsu (1979, pp. 578-579). The reader is referred to

Appendix D.

Let T1 denote the power of the test for a2. Then from (3.56)





3/[(ris1) Fp] FT,r-1,(r-1)(s-1)-p Ha (3.78)



where H:a~ 0. Under Ha, w1wl is distributed like a random variable

of the form (see Box (1954b))


r-1
T= E (j 61)j, (3.79)
j=1


where the Xj's are independent chi-squared variates with one degree

of freedom, and A is the jth diagonal element of A* (j=1,2,---,r-1);

A* is defined in (3.35) and 61 in (3.51). The exact distribution of T

is complicated. To obtain approximate power values let



MSz3= V3*3 (3.80)
3= (r-1)(s-l)-p


H = yl/cf (3.81)
MS3/1b






-46-


h = cf 7,r-l,(r-1)(s-1)-p


where in this case
r-1
S(j+ 1)2
c j=1 (3.83)
c=1 r-1
E (j.+ 1)
j=l


Fr-1

f = L (3.84)
(A+1)2
j=1

2 2
= c (3.85)
=1- 02 2
1 a p A1, maxe


Then


L = P(H > hHa) P(Fff2 h)



+[p/{3(f +2)(f +4)B(lf,f2)}]


x(1 fh-(f + f2)/(fh) f


x[( +)( +) 2(f +f2)(f +4) (f + f2 + 2)(f+f2) (3.86)
x[(f+2)(f+4)- + 2
S1 + f/(fh) {1 + f/(fh)}2


where


(3.87)


f = (r-1)(s-1)-p






-47-


[r-1 )r-1 3
p + 1= -1) I (3.88)
"r-1 +)2
j=1


and B(ml,m2) is the beta function. We note that from (3.86) the power

of the exact test depends on the design parameters Al,max',l*,' r-

the level of significance; and the ratios a /a. and ,a2 of the

variance components.



3.1.8 Concluding Remarks



The key to the construction of the exact tests for the main

effects' variance components were the random vectors W1:3 and W2:3 in

(3.49) and (3.67), respectively. Both of these vectors represent a

linear combination between a function of y and a portion of the error

vector.

Ve note that C, in (3.49) and C* in (3.67) are not unique. Thus

the values of the test statistics F1 and F, (see (3.56) and (3.72),

respectively) will depend on the choice for C1 and C*, respectively.

However, the distributions of W1:3 and Y2:3 are invariant to these

choices. It is in this sense that our methodology is similar to

techniques involving resampling.

We also presented an exact test for the variance component

associated with the interaction effect. This test is similar to

Thomsen's (1975) procedure. Seely and El-Bassiouni (1983) point out






-48-


that Thomsen's (1975) test for a 2 is a Vald's test. Since our test

is based on the same degrees of freedom as Thomsen's test, we

conclude that our test for apl is also a Wald's test. Consequently,

F3 in (3.76) can be equivalently expressed as


R (a() |itsg,)/[(r-1)(s-1)-p]
F ( ) (3.89)
3- SSE/(n-(rs-p))


where R(aP) Iu,a,)/) represents the sum of squares for (a/3) adjusted

for p, a, and f/. Here, a and # are treated as if they were fixed

effects (see Seely and El-Bassiouni (1983, p. 198)).

Consider again wl:3 defined in (3.49). It is easy to verify that

Yl:3 can be written as

1
Y1:3 = Aa+ (3.90)


where


A =(A 0) (3.91)
(0 0


(see (3.35)), and


E~N(O, 61Is(r-1)p). (3.92)



b6 is defined in (3.51). Now, we have

1
rank(A2) = r-1






-49-


and (using similar notation to that of Seely and El-Bassiouni),



f= s(r-1)-p-(r-1) = (r-1)(s-1)-p



kW = r-1.



Thus, the test statistic F1 is a Wald's test based on the y1:3 model.

Likewise, we can also show that F2 is a Wald's test based on the W2:3

model.



3.2 The Unbalanced 3-Fold Nested Random Model



3.2.1 Introduction



The unbalanced 3-fold nested random model can be expressed as



Yijk = +ai 3ij +Tijk+Eijke (3.93)



(i=1,2,-. .,a; j=l,2,...,bi; k=1,2,...,cij; e=1,2,-. ,nijk),

where p is an unknown constant parameter; ai, ,ij' 7ijk are random-

effects associated with the first-stage, second-stage and third-stage

nested factors, respectively; and Eijke is the random error component.

We assume that ai, /ij, 7ijk, and cijke are independently distributed

normal random variables with zero means and variances a2, ao, 2a, and

2 respectively. We assume there are no missing cells. If n is the






-50-


total number of observations and c and b the total number of levels

of the third- and second-stage nested factors, respectively, then



n= E nijk
i,j, k


c= Ecij (3.94)
1,j


b= Ebi.
i


We are concerned with constructing exact tests for a2 and a2.

We again need to use resampling to achieve our goal. The results

given in Seely and El-Bassiouni (1983) will allow us to conclude that

our tests are Wald's tests. We note that an exact test for a2 is

known to exist based on comparing R(71p,a,fl) to SSE where R(7|p,a,13)

is the sum of squares for 7 adjusted for p,a, and P (treating a,6 and

7 as fixed effects); and SSE is the error sum of squares. That is, to

test Ho:a2=O vs. Ha: 2 O we would reject Ho for large values of Fy

where



SR(7l p,a,)/(c-b)
F7 SSE/(n-c)



Under Ho, Ft Fcb,nc (see, e.g., Searle (1971)). Resampling was not

needed to develop this test.

Now, in matrix notation, model (3.93) becomes


y = Pln+X -+Xt /XX37+ (


(3.95)






-51-


where y is the nxl vector of observations; in is a vector of ones of

dimension n; X1, X2, and X3 are known matrices of one's and zero's (see

(3.97) (3.99)); a, 0, and 7 are random vectors whose elements are

the ai, fij, and 7ijk, respectively; and e is the nxl vector of

errors. If ni and nij are defined as


ni .= nijk
j,k


nij= knijk
k


a
Xl= (D in.
i=1 1

b.
a 1
X,2= E in..
i=1 j=l ij


a b1 cij
3 -- $ 1T !nijk
X i=l je ke in
i=1 j=1 k=1 ijk


(i =1,2, -. ,a)

(3.96)

(i=l1,2, ,a; j=1,2, -- ,bi)


(3.97)



(3.98)



(3.99)


From (3.95) we have that


where


S 1(3Jn.)C+( Jn.j 2 +(i kJn ijk)2 + n .
i 1 i, j 1 aJk ijk


then,


(3.100)


y~N(pln, 0)






-52-


Now, define yijk as


nijk
Yijk=ifi E Yijkt


(i=1,2, .- ,a; j=1,2, ---,bi; k=1,2, -,cij).


(3.101)


Then from (3.93) we


(3.102)


Yijk = P ai +ij +ijk +ijk


(i=1,2,. .,a; j=1,2,...,bi; k=1,2,.--,cij), where


1j
Zijk- nijk


nijk
E ijk'"
9=1


In matrix form, (3.102) becomes


y =pc + + +A2 + Ic2 +


(3.103)


where y is the cxl vector of means; T is the cxl vector

Y = (En l,112',... abacaba)'; and A1 and A2 are defined as


a
Al =. 1
i=l i



a bi
A2= l 1c
i=1 j=l 1ij


(3.104)


(3.105)


have







-53-


where


ci= .cij
c..J
13


(i =1,2,-.-,a).


(3.106)


We note that


range(1c) C range(A1) C range(A2).


In fact,


Ic= Alla = A2lb




Al = A(.ilbi)


(3.107)


From (3.103) we have


where


(3.108)


and


(3.109)


a
A1 = AA'- E Jc.
1i= 1


y~N(plc, E)






-54-


a bi
A2=A2A/= = Jc. (3.110)
i=l j=l 1i




a bi Cij -1
K= $ Enik. (3.111)
i=1 j=l k=1


Besides the assumptions concerning the random effects, we will also

assume:



(i) The model used is model (3.93) (or model (3.95))

(ii) b-a>0 (b is defined in (3.94)) (3.112)

(iii) n>2c-1 (n and c are defined in (3.94))

(iv) c > 2b-1.



The last two assumptions (along with the others) are needed to insure

the validity of our procedures. Neither is overly restrictive.

Assumption (iii) can be satisfied if, for example, nijk>2 for all

(i,j,k) while Assumption (iv) will hold if, for example, cij22 for

all (i,j).



3.2.2. Preliminary Development



In this section we use the last c-1 rows of the cxc Helmert

matrix (see Searle (1982), p. 71) to transform model (3.103) into one

that has a zero mean vector. We also present some results concerning

the ranks of the model matrices in the transformed model.






-55-


Let HI be the (c-1)xc matrix resulting from the deletion of the

first row of the cxc Helmert matrix. HI has orthonormal rows and is

such that Hi1c=0c-l. Define the (c-l)xl vector u as


U = Hly.


(3.113)


Then,


E(u) = HIE(y) = pHl1c = Qc_1


Var(u) = H1A1Hlo + HlA2H O + fl c-1 + L


(3.114)


where


L = H1KH .


(3.115)


Recalling (3.104) and (3.105), it is easily seen that



rank(A,) = a


(3.116)


rank(A2) = b.


The next lemma provides information on the ranks of HIA1H' and H1A2H .






-56-


Lemma 3.3



(a) rank(HI iH') =a-l



(b) rank(Hix2H') =b-1



(c) rank(HliHA+H liH') = b- 1.



Proof:



(a) rank(HA1H,') = rank(H1AA'H')


= rank(HIA1)



< rank(A1) = a.


(see (3.116))


However, from (3.107), c =A1a. Since Ic = Qc-1 it follows

that the a columns of HIA1 are linearly dependent. Thus

rank(H1A1)
Frobenius Inequality (see Lemma A.1), we have



rank(H11Hi) = rank(H1Aj)


> rank(H1) +rank(A1) -rank(Ic)






-57-


=c-l+a-c=a-1.



Thus, rank(H1A1Hi ) = a-1.



(b) Similar to (a).



(c) Since range(A1)Crange(A2) we have that range(HiA1)Crange(H1A2).

Consequently,



range (HA1AAH') = range(H1iA) C range(H1A2) = range(HA2An H) .



Thus there exists a matrix B such that



HiAIH' = (HiA2H))B.


(Since both H1A1HA and H1iA2H are n.n.d.


B must be n.n.d.)


Hence,


rank(H1A1Hl + H1A2H) = rank[(HHA2H) B + HiA21H


= rank[Hi2HA (B+I)]


= rank(H1A2HI)


(B+Ihas full rank)


(see (b)).


=b-1






-58-


3.2.3 An Exact Test Concerning a2



Introduction. In this section we construct an exact test for

the variance component associated with the second-stage nested

factor. The idea is to first make a transformation of model (3.113)

based on resampling from the error vector. This transformation will

result in a model that is a special case of the model described in

Seely and El-Bassiouni (1983, p. 198). We will then apply the

results in Seely and El-Bassiouni's Equation 3.2 to develop the exact

procedure.

Utilizing resampling. Recall model (3.95). Let W be the

partitioned matrix



W=[1in:Xi:X2:X3] (3.117)



and let R be the matrix



R = I W(W'W)'. (3.118)



Since



range(1n) Crange(X) C range(X2)C range(X3),



(3.118) can be expressed as


R = In X3(XX3)x. 3-


(3.119)






-59-


The error sum of squares associated with model (3.95) is given by



SSE=y'Ry. (3.120)



It is known that R is symmetric, idempotent, and has rank n-c.

Furthermore, SSE/a. has the chi-squared distribution with n-c degrees

of freedom independently of y (see, e.g., Lemma B.6). We can express

R as

R = CrC' (3.121)



where C is an orthogonal matrix and r is a diagonal matrix whose

first n-c diagonal elements are equal to unity and the remaining c

entries are equal to zero. By assumption (3.112)(iii), we can

partition C and r as



C = [C1:C2:C3]

(3.122)

r =diag(I1,I,2,0)



where



I =c-1

(3.123)


'12 = n-2c+l > 0






-60-


and C1, C2, and C3 are matrices of orders nxI1, nxy2, and nxc,

respectively. Note that 71+72 = n-c = rank(R).

From these partitionings and the fact that (recall C is

orthogonal)



CiCi =I (i =1,2,3)

(3.124)

C'Cj =0 (i# j),



we obtain



R= CC' + CC'. (3.125)



Now, define the (c-1)xl random vector w as


1
L= +(AmaxI,1- L)2C'y (3.126)



where Amax is the largest eigenvalue of L (see (3.115)) and C' is the

nx~l matrix defined in (3.122). The matrix (AmaxI 1 -L) is p.s.d.

Hence (AmaxI 1-L)2 is well-defined with eigenvalues equal to the

square roots of the eigenvalues of (AmaxII1-L), which are

nonnegative. The first major result of this section is given in the

next theorem which provides some distributional properties for w.






-61-


Theorem 3.3



The random vector w is normally distributed with



E(y)= c-i


(3.127)


Var(y) = HiAH, 2 + H1A2H+2 6Ic_-


where 6 = + Amax27



Proof:



The fact that w is normally distributed is clear. We know that

E(u)=O (see (3.114)). From Lemma B.6 (ii) we have that Rln=O.

Since range(C1)C range(R), it follows that C(1n=Q (recall that R is

symmetric). Hence



E(w) = E(u) + (Amaxl1 L)2CE(y)


1
= O+ (AmaxI1 L)Yi (ln)p



=0.



Now, we claim that u and C y are independent in the expression given

for y in (3.126). To verify this recall that SSE is independent of y






-62-


(see Lemma (B.6)(v)). Consequently,


DOR = 0



where D is such that Dy=y, and 0 is defined in (3.100). Since

range(C1) Crange(R) it follows that



DQC1 = 0.



Thus y is independent of C'y. Since u is a function of y, u is also

independent of C'y as claimed. Hence,


1 1
Var(w) = Var(u)+(Amax1 L)2CQCi (AmaxI771 L)2.


(3.128)


From Lemma B.6(ii), RR=R2a2. Thus since range(C1)C range(R) and R

is idempotent we have



C, CI = CcClo2 = a11.



The last equality follows from (3.124). Therefore we can express

(3.128) as



Var(w) = Var(u) + (Amaxy L)O


= HI1AHl + H"a Hll + Ica + La + (Amaxli1 -L
I1 1 2 C- -t77






-63-


= HIA1Ha ~ + HiA2H 61 + 61_1



where 6= + Amaxa O



From Theorem 3.3 we note that model (3.126) can also be written as



w=B B1+B2P+ (3.129)


1
where B1= HA1; B2 = HA2; and = HI7+H + (AmaxI1 L)2C'i ~N(O,b6Il)

independently of a and /. This model in the form required by Seely

and El-Bassiouni (1983, p. 198). We are now in position to construct

an exact test for testing Ho:uO O= vs. Ha:0 0O.

The exact test. Let us start with some notation. SSE, will

denote the error sum of squares associated with model (3.129). That

is,



SSE= w'(Ic-1- [Bi:B2]([Bi:B2]'[B :B2])-[B1:2]', (3.130)



Also, let R,(Pla) denote the sum of squares for # adjusted for a

(treating a and 6 as fixed) in model (3.129). Thus



Rw( g) = '(I1- B(B'B)-B) -SSEw. (3.131)



Since range(B1) Crange(B2), (3.130) and (3.131) can be equivalently

expressed as






-64-


SSER = 'wl(I_ -B2(B'B,)-B



R( a) = wy B2(B(B2)-B B,1(B B1-B'.


(3.132)



(3.133)


The second and last major result of this section is given in the next

theorem.



Theorem 3.4



a) A g-inverse of B'Bj = A'.HH1Aj is given by (A'.HH1A')-= (A.Aj)-1

(j=l,2).



b) RV(PI Q)/6 _Xb-Xa under Ho- =0O



c) SSEW/6 ~ x2
Xc-b



d) R.(flIg) is independent of SSEW



Proof:


a) Since HiH'i=Ic_1 and H1lc=0c_1. It follows that H'HI=Ic- c

where Jc= Ic'. Also recall that 1 = Ala = A2b (see (3.107)).

Thus


A'H'HlAj = A'(Ic-Jc)Aj






-65-


= A'-Aj A' JA


= A'Aj -A' AjJ A'jAj







t a if j=1
t b if j=2.


(j=1,2),


(3.134)


Therefore,



A'HHA (A' jA )A'H'HA



=(A'-A A'jJtA'.Aj)(A'jAj)-'(A'Aj A'jAjJ tAAj)



=(It -A' jAJt)(A'Aj A'AjJtA jA


= A'A -A' A J A' A + A'. A J A' A J A'. A
=AjAj-JJ it JA J c AjJ J AjJ J


=A'Aj -JAjJ +A A'JcA JA'Aj


= A'. A -AJcA' +A'.Jc cA
=j A Cj C 2 A c JJ


= A'.Aj C
Si i i


= Aj(Ic- J)Aj = A' H'Aj.


where






-66-


Although (b), (c), and (d) follow from Seely and El-Bassiouni's

(1983) Equation (3.2), we nevertheless prove these results for

the sake of completeness.



b) First note that (3.133) can be expressed as


(3.135)


w'(P2 P)


where


(j = 1,2).


(3.136)


We need to verify that (P2-P1)(Var(w))/6 is idempotent of rank b-

a when = 0 (see Lemma (B.4)). Recall from Theorem 3.3 that



Var(w) = H1A1AH la + H1A2AH + 6c-1



Since Pj is idempotent of rank t-1 (see part (a) of this theorem,

Lemma (3.3), and (3.134)) and range(P1) Crange(P2) it follows

that there exists a matrix 0 such that



P1 = P20


==0'P2. (by symmetry)


Pj =HAj(A'Aj)-1'A'H'






-67-


Thus


P1P2= 0'PP, = OP2 = PI.


Consequently, P2-P1 is idempotent and has rank (b-1) -(a-1) =b-a.

Now, since range(H1A1,H') C range(HIA2AgAH) = range(P2), we have


(P1P2 = P1)


=(I- P ,) [HAA'Hoa + H1A2A He' + 6P2]


= (I -P1)H1A2AH'I +6(aI-P1)P


(3.137)


= 6(P2 P1)


(when a = 0).


Formula (3.137) follows since range(H1A1A'H')=range(P1) and

(I-PI)PI=O0. Consequently, (P2-P1)(var(y))/6 is idempotent of

rank b-a. Thus from Lemma B.4 we have


R( l)/6~Xb-a under Ho:92 =0.


c) SSEW in (3.132) can be expressed as


(P2- P1)(Var()) = (I -P)P2Var(yw)


SSEW = w'(Ic-1 P2)"






-68-


In the expression for Var(y), range(H1A1AH1) C range(HIA2AH')

=range(P2). Thus since (I-P2)P2=0 it follows that

(I-P2)(Var(w))/6=I-P, and therefore has rank (c-1)-(b-1)=c-b.
Thus from Lemma B.4,



SSE/~ c-_b



d) To show that R(/PIa) is independent of SSEw we need to verify

that

(P2-P )(Var(w))(Ic -P2)=0



(see Lemma B.2). From (c) we have



(Var(w))(I-P2) = (I P2)


Thus


(P2-P)(Var(w))= 6(P2-P)(I-P2) = 0.


The last equality follows since range(P1)Crange(P2).

Consequently R(/PI|) and SSEW are independent.



We can conclude from Theorem 3.4 that


Rw(1q)/(b-a)
F SSE/(c-b) b-ac-b
/3 SSE,/(c-b) b-a,c-b


(3.138)






-69-


under IHO:e2 =O. We would reject Ho at the v level of significance if

Ff>Fv,ba,c-b where Fu,b-a,c-b is the upper 100%l point of the

corresponding F distribution. The fact that we reject HO for large

values of our test statistic follows from the next lemma.



Lemma 3.4



a) tr[(I -P1)H1A2A'H'] >0



b) rank[(I -P)HAAH] = b-a, where (I-P1)H1A2A2H1 is part of the

expression for (P2-P)(Var(w)) in (3.137).



Proof:



a) Both (I-P1) and H1A2A H' are n.n.d. Thus tr[(I-P1)H1A2A H] >0

(Graybill (1983, p. 397)). However, since range(P1) Crange(P2)

=range(H1A2A''H) it follows that (I-P1)H1A2AH 0 if it were

zero then range(H1A2AAHl) would be contained in range(P1), a

contradiction). Hence tr[(I-P1)H1A2A H']>0 (see Graybill (1983,

p. 397)).



b) First note that



P2HIA2A'H' = HIA2AHi.


Thus, since P1P2 =P1






-70-


(I P1)H1A2A'H = (P2 PI)H1A2AH.



Consequently,



rank[ (I PI)HIA2AH'] = rank[ (P2 P )HiA2AH'l]



< rank(P2-Pi) = b-a.



Conversely, by the Frobenius Inequality (see Lemma A.1),



rank[(I PI)HIA2A'Hl] > rank(I P1) + rank(H1A2A'H') rank(Ic_ )



= c-a+b-1- (c-1)



= b-a.



Hence, rank[(I -P1)H1A2A~H'] = b-a. 1



If 0P3(>0) denotes the average of the nonzero eigenvalues of

(I-P1)HIA2A'H' then from (3.137)


b1-aI) = tr[(P2- P)Var(w) (b-a) = Bp/2 +6.



We note that 0 is an average since rank[(I-P1)HA2AH']= b-a by (b) of

this lemma. We now develop an exact test for ac.






-71-


3.2.4. An Exact Test Concerning ar



Introduction. In this section an exact test is constructed for

the variance component associated with the nesting factor. Similarly

to Section 3.2.3, we would like to use model (3.129) to develop our

exact procedure. Unfortunately, because range(B1)Crange(B2) in

(3.129), we have



R( I ) = wIc- B(B2 B2)-B SSEw



=0.



Thus another approach is necessary. The problem here is similar to

the problem we would have encountered if we had tried to use the

original model (3.95) to construct an exact test for ao. Namely

because range(X2) Crange(X3), R(31p,g,7)=0.

We overcame this problem by first averaging over the last

subscript, then transforming the resulting model into one that had a

zero mean vector. Finally, another transformation was made based on

resampling from the error vector to arrive at model (3.129). From

here we were able to derive an exact test for a .

The same idea can be used again to develop an exact test for oa.

Now, however, our starting model is model (3.129). This model is

similar to an unbalanced random two-fold nested model (without an

intercept term) in the sense that it enjoys all the same mathematical

and statistical properties of such a model.






-72-


Initial development. Our first step will be to "average over

the last subscript". This can be accomplished by considering the bxl

random vector w defined as



D= (BAB22)-B



= (A H'HIA2)A2Hy


(A~'A2)A Hl .


(3.139)


The last equality follows from Theorem 3.4(a).

Next we transform model (3.139) into one that has a "zero mean

vector." Although E() )=O, it is still convenient (for mathematical

reasons) to make this transformation. Let H* be the last b-1 rows of

the bxb Helmert matrix. Then H*H*'= Ib and Hlb= b-1 Define y

as the (b-l)xl vector


y= H*.


(3.140)


The mean and variance of y is provided in the next lemma.


Lemma 3.5



a) E(v) =Qb-1



b) Var(y) = Ha Jb.1 + b-1 +HI(A2Az)- H'6


(3.141)






-73-


Proof:


a) From Theorem 3.3, E(w) =Q_1.


Thus


E(y) = HE(D)


= H(A A2) -1AH'E(y) = .



b) Also from Theorem 3.3,



Var(w) = HlA c1a + HI A2Ha 1 + -1


Thus,


Var(y) =H*[Var(g)]H*'


= H(AA2)-'A H'[Var( )]HiA2(A A2)-'H '



= H*(A'A,) -1A [HIAH1H "a +A, 2H'0 + 6Ic-1_

x HiA2(AA )- H1


Now, from (3.107),


A1 = A2( i Jbi Therefore


HiAIH' = HIA2 1 bi H'I.
eJb)A''1


(3.142)






-74-


Also,

) A2H1A 2 1 1 I(A2A2) 2Ic c)A2



= H (A A2) -'[A A2 -A JcA21


= H(AA ) A -A JbA (from (3.107))


= H[Ib- JbA A2]


=H (recall that Hb = 0) (3.143)


Consequently, from (3.142) and (3.143),


Var(v) = [H i1Jbi HlJa+HA H*r +H *(A2A2)'A2H16]HA2(A2A2)1 H


= ( H Jb H'a + Hb -- + l(AA A)-1H *6.
'i=1 i *


:( i=1 b )H +b-l P T122) -1H6



Our last step will be to resample from the error vector. In

this case, the error vector is R*w (R*=Ic1 -P2 ) since the error sum

of squares for the w model (see (3.129)) is given by SSE =w'(Icl-P2)w .

Utilizing resampling. Noting that R* is symmetric, idempotent,

and has rank c-b, we can write it as






-75-


R* = C*F*C*'


(3.144)


where C* is orthogonal and F* is diagonal whose first c-b diagonal

entries are equal to unity and the remaining entries are equal to

zero. Furthermore C* and F* can be partitioned as



C* = [C:C :C C*]


(3.145)


r*=diag(I *,I *,0),
71 '72


where


'7 =b-1


(3.146)


2 = c-2b+ 1.



We note that +' =+ = rank(R*) and 9 >0 by (3.112)(iv). Also, C*, C*,

and C* are matrices of orders (c-l)x*, (c-1)x7*, and (c-1)x9,

respectively, satisfying


CIyCi = I


(i = 1,2,3)


(3.147)


CiC= 0 (i j).
1 3






-76-


Thus,



R* = C*C*' +C (3.148)



Define the (b-1)xl random vector r as


1
S= Y + (Amax L*)iCw (3.149)



where Amax is the largest eigenvalue of L* = H(A2A2)-'H' and
1
(A*axI *-L*)2 is a well-defined matrix with eigenvalues equaling the

square roots of the eigenvalues of (AmaxI *-L*), which are
771
nonnegative. The major result of this section is given in the

following theorem.



Theorem 3.5



The random vector r is normally distributed with a zero mean

vector and a variance-covariance matrix given by



Var(r) =H*( a Jbi)H' + 6*Ib-1 (3.150)
1i=1 E


where 6* = +Aax5.






-77-


Proof:



Clearly r is normally distributed. From Lemma 3.5 we know that

E(v)=Qb-1. E(w)=OQc_ from Theorem 3.3. Thus


1
E(r)=E(y)+(Amai -L*)C'E(W) (see (3.149))



=0.



We claim that v = HI = (A'A )'AH'i (see (3.140) and (3.139)) is

independent of C*'w in (3.149). To verify this consider


(A A2)-'AH':[Var(w) R* = (A'A2)-'A H'[Var(w)](I-P2)



=6(AA2)1'A'H(I-P2) (recall (3.127) and (3.136))



=0.



The last equality follows since range[(A/A2)l'A'H] = range(P2). Since

range(C*)C range(R*) it follows that



(AA2)-AWHI [Var(w) ]C= 0.



Thus v is independent of Cu'w as claimed. Consequently, from (3.149)


1 1
Var() =Var(v) + (A*axI L*)2C'[Var() ]C(A*axI L*)2
91 91






-78-


Now, recalling that [Var(y)]R*=6R* (see (3.127) and (3.136)),


C*'[Var(w) ]C* = C*'R*[Var(w) ]R*C*



= 6C*'R*C*


(from (3.148))



(R* is idempotent)


= 6C*ICb



=6Ib-1


Hence, from Lemma 3.5(b) and the fact that L* =H*(AA2) -'HI,



Var(r) = Var(y) + (AaxI L*)S
771


= H 1bi)i ) + Ibl + L5* + (AmaxI -L*)6


=H Jb a b-16
i=1 77






where 6* = o + max6.



We can infer from Theorem 3.5 that model (3.149) can be re-

expressed as


(3.151)


1
where ?=H 1 +H *+(AmaxI *-L*)2C *1* N(Q,6*Ibl)


independently of a


and T* = (AA2)- A'H'i*. The vector is defined in (3.129). Model


r=(H *ilb i+
-b1






-79-


(3.151) is similar to an unbalanced one-way random effects model with

zero intercept. We note that



rank(H H lbi )< rank( i bi)


But, since ( i 1bi a=ib and H b=Ob-1, it follows that

rank(H i=lb ) Si=1 i
Lemma A.1),



rank (H* ilb ) rank(H*) +rank( i b )-rank(Ib)
1 i i= l" / b)


=b-1 +a-b



=a-1.



Hence, rank(H i bi)=a-l. We have therefore just proven Lemma 3.6.



Lemma 3.6



In (3.151), rank(H i=(lb )=a-1.



We can now develop the exact test for ra.

The exact test. We now construct an exact test for HO:a=2=0

vs. Ha:2a~#O. To start with let F be the (b-l)xa matrix


a
F=HHJ Dlb.b (3.152)
ii1 1






-80-


Also, let SSET denote the error sum of squares for model (3.151).

That is,


SSET = '(Ib-1 -F(F'F)-F')r.


(3.153)


R,(q) will denote the sum of squares for the a effect in model

(3.151). That is


Rr(q) = r'F(F'F)-F'.


(3.154)


In the next theorem we find a generalized inverse for F'F and show

that R(go) and SSE, are independent and have scaled chi-squared

distributions (the former having a chi-squared distribution only

under HO: 2=0).



Theorem 3.6



(a) A g-inverse of F'F=( i 1Ib.i)1'H1 1bi.) is


a -
(F'F)- = b1.
i=1


(b) R(aQ)/6*a~a_-1, under Ho:a* =0



(c) SSE/6l* 2b
Xb-a


(d) r(qa) is independent of SSE,.






-81-


Proof:


The proof of Theorem 3.6 follows along much the same reasoning

as that for Theorem 3.4 and is thus omitted.


A test statistic for testing HO:oa2=0 vs. Ha:a2 0O is given by


SR(()/(a-1)
F SSE/(b-a)
a SSE,/(b-a)


(3.155)


Under HO, Fa- Fa-l,b-a Note that


(3.156)


E =-i] +6*-


where 0a is the average of the eigenvalues of FF'=Hi=~l(Jbi)t'' If

H* denotes bxb Helmert matrix, then


(3.157)


Consequently,


b i= i bi


= tr


b i 1 bi)b

b i=1 bi)lb


4i-b = Jb H* '
i ( i= b
Hi 1bbi


=H 1 11 b
-H*
Hi n






-82-


lit 1 i +( ( i ) i




= br tE)




Hence,



ba 1 b- i >0. (3.158)



Therefore we would reject HO:aO==0 for large values of F.a



3.2.5. Reduction of the Exact Tests when the Design is Balanced



Introduction. We now show that the exact tests for a2 and ao

given in (3.155) and (3.138), respectively, reduce to the usual ANOVA

tests when the design is balanced. The test statistic F7, given in

Section 3.2.1, is the usual ANOVA test for testing HgO:2=0O vs.

Ha: u' O.

When the design is balanced we have



b b* Vi



cijc* V(i,j) (3.159)


nijkn* V(i,j,k).






-83-


Consequently,


b= bi = ab*
i
1


c = c j =ab*c*
1,i


(3.160)


n =E nijk = ab*c*n*
i,j,k




Also, recalling (3.104), (3.105), and (3.111)



A1 = b*c* = Ia b* c*
1


A2=. 1* =Ia Ib* lc
1,Jc


K = E n*-1 =-- (Ia I (
i,j,k n b* c*)


Thus the matrix L defined in (3.115) becomes



L= HIKH' =1 1c-



It's largest eigenvalue is therefore Amax= The matrix L*,
(3.149), was defined as
(3.149), was defined as


(3.161)


(3.162)



used in


L*= H*( AA)-1 '.
1 2 -'H*/






-84-


When the design is balanced, L* becomes (see (3.161))



L* = H(c*Iab*)-1H'



= lb-1 (3.163)


The largest eigenvalue of L*, A*ax, is therefore -. The matrices P1
c
and P2 were defined in (3.136). When the design is balanced they

reduce to



P1 = H(IaJb* Jc*)H'

(3.164)

P2 = Hl(Ia Ib* Jc*)



where Js= s1s*

Reduction of the exact tests. Consider first the vector w

defined in (3.126). When the design is balanced w becomes


1
W= + (AmaxI- L)2CIy



=u (from (3.162))



=HI (see (3.113)). (3.165)





-85-


Using (3.163), the vector r, defined in (3.149), reduces to

1
S= y + (A*axI -L*)C /,
'r v-,:Y+ ma 1 1


(from (3.163))


(see (3.140))


= (AA )-'A'H
=H(A'A2) A2 1W


= -H (I,( I* 1*)H-w
C* b* & c


(see (3.139))


(from (3.161)). (3.166)


Since the design is balanced,


Ic- P2 H[Ia Ib* (c* -Jc*


and


P2-P P = H[Ia (Ib* -Jb*) Jc*


Consequently,


(see (3.135))


= y'IHI[Ia (Ib b*) -Jc* IHI


Rw(fl a) = w'(P2 P)w





-86-


=y{Ia (Ib*-Jb*) Jc*


(3.167)


The last equality follows since H'Hl=Ic-Jc and Jb*(Ib*-Jb) =0.
Also


SSEW = w(Ic1 P2)y


(see (3.132))


(3.168)


The last equality follows since Jc (I* -Jc) =0.
(3.153) and (3.154) we have


SSE, = r'(Ib-1 -F(F'F)F')r


='rb(Ib-l- -FF')r
b b*


Furthermore, from


(by Theorem 3.6(a))


= '(Ib-l-H*(IaJb*,)H')r


(see (3.152))


= r'H[Ia (Ib*- Jb*)1~'T


From (3.166) we can further reduce SSE, to


SSEr= i H lQ'H 'H Ia 00(b* -b*)I'H*QH'
C1w-_ n**, 711 b* -_jbP,) 1 'HrIW
SSEr -2- c,~1~~ -La (b


= y'HHI[Ia b* & )(Ic* c*)H


=y Ia Ib* ((Ic -Jc*) 4






-87-


= 'H2- ,IaS b* b b*)]Q.Hl


where, from (3.166),


r = Ia Ib* lc*


Since


Ia (Ib* -Jb*) Ia Ib-J*b*) c*


we have


SSEr = /'Hla ( )


= -i: (P2 P1) w.
c


Thus,


c*SSE' = R('3 g).



Similarly, we can show that


R7() = r'F(F'F)-F'r





C*- l
= c, PI w


(3.169)






-88-


Thus,


c*Rr(a) =y'H'P1Hly.



Since the design is balanced,



Y = 1(la I
n* Ib*( Ic* "n*Y


(3.170)






(3.171)


Thus (3.167), (3.168), and (3.170) can be expressed as, respectively,


and


R,( Ia) = 1*Ya (Ib -Jb*) Jc* & Jn*-



SSEWY = l aI* la c* Jc*) J*







Rr(a) = n c (Ia a)Jb* Jc* Jn*f


(3.172)



(3.173)







(3.174)


Therefore, from Khuri (1982), it follows that n*c*R,(g), n*R(/l|g)

=n*c*SSE,, and n*SSEw reduce to the usual ANOVA sums of squares for

balanced data.



3.2.6. A Numerical Example



The following data are taken from Bennett and Franklin (1954, p.

405) and has been made unbalanced for illustrative purposes. The






-89-


data represent a part of a statistical study on the variability of a

number of properties of crude smoked rubber. The figures given in

Table 3.2 represent measurements of the modulus at 700% elongation

made using 24 samples. On each of these samples at most 3

determinations of the modulus at 700% elongation were made, resulting

in a total of 53 observations. The three factors of interest are

supplier, batch, and mix. We assume each factor to be random. Yijki

will represent the Ith replicate determination from mix k, which is

derived from batch j of supplier i. The linear model in this

experiment is




YijkE =p+ai +I3ij +ijk+ ijkE'



where i=1,2,3,4; j=1,2,-..,bi; k=1,2; E=1,2,. *,nijk. In this

case b =2, b2=4, b3=3, and b4=3. Notice that cij =2 for all (i,j)

so that the design associated with this experiment is partially

balanced.

We start our analysis by testing for a mix effect. That is, we

test for Ho0:0=0 vs Ha:a2#0. The appropriate test statistic is F

given in Section 3.2.1. In the present case we find



F =1.3110

numerator d.f. =12

denominator d.f. =29

p-value =0.2650.






-90-


Thus there does not appear to be any significant variation due to the

mix effect.

To test for a significant batch effect we use Fp, the test

statistic developed in Section 3.2.3. For these data we find



Fp =0.3800

numerator d.f. =8

denominator d.f. =12

p-value = 0.9114.



No significant variation in the moduli of the 24 samples exist

among batches.

The test statistic Fa (see (3.155)) is used to test for a

significant supplier effect. Ve find



Fa =39.0024

numerator d.f. =3

denominator d.f. =8

p-value <0.0001.



There is highly significant variation in the moduli of the samples

among the four suppliers.

All calculations were performed using S-plus.






-91-


Table 3.2


An example of an unbalanced random 3-fold nested model


Supplier A B C D

Mix 1 2 1 2 1 2 1 2


Batch I








Batch II








Batch III








Batch IV


196

200


250

238


226


248

249


193








204

165

194



209

221


196

156


186

196





174

172





202

211

204


323

279

251


238

250





273

221


262

272





223

256

230


(Source: Bennett and Franklin (1954, p. 405))




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EXACT TESTING PROCEDURES FOR UNBALANCED
RANDOM AND MIXED LINEAR MODELS
By
ROBERT CALVIN CAPEN
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1991

To my mother and the memory of my father

ACKNOWLEDGMENTS
I would like to express my most sincere appreciation to Dr.
Andre' I. Khuri for his patience, understanding, and guidance during
the writing of this dissertation. Also, I would like to thank Dr.
Malay Ghosh, Dr. Ramon Littell, Dr. P.V. Rao, and Dr. Timothy White
for serving on my committee.
I am grateful to my fiancée, Debra Thurman, for all the love
and support she has provided me during these past two years. She
gave me the strength to continue.
I would also like to thank Jeff, Joe, Amber, and Julie for
their friendship and moral support.
Finally, I wish to acknowledge Leslie Easom for her superb
typing job — thank you very much.

LIST OF TABLES
Table Page
3.1 An example of an unbalanced random two-way model
with missing cells 44
3.2 An example of an unbalanced random 3-fold nested
model 91
3.3 Information needed to determine the vector y in
(3.257) for a three-way cross-classification model
in which Factors A and B are fixed while Factor C
is random 129
3.4 Information needed to determine the vector x in
(3.288) for a three-way cross-classification model
in which Factor A is fixed while Factors B and C
are random 139
3.5 A comparison of the powers of the exact tests for
in an unbalanced mixed two-way cross-classification
model 162
3.6 Numerical scores resulting from a comparison between
dyed cotton-synthetic cloth and a standard 186
3.7 Some useful quantities used in the construction of the
exact tests for the variance components utilizing
the methodology developed in Section 3.3.3 188
3.8 Expected mean square values for the cloth example ... 189
3.9 Results from the analysis of the variance components
for the cloth example 190
3.10 Some useful quantities used in the construction of the
exact tests for the fixed-effects utilizing the
methodology developed in Section 3.3.4 191
3.11 Summary of results for the cloth example 192
3.12 Performance values for three machines 194
- iv-

Table
Page
3.13 Results from the analysis of the variance components
for the machine example 195
3.14 Some useful quantities needed in the development of
the exact confidence intervals for the machine
example 196
3.15 Exact and approximate 95% confidence intervals for
the machine example 198
3.16 Some useful quantities used in the construction of
the exact tests for the four variance components
in the example 262
3.17 Results from our testing procedures 264
-v-

TABLE OF CONTENTS
Page
ACKNOWLEDGMENTS iii
LIST OF TABLES iv
ABSTRACT viii
CHAPTERS
ONE INTRODUCTION 1
TWO LITERATURE REVIEW 5
THREE EXACT TESTING PROCEDURES 13
3.1 The Unbalanced Random Two-Way model
with Some Cells Missing 13
3.1.1 Introduction 13
3.1.2 Preliminary Development 19
3.1.3 An Exact Test Concerning a2a 26
3.1.4 An Exact Test Concerning a2^ 36
3.1.5 An Exact Test Concerning 3.1.6 A Numerical Example 42
3.1.7 The Power of Exact Tests 45
3.1.8 Concluding Remarks 47
3.2 The Unbalanced Three-Fold Nested Random Model .... 49
3.2.1 Introduction 49
3.2.2 Preliminary Development 54
3.2.3 An Exact Test Concerning a^ 58
3.2.4 An Exact Test Concerning a2a 71
3.2.5 Reduction of the Exact Tests when the
Design is Balanced 82
3.2.6 A Numerical Example 88
3.2.7 The Power of Exact Tests 92
3.2.8 Concluding Remarks 98
-vi-

3.3 The General Unbalanced Mixed Model with Imbalance
Affecting the Last Stage Only 99
3.3.1 Introduction 99
3.3.2 Preliminary Development 100
3.3.3 Analyzing the Variance Component 110
3.3.4 Analyzing the Fixed Effects Parameter 119
3.3.5 Reduction of the Tests When the
Design is Balanced 141
3.3.6 The Power of the Exact Tests 143
3.3.7 An Alternative Approach to the Analysis of
the Unbalanced Mixed Model with Imbalance
Affecting the Last Stage Only 151
3.3.8 Numerical Examples 184
3.3.9 Concluding Remarks 199
3.4 The Random Unbalanced Nested Model with Imbalance
Affecting the Last Two Stages Only 201
3.4.1 Introduction 201
3.4.2 Notations and Preliminary Development 201
3.4.3 An Exact Test for 204
3.4.4 An Exact Test for 211
3.4.5 An Exact Test for cr\, o\, • • • , a\-2 227
3.4.6 Reduction of the Exact Tests When the
Design is Balanced 247
3.4.7 The Power of the Exact Tests 254
3.4.8 A Numerical Example 261
3.4.9 Concluding Remarks 261
FOUR CONCLUSIONS AND DIRECTIONS FOR FUTURE RESEARCH 265
APPENDICES
A MATHEMATICAL RESULTS 269
B STATISTICAL RESULTS 280
C AN ILLUSTRATION OF THE UTILITY OF RESAMPLING 286
D HIROTSU’S APPROXIMATION 292
REFERENCES 294
BIOGRAPHICAL SKETCH 298
— vi i-

Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
EXACT TESTING PROCEDURES FOR UNBALANCED
RANDOM AND MIXED LINEAR MODELS
By
Robert Calvin Capen
August 1991
Chairman: Dr. André I. Khuri
Major Department: Statistics
Although much attention has been given to the problem of
estimating variance components in unbalanced random and mixed linear
models, the same cannot be said about the development of exact testing
procedures for the aforementioned quantities. One reason for this is
the mathematical complexity of the problem. Even in some balanced
random or mixed models, no exact analysis of variance tests exist for
testing certain variance components. Consequently, approximate
methods, such as Satterthwaite’s procedure, are typically relied upon
to furnish the required tests. Unfortunately, the exact distributions
of these approximate procedures are generally unknown, even under the
null hypothesis.
-vi11-

Recently, some progress has been made in the construction of
exact tests for the variance components and for estimable linear
functions of the fixed-effects parameters in unbalanced random and
mixed models. In this dissertation we extend some of these known
results and generate some new ones. Specifically, exact procedures are
developed to test hypotheses concerning the variance components in an
unbalanced random two-way model with some cells missing, an unbalanced
random 3-fold nested model, and an unbalanced random s-fold nested
model where the imbalance affects the last two stages only. Also,
methods are derived to test hypotheses about the variance components
and about estimable linear functions of the fixed-effects parameters in
a general unbalanced mixed model with imbalance affecting the last
stage only.
-IX-

CHAPTER ONE
INTRODUCTION
Although much attention has been given to the problem of
estimating variance components (see, e.g., Rao and Kleffe (1988)),
the same cannot be said about the development of exact testing
procedures concerning the variance components in unbalanced random
and mixed linear models. Perhaps one reason for this is due to the
mathematical complexity of the problem. Even for some balanced
models no exact analysis of variance (ANOVA) tests exist for testing
certain variance components. For example, in a three-way cross-
classification random-effects model, no exact procedures exist for
testing hypotheses concerning the variance components associated with
the main effects of the model. Also, when exact tests do exist for
balanced models, they generally cannot be used to test the
corresponding hypotheses in unbalanced models. The reason for this
is that for unbalanced models, the traditional test statistics used
with balanced data are, in general, no longer based upon independent,
chi-squared-type sums of squares.
Approximate methods, such as Satterthwaite’s (1946) procedure,
are typically relied upon to furnish tests for the variance
components when no such exact procedure exists. However, approximate
methods are just that -- approximate. Some may be relatively easy to
implement, and perform reasonably well in certain situations, but
-1-

-2-
they cannot be expected to always produce adequate solutions (see,
e.g., Cummings and Gaylor (1974), Tietjen (1974), and more recently,
Ames and Webster (1991)). Thus there is a clear need for the
development of exact testing procedures.
In this dissertation we have concentrated on developing exact
methodology for some unbalanced random and mixed linear models. Much
of this work has been inspired by the derivations given in Khuri
(1984, 1987, 1990), Khuri and Littell (1987), Gallo (1987), and Gallo
and Khuri (1990). Their results and those obtained in this
dissertation are based on certain transformations which reduce the
analysis of the unbalanced model to that of a balanced one. A
crucial theme in the construction of these transformations is a
scheme for resampling from the error vector. In this scheme, only a
portion of the error vector is utilized in making the transformation.
Consequently, the values of the resulting test statistics will not be
unique. However, the distributional properties of these statistics
will remain unchanged no matter what portion of the error vector is
used in the transformation. It is in this sense that our methodology
is similar to procedures involving resampling. Appendix C provides
an illustration of the utility of resampling as a means of
“counteracting” the imbalance in a design.
This thesis is divided into four chapters. Chapters one and two
give a brief introduction to the area of exact testing procedures in
unbalanced random and mixed linear models and a review of the
pertinent literature. The main contributions of this study are found
in Chapter three. This chapter is divided into four sections. In

-3-
the first section we develop exact tests for the variance components
in a random two-way model with some cells missing. Section two deals
with the unbalanced random 3-fold nested model. Here the imbalance
is assumed to affect all stages of the design. In the last two
sections we concentrate on more general models. The general mixed
model with imbalance affecting the last stage only is analyzed in
Section three. While in Section four we obtain exact procedures for
testing the variance components in a random s-fold nested model with
imbalance affecting the last two stages only. Finally, Chapter four
summarizes our findings and offers directions for future research.
Throughout this work we will try to motivate, as much as
possible, the reason behind each transformation we use. At times,
the mathematical derivations may become “intense” in the sense that
there may be pages containing nothing but equations. Unfortunately,
there seems to be no way to avoid this. A numerical example is given
at the end of each section in Chapter three to illustrate the
proposed methodology. These examples also serve as useful (and
probably needed) “rest stops” between pages full of formulas.
The following notation will be used throughout this thesis:
• rank(A) will denote the rank of the matrix A
• range(A) will denote the range space of the matrix A, that
is, the vector space spanned by the columns of A.
• tr(A) will denote the trace of the matrix A.
• A- will denote a generalized inverse of A, that is, any
matrix which satisfies AA A = A.

-4-
A O
O B
This can also be
diag(A, B) will denote the matrix
represented by A®B where “©” is the direct sum operator.
A®B will be used to represent the (right) direct (Kronecker)
product of two matrices. Briefly, if A is nxm and B is pxq
then A®B is the np x mq partitioned matrix whose (i,j)
• \ th
matrix is a—B where a^j is the (i,j)^ element of

CHAPTER TWO
LITERATURE REVIEW
The first paper that addresses the issue of exact tests for
variance components in unbalanced models is due to Wald (1940). Wald
constructed an exact test for in the random one-way model:
yij=P + ai+cij (2.1)
(i=1,2,•••,p; j=1,2,•••,m.), where p is an unknown constant
parameter; o> and • are independent, normally distributed random
variables with zero means and variances u2a and respectively. He
later extended this result to the general random-effects model
without interactions. Wald (1947) also considered the mixed model
y = Xg + Zb + £
(2.2)
where q is a vector of fixed effects; b~N(Q, tr^I) independently of
f~N(0, construct an exact test for
As pointed out in Spjptvoll (1968), Wald’s idea cannot be used to
test variance components in all linear models. To illustrate this
phenomenon, Spjptvoll considered the random two-way model. Based on
-5-

-6-
a modification of Wald’s method he was able to derive an exact test
for the variance component associated with the interaction effect.
However, only under the assumption of additivity could he construct
exact tests for the main effects variance components.
By modifying an approach suggested by Graybill and Hultquist
(1961), Thomsen (1975) was able to construct exact tests for the
variance components in an unbalanced two-way random model. Thomsen
did this by first making an orthogonal transformation as outlined
below. Let y be the vector of cell means, then in matrix notation we
have (see Thomsen (1975, p. 258)):
y — Plrs + + B2/? + Irs(o^) + £ (2-3)
where lrs is the vector of ones of dimension rs; Irs is the rsxrs
identity matrix; Bj and B2 are defined as
Bj — Ir ® ls
= lr ® ^s
(2.4)
and r is the number of rows and s is the number of columns in the
design. Since Aj = BjBj7 and A2 = B2B27 commute, there exists an
orthogonal matrix P which simultaneously diagonalizes Aj and A2.
Further, the first row of P may be chosen to be (l/VTs)lpS. Making
the transformation

-7-
z = Py (2.5)
results in Thomsen’s “semi-canonical” form. Thomsen used the last
rs-1 elements of z to derive exact tests for (the variance
component associated with the interaction effect) and for (the main effects variance components). However, as in Spj0tvoll
(1968), the development of the latter two tests rested on the
assumption that <7^ = 0. Also, Thomsen assumed the layout contained
no missing cells. (In a later section he dropped this assumption,
but still assumed the design was connected.)
In 1983, Seely and El-Bassiouni returned to Vald’s variance
component test. In their paper, necessary and sufficient conditions
were given under which the Wald test can be used in mixed models.
They also developed a uniqueness property of Wald’s test. Thus based
solely on matrix ranks, one can determine if a particular variance
component test is a Wald’s test. They applied their notion to the
tests developed by Spjptvoll (1968) and Thomsen (1975). They
concluded that the tests derived by Thomsen are indeed Wald’s tests,
while only the test for interaction developed by Spj0tvoll is a
Wald’s test.
Although both Spj0tvoll and Thomsen needed to assume additivity in
their development of exact tests for the main effects variance
components in an unbalanced random two-way model, Khuri and Littell
(1987) demonstrated that this assumption was unnecessary. Initially,
their work proceeded similarly to Thomsen (1975). However, they also
made a particular transformation that reduced the analysis of the

-8-
unbalanced model to that of a balanced one. Their idea was based on
a scheme for resampling from the error vector. After making this
transformation they were left with a model of the form ^see Khuri and
Littell (1987, p. 548)),
lj — PjBjO -t P^B2/? "t £
(2.6)
where g~N(0, + Amax ae and ^max Is largest eigenvalue of the matrix PjKPj' where
K = diag(n^j,nj2 > • • • ?nrs) an (see (2.5)). Furthermore, a, /? and c* are mutually independent. Bj
and B2 are defined in (2.4).
After partitioning w into u>=(u'Q, u/a, y'ap)'> Khuri and Littell
developed exact tests for a2a and <7^ based on (u'a, v'ap)' an
v'Qp)', respectively. In particular, their test statistic for testing
H0: F =
VgVa/(r-l)
v'a^a^i(r-l)(s-l)y
(2.7)
It should be noted that
rank(PjBj: P1B2) = r + s — 2
rank(PjBj) = r — 1
rank(P1B2) = s — 1.

-9-
Consequently, according to Seely and El-Bassiouni (1983), the tests
derived by Khuri and Littell (1987) for the main effects variance
components are Wald’s tests based on the w model in (2.6).
One attractive feature of a Wald’s test is that when applied in an
unbalanced one-way-random effects model, it may be interpreted as
being “almost equal to the most powerful invariant tests against
large alternatives” (Spjptvoll (1967, p.422)). This result is also
hinted at in Mostafa (1967). Also, under certain conditions, Mathew
and Sinha (1988) show that the Wald test for testing the variance
component associated with the unbalanced random two-way mixed model
without interaction is the uniformly most powerful invariant test
under a certain group of transformations. This is more fully
discussed in chapter four.
Besides the above mentioned results, other exact procedures do
exist in the literature. Khuri (1987) developed an exact test for
the nesting effect’s variance component in an unbalanced random 2-
fold nested model. The idea of resampling from the error vector was
also utilized in this paper. Verdooren (1988) also analyzed the
unbalanced random 2-fold nested model. He was able to derive an
exact test for pj for each given value of p2- Here pl is the ratio of
the nesting effect’s variance component to the error variance
component and p2 is the ratio of the nested effect’s variance
component to the error variance component.
Exact tests for the random and fixed effects in some unbalanced
mixed models were derived in Gallo (1987) and Gallo and Khuri (1990).
Hocking (1988) also considered the mixed linear model. Although he

-10-
was more interested in estimation than in testing, he did note that
his estimators could be used to develop tests of hypotheses for
designs with moderate imbalance. These tests, however, are not
exact. Ve only mention them since they may represent reasonable
alternatives to the methodology presented in Section three of Chapter
three.
Turning to more general models, Khuri (1990) developed exact tests
for random models with unequal cell frequencies in the last stage.
These designs include all cross-classification random effects models
with no missing cells. Utilizing resampling, Khuri was able to
construct a set of mutually independent sums of squares that are
distributed as scaled chi-squared variates. These sums of squares
were then used to test hypotheses concerning the variance components.
It should be noted that not all variance components can be tested
exactly using Khuri’s procedure. This should not be too surprising,
however, since this phenomenon occurs even for some balanced random
(or mixed) models (cf. Kendall and Stuart (1968), p. 83).
Using a different approach, Ofversten (1989) was also able to
construct exact tests for the variance components in general mixed
linear models. His idea was to use a result found in Allen (1974) to
orthogonally transform the model matrices into so-called layer
triangular form. The nice thing about this type of transformation is
that when combined with orthogonal row permutations it can be used to
reduce matrices to ones with full row rank. This technique is
repeatedly utilized in Ofversten’s work to produce exact tests for
the variance components in certain mixed models. However, when

-11-
certain rank conditions are not satisfied this approach cannot be
used to produce the exact procedures. When this occurs, Ofversten
relies on resampling to develop his exact tests. Ofversten’s
methodology can be used in any unbalanced mixed model to construct
tests for the variance components. However, it is not altogether
clear as to how to adapt his methods to designs not examined in his
dissertation. More is said about Ofversten’s results in Chapter four.
Partial motivation behind the need to develop exact testing
procedures can be found in Tietjen (1974) and Cummings and Gaylor
(1974). Both of these papers considered the random 2-fold nested
model. In his simulation study, Tietjen showed that the approximate
test based on Satterthwaite’s (1946) procedure for testing the
nesting effect’s variance component can be markedly off in
approximating the true p-value. This led him to recommend the use of
the conventional F-test (that is, the test used when the data are
balanced) as an approximate test for this hypothesis.
Cummings and Gaylor (1974) considered various 2-fold nested
designs in their simulations. These designs were specifically chosen
so that the mean squares involved in the approximate (Satterthwaite)
F-test were either dependent, failed to be distributed as scaled chi-
squared variates, or both. They demonstrated that when the mean
squares were dependent, the estimated test size was less than the
stated test size and thus, the approximate test was conservative.
When the mean squares were independent but did not have chi-squared
type distributions, the opposite occurred. The estimated test size
was greater than the stated test size. When the mean squares were

-12-
both dependent and not distributed as scaled chi-squared variates, a
“counter-balancing” effect was observed. The approximate F-test in
this case seemed to perform well. However, this counteracting effect
may not occur in higher level designs. As Cummings and Gaylor (1974,
p. 771) state “there is a possibility the test size disturbances due
to mean square dependence and nonchi squaredness will provide some
reinforcement.”
In the next chapter we extend the work of Khuri (1987, 1990),
Khuri and Littell (1987), and Gallo and Khuri (1990). The idea of
resampling from the error vector is vital in our development and is
used repeatedly throughout this dissertation. The reader is referred
to Appendix C where the utility of this approach is illustrated.

CHAPTER THREE
EXACT TESTING PROCEDURES
3.1. The Unbalanced Random Two-Way Model with Some Cells Missing
3.1.1. Introduction
In this section we will consider the unbalanced random two-way
model with some cells missing. Specifically, we will concentrate on
the following model:
yijk = p + ai+/?j + (a^)ij+fijk
(3.1)
((i,j)€f; k = 1,2, • • ■ ,ni j); where /j is an unknown constant parameter;
, ¡3j, and are independently distributed normal random
variables with zero means and variances er^, respectively; f is the set of subscripts, (i,j), which label the
nonempty cells of (3.1). That is,
?={(i,j): l 0}
(3.2)
where r is the number of rows and s the number of columns of the
design. Note that J is a proper subset of IXJ, where I =
{i : i = 1,2,•••,r}; J = {j: j = 1,2,••• , s} ; and X denotes the Cartesian
product.
-13-

-14-
In matrix form, (3.1) becomes
y — ^2^ + ^3(^) + i
(3-3)
where y is the vector of observations of dimension n= n- •; ln
(i,j)€J J
is a vector of ones of dimension n; Xj, X2, and X3 are known matrices
of zeros and ones of orders nxr, nxs, and nx(rs-p), respectively,
where p is the number of empty cells; g, /?, and (a/3) are column
vectors whose elements are the a^, /3j, and (a/3) — , respectively
^(i,j)6Í)^; c is the nxl vector of errors. Define
1 1J
y i i = nT-r £ y? ik
J ij k=i J
(3.4)
for all (i,j)Gf. From (3.1) we obtain
yij = /i + «i + ^j + (^)ij + fij
(3.5)
((i , j) G ¡f), where
(3.6)
for all (i>j)GÍ. In matrix form, (3.5) becomes
y — /^Irs-p + \-20-\- Irs-pí*^) “I" £
(3.7)
where y is the (rs-p)xl vector of (filled) cell means; Irs_p is the

-15-
identity matrix of order rs-p; Aj and A2 are known matrices of zeros
and ones of orders (rs-p)xr and (rs-p)xs, respectively. A1 and A2
can be expressed as
Ai
P
(3.8)
(Is-p1 • °(s-p1) xPl)Ei
(IS-p2: °(s-p2) Xp2^E2
(Is-Pr- °(s-pr)xpr)Er
(3.9)
where p, is the number of empty cells in row i (i = 1,2,•• • ,r);
0/ \ is a matrix of zeros of order (s-p-)xp- ( i = 1,2, • • • ,r) ;
(s-Pi)xPi v Ki
and Ej is a product of interchange matrices (i = 1,2,• • • ,r) . We take
(IS-Pi- 0(s_pi)xpi) = Is when pi=0 ( i = 1,2, • • • , r) .
Example 3.1
Consider the following two-way table, an “X” represents a filled
cell.

-16-
Then,
r = 3
s = 4
Pi = 0
P2 = 1
p3 = 2
P = 3
furthermore,
Ej — I4 — Eg
1
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
1
0
0
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
1
0
0
consequently,
Aj = diag(l4,l3,l2)

-17-
U
1
0
0
0
0
0
1
0
0
0
0
1
-*
( ^2:X 2)
Besides the usual assumptions concerning the random effects, we will
also assume:
(i) The model used is model (3.1) (or model (3.3))
(ii) Both r and s are at least two
(iii) rank(Aj: A2) = r + s — 1 (3.10)
(iv) P <(r-l)(s-l)
(v) n >2(rs-p)—min(r,s)
Assumption (iii) implies that the design is connected ^Gately (1962)
p. 46^, while the last assumption (along with the others) is needed to
insure that the exact tests for the main effects’ variance components
are valid.
The inclusion of missing cells into any design adds a degree of
difficulty to the analysis. The two-way random model is no
exception. One problem is that the matrices Bx = AjAj and B2 = A2A2 do
not, in general, commute. Bj and B2 will commute if all cells are
filled ^see Thomsen (1975) or Khuri and Littell (1987)^.

-18-
Eeconsidering Example 3.1, we have
Bj—diag(J4, J3, J2)
where >Ja=lalá• While
I4
(o?)
b2 =
(I3:0)E2
i3
(i3-9) e, („?)
( 12 : 0 )
(I2:0)E' (0v)
hence
J4
« ( P)
4 P)
BjB2 =
J3(I3:0)E
2 ^3
â– Ej( ^3 â–  Q)E-jf q' j
J2(I2:0)
J2(I2:°)E2 (
P) *
Since, for example,
while J2(I2:0)
=(J2:0), we see
is not symmetric. Thus Bj and B2 do not commute. Consequently no
orthogonal matrix exists that will simultaneously diagonalize B4 and
B2.

-19-
The next section lays the groundwork necessary for the
construction of the exact tests.
3.1.2 Preliminary Development
Recall (3.7). From the distributional properties of the random
effects, we can conclude that
y ~ N(plrs-p? £)
where
E — BjCTjj + B2 (3.11)
and
(3.12)
(3.13)
B2 — a2a2
The following lemma provides the ranks of Bj and B2.

-20-
Lemma 3.1
(a) rank(B1)= r
(b) rank(B2)= s
proof:
(a) rank(B1) = rank(A1A^) = rank(A1) . Since
rank(Aj) =(© ls-p.) (see (3.8)j
r
= 53 (1)= r, the result follows.
i = l
(b) Without loss of generality, we can assume that each column (and
row) in the layout includes at least one cell containing
observations. Thus, each column in A2 contains at least one “1”.
Hence there exists a permutation matrix II such that
ha2
I
s
$
for some (rs-p-s) xs matrix $. Consequently,
rank(B2) =rank(A2A2)
=rank(A2)

-21-
= rank(nA2)
= rank
xs
$
â–¡
Now, let
u=H1y (3.14)
where Hj consists of the last rs — p— 1 rows of the (rs — p) x (rs — p)
Helmert matrix ^see Searle (1982), p. 7lj. We note that
Hjlrg-p=Qrs_p_i• Then, it is easily verified that
E(H)=Qr8.p.1
(3.15)
Var(u) = + UfoJi'rf + lrs_p_^l0 + HXKH\a\.
The purpose of this transformation is to construct a vector which has
a zero mean vector. The ranks of and are examined in the
next lemma.
Lemma 3.2
(a) rank(H1B1Hj) = r — 1

-22-
(b) rankCHjBjHj) = s — 1
(c) rankiHjBX+HjBX) =r+s-2
Proof:
(a) rankiHjBjHj) = rC^AjAjHÍ)
= r(H1A1)
However,
^llr — ^®^-s-p^)ir — lrs-p‘
Thus,
HjAjlr — Hilrs_p — Qrs_p_i •
Therefore,
rank(H1B1Hj) Conversely, by the Frobenius Inequality (see Lemma A.l), we have

-23-
rank(H1B1H/1) =rank(H1A1)
> rank(Ha) + rank(A1) — rank(Irs_p)
= rs — p—1+r—(rs — p) = r — 1.
Consequently, rank(H1B1H,1) = r — 1.
(b) Similar to (a).
(c) rank(H1B1H,1 + HjBjHj) < rank(H1B1H'1) + rank(H1B2H,1)
= r + s - 2
^from (a) and
Conversely, the Frobenius Inequality gives us
rankíHjBjH'! +HJB2H;) = rank^Bj + B2)Hi)
= rank^Hj (Aj A^ + A2A2
= rank! H
(Hi[Ai:A2]
K
a;
= rank^H1[A1:A2])
> ran k(Hj) +rank^[Aj:A2]^ — rank(Irs_p)

-24-
= rs — p — 1 + r + s — 1 — (rs — p)
= r + s — 2
^by assumption, rank ^[Aj: A2]j = r + s — 1 ^see (3.10)jj.
â–¡
The next step in our analysis is to simultaneously diagonalize
HjBjHJ and HjB^i. Because Bx and B2 do not, in general, commute,
neither will HjBjHj and H^Hj. So we cannot achieve simultaneous
diagonalization by way of an orthogonal transformation. However,
both HjBjHj and are n.n.d. Consequently a nonsingular matrix,
M, of order (rs — p — 1) x (rs — p — 1) , will exist that will diagonalize
both HjBjHj and HjB2Hj (see Lemma A.4). Define
Aj = MHjBjHjM'
(3.16)
a2 = mh1b2h/1m/
(3.17)
Both and A2 are diagonal matrices of order rs—p—1. By Corollary
A.4.1 of Lemma A, we can express Aj and A2 as
Ai - diag(lr_1, 0S_15 °(r_i)(s_i)_p) (3.18)
A2 = diag^Or_1, Is_l5 °(r_i)(s_i)_p)* (3.19)
Consider, then, the (rs-p-l)xl vector y defined as

-25-
v = Mu.
It is easily shown that
E(v) = 0
(3.20)
(3.21)
Var(v) = + A2 where
F = MM'
(3.22)
G = MHjKHjM'.
Ve can partition y into
Y = (Yi ? Y2> Yá)'» (3.23)
where v. consists of the r—1 elements of y whose covariance matrix is
free from matrix is free from and y3 consists of the (r —l)(s —1)—p
elements of y whose covariance matrix is free from both and cr^.
Hence

-26-
E(v) = O
(3.24)
Var(v) =diag(Ir l<7^, Is_1 This result follows from (3.18), (3.19), and (3.21).
We are now ready to develop our exact tests. To begin with, we
# o
develop an exact test concerning aa.
3.1.3 An Exact Test Concerning a2a
Initial development. In this section an exact test for testing
Ho *4
= 0 vs. Ha
4#
0 is
derived by ut
ilizing the
subvectors vx and y3
of V
in (3.23).
To
begin
with,
partit
ion F and G
^see (3.22)^ as
F11
F12
F13
F =
F12
F22
F23
(3.25)
and
F13
F23
F33
G11
g12
G13
G =
G12
G22
G23
(3.26)
G'l3
G23
G33
where Fu(Gn), F22(G22), and ^33^33) are, respectively, (r —l)x(r —1),
(s-l)x(s — 1), and ^(r - l)(s - 1) - p)x^(r - 1) (s - 1) — p) matrices. Also,
let y1.3 be the (s(r —1)—pjxl vector defined as

-27-
-1:3 =
(3.27)
where Vj is (r-l)xl while v3 is ^(r — 1 (s — 1) — pjx 1. Then, from
(3.21) we have
E(y1:3) = Q (3-28)
Var(Y1;3) = Aj where
0
0(r-l)(s-l)-p
(3.29)
^11 ^13
^13 ^33
(3.30)
:3 —
Jll
GÍ
13
J13
J33
(3.31)
From (3.22) notice that since both F and G are positive definite
(p.d.), F1;3 and G1:3
are also both p.d.

-28-
Now, there exists a nonsingular matrix N or order s(r —1)—p such
that (see Lemma A.3)
and
NF1:3N' = I
s(r-l)-p
(3.32)
NAjN' = A15
(3.33)
where A1 is a diagonal matrix whose diagonal entries are solutions 6
to
|Ai-0F1:3| =0. (3.34)
Ve note that if Fj1.3 represents the first r —1 rows and columns of F1;^3
then the nonzero solutions to (3.34) are exactly the eigenvalues of
T?ll
M :3*
Without loss of generality we may express Aa in (3.33) as
Ai =
Aí
0
0
(3.35)
where A* is a diagonal matrix of order r—1 whose diagonal entries are
the eigenvalues of Fj1^. (If Ax is not in this form then we can find
an orthogonal matrix P such that
PAjP7 =
A1
0
0
0

-29-
Thus, letting
N* = PN
we see that
N*AXN*' = PNAjN'P'
= PAjP' (see (3.33))
A* 0
0 0
â– )
The purpose of introducing the matrix N was to simultaneously
diagonalize and F1;3 in the expression for Var(y1;3) given in
(3.28).
Define the s(r—1)—p random vector x1;3 as
-3
= N
= Ny1:3
(3.36)
where Xj is (r-l)xl and x3 is ((r — 1)(s — 1) — p)x 1. Then
E(*1;3) = Q
(3.37)
Var(x1;3) = A1^+Is(r_1)_p^ + L1
-30-
This follows from (3.28), (3.32), and (3.33). Lj is the ^s(r-l)-p^
xis(r-l)-p) full rank matrix
l1 = ng1:3n'.
(3.38)
Ve can equivalently express Var(x1:3) in (3.37) as
Because L1 is not diagonal, x1 and x3 are not independent. To achieve
independence between xx and x3, we resample from the error vector.
The results of Lemma B.6 are utilized heavily in the forthcoming
development.
Utilizing resampling. Because model (3.1) or (3.3) is a special
case of model (B.l) the error sum of squares, SSE, associated with
this model can be expressed as
SSE = y'Ry
(3.40)
where R is the nxn symmetric, idempotent matrix
R=In-V(V'V')-V'
(3.41)
where
(3.42)

-31-
and Xj, X2, and X3 are described in (3.3). Since
range(ln)C range(X3)
range(X-)C range(X3) (i = 1,2),
(3.41) can also be written as
R = In-X3(X'X3)-1X'
In (i,f)e/Jnij/nij)
(3.43)
This last equality follows since
X3 =
(i,j)€í" Ü
(3-44)
Following Lemma B.6 (iv), there exists an orthogonal matrix C and a
diagonal matrix T such that
R = crc'.
(3.45)
Partition C and T as
C — [Cj:C2:C3]
(3.46)

where
jll = s(r — 1) —p
(3.47)
772 = n — 2(rs — p) + s.
We note that r]l + rj2 = n — (rs — p) = rank(R) and r]2>0 by (3.10)(v).
Cj,C2, and C3 are matrices or orders n x rj1, n x r/2 and nx(rs-p),
respectively. From (3.46), R can be written as
R = CjCj + C2C2. (3.48)
Now, consider the ^s(r —1)—pjxl random vector y1;3 defined to be
^1:3 = ^1:3 + (^l, max 177 j -Li)^iy (3.49)
1
where Aj5max is the largest eigenvalue of Lj and (Aj ,maxl;7 ~ Iq)2 a
symmetric matrix with eigenvalues equal to the square roots of the
eigenvalues of (A1>maxI,j — Lj), which are nonnegative.
Let w1;3 be partitioned just like x1:3 in (3.36), that is
(3.50)
where Wj and w3 are of orders r —1 and (r — 1)(s — 1) — p, respectively.

-33-
The major result of this section is given in the following
theorem.
Theorem 3.1
(a) E(wj)=0; E(w3)=0
(b) Wj and w3 are independent, normally distributed random vectors
and have the following variance-covariance matrices:
Var(wj) = +¿1Irl
Var(y3) =¿1I(r_1)(s_1)_p
where — aafí + Aj ? max <7f.
Proof:
(a) From (3.49) we have
^(^1:3) — :3) + (Aj ^1)2C1E(y)
(3.51)
From (3.37) we know that E(x1;3)=0. Since Rln = 0 ^see Lemma
B.6(ii)^ and range(C1) C range(R), we have that 1 n = Q. Thus the
result follows after noting that E(y)=pln ^see (3.3)j.

-34-
(b) Clearly Wj and w3 are normally distributed. From Lemma B.6(v) we
have that y is independent of SSE. Consequently
DfiR = 0
where D is such that Dy = y, and
fi = Var(y) =X1X'14 + X2X'^ + X3X'^/j + ^In (3.52)
^see (3.3)^. However, since range(C1) Crange(R) it follows that
DfiCi = 0.
Thus y is independent of Cjy. Since x1;3 is a function of y, it
must also be independent of C^y. Hence
Var(yi:3) =Var(il:3) + (Al,maxI7?1
Li)2 C1fiC1(A1 ;inaxI^i
(3.53)
From Lemma B.6(ii), RQR = R2ct2. Thus since range(Ca) C range(R)
and R is idempotent we have
CjnCi = C/1C1(T2 = The last equality follows from the fact that C = [Cj:C2:C3] is
orthogonal. Therefore, (3.53) can be expressed as

-35-
Var(y1:3) = Var(x1:3) + (Aj ,maxI^ - Iq)^
- Ai°a + Is(r-l)-p - Aiira +Is(r-i)-p( Consequently, from (3.35),
Var(wt) = A^a + Vr-l
Var( w3) = M(r_i)(s_i)-p
where 6^
+ Al, max17
2
e •
â–¡
As a result of Theorem 3.1 we obtain
(i) SS^yiCA^ + Mr-ir1“1~Xr-1 (3‘54)
(ii) SS3 = w^3/¿1~X(r.1)(s_1)_p (3.55)
(iii) SSj and SS3 are independent.
A test statistic for testing Ho: V3V3/((r-1)(B~1)-p)
(3.56)

-36-
Under Hn, F1~F , / w . Note that
0’ i r-1,(r-1)(s-1)-p
E(|^1) = ^l'Ta + (5i (3.57)
where 61 is the average of the eigenvalues of F^.-j ^see (3.35)^. Since
Fj;3 ^see (3.30)j is p.d. it follows that Fj'i-j is p.d. Thus F^.g is
p.d., implying that 0X>O. Consequently, we would reject HQ at the 7-
level of significance if
Fi >F7,r-l,(r-l)(s-l)-p
where F % is the upper 1007% point of the
7,r-l,(r-l)(s-l)-p
corresponding F-distribution.
Ve now turn our attention to the construction of an exact test
concerning a
3.1.4 An Exact Test Concerning <7^
Initial development. This section is concerned with the
construction of an exact test for <7^. Much of the development in
this section parallels that of Section 3.1.4 except that we now focus
on the subvectors y2 and v3 of y in (3.23).
Let y2:3 be the (r(s-l)-pjxl random vector
(3.58)

-37-
where y2 is (s-l)xl and v3 (recall) is ((r-1)(s-1)-pJx1. Similarly to
Section 3.1.3 we can make a nonsingular transformation, T, to obtain
a random vector x2. 3 = [x2: x3,]/ = Tv2; 3 with
^(-2:3) — 0
Var (x2.3) = A2^ + Ir(s_i ) _p*a/? + L2ae
(3.59)
where A2 = diag(A|, 0) and A2 is a diagonal matrix of order s-1 whose
diagonal entries are the eigenvalues of F^.g. F2*3 represents the
first s-1 rows and columns of F2?3 where
^2:3 ~
22
F'
r23
23
33
(3.60)
|see (3.25)^. Also, L2 is the ^r(s-l)-pjx^r(s-l)-p^ full rank matrix
L2 = tg2;3t/.
(3.61)
T is the nonsingular matrix that simultaneously diagonalizes F2;3 and
A2 where
A2 —
xs-l
0 0
(r-1)(s-1)-p
(3.62)
^see(3.24)j; and G
2:3
is defined to be

-38-
(3.63)
^see (3.26)).
Ve note that x3, defined in x2:3 above, is not the same as x3 in
(3.36) since a different nonsingular transformation was necessary to
simultaneously diagonalize F2;3 and A2.
Because x2 and x| are not independent
(see
(3.39))), we will again need to employ resampling to aid in the
construction of the exact test.
Utilizing resampling. Consider again the matrix R defined in
(3.41) or (3.43). Unlike the partitioning of C and T described in
(3.46), we now partition C and f as
C — [C*:Cj:C3]
(3.64)
T = diag(I *,I *,0)
V\ U2
where
V1 =r(s-l)-p
(3.65)
t]2 — n-2(rs-p) + r.
Note that Vi + *?2 = n-(rs-p) = rank(R) and tj2 > 0 by (3.10)(v). C*, C2,

-39-
and C| are matrices of orders nxi)¡, nx?72, and nx(rs-p),
respectively. From (3.64), we can express R as
R = CjCÍ'+ CjC^ (3.66)
see also
(3.45)).
Now define the ^r(s-l)-p)xl random vector w2:;
^2:3 — x2:3+(A2 max1^*
"l
(3.67)
2 max the largest eigenvalue of L2 and (A2 maxI * — L2)2 is a
’ ’ Vi
where A
symmetric matrix with eigenvalues equal to the square roots of the
eigenvalues of (A2 maxI * — L2), which are nonnegative.
’ ' 'h
Let w2;3 be partitioned just like x2;3, that is
^2:3 =
w,
Wl
(3.68)
where w2 and are of orders s-1 and (r-l)(s-l)-p, respectively.
Theorem 3.2 represents the major result of this section.
Theorem 3.2
(a) E(w2)= 0; E(w3) = 0
(b) y2 and are independent, normally distributed random vectors
and have the following variance-covariance matrices:

-40-
Var(w2) = A|ct^ + ¿2Is_1
Var(y*)=¿2I(r_1)(s_1)_p
where ¿2 = ffa/? + A2, max'7? •
(3.69)
Proof:
The proof of Theorem 3.2 is very similar to the proof of
Theorem 3.1 and is thus omitted.
â–¡
As a result of Theorem 3.2 we obtain
(i) SS2 = y'(A2V^ + ¿2Is_1)-1 (ii) SS3 = ~ (3-71)
(iii) SSS2 and SS3 are independent.
A test statistic for testing H0: F ‘¿'2^2/(s~1)
y3/y3/((r-1)(s"1)-p)
Under H0, F2~Fs_1,(r_1)(s_1)_p* Note that
— "F ^2
(3.72)
(3.73)
where 02 is the average of the eigenvalues of Fj1^.
Since F2;3 (see

-41-
(3.60)) is p.d. it follows that F2Í3 is p.d. Thus Fj1^ is p.d.,
implying that 02>O. Consequently we would reject H0 at the 7-level
of significance if F, >F , .w
& 2 7,s-l,(r-l)(s-l)-p
A test for the variance component associated with the
interaction effect is developed in the next section.
3.1.5 An Exact Test Concerning
In this section we develop an exact test for (3.23)). We note at the onset that this test is equivalent to
Thomsen’s (1975) test.
From (3.24)—(3.26) we have
E(y3)=0
Var(y3) = + G33 (3.74)
Since y is a function of y ^see (3.14) and (3.20)J and y is
independent of SSE ^Lemma B.6(v)), v is also independent of SSE. Thus
y3 is independent of SSE as it is a subvector of v. Also, it clearly
follows from (3.74) that
Y3(F’33^/? + G33^)'1Y3~X2(r_1)(s_1)_p.
Because SSE/cr^
Ha: ^n-(rs-p)
a test statistic for testing H0: (3.75)

-42-
-3^33 ¥3 /(
'(r-l)(s-l)-p)
SSE /
(n-(rs-p))
1
Under H0, F3
(r-l)(s-l)-p,n-(rs-p)'
Note that
v;g
-i.
3U33 X3
(r-l)(s-l)-p
= V2
a/? + (3.76)
(3.77)
where 63 is the average of the eigenvalues of G33F33. Since both G33
and F33 are p.d. 03 > 0 ^see Graybill(1983, p. 397)^. Hence we would
reject H0 for large values of F3. In this case, we did not need to
resample to construct the exact test.
3.1.6 A Numerical Example
An automobile manufacturer wished to study the effects of
differences between drivers and differences between cars on gasoline
consumption. Four drivers were selected at random; also five cars of
the same model with manual transmission were randomly selected from
the assembly line. Each driver drove each car a maximum of three
times over a 40-mile test course and the miles per gallon were
recorded. For technical reasons the drivers did not drive the cars
the same number of times. Unfortunately, unforeseen mechanical
problems forced the cancellation of some of the test drives. The
data is recorded in Table 3.1
We first test for a significant interaction effect using (3.76).
Ve find that

-43-
F3 = 0.7167
numerator d.f. =9
denominator d.f. = 22
p-value = 0.6885.
Thus there does not appear to be a significant interaction effect.
The results concerning the main effects variance components are now
presented.
For testing H0: use (3.56). For this data we find that
Fj = 261.1971
numerator d.f. =3
denominator d.f. =9
p-value < 0.0001
The test for the car variance component resulted in ^see (3.72)^
F2 = 71.2372
numerator d.f. =4
denominator d.f. =9
p-value < 0.0001
All calculations were performed using S-plus.

-44-
Table 3.1 An example of an unbalanced random two-way model with
missing cells
B(car)
A(driver)
1
2
3
4
5
25.3
28.1
24.8
28.4
1
25.2
28.9
25.1
27.9
30.0
26.7
33.6
36.7
31.7
35.6
33.7
2
32.9
36.5
31.9
35.0
33.9
31.8
32.9
27.7
30.7
29.7
29.2
3
28.5
30.4
30.2
28.9
31.5
4
29.2
32.4
31.8
30.3
29.3
32.4
30.7
29.9
(source: Dr. André I. Khuri)

-45-
3.1.7 The Power of the Exact Tests
The approximate power of the exact test concerning a2a is derived
in this section. Similar expressions can also be developed for the
test for cr2p and the test for Our development is based on the
results in Hirotsu (1979, pp. 578-579). The reader is referred to
Appendix D.
Let denote the power of the test for a2a. Then from (3.56)
= P
tfitfi/(r-l)
w^w3/[(r-l)(s-l)-p] - 7,r-l,(r-l)(s-l)-p' a
(3.78)
where Under Ha, WjWj is distributed like a random variable
of the form ^see Box (1954b)j
r-1
= E (AV*
.1=1 J
+ *l)x
j’
(3.79)
where the Xj’s are independent chi-squared variates with one degree
of freedom, and Aj is the j*" diagonal element of A* (j = 1,2, • • • ,r-l);
A* is defined in (3.35) and S1 in (3.51). The exact distribution of T
is complicated. To obtain approximate power values let
ms3 =
W3W3
(r-1)(s-1)-p
(3.80)
tt ... vivi/cf
MS3/¿x
(3.81)

-46-
cf t7,r-l,(r-l)(s-l)-p
(3.82)
where in this case
eVV + 1)2
j=l J
1 r-l
E (>V + 1)
j=i J
eV^ + i)
>1 J
E^A^ + l)2
j=l J
2 r \ 2
^a/} + Al ,maxaf
(3.83)
(3.84)
(3.85)
Then
tf1 = P(H>h|Ha)=P(Ff>f2>h)
+
p/{3(f + 2)(f + 4)B(if,If2)}
x
(^tr(f+f!)(tf
X
(f + 2) (f + 4) —
2(f + f2)(f + 4)
l+f2/(fh)
(f + f2 + 2)(f + f2)
{l + f2/(fh)}2
(3.86)
where
i2 =(r-l)(s-l)-p
(3.87)

-47-
P =
eVv + 1)
eVv + i)3
u=1 J
[j-1 J
eVv + i)2
j=l J
(3.88)
and B(m1,m2) is the beta function. Ve note that from (3.86) the power
of the exact test depends on the design parameters ^i}max’^l’‘ '‘ ’^r-1’
the level of significance; and the ratios variance components.
3.1.8 Concluding Remarks
The key to the construction of the exact tests for the main
effects’ variance components were the random vectors w1;3 and w2;3 in
(3.49) and (3.67), respectively. Both of these vectors represent a
linear combination between a function of y and a portion of the error
vector.
We note that Cx in (3.49) and C* in (3.67) are not unique. Thus
the values of the test statistics Fj and F2 ^see (3.56) and (3.72),
respectively^ will depend on the choice for Cj and C*, respectively.
However, the distributions of w1:3 and w2:3 are invariant to these
choices. It is in this sense that our methodology is similar to
techniques involving resampling.
We also presented an exact test for the variance component
associated with the interaction effect. This test is similar to
Thomsen’s (1975) procedure. Seely and El-Bassiouni (1983) point out

-48-
that Thomsen’s (1975) test for is a Wald’s test. Since our test
is based on the same degrees of freedom as Thomsen’s test, we
conclude that our test for <^aa is also a Wald’s test. Consequently,
F3 in (3.76) can be equivalently expressed as
Rita/?) I/*>«>/?)/[(r-l)(s-l)-p]
F3 = -^ -/ r (3.89)
SSE/(n-(rs-p)J
where R^(a/?) | represents the sum of squares for (a/?) adjusted
for p, a, and (3. Here, a and ¡3 are treated as if they were fixed
effects ^see Seely and El-Bassiouni (1983, p. 198)^.
Consider again w1;3 defined in (3.49). It is easy to verify that
w1;3 can be written as
l
w1;3 = A^g + ij (3.90)
where
|see (3.35)V and
(3.91)
lí~N(0, Ms(r_i)_p)- (3.92)
is defined in (3.51). Now, we have
2
rank(Aj)= r-1

-49-
and (using similar notation to that of Seely and El-Bassiouni),
fv= s(r—1)—p—(r-1) = (r-l)(s-l)-p
ky = r-l.
Thus, the test statistic Fx is a Wald’s test based on the w1;3 model.
Likewise, we can also show that F2 is a Wald’s test based on the w2:3
model.
3.2 The Unbalanced 3-Fold Nested Random Model
3.2.1 Introduction
The unbalanced 3-fold nested random model can be expressed as
Yijk£-^ + ai +^ij + Tijk + eijki
(3.93)
(i - 1,2,•••,a; j - 1,2,• • • ,bi; k — 1,2,• • • ,cij; Í— 1,2,•••,n¿jk),
where p is an unknown constant parameter; , P^y "^ijk are ran<^om_
effects associated with the first-stage, second-stage and third-stage
nested factors, respectively; and is the random error component.
We assume that , P±y ^ijk’ an<^ ei jk£ are in normal random variables with zero means and variances <7^, respectively. We assume there are no missing cells. If n is the

-50-
total number of observations and c and b the total number of levels
of the third- and second-stage nested factors, respectively, then
n = £ n
i.j.k
ijk
c
b = £bi .
i
(3.94)
Ve are concerned with constructing exact tests for cr^ and cr^.
Ve again need to use resampling to achieve our goal. The results
given in Seely and El-Bassiouni (1983) will allow us to conclude that
our tests are Wald’s tests. Ve note that an exact test for a^ is
known to exist based on comparing R(y\ n,a,(3) to SSE where R(j\ [i,a,¡3)
is the sum of squares for 7 adjusted for fi,a, and f3 (treating a,¡3 and
7 as fixed effects); and SSE is the error sum of squares. That is, to
test Ho:cr^, = 0 vs. Ha: where
p _R(7lA*»tt»fl)/(c-b)
1~ SSE/ (n — c)
Under H0, F7 ~ F(,_jJ n c ^see, e.g. , Searle (1971)J. Resampling was not
needed to develop this test.
Now, in matrix notation, model (3.93) becomes
y = Pin + Xi “ + X2^ + X3T + £
(3.95)

-51-
where y is the nxl vector of observations; ln is a vector of ones of
dimension n; Xj, X2, and X3 are known matrices of one’s and zero’s ^see
(3.97) - (3.99)); a, P, and 7 are random vectors whose elements are
the and Tjjk’ respectively; and c is the nxl vector of
errors. If n^ and n-j are defined as
then,
ni = £>ijk (i=l,2,---,a)
J 5 K
(3.96)
n|.= ^n—^ (i = l,2,---,a; j = l,2,-<-,b^)
j k J
^1 ~.® In •
1=1 1
(3.97)
X2 =
a bi 1
© ® ln. .
i=l j=l “ij
v a bi Cij ,
X3 = © © © ln.
i=lj=lk=l ijk
(3.98)
(3.99)
From (3.95) we have that
y ~ N (/il n 5 Í1)
where
^ — ( © dn • ) 1 1 i,j H 1, J, k ijk
(3.100)

-52-
Now, define as
1 "ijk
yijk - nijk 53 yi jk£
(3.101)
(i=1,2,•••,a; j=1,2,•••,b±; k=1,2,•••,cij). Then from (3.93)
have
yijk = M + «i+/?ij+7ijk + ^ijk
(3.102)
(i=1,2,•••,a; j = 1,2,•■•,bj; k=1,2,••• ,Cjj), where
1 nijk
cijk — nijk £ fijk«-
In matrix form, (3.102) becomes
y = nlc + Ajg + \2/3 + Ic7 + e
(3.103)
where y is the cxl vector of means; e is the cxl vector
7 = (7in ,7112, • • • ,^ab c ),j and Aj and A2 are defined as
a aba
a
A, = © lc.
1 i=l ci
(3.104)
. a bi
Ao — ® © If-
i=l j=l~Cij
(3.105)

-53-
where
c¡ = S°ij
(i — 1,2,•• • ,a).
(3.106)
We note that
range(lc) C range(A:)C range(A2) .
In fact,
le — ^íla — ^2lb
Al = A2(_e Ib.)-
1=1 1
(3.107)
From (3.103) we have
y~N(/4c> E)
where
E = Aao-^ -f A2^ + Ic<7^ +
(3.108)
and
Aj — AjAj — .0 J
i = l
(3.109)

-54-
^2 — ^2^2 —
bi
i—1 j=l 1J
(3.110)
a i 1J _i
K = © 0 © n- .
i = l j=l k=l JK
(3.111)
Besides the assumptions concerning the random effects, we will also
assume:
(i)
The model
used is model (3.93) (or model (3.95)^
(ii)
b-a > 0 (b
is defined in (3.94)j
(3.112)
(iii)
n > 2c — 1 (
n and c are defined in (3.94)j
(iv)
c > 2b — 1.
The last two assumptions (along with the others) are needed to insure
the validity of our procedures. Neither is overly restrictive.
Assumption (i i i) can be satisfied if, for example, nijk —^ for all
(i,j,k) while Assumption (iv) will hold if, for example, c^j>2 for
all (i,j).
3.2.2. Preliminary Development
In this section we use the last c —1 rows of the cxc Helmert
matrix ^see Searle (1982), p. 7lj to transform model (3.103) into one
that has a zero mean vector. Ve also present some results concerning
the ranks of the model matrices in the transformed model.

-55-
Let Hj be the (c-l)xc matrix resulting from the deletion of the
first row of the cxc Helmert matrix. Ha has orthonormal rows and is
such that H11C = 0 j. Define the (c-l)xl vector u as
u = Hjy.
(3.113)
Then,
E(u)=H1E(y)=/iH1lc = Oc_1
Var(u) =H1A1H'1^ + H1A2H,1^ + Ic_1^ + L where
L = HjKHj.
(3.115)
Recalling (3.104) and (3.105), it is easily seen that
rank(A^) = a
rank(A2) = b.
(3.116)
The next lemma provides information on the ranks of HjAjHÍ and HjAjH^ .

-56-
Lemma 3.3
(a) rank(HjAjHj) = a — 1
(b) rank(H1Á2H^) = b — 1
(c) rankCHjÁ^i + HjÁjHj) =b-l.
Proof:
(a) rank(HjAjHj) = rank(H1A1AjHj)
= rank(H1A1)
< rank(Aj) =a.
However, from (3.107), lc = A1la. Since H11C = Qc_i it follows
that the a columns of HjAj are linearly dependent. Thus
rank(H1A1) < a — 1 => rank(H1A1H^) < a — 1. Conversely, by the
Frobenius Inequality (see Lemma A.l), we have
rank(H1A1Hj) = rank(H1A1)
> rank(Hj) +rank(A1) —rank(Ic)

-57-
= c — 1+a — c = a—1.
(b)
(c)
Thus, rank(H1A1Hj) = a — 1.
Similar to (a).
Since range(A1) C range(A2) we have that range(H1A1) C range(HaA2) .
Consequently,
range (H1A1AjH^) = range(H1A1) C range(HjA2) = range(H1A2A2H^) .
Thus there exists a matrix B such that
HjÁjHj = (HjÁ2Hj)B.
(Since both HjAjHj and HjAjHj are n.n.d. B must be n.n.d.)
Hence,
(b)
(c)
ran
k (Hj AjH; + H^H'j) = rank[( HjA2Hj ) B + H^Hj]
= ran
k[H1A2Hj(B+I)]
= rank(HjA2Hj) ( B+I has full rank)
â–¡

-58-
3.2.3 An Exact Test Concerning
Introduction. In this section we construct an exact test for
the variance component associated with the second-stage nested
factor. The idea is to first make a transformation of model (3.113)
based on resampling from the error vector. This transformation will
result in a model that is a special case of the model described in
Seely and El-Bassiouni (1983, p. 198). We will then apply the
results in Seely and El-Bassiouni’s Equation 3.2 to develop the exact
procedure.
Utilizing resampling. Recall model (3.95). Let W be the
partitioned matrix
(3.117)
and let R be the matrix
R = I - V(W'V)_W/.
(3.118)
Since
range(ln) C range(Xx)C range(X2) C range(X3),
(3.118) can be expressed as
R=In-X3(X'X3)-1X'.
(3.119)

-59-
The error sum of squares associated with model (3.95) is given by
SSE = y'Ry. (3.120)
It is known that R is symmetric, idempotent, and has rank n-c.
Furthermore, SSE/cr^ has the chi-squared distribution with n-c degrees
of freedom independently of y (see, e.g., Lemma B.6). We can express
R as
R = CrC' (3.121)
where C is an orthogonal matrix and T is a diagonal matrix whose
first n-c diagonal elements are equal to unity and the remaining c
entries are equal to zero. By assumption (3.112)(iii) , we can
partition C and T as
C — [C1:C2:C3]
r = d i ag (177x, 1772,0 )
(3.122)
where
T11 = c-1
(3.123)
í72 = n-2c+l > 0

-60-
and Cj, C2, and C3 are matrices of orders n x , nxi¡2, and nxc,
respectively. Note that rj1 + r;2 = n — c = rank(R) .
From these partitionings and the fact that (recall C is
orthogonal)
(^ = 1 (i = 1,2,3)
C'iCj = 0 (i#j),
(3.124)
we obtain
R = C1C;+C2C^. (3.125)
Now, define the (c-l)xl random vector w as
y = y + (W^1-L)^,1y (3.126)
where Amax is the largest eigenvalue of L ^see (3.115)^ and is the
nxrjj matrix defined in (3.122). The matrix (Aj^xl,^ — L) is p.s.d.
l
Hence (AmaxIj7 — L)2 is well-defined with eigenvalues equal to the
square roots of the eigenvalues of (AmaxIjj — L) , which are
nonnegative. The first major result of this section is given in the
next theorem which provides some distributional properties for w.

-61-
Theorem 3.3
The random vector w is normally distributed with
E(y)=Qc_1
(3.127)
Var(w) = HjÁjHÍ^ + H1A2H,1 where 6 = a* +Amaxa^.
Proof:
The fact that uj is normally distributed is clear. We know that
E(u) = 0 ^see (3.114)j. From Lemma B.6 (ii) we have that Rln = 0.
Since range(Cj)Crange(R), it follows that Cjln=0 (recall that R is
symmetric). Hence
E(«) = E(u) + (AmaxI^ -L)*CjE(y)
= 0 + (AmaxIJji — L)^C1(ln)/j
= 0.
Now, we claim that u and C^y are independent in the expression given
for y in (3.126). To verify this recall that SSE is independent of y

-62-
^see Lemma (B.6)(v)^. Consequently,
DfiR = 0
where D is such that Dy = y, and fi is defined in (3.100). Since
range(C1) Crange(R) it follows that
=0.
Thus y is independent of C^y. Since u is a function of y, u is also
independent of C^y as claimed. Hence,
Var(y) = Var(u) + (AmaxI^ - L^ilC^ - L)*. (3.128)
From Lemma B.6(ii), RilR = R2 is idempotent we have
Cine, = 050,»? = <#,,.
The last equality follows from (3.124). Therefore we can express
(3.128) as
Var(y) = Var(u) + (A^I^ - L) = H1A1H1(ra + HjAjHjcr^ + + L^f + (^max^j — ^')cre

-63-
+ ¿I
C-
1
where 6 = + Amax From Theorem 3.3 we note that model (3.126) can also be written as
u — BjO -1 B2/i c* (3.129)
1
where Bj = HjAj ; B2 = H1A2; and e* = Hj7 + HjJ + (AmaxI,j — L)2Cjf ~ N(0,¿IC)
independently of a and /?. This model in the form required by Seely
and El-Bassiouni (1983, p. 198). We are now in position to construct
an exact test for testing H0:cr^=:0 vs. Ha:u"^^0.
The exact test. Let us start with some notation. SSEW will
denote the error sum of squares associated with model (3.129). That
is,
SSEy = 0/(1^ - [Bj: B2]( [B,: B2],[B1: B2] ■ B2]')y • (3.130)
Also, let Rw(/?|g) denote the sum of squares for /? adjusted for a
(treating a and /? as fixed) in model (3.129). Thus
Ry(^l«)=y,(lc_1-B1(B/1B1)-B'1)y-SSEy. (3.131)
Since range(Bx)Crange(B2) , (3.130) and (3.131) can be equivalently
expressed as

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SSEy = y'(l c _ 1 - B2 ( B' B2 )“Bi)y (3.132)
Ry(^ls) =y'(B2(B'B2)-B' -B1(BiB1)-Bi)w. (3.133)
The second and last major result of this section is given in the next
theorem.
Theorem 3.4
a) A g-inverse of BjBj = AjHjHjAj is given by (AjHjHjAj) = (^j^j) 1
(j = l,2).
b) \,(/?|a)/<5 ~ Xb_a’ under H0: c) SSEy/5~x2c_b
d) Rw(/?|g) is independent of SSEW
Proof:
a) Since HjHj = Ic_i and H11C = QC_^. It follows that H^H1 = IC—ijQ
where Jc = lcl(:. Also recall that lc = ^lla = ^2lb (see (3.107)^.
Thus
A'jH/1H1Aj=A/j(Ic-¿Jc)Aj

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= A'jAj-lA'jAjJtA'jAj (j=l,2),
where
t =
£
if j=i
if j=2.
Therefore,
(3.134)
= (A'jAj-¿A'jAjJtA,jAj)(A'jAj)-1(A,jAj-ÍA,jAjJtA,jAj)
= (It-ÍA'jAjJt)(A'jAj-¿A'jAjJtA'jAj)
= AjAj -?AjAjJtAjAj + ¿AjAjJtAjAjJtAjAj
= AjAj _§AjJcAj +^AjJcAjJtAjAj
= A jA j “^A j J°A j + ^2A jJCJCA j
= AjAj-^AjJcAj
= Aj(Ic-ijc)Aj = A'jHiH1Aj.

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Although (b), (c), and (d) follow from Seely and El-Bassiouni’s
(1983) Equation (3.2), we nevertheless prove these results for
the sake of completeness.
b) First note that (3.133) can be expressed as
y'(P2-P l)^ (3.135)
where
Pj=HlAj(AjAj) AjHi =
(3.136)
We need to verify that (P2 — Pj)^Var(w)j/(5 is idempotent of rank b-
a when cr^ = 0 ^see Lemma (B.4)^. Recall from Theorem 3.3 that
Var ( w) = HjAjAiH'a* + H^A'H'aJ + Hc_1
Since Pj is idempotent of rank t-1 (see part (a) of this theorem,
Lemma (3.3), and (3.134)^ and range(Pj) Crange(P2) it follows
that there exists a matrix 0 such that
P1 = P20
= 0,P2. (by symmetry)

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Thus
P1P2 = 0/P2P2 = 0,P2 = P1.
Consequently, P2 — Px is idempotent and has rank (b-1)—(a-1) = b-a.
Now, since range(H1A1A^Hj) C range(H1A2A2Hj) = range(P2) , we have
(P2 - Pj)(Var(w)) = (I - Px)P2Var(u) (?{P2 = Px)
= (I - Pj) [HjAjAiH'or» + H1A2A'H'1^ + 5P2]
= (I - P1)H1A2A2Hj = ¿(P2 —Pj) (when = 0).
Formula (3.137) follows since range(HjAjA^Hj)= range(Px) and
(I—P1)P1=0. Consequently, (P2 - Pj)^var(w)^/<5 is idempotent of
rank b-a. Thus from Lemma B.4 we have
Rw(/?l“)/¿ ~ Xb-a under H0:^ = 0.
c) SSEW in (3.132) can be expressed as
SSEy = y'(Ic_1-P2)y.

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In the expression for Var(w), range(H1A1A^H1)Crange(HjA2A2HÍ )
= range(P2). Thus since (I— P2)P2 = 0 it follows that
(I — P2)^Var(ui)'j/S = I — P2 and therefore has rank (c-1) — (b-1) = c-b.
Thus from Lemma B.4,
SSEy/<5 ~ x
2
c-b ‘
d) To show that Rw(/?|a) is independent of SSEW we need to verify
that
(P2-P1)(Var(y))(Ic_1-P2)=0
(see Lemma B.2). From (c) we have
(Var(w))(I — P2) = ¿(1— P2).
Thus
(P2-P1)(Var(^)) = ¿(P2-P1)(I-P2)=0.
The last equality follows since range(Pj)Crange(P2) .
Consequently Rw(/?|qO and SSE^ are independent. â–¡
Ve can conclude from Theorem 3.4 that
„ Ry(/?|g)/(b-a) ^
P SSEy/(c-b) ~*b-a,c-b
(3.138)

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under Hg:a^ = 0. Ve would reject Hg at the v level of significance if
F/?>Fi/,b-a,c-b where Fu, b-a,c-b is the upper 100"% point of the
corresponding F distribution. The fact that we reject Hg for large
values of our test statistic follows from the next lemma.
Lemma 3.4
a) tr[(I-P1)H1A2A'H/1]>0
b) rank[(I — P1)H1A2A2Hj] = b-a, where (I — Pj)H1A2A2H^ is part of the
expression for (P2 — Pj )^Var(w)) in (3.137).
Proof:
a) Both (I— Pj) and HjA2A2Hj are n.n.d. Thus tr[(I — P1)H1A2A2Hj] > 0
^Graybill (1983, p. 397)). However, si nee range(Pj) Crange(P2)
= range(HjA2A2Hj) it follows that (I — P1)H1A2A2Hj ^ 0 ^if it were
zero then range(H1A2A2Hj) would be contained in range(P1), a
contradiction). Hence tr[(I — P1)H1A2A2Hj] >0 ^see Graybill (1983,
p. 397)).
b) First note that
P2H1A2A'H,1 = HjAjAjHj .
Thus, since PjP2 = P2

-70-
(I - P,)Hja2a'h; = (P2 - Pj )HjA2A'h; .
Consequently,
rank[(I -PJHjAjA^H;] = rank[(P2 - PJHjA^H'J
< rank(P2 — Pj) = b-a.
Conversely, by the Frobenius Inequality (see Lemma A.l),
rank[ (I — Pj JHjAjAjHÍ ] > rank( I — Px) + ran^H^A^) — rank( Ic_^ )
= c-a + b-1 — (c-1)
— b-a.
Hence, rank[(I — P^HjAjAjHj] = b-a.
â–¡
If 0>0) denotes the average of the nonzero eigenvalues of
(I — P1)H1A2A2Hj then from (3.137)
E
b-£
= tr[(P2 - Pj)Var(w)]/(b-a) = 0p Ve note that 6p is an average since rank^(I — Pj)HjA2A2H^] = b-a by (b) of
this lemma. Ve now develop an exact test for

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3.2.4. An Exact Test Concerning a^
Introduction. In this section an exact test is constructed for
the variance component associated with the nesting factor. Similarly
to Section 3.2.3, we would like to use model (3.129) to develop our
exact procedure. Unfortunately, because range(Bx) Crange(B2) in
(3.129), we have
Ry ( 9 I 2 ) = y'(í c.-1 - B2 0B2B2 ) - SSEy
= 0.
Thus another approach is necessary. The problem here is similar to
the problem we would have encountered if we had tried to use the
original model (3.95) to construct an exact test for Namely
because range(X2) C range(X3) , R(/?| n,a,y) = 0.
We overcame this problem by first averaging over the last
subscript, then transforming the resulting model into one that had a
zero mean vector. Finally, another transformation was made based on
resampling from the error vector to arrive at model (3.129). From
here we were able to derive an exact test for The same idea can be used again to develop an exact test for
Now, however, our starting model is model (3.129). This model is
similar to an unbalanced random two-fold nested model (without an
intercept term) in the sense that it enjoys all the same mathematical
and statistical properties of such a model.

-72-
Initial development. Our first step will be to “average over
the last subscript”. This can be accomplished by considering the bxl
random vector w defined as
V — (^2^2) ®2^
= (A'HiHjA^-A'H'y
= (A'A^-U'Hiy. (3.139)
The last equality follows from Theorem 3.4(a).
Next we transform model (3.139) into one that has a “zero mean
vector.” Although E(w)=0, it is still convenient (for mathematical
reasons) to make this transformation. Let H* be the last b-1 rows of
the bxb Helmert matrix. Then HjH*7 = 1^and H*!.^ = 1 * Define v
as the (b-l)xl vector
y = HjO. (3.140)
The mean and variance of y is provided in the next lemma.
Lemma 3.5
a) E(v)=0bl
b) Var(v) =H^©iJbi)l1+'4+ Ib_1^ + H*(A'A2)-1H1+/6
(3.141)

-73-
Proof :
a) From Theorem 3.3, E(w)=Qc_^. Thus
E(v) =H*E(w)
= HÍ(A'A2)-1A'H'1E(y) =0.
b) Also from Theorem 3.3,
Var(w) = HjÁjíTq + HjAjHjcr^ + 6IC_j .
Thus,
Var(v)= H*[Var(w)]H*'
= H*( A2A2)_1A2Hj[Var(w) ]HaA2( A2A2)~1H*/
= HÍ(A'A2)-1A'H;[H1A1H'4 + H1Á2H/1^ + éIc_1]
x H1A2(A2A2)_1H*/
Now, from (3.107), Áj = A2^ _© ^A2. Therefore
HjA^HjA^Jh.y'H;.
(3.142)

-74-
Also,
Hí(A2A2)”lA2HíHlA2 = ^1 ( A2A2 ) _1 A2 ( 1C — ÜJ C ) A2
= H*(A2A2) ^L A2A2 — ¿A2JcA2J
= H1*(A'A2)-I[A,2A2-lA5A2JbA;A2] (from (3.107))
= Hf[Ib-ÍJbA'A2]
= Hj. (recall that Hjl^ = 0)
Consequently, from (3.142) and (3.143),
(3.143)
Var(v) = [H*( © J, Wa* + H^'H^ + H*(A'A2)-1A'H'1¿]H1A2(A'A2)-1H
xi=l i' H
.©idb.^M + Ib-ik + H^A'A,)-^.
â–¡
Our last step will be to resample from the error vector. In
this case, the error vector is R*w ^R*=Ic_j— P2 j since the error sum
of squares for the w model ^see (3.129)j is given by SSEW = w'(Ic_j-P2)w.
Utilizing resampling. Noting that R* is symmetric, idempotent,
and has rank c-b, we can write it as

-75-
r* = c*r*c
*7
(3.144)
where C* is orthogonal and T* is diagonal whose first c-b diagonal
entries are equal to unity and the remaining entries are equal to
zero. Furthermore C* and T* can be partitioned as
C* = [CÍ:C2*:C3*]
r* = diag(I *,I *,0),
Vl 72
(3.145)
where
rj* = b-1
r¡2 = c-2b + 1.
(3.146)
Ve note that rj* + = rank(R*) and > 0 by (3.112)(iv). Also, C*, C2,
and C| are matrices of orders (c-l)x7?*, (c-l)x7j2, and (c-l)x^*,
respectively, satisfying
C|'C? = I (i = 1,2,3)
(3.147)
C?'Cj = 0
(i ¿ J')-

-76-
Thus,
R* = C*C*' + CjCj'. (3.148)
Define the (b-l)xl random vector r as
r = Y + (A*axI +-L*)^C*'y (3.149)
ui
where A^ax is the largest eigenvalue of L*=H*(A2A2)_1H*/ and
l
(A*ax-^ * —L*)2 is a well-defined matrix with eigenvalues equaling the
’ll
square roots of the eigenvalues of (Aj^axI * — L*), which are
’ll
nonnegative. The major result of this section is given in the
following theorem.
Theorem 3.5
The random vector r is normally distributed with a zero mean
vector and a variance-covariance matrix given by
Var(r) = H*( (3.150)
where 6* = + A*ax$.

-77-
Proof :
Clearly r is normally distributed. From Lemma 3.5 we know that
E(v)=Qk_j. E(w)=Qc_-^ from Theorem 3.3. Thus
E(r)=E(y)-KA¡UxIv*-L*)2Cl*'E(y) (see (3.149))
= 0.
We claim that y = H* w = (A2A2) (see (3.140) and (3.139)) ii
independent of in (3.149). To verify this consider
(AiAa)-1AiH;[Var(tf)]R* = (A¿A2)-1AiHiCVar(tf)](I-P2)
= ¿(A^A2)_1A^H;(I-P2) (recall (3.127) and (3.136))
= 0.
The last equality follows since range[\A2A2)-1A2HjJ = range(P2) . Since
range(C*)Crange(R*) it follows that
(A2A2)-1A2H1[Var(w) ]C* = 0.
Thus y is independent of C*'w as claimed. Consequently, from (3.149)
Var(r) = Var(v) + (A*axI +-L*)2CÍ',[Var(y)]Cjt(A*axI +-L*)2
u l ui

-78-
Now, recalling that [Var(w)]R*=6R* (see (3.127) and (3.136)),
C*'[Var(w)]C* = C*'R*[Var(y)]R*C* (from (3.148))
= ÍCj'R*^
(R* is idempotent)
= 6 Cf'C?
b-1'
Hence, from Lemma 3.5(b) and the fact that L*=H*(A2A2) 1H*/,
Var(r) = Var(v) + (A*axI * - L*)<5
vi
= Hrf .© Jb.+ Ib_i& + L*¿ + (■Amaxl * - L*)«
' 1 = 1 i7 r '1
= Hi( .© Jb-)Hí, v i = l i7
where S* = a2a + X
max'"
â–¡
We can infer from Theorem 3.5 that model (3.149) can be re¬
expressed as
t — (h* © 1. V + I (3.151)
' i=l~Di7
where £ = H*/l + H*e* + (A*axI * -L*)2C*'e* ~ N(0,<5*I> -. ) independently of a
~ *71 ~ - D-l
and J* = (A^Aj)-^^^*. The vector e* is defined in (3.129). Model

-79-
(3.151) is similar to an unbalanced one-way random effects model with
zero intercept. We note that
rank^H* _® 1^ ^ But, since ^ ® 1^ ^la
rank^H* ® 1^ ^ Lemma A.1),
= 1^ and H*lk = Qk_i5 it follows that
Conversely, by the Frobenius Inequality
(see
ran
•® lb-)> rank(H*)+rank^
Albi)~rank(Ib)
= b-1 + a-b
= a-1.
Hence,
ranklH
*
i
i=r
.,)=
a-l,
Ve have therefore just proven Lemma 3.6.
Lemma 3.6
In (3.151), rankÍH* ® 1, j = a-l.
' i = l~Di'
We can now develop the exact test for
The exact test. We now construct an exact test for Hg:cr^ = 0
vs. Ha:o-^^0. To start with let F be the (b-l)xa matrix
(3.152)

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Also, let SSEr denote the error sum of squares for model (3.151).
That is,
SSEr = r,(Ib_1-F(F'F)_F,)r- (3.153)
Rr(g) will denote the sum of squares for the a effect in model
(3.151). That is
Rr(g) = r/F(F'F)Tr. (3.154)
In the next theorem we find a generalized inverse for F'F and show
that Rr(a) and SSEr are independent and have scaled chi-squared
distributions (the former having a chi-squared distribution only
under Hq : Theorem 3.6
(a) A g-inverse of F'F = f ® 1Í © 1. ] is
' i=l~Di' ' i=l-Di/'
(F'F)- =.® bj1.
(b) Rr(g)/¿* ~ Xa_i> under HQ:cr^ = 0
(c) SSEr/¿+~X2b_a
(d) Rr(a) is independent of SSET.

-81-
Proof :
The proof of Theorem 3.6 follows along much the same reasoning
as that for Theorem 3.4 and is thus omitted. â–¡
A test statistic for testing IIq:cTq = 0 vs. Ha:<7a^0 is given by
Rr(g)/(a-l)
Fa_ SSEr/(b-a) •
(3.155)
Under Hq, Fa~ Fa_1>b_a.
Note that
Rr(a)
a-1
= 6aal + 6*
(3.156)
where 6a is the average of the eigenvalues of FF7 = H*f © J. jH*'. If
'i=l iy
H* denotes bxb Helmert matrix, then
H*
J_ 17
(3.157)
Consequently,
a T
V»
,)=trK
H*'J
E!b(
itijbi>b
Hí(
= tr

-82-
Hence,
e~ =
a a-1
! a
i = l
>0.
(3.158)
Therefore we would reject Hg:cr^ = 0 for large values of Fa.
3.2.5. Reduction of the Exact Tests when the Design is Balanced
Introduction. We now show that the exact tests for a^a and a^
given in (3.155) and (3.138), respectively, reduce to the usual ANOVA
tests when the design is balanced. The test statistic F^, given in
Section 3.2.1, is the usual ANOVA test for testing = 0 vs.
When the design is balanced we have
bi = b* Vi
cij = c* V(i,j)
nijk = n+ v(i>j>k)-
(3.159)

-83-
Consequently,
b = b^ = ab*
i
c
= £ <
i, j
ij
ab*c
*
n = £ ni jk = ab*c*n*
i, j>k
Also, recalling (3.104), (3.105), and (3.111)
Ai - -b*c* ~ Ia® -b* ® ~c*
A2 = .© .1 * = Ia ® 1^* ® 1 *
1 5 J
K= ® n*“1=Ji(Ia®Ih*®Ir*)-
i, j , k n be
Thus the matrix L defined in (3.115) becomes
L = HjKHj
It’s largest eigenvalue is therefore -^max^"^?' The matrix L
(3.149), was defined as
(3.160)
(3.161)
(3.162)
, used in
L* = H*(A'A2)-1H*'-

-84-
When the design is balanced,
L* becomes (see
(3.161))
L*
)-1H
*i
l
b-r
(3.163)
The largest eigenvalue of L*, Aj^ax, is therefore The matrices Px
and P2 were defined in (3.136). When the design is balanced they
reduce to
= (la ® Ijj* ® ^C*)®1
(3.164)
where Js=^lsl's.
Reduction of the exact tests. Consider first the vector w
defined in (3.126). When the design is balanced u> becomes
~ = - + (^max^j ~ L)5Ciy
^from (3.162))
= Hxy
^see (3.113)).
(3.165)

-85-
Using (3.163), the vector r, defined in (3.149), reduces to
r = Y + (A*axi *-L*)5cry
'h
=v (from (3.163))
=H*w (see (3.140))
= II*(A^A2)_1A^H'1w (see (3.139))
= ^*H*(Ia® Ib*® l'c*)Hiy (from (3.161)). (3.166)
Since the design is balanced,
Ic-1-P2 = Hi[!a ® Ib* ® ( Ic* - Jc*
and
P2 — Pa — Hj(la (1^* — J^*) ® •
R„(^|ff) =y/(P2-P1)y (see (3.135))
= y'HiHi[la ® (Ib* - Jb* ) ®
Consequently,

-86-
(3.167)
The last equality follows since H'1H1 = IC — Jc and = 0.
Also
SSEy = y,(Ic_1-P2)y (see (3.132))
= y'HiHi[la ® Ib* ® (Ic* "
= y'[Ia®Ib*® (Ic*_Jc*%-
(3.168)
The last equality follows since J *(I * — J *)=0. Furthermore, from
c c c
(3.153) and (3.154) we have
SSEr = r/(Ib_1-F(F'F)-F')r
= r,(Ib_i — •^■FF,)r (by Theorem 3.6(a))
= r/(Ib_1-Hf(Ia®Jb,)Hjt/)r (^e (3.152))
= r'HÍ[ia®(ib,-Jb0]HÍ/r.
From (3.166) we can further reduce SSEr to
SSEr = ® (Ib* - Jb* jjHÍ'HÍQrH;y

-87-
= ® (Ib* - Jb*)]QrHiy
where, from (3.166),
qr = Ia®Ib*®l/c*-
Since
^Q'rfla ® ( V - Jh*)]Qr = *a ® (*h*
—Jb*)®Jc*
we have
SSEr = (Ib* - Jb+) ® Jc*]H,1y
= ^±',(P2-Pi)y-
Thus,
c+SSEr = Ry(£|g).
(3.169)
Similarly, we can show that
Rr(g) = r'F(F'F)~F'r
a ® ^b*
®J *
c
)Hjy

-88-
Thus,
c*Rr(g) = y/HjP1H1y.
(3.170)
Since the design is balanced,
y ^Ua®1^®1
c
(3.171)
Thus (3.167), (3.168), and (3.170) can be expressed as, respectively,
(3.172)
(3.173)
and
(3.174)
Therefore, from Khuri (1982), it follows that n*c*Rr(g), n*Rw(/?|g)
=n*c*SSEr, and n*SSEw reduce to the usual ANOVA sums of squares for
balanced data.
3.2.6. A Numerical Example
The following data are taken from Bennett and Franklin (1954, p.
405) and has been made unbalanced for illustrative purposes. The

-89-
data represent a part of a statistical study on the variability of a
number of properties of crude smoked rubber. The figures given in
Table 3.2 represent measurements of the modulus at 700% elongation
made using 24 samples. On each of these samples at most 3
determinations of the modulus at 700% elongation were made, resulting
in a total of 53 observations. The three factors of interest are
supplier, batch, and mix. We assume each factor to be random.
will represent the replicate determination from mix k, which is
derived from batch j of supplier i. The linear model in this
experiment is
yijk£-^ + ai + /?ij+7ijk + €ijk£’
where i = l,2,3,4; j = 1,2, • • • ,b^ ; k = l,2; £ = 1,2, • • • ,n^In this
case bj = 2, b2 = 4, b3 = 3, and b4 = 3. Notice that c^j = 2 for all (i,j)
so that the design associated with this experiment is partially
balanced.
Ve start our analysis by testing for a mix effect. That is, we
test for Hq: given in Section 3.2.1. In the present case we find
FT = 1.3110
numerator d.f. =12
denominator d.f. =29
p-value = 0.2650.

-90-
Thus there does not appear to be any significant variation due to the
mix effect.
To test for a significant batch effect we use F^, the test
statistic developed in Section 3.2.3. For these data we find
Fp = 0.3800
numerator d.f. =8
denominator d.f. =12
p-value = 0.9114.
No significant variation in the moduli of the 24 samples exist
among batches.
The test statistic Fa ^see (3.155)j is used to test for a
significant supplier effect. We find
Fa = 39.0024
numerator d.f. =3
denominator d.f. =8
p-value < 0.0001.
There is highly significant variation in the moduli of the samples
among the four suppliers.
All calculations were performed using S-plus.

-91-
Table 3.2
An example of an unbalanced random 3-fold nested model
Supplier
A
B
C
D
Mix
1
2
1
2
1
2
1
2
Batch I
196
226
196
196
189
196
323
262
200
186
156
186
175
279
234
190
251
Batch II
250
248
193
186
211
196
238
262
238
249
196
197
190
250
272
210
Batch III
204
174
196
180
273
223
165
172
197
166
221
256
194
186
230
Batch IV
209
202
221
211
204
Source: Bennett and Franklin (1954, p. 405

-92-
3.2.7. The Power of the Exact Tests
Introduction. The approximate power of the exact test
concerning factor, is derived in this section. Similar expressions can also be
developed for the other two variance components. Ve again use
Hirotsu’s (1979) results to develop our expression for the power (see
Appendix D).
Let 'Lq denote the power of the test for cr^. Then from (3.155)
a
MoO/(a-l)
SSEr/(b-a)
> F
v,a-l,b-a
(3.175)
where HaiO’^^O and Rr(g) and SSEr are defined in, respectively,
(3.154) and (3.153). Under Ha, Rr(a) is distributed like a random
variable of the form (see Box(1954b))
a-1
E/ajXj
j=i J J
(3.176)
where the Xj’s are independent
of freedom, and 9 • is the j*
’ «J J
chi-squared variates
nonzero eigenvalue of
with one degree
U = FF'<4 + F(F,F)_F,<5*, (3.177)
where F is defined in (3.152). Equation (3.177) follows from Theorem
3.5. Clearly FF' and F(F'F)_F' commute and rank(FF':F(F'F)_F')

-93-
=rank(FF') =a-1. Thus there exists an orthogonal matrix P such that
PUP' =
(3.178)
where Aa is the (a-1)x (a-1) diagonal matrix whose diagonal entries
are the nonzero eigenvalues of FF'. (If PUP' is not in the form
indicated in (3.178) then pre- and post-multiply by permutation
matrices until this form is achieved. The locations along the
diagonal of the nonzero eigenvalues of FF' and F(F'F)~F' must match
exactly.) The nonzero eigenvalue of U, is therefore
2
a
+ 6*
where Aftj is the diagonal entry of Aa (j = 1,2, • • • ,a-l) .
The approximate power. To obtain the approximate power (see
Appendix D) let
MSEr = SSEr/(b-a)
(3.179)
Rr(g)/df
MSEr/<5*
(3.180)
■ ¿*(a-l)F
(-}*-£ rj/,a-l,b-a’
(3.181)

-94-
lí ía =
then in this case
a¿Va/a+l)2
j=l J
E (Aajía + *)
j=l
E (AQjía+ !)
>1 J J
E ('*aj£a + 1)2
J=1
(3.182)
(3.183)
Furthermore, define
f2 = b-a
a-1
a-l ,
E (^ají« + 1)
J=1
E (Wa+l)3
U=1 J
jía+l)2
j=l J
(3.184)
Then (^in (3.175)J can be approximated by using the formula given in
(3.86) and making the appropriate substitutions for H, h, d*, f, f2,
and p. We note that “d*” in (3.182) plays the roll of “c” in (3.86).
Since 6 = + Amaxcrj we see that the power of the exact test will
depend on the design parameters Amax, A*ax, Aal,-*-,Aa a_^; the level
of significance; and the ratios variance components.

-95-
The design parameters Amax and A*ax. Bounds for Amax and A^ax
are given in the next lemma.
Lemma 3.7
(a) The largest eigenvalue, Amax, of the matrix L in (3.115)
satisfies the double inequality
I V" 1 < \ < 1
Ci,j,k "ilk" maX-n(0
(3.185)
where n^1^= min (n**v.).
i , j, k ^
(b) The largest eigenvalue, A*ax, of the matrix L* = H*(A2A2)-1H*' isee
(3.149)j satisf ies the double inequality
I V' _JL_ < a* < 1
bi,jcij max c(0
(3.186)
where — mi
= min (c- •).
1 ? J J
Proof:
(a) If emax(A) denotes the largest eigenvalue of the matrix A, then
^max — emax(b) — emax(^i^i)
— emax(^bl^l)

-96-
— emax(^)emax(^l^l)
— emax(^)emax(^i^l)
= emax(K) (Hj has orthonormal rows)
It is also true that Amax is greater than or equal to the
eigenvalue of L. That is,
Amax > ¿Ttr ( L ) = ¿Ttr ( HiKHi )
= ¿ítr[K(Ic-Jjc)]
= ^iy[tr(K) -itr(KJc)]
= ¿T[tr(K)-Il'cKlc]
1
c— 1
E
i» j>k
1
nijk
i E
i, j,k
1
nijk
E
i j J5k
1
nijk
Thus, from (3.187) and (3.188),
i E
i > j,k
n
— < A
i jk
max —
(3.187)
average
(3.188)

-97-
(b) Recalling that (AjAj)-1 = © c-1- and H* has orthonormal rows (see
i, j J V
(3.105) and (3.140)^, we see that the proof of (b) will follow
similarly to that of (a). D
The value of Amax will not depend on which (c-l)xc suborthogonal
matrix is used in (3.113) ^see also (3.115)J. To see this let H be
any other cxc orthogonal matrix whose first row is l'c. Let Hj be
the (c-l)xc matrix resulting from the deletion of the first row of H.
Then there exists a (c-l)x(c-l) orthogonal matrix M such that
Hj = MHj.
Consequently,
emax(H1KH,1)=emax(MH1KH'1M')
= emax(H1KH'1) (M'M = IC_1)
= A
max •
Similarly, AJ^ax is invariant to the choice of which (b-l)xb
suborthogonal matrix is used in (3.140).

-98-
3.2.8 Concluding Remarks
As long as two rank conditions are satisfied, the results given
in Seely and El-Bassiouni’s (1983) Section 3, allow us to conclude
that an exact test for the second to last stage effect in unbalanced
nested designs can be constructed by comparing the mean square for
that effect (adjusting for all the other effects in the model) to the
error mean square for the model. Such a test would be a Wald’s test.
Thus, the test for given in Section 3.2.1 is a Wald’s test.
The above observation provided the motivation for the two
transformations introduced in (3.176) and (3.149). Each of these
transformations was based on resampling from the error vector. The
first reduced the analysis of the unbalanced 3-fold nested random
model to that of an unbalanced 2-fold nested random model. While the
second reduced the problem to that of analyzing an unbalanced 1-fold
nested random model. In either case, the exact test was based on
comparing the mean square associated with the second to last stage
effect in the new (transformed) model to the error mean square
associated with the new model. Consequently, both of these exact
tests are Wald’s tests.
Although not pursued in this dissertation, the extension to more
general unbalanced nested models should offer no resistance. A
partial extension is examined in Section 3.4. There we study a
general unbalanced random nested model where the imbalance affects
the last two stages only.

-99-
3.3 The General Unbalanced Mixed Model with Imbalance
Affecting the Last Stage Only
3.3.1 Introduction
Khuri (1990) developed exact tests for testing hypotheses
concerning the variance components in general unbalanced random
models where the imbalance affected the last stage only. This
extended the work of Khuri and Littell (1987). Gallo and Khuri
(1990) constructed exact tests for both the random- and fixed-effects
in an unbalanced mixed two-way cross-classification model. In this
section we extend the work of Gallo and Khuri (1990) by considering a
general unbalanced mixed model. We will require that the imbalance
affects the last stage only.
After some preliminary development we start our analysis by
studying the random effects in the model. Next we construct
procedures for the fixed-effects. Finally, we reexamine our model
assuming only a few cell counts are large relative to the smallest
cell frequency. Under this scenario we will employ the results in
Khuri (1984) to construct exact simultaneous confidence intervals on
estimable linear functions of the fixed-effects parameters.
In our construction of the exact procedures we will closely
follow the terminology and notation laid out in Zyskind (1960), Smith
and Hocking (1978), and Khuri (1982, 1984, 1990).

-100-
3.3.2 Preliminary Development
Introduction. A mixed model for a general unbalanced design
whose imbalance arises from unequal cell frequencies in the last
stage only can be written in the form
y*~i5o7*i(*i)+£*
(3.189)
where 0= (k15k2, • • ■,ks) is a complete set of subscripts that identify
a typical response, y. The sets 0^ and 9^ are the sets of
nonrightmost and rightmost bracket subscripts ^see Khuri (1982)^,
respectively, associated with the i^ effect 7. ,-x . (i = 0,1, • • • .
0i^i)
When i = 0 we define 0q = 0q = <¿>, the empty set, and the corresponding 7
is the grand mean. We assume that 7. for i = 0,1, • • - ,v-p
(0 < p , for i= u-p + 1,
and eg are independent, normally distributed random variables with
zero means and variances 0^ ,0^, and Since the design is balanced except for its last stage, the
ranges of subscripts kj,k2,•••,ks can be expressed in the form
(1,2,
kj “i 1,2,
’aj
for j=l,2,
for j=s,
, s—1
(3.190)
where r= (kj ,k2, • • •,kg_^) consists of the first s-1 subscripts of
Ve assume that the data set contains no missing observations.

-101-
Let us define the set T by
Then c, the number of elements in T, is
s — 1
c = n -
(3.192)
Also, let = be the set of subscripts associated with the i^
effect which results from concatenating the sets of 9j and 9^
(i = 0,1, • • • ,u) ; V’j is the complement of ip^ with respect to r
(i=0,1,• • • ,v). The complement is with respect to r and not 8 since
ks only appears in the Cq term of (3.189).
Matrix formulation. In matrix notation model (3.189) can be
expressed as
(3.193)
i=0
where y is the nxl vector of observations (n = total number of
observations= nr); is an n x c • (i =0,1, • • • v) matrix of ones and
r€T
zeros; /?• is a vector consisting of the c- elements of yQ
~ 1 I#
(i=0,l, •••,!/); and e is the nxl vector of random errors. The
integer c- is given by
for i = 0
for i = 1,2, • • • ,u.
(3.194)

-102-
The assumptions made earlier concerning the 7. . (i = 0,1, • • • ,v)
now be restated as
(i) /Jqj/Jj, • • • ,are unknown parameter vectors for
0 < p < v.
(ii) The vectors /?¡,_p+i > ' ' • > Py > and Í are independent and
normally distributed random vectors with zero mean
vectors and variance-covariance matrices given by
Var(£b ) = cr|lc. for i = u-p+1,
Var(e) = <7fln.
Besides these assumptions we will also assume
(iii) the model used is model (3.189) ^or model (3.193)^
/ "~p \
(iv) n > max(2c — £) m-, 2c— J2 m-, 2c— £) m-), (3
V i=0 i £ fc ieTc ;
where m- , fc, and Assumption (iv) (along with the other assumptions) are needed to
insure that our procedures are valid. Later we will strengthen
Assumption (iv) by assuming that n > 2c ^see Section (3.3.7)^.
From Assumptions (i) and (ii) it follows that model (3.193)
can
.195)
can
be written as

-103-
y = Xg + Zh + e
(3.196)
where X = (XQ:Xj: • • • :X„_p) ; Z = (X^^ :X^^: • • • :X„) ;
g = (/Jq/?^,_p)' is the vector of fixed effects and
h = (/?^,_p+j :^i/_p+2: ' ’ ‘ : t^ie vector of random effects. It is
easy to verify that
E(y) =Xg
(3.197)
Cl = Var(y) = £ «ríXjX'. + *2eIn.
i=i/-p+l
Cell means. If it were not for the e term, which consists of the
n elements of Cq, model (3.193) or (3.196) would represent a general
balanced mixed-effects model. Various results concerning general
balanced mixed models can be found in, for example, Smith and Hocking
(1978) and Khuri (1982, 1984). The idea then is to reduce the
analysis of the unbalanced mixed model to that of a balanced model.
To that end let
ii T
E Ygt rGT
kE=l
(3.198)
if y is the vector consisting of the c values of yr, r£T, then, from
(3.198), we have
y = Dy
(3.199)

-104-
where
D = (V'V)"^'
and W is the nxc block diagonal matrix
V =
T GT
Clearly,
® t nr - nr'
(3.200)
Now, from (3.189) and (3.198) we have
yT-£'V (Q. ) + £r> reT
1=1 lv 1'
(3.201)
where
1 nr
GT.
ks=1
Since 53 7a . in (3.201) is of the same form as in a balanced
i=0
model, we can express (3.201) in matrix form as ^see Khuri (1982)^
yT = É Hí/?í +í
i=0
(3.202)

-105-
where the cxcj matrix can be expressed as
Hi=®L.£, i =0,1,
1 C= 1 xt
(3.203)
and is given by
Li£ 11=
fXa£ for
for
(3.204)
(i=0,1,•••,v; l — 0,1,•••,s-l) . Furthermore, if
= (HqiH, : - - --H„_p)
F=(lt-p+l:lt-p+2!"-:B*'>
then from (3.202) we obtain
(3.205)
y = Hg + Fh + e .
(3.206)
Define the cxc matrices (i=0,1,•••,i/) as
A^H^, i =0,1, •••,!/.
(3.207)
Then from (3.203) and (3.204) we have
s-1
A: = ® M- /,,
1 £=1 xi
i = 0,1, • • • ,i/
(3.208)

-106-
where
Mi
L
for kg 6 V’i
for kg € V»i
(3.209)
(i = 0,1, • • • ,i/; i— 1,2, • • • ,s-l) . Hence
E(y)=Hg =£%/?!
~ i=0
(3.210)
£ = Var(y) = £ i=i/-p+l
where
K = ® n-1.
re T T
(3.211)
It can be easily verified that A^j = A jAi for all i,j = 0,1, • • • ,^ (see
Khuri (1982, p. 2910)j. Consequently, there exists an orthogonal
matrix Q of order cxc such that
QAiQ' = Ai? i =0,1, •••,!/
(3.212)
where Ai is a diagonal matrix (see Lemma A.5). The construction of Q
is described below.
v
The matrix Q. Recall that J2 In /~ñ \ in (3.201) is not affected
i=0
by the imbalance in the last stage of the design. Thus it can be
thought of as representing a balanced mixed model with one
observation in each r-cell. Let Pi be the cxc matrix associated with

-107-
the sum of squares for the effect in the derived model
Zr=i?o\('i>- (3'213)
Some properties of P.(i =0,1,■•■,v) are provided in Khuri (1990,
p. 181).
Let rank(P^)=m^ ( i = 0,1, • • • , v) , then
£ mi = £ rank(Pi)
i=0 i=0
V
= £ tr(P-)
i=0
(P^ is idempotent for
i = 0,1,• ••,v)
-*&*)
= tr(Ic) ^see Khuri (1990) Lemma 3.1(iii)j
= c.
Let be a matrix of order m^xc whose rows are orthonormal and span
the row space of P^ (i=0,l,- ••,!/). The rows of can be obtained
by orthonormalizing any m^ linearly independent rows (or columns) of
P^ (i = 0,1, • • • ,i/). The following lemma is proved in Khuri (1990).

-108-
Lemma 3.8
The matrices Qg, Qj, • • • , Qj, have the following properties:
(a) Q0=^lc
(b) i = 0,l,
Q.Q'j=0, i#j
TO,
(c) A.q'i= 1 J i = o, i,
^ U5 j^i ’ ^i ^ j ’
,i/; j —0,1,
where bj is the number of ones in a typical column of Hj,
j = 0,1, • • • ,u ^see (3.203)j.
A consequence of Lemma 3.8 is that if ip^ represents a set of
subscripts associated with a random effect and ip ^ represents a set of
subscripts associated with a fixed effect, then Q^Aj = 0 which implies
QiHj=0 (i = y-p+1, • • ■ ,u; j = 0,1,••• ,v-p). Hence from (3.205),
Q^H = 0 (i = i/-p+l, • • • ,v) . The matrix Aj was defined in (3.207).
Write the cxc matrix Q as
Q = [Qq:Qi: * * *;Qj/]/,
(3.214)
then from Lemma 3.8, Q is orthogonal and simultaneously diagonalizes
Ag,Aj, • ■ •
Furthermore, from (3.210),

-109-
Var(Q-y) = £ ^.Q.A •Q,i + ^KQ'. (3.215)
j=^-P+l
(i=0,l,**',i/). However, from Lemma 3.8
E ^QiAjQ'i = E b/.i
j^-p+l J 1 J 1 jeVj J J 1
^i-bm^ > i — 0,1,
(3.216)
where
6- = E b-o-2-, i = 0,1, • • • ,r
jevi J J
(3.217)
vi = { j:r-p+l < j < C V’j} > i = 0,1, ■ • • ,v.
Thus from (3.215) we have
Var(Qjy) =¿iIm.+ Also, for i ^ i'
CoviQjyjy'Q'.,) = Qj
v
E
j=r-p+l
Q',
/
= Qi
£ ^A.Q^ + ^KQ',
j^-p+1 J J 1 1
(3.219)

3.3.3 Analyzing the Variance Components
Introduction. In this section we develop some independent, chi-
squared distributed sums of squares that can be used to test
hypotheses concerning the variance components. These sums of squares
are analogous to the sums of squares associated with the random-
effects in an ANOVA table for a balanced mixed model. In fact these
sums of squares will reduce to the usual ANOVA sums of squares
(associated with the random effects) when the design is balanced.
The key to the construction of these sums of squares is the idea
of resampling from the error vector (see Appendix C). It should be
noted, however, that resampling is not a “cure-all”. Resampling does
not guarantee that we will be able to produce an exact test for all
the variance components of the model. This should not be too
surprising, though, since even in certain balanced models the same
phenomenon occurs.
In designs with imbalance affecting the last stage only,
resampling, in some sense, counteracts the imbalance by reducing the
analysis of the original unbalanced model to that of a balanced one.
Consequently we should not expect to do any better (in terms of
constructing exact tests) when utilizing resampling in unbalanced
models than if the design was balanced to begin with.
Initial development. Recall the definition of Q in (3.214).
Let Qr (r for random) be the (c—J)m-jxc matrix
V i=0 '

-111-
Qr ~ [Q^-p+l: Qi/-p+2:
â– xr-
(3.220)
( "'p \
Also, define the Ic— £m-Jxl vector u.
V i=0 '
yr = Qry-
(3.221)
The use of the matrix Qr causes ur to have a zero mean vector. In
fact, it is easily verified that (recall that Q^H = 0, i = u-p+1, • • • , u)
E(ur) = QrHg = 0
Var(ur) = © ¿-Im.+^Gr
i=j/-p+l i
(3.222)
where
Gr = QrKq/r. (3.223)
Utilizing resampling. The error sum of squares for the original
unbalanced model in (3.189) can be expressed as
SSE= ^(y0-yr)2= £ £ (y^-y,.)2 (3.224)
6 reT ks=l
where, recall, 0=(r,ks) and yr is defined in (3.198). In matrix form
(3.224) becomes
SSE = y'Ry
(3.225)

-112-
where
R = In — V(V,W)-1V/
(3.226)
and W is the nxc matrix
(3.227)
In Khuri (1990), the following lemma is proved (see also Lemma B.6).
Lemma 3.9
(a)R is symmetric, idempotent, and has rank n-c.
(b)DR = 0, where D is defined in (3.200).
(c) RX^=0, i=0,where Xq ,X1, • • • ,X„ are the matrices in
(3.193).
From (3.197) and Lemma 3.9 we can conclude that y and SSE are
independent. Furthermore
R tl/al = R
(3.228)
thus
SSE/tr* ~ X
2
n-c *
(3.229)

-113-
Now, since R is symmetric, idempotent, and has rank n-c, we can
express it as
R = CrArC
/
r
(3.230)
where Cr is an orthogonal matrix and Ar is a diagonal matrix with n-c
ones and c zeros. Without loss of generality, we can partition Cr
and Ar as
Cr — [Crl: C
Jr2-'-'r3]
Ar — diag( , I;yr2 5 0)
(3.231)
where
u-p
Vrl =c - £ m-
i=0
v-p
,?r2 = n"2c + £
i=0
(3.232)
and Crl, Cr2, and Cr3 are matrices of orders nX7jrl, nx^r2, and nxc,
respectively. Note that r]rl + r/r2 = n-c = rank(R) . Also r¡r2 > 0 by
Assumption (iv) ^see (3.195)^). Without this assumption, rjr2 may be
nonpositive which would invalidate our methodology.
Since Cr is orthogonal
CriCri = I (i = 1,2,3)
(3.233)
CriCrj=° Ci # J).

-114-
Consequently
R — CrlC^j + Cr2Ci-2
and
SSE = y'Ry = y'CrlC'rly + y'C^C^y
= SSErl+SSEr2
where
SSEri =y/cric^iy, i = l,2.
Clearly y is independent of both SSErl and SSEr2. Also since
C^R = C^ (i = l,2), it follows that
CriCrin/^ = Cri(C^R)il/^
= CriCri(Rii)/ (see (3.228))
i = 1,2.
Thus Cr-C^ft/(7^ is idempotent (i = l,2) and = 0.
SSErl and SSEr2 are independent with
= C -C' -R
r i r i
= c . c'.,
rrn ’
(3.234)
(3.235)
Hence

-115-
SSErl/cr^ ~ XJrl
SSE.-j/o'f ~ X^r2'
(3.236)
The exact tests. Define the rjrl x 1 random vector wr as
wr — ur + (Ar jmax^r;rl Gr)2Crly (3.237)
where Gr was defined in (3.223); Ar>max is the largest eigenvalue of
Gr; ur was defined in (3.221); and C^i was defined in (3.231). The
1
matrix (Ar >inaxIr?ri-Gr)2 is well-defined since Ar >maxI,rl - Gr is
positive semidefinite (p.s.d.).
Partition ur in (3.221) as
yr = (a'r-p+i1 • • • !si/)#
= (ftp+l:'-':!X)' (3-238)
where Uj is of dimension m^; i =r-p+1,• • • ,u. Partition wr just like
ur. That is
= 04-
p+i1
s«4)#
(3.239)
where is of order m. xl, i = i/-p+l, • • • , u. The major result of this
section is now presented.

-116-
Theorem 3.7
(a) E^^O, i = I/-P+1, • • • ,v.
(b) p+1 > ‘ ‘ ' >Vj/ are independently distributed as normal random
vectors with y^ having variance-covariance matrix
Var(yi) = («. + Ar jmax^)Im. , i = v-p+1, • (3.240)
(c) yr is independent of SSEr2.
Proof:
(a) From (3.222), E(ur)=0. Also from Lemma 3.9(c), RX
= R[Xq: • • • :XJ/_p] = 0. Thus
C'rlX = (C'rlR)X = C'rl(RX) =0.
Consequently the result follows after noting that E(y)=Xg ^see
(3.197)^ and recalling (3.237).
(b) Clearly yr is normally distributed. Ve claim the ur is
independent of C^y. Now, since DÍ2R = 0 (y is independent of
SSE) it follows that DQCrl=0 ^range(Crl) C range (R)). Th us y is
independent of C^y. Therefore ur is also independent of C^y as
it is a function of y (see (3.221)1. Consequently

-117-
l
Var(wr) = Var(ur) + (-^r,max^rjrl — ^r)2^ri^^ri (^r, max^i7rl
But, from (3.228),
C'rlfiCrl =CriRfiRCrl = Cr1(o-2R)Crl = 0^1^.
Hence, from (3.222),
Var(wr) = © ^i^m- + ^e^r + (-V ,max^77rl — ^r) i=i/-p+l i ,rl
1/ 2
= . ® -.^i^m- +^r,max°re^i7r-1 •
l—V—D+1 1 'rl
Consequently yl/_p+\> ' ' ’ tVu are independent and
Var(yi) = (¿i +'V,max<^)Imi, 1 = "-P+1
5 5 1
(c) Since ur is independent of SSE, it follows that it is
independent of SSEr2 ^range(Cr2) Crange(R)j. Also, since SSErl
and SSEr2 are independent, it follows that C^y is independent of
SSEr2. Thus ur and SSEr2 are independent. â–¡
Let us define SS • as
r i
SSri = ^Wi ’
i = r-p+1 ,•••,!/.
(3.241)

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Then, from Theorem 3.7,
(i) SSr j/_p+j , • • • ,SSr„, and SSEr2 are mutually independent.
(ii) SSri/(5i+Ar5maxir2) ~x2 _ ^ i =„-p+l, . . .
Thus, analyses concerning the variance components 5 can
proceed using SSr l/_p+i, • • • ,SSrt,, and SSEr2. To illustrate the idea
let us suppose we wish to test Hg: Also, suppose that there exists a j( ^i)€{v-p+1,•• • ,r+l) such that
under Hq: SS
r 1
-éj+Ar,i
(3.242)
(For notational convenience take SSr J/+^=SSEr2, 6¡/+^=0, and
= r?r2/Ar ^max. ) Then a test statistic for this hypothesis is
_ ssri/mi
_ SSrj/mj"
Under Hg, Fr^~Fm.jm.. From (3.217) we would reject Hg for large
values of F j .
If no such j exists satisfying (3.242) then no exact test can be
developed based on our methodology. In such a situation we must
resort to Satterthwaite’s procedure to construct an approximate test
for testing this hypothesis.

-119-
3.3.4 Analyzing the Fixed-Effects Parameters
Introduction. In this section we concentrate on the fixed
portion of the model (3.189). When possible, exact tests will be
developed to test hypotheses about the unknown parameters in the
model. When no such exact test can be developed we will utilize
Satterthwaite’s procedure (after making an appropriate transformation)
to develop an approximate test.
Initial development. To begin with, let u be the cxl random
vector defined as
(3.243)
u = Qy
where Q is the cxc orthogonal matrix defined in (3.214). If we let
r-p
(f for fixed) be the xc matrix
(3.244)
then, recalling (3.220), we can write (3.243) as
(3.245)
where u^p = (uq: u^ : • • • : u^p)'. It is easily verified that
(3.246)

-120-
where
G = QKQ'
(3.247)
and H and K are the matrices defined in (3.205) and (3.211),
respectively. Formula (3.246) follows from (3.215)-(3.219) and the
fact that Q^H = 0, i = i/-p+l, • • • ,i/ ^see Lemma 3.8(c)j. Now, from
(3.246) it follows that
Var(u^) + crjG^ (3.248)
where
Gf = QfKq'f.
(3.249)
Also,
Cov(uf ,ui.) = n-jq^Kq'r.
(3.250)
Thus and ur are not independent. ^They would be if the design was
balanced (see Lemma 3.8(b))j.
Now, recall that m^ =rank(P^) = rank(q^), i =0,1,• • • ,u ^see the
discussion after (3.213)^. Also, let z be the cxl vector consisting
of the c zr’s from the derived model (3.213). Then, from Khuri’s
(1990) Lemma 3.1 (iii), we can express the total sum of squares
associated with model (3.213) as

-121-
z z = z
/ V \ v
(EPi>=E5'Pi5-
\=o J i=0
(3.251)
Since Q is orthogonal, we can equivalently express (3.251) as (see
(3.214))
Consequently, since is idempotent and range(P-)=range(Q^),
(3.252)
'i q i, i=o,i,•••,!/.
Lemma 3.10
(a) rank(Q^H) = m^ , i =0,1, • • • ,u-p,
u-p
(b) rank(H) = ^2 m- .
i=0
(c) If li=Q'iQiHg, i =0,1, ■ • • ,I/-P, then rQ, r1, ■ • ■ ,Tu_p are linearly
independent and span the space of all linearly estimable
functions of g, the vector of fixed effects in model (3.196).
Proof:
(a) By Lemma 3.8, (3.205), and (3.207)

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= ( E bjK,
vjeUj 1
= ¿iImi» i =0,1, • • • ,i/-p
where
Uj = { j : 0 < j < u-p, V’iCV’j}, i =0,1, • ■ • ,i/-p
Thus
rank(Q^H) = rank( Q^HH'Q'^ )
= rank(¿ÍImJ = mi , i = 0,1, • • • , i/-p.
(b) First note that
QfHH'q'f
Q0
Q„-p
From Lemma 3.8 we have
i = i' = 0,l, • • • ,u-p
i#i'-

-123-
Consequently,
QfHH'Qf =i®Q5iImi
Thus
i/-p
rank(H) > rank(Q^H) = rank(Q^HH/Q^) = Y .
Conversely, from Lemma 3.8 and the fact that Q is
have
H = Q'QH =
(
(
(
E
i=t/-p+l
Since
/ v-P v ""P \ ""P
( EQ'iQiX EQ'jQj^EQ'iQi + E Q'iMjQj
v i=0 A j=0 J i=0 i ¿ j J J
i/-p
- E Q'iQi,
i=0
r / "
we see that Y Q';Qi is idempotent and thus rank! J
i=0 v i=
/ v-P \ v-p v-p
= tr( E Q'iQi) = E tr(Q'iQi) = E mi- Hence
(3.253)
orthogonal, we

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/ "_p \
rank(H) = rank ( 53Q-Q-JH
' ; -n 1 x'
— rank( E Q^Qj) = E i"i •
i/-p . i/-p
(3.254)
i=0 ' i=0
i/-p
Therefore, (3.253) and (3.254) imply that r(H) = E mi•
i=0
(c) Recall that r- = QjQjHg, i = 0,1, • • • ,i/-p. Suppose that Tq,
Il> • • • >Ii/-p were linearly dependent, then there would exist
constants ajj,a*,•••,a*_p, not all zero, such that
t/-p
E a*Q^QiHg = 0.
i=0
r-p
This implies that E atQ^ QiH = 0. Premultiplying by Q -(0 < j < t'—
i=0 J
p) results in
j/-p
Hence we conclude that aj=0. Consequently a* = 0 for
i =0,1,•••,r-p which is a contradiction. Therefore
Tq,t1,•’•,r^-p must be linearly independent.
Now,

-125-
Thus, any estimable linear function, |'Hg, of g can be written as
a linear combination of the elements of Iq ,Z\ > * • • ,l„_p °f the
v-p
form Therefore, TqiZ\, • • • span the space of all
i=0
linear estimable functions of g. â–¡
From Lemma 3.10 we can restrict our attention to developing
methodology concerning Zq>Z\ , • • • ,T¡/-p‘ Before we proceed, let us
recall that ^see (3.246)j
ui~N(QiHg, ^ilm.+^QiKQi), i = 0,1, • • • ,i/-p
yi0~N(9>¿i0Imio + ^Qi0KQi0), io = «'-P+1» • • •
(3.255)
In (3.255) notice that
Qi^g = Q i - i » i = 0,1, • • • ,i/-p. (3.256)
The vector of “proper error terms”. If the design was balanced
and we wished to test a testable hypothesis concerning an estimable
linear function of the fixed effects in the model, we would do so by
scanning the “E(MS)” column of the corresponding ANOVA table for a
mean square which could serve as an “error” term (if one existed) for
the test statistic used in testing our hypothesis. This mean square
would be chosen in such a way that under the null hypothesis our test
statistic would have an F-distribution. If no such mean square
existed then, typically, we would utilize Satterthwaite’s procedure
to develop an approximate test.

-126-
When the design is unbalanced there generally does not exist a
mean square that can serve as a proper error term. Consequently
exact tests (based on the ANOVA sums of squares) do not usually exist
for the above mentioned hypothesis testing problem. The most common
approximate test is based upon Satterthwaite’s procedure.
Unfortunately the sums of squares involved in Satterthwaite’s quasi
F-ratio are typically neither independent nor chi-squared
distributed. Either of these “failures” is enough to invalidate
Satterthwaite’s methodology. This has led some authors to advocate
the use of traditional (balanced) methods even in the presence of
imbalance ^Bliss (1968) or Tietjen (1977), for example^.
Ve now present a technique that will allow us to construct exact
tests (when such exist) for estimable linear functions of the fixed
effects in model (3.196). This will be done by using resampling to
make a transformation that will result in a set of mutually
independent, (scaled) chi-squared distributed sums of squares. The
idea will be illustrated by considering a three-way cross¬
classification design in which factors A and B are fixed while factor
C is random.
Let the random vector v be defined as
Xf
(3.257)
where v^ and yr are determined as follows:
(i) Put u^ (i = 0,1, • • • ,i/-p) in v^ if and only if there exists
an ift £ {i/-p+l for which 6- — 6- .
1 lQ

-127-
(ii) Put the corresponding u- (ig = j/-p+l, • • • ,u) in
Thus, consists of those vectors for which a proper error term
exists and yr consists of the corresponding vectors used in those
proper error terms. Recall that u^ =where is defined in
(3.214), k = 0,1, • • • ,i/. Also <5^ (k = 0,1, • • • ,j/) is defined in (3.217).
It is also convenient to introduce the following notation:
= {0 < i < i/-p: u^ is part of v^}
(3.258)
fr = {i^-p+1 < ig < u:
is part of yr}
(3.259)
f = íf Ufr.
We note that and fr (and hence f) may be empty. In such cases we
cannot apply our procedure.
Example 3.2
Consider a three-way cross-classification design in which
factors A and B are fixed while factor C is random. The model can be
expressed as
yijki = ^+ ai+^j + (a^)ij+7k+(a7)ik+(/?7)jk+(«^7)i jk + fijk£ (3.260)
(i=l,2,..-,a; j = 1,2, • • • ,b; k = l,2,---,c; Í = 1,2, • • • ,n-jk) .

-128-
Following the notation in Section 3.3.2: v = 7, p = 4, and 7^, (<*7)^5
(/?7)jk, (Qr/?-y) i jj^» and fjjkg are independent, normally distributed
random variables with zero means and variances respectively. Table 3.3 summarizes the pertinent information needed
in the determination of v.
From Table 3.3 we see that
Yf = [yó^y^aá]7
Yr = [u^y^Ue^?]'-
(3.261)
Thus, in this example, the vector y is exactly the vector u. It is
interesting to note that if we had instead assumed factor A fixed and
factors B and C random then y would not consist of any elements and
our procedure could not be applied. That is, no exact tests (based
on our methodology) could be constructed for any estimable linear
function of the fixed effects. We would thus need to apply
Satterthwaite’s procedure. More is said about this in the sequel.
Utilizing resampling. Recall (3.225) and (3.226). Instead of
the partitioning of R as described in (3.230)-(3.232), we now express
R as
R = CfAfC!f (3.262)
where is an orthogonal matrix and A_p is a diagonal matrix
consisting of n-c ones and c zeros. We can partition and A^ as

-129-
Table 3.3 Information needed to determine the vector v in (3.257)
for a three-way cross-classification model in which
Factors A and B are fixed while Factor C is random
i
V-i
bi
V^see (3.217))
¿¿(see (3.217))
0
4>
abc
{4,5,6,7}
abo^ + b<7g + a i
{i}
be
{5,7}
b 2
{ j}
ac
{6,7}
2 2
a<76 + al
3
{i> j>
c
{7}
A
4
{k}
ab
{4,5,6,7}
aber^ + b cr\ + acr| + <77
5
{i»k>
b
{5,7}
b 6
{j5k>
a
{6,7}
aa6 + A
7
{i,j>k}
i
{7}
A
Cf - tCfl:Cf2:Cf3^
Af = diag(I7?fi,Ir?f2,0),
(3.263)
where
7fi =
c- E mi
i G fc
(3.264)
7f2 = n — 2c + E mi
i G fc
and ¡fc is the complement of Í with respect to {0,1, • • ■ ,i/}. If fc =
then r]_£l = c and r¡^2 = n—2c. Note that + i?f2 = n — c = rank(R) . Also,
by (3.195), 77i2>0.

-130-
As v (in (3.257)) consists of those for which i6Í, let Qv
consist of all those for which iGf. In Example 3.2
Qv = Q = [Qg: Qi: ’ • • : Qj/]7. If Y is void of elements (or, equivalently
'S = ) then Qv will not contain any elements and, as mentioned
previously, our procedure cannot be applied.
Now, define the random vector as
«f = V+(-'f,maxV"Gf)H.Í <3'265>
where max is the largest eigenvalue of and is the matrix
Gf = QyKQy- (3.266)
l
The matrix (Af ? maxI^f ^ - Gf )2 is well-defined since (Af >maxI,fi-Gf) is
p. s . d .
Before we proceed we need the following notation: Suppose
S = {/3q,Pi , • • • >/>t) where each p. is an integer satisfying
*0 < pj < • • • < and t + 1 is the number of elements in S. Then
w = (wj: j G S) = (w' :w^:
•W-
Now, partition v as
Y=(Yi:i€f), (3.267)
where each v^ is of dimension m^, ief. We note that {v^cieí}
represents a subset of {uq,u1} • • •,u„}. Partition just like v.

-131-
That is
— (w_£i ; i 6 Í)
(3.268)
where is of order m. xl, i G Í. The key result of this section is
given in the next theorem which provides some distributional
properties for {w^:i€f} and justifies the transformation made in
(3.265).
Theorem 3.8
(a) E(wfi) =
i £
i £ fr
(3.269)
where (i=0,1,•••,r-p) is defined in Lemma 3.10(c).
(b) {w_f^:iGf} consist of independently distributed, normal random
vectors with having variance-covariance matrix
Var(yfi) = (^+Af)[nax Proof:
(a) From Lemma 3.9(c) , RX^=0, i = 0,1, • • • ,u. Therefore, since
range(C.^) C range(R) and R is symmetric, it follows that
C'flXi=0, i = 0,1, • • • ,v. Consequently
E(C;fly) =C#flXg = 0.

-132-
(recall that X = [Xg :Xj: • • • :Xi/_p]) . Since
i= 0,1,• • • ,r-p
i = r-p+1
(see (3.255)), it must be that
Thus the result follows after recalling the definition of u>^
given in (3.265).
(b) Clearly is normally distributed for each i 6 Í. Now recall
that Uj=Q-y, i = 0,1, • • • ,v, (see (3.243)). Since v consist of
exactly those u- for which i £ f and Qv consists of the
corresponding it follows that
y = Qvy-
(3.271)
Since y is independent of SSE, Df2R = 0. Since range(C:p1)
C range(R), DQC^1 = 0. Thus y is independent of C^y.
Consequently y is also independent of C^y as it is a function of
y. Therefore
i
i
Var((!,f)=Var(v) + (AfiBaxI,fi-Gf)2c'finCfl(AfjJ,axI,fi-Gf)5.
By the now familiar argument, it is easy to show that

Also, from (3.217) and (3.218),
Var(v)
i|/iImi + Hence, after some simplification, we obtain
Var(y.p) =
f ) = . w +
T i £ Í 1
-2A
e f,max
)I«
Thus
Var(yfi) = (¿i+ i € J.
â–¡
From Theorem 3.8, we can conclude:
and
~ N(0,
*0
)
for
x0
eJT
(3.272)
where
*fk
-¿k +
kef.
(3.273)
The exact procedures. First recall, from Lemma 3.10(c), that
I^Q'iQiHg, i =0,1, • • • ,i/-p. Consequently,

-134-
Q'iQiii =ii, i = 0,1, • • • ,i/-p.
(3.274)
Thus, constructing exact simultaneous confidence intervals for all
linear combinations, d'r^, of (d6Rc) is equivalent to constructing
exact simultaneous confidence intervals for all r'Q^r^, where r'= d'Q^
and hence is in the row space of , i = 0,1, • • • ,i/-p.
From (3.272) we can conclude that
(i) (tffi-Qiri)/(«fi-Qiri)/ífi~x2i. f ieJf
(ii) yfioyfi0/ifi0~Xmio> ioe;fr-
(3.275)
By our construction of y in (3.257), for each i€there exists an
ig G fr such that 6^ = 6- (implying —ífj)* Therefore if i 6 and
inGfr are such that £p. = • then, for MS_p • = u/p- w_p- /m- ,
u rig xi tiq -ti0-ti0' i0
E[MSfi|
(3.276)
and
(^fj -Qili)/(yfi -Qili)/mj
MSf-
tl0
mi,mi.
(3.277)
Hence, simultaneous confidence intervals on all linear functions,
~^i~i’ of ^r 6 range(Q^ )j can be obtained from the formula

-135-
r'Qin e r'yfJ ± (m.MSfi Ftt m >ra r'r)
U i ig
= 1 — a
(3.278)
(i Gíp ig € fr such that = We note that and
thus since is idempotent, d'Q'^w^ is the least squares estimator
of d'r- i |see (3.272)^. Formula (3.278) is based on Scheffé’s S-
method ^Sheffé (1959, pp. 68-72)^.
To test Hq:L,t;^=| for i G^, where L' has full row rank r^, we
proceed as follows: Recall (3.272), let r- =^i^fi the least
squares estimator of !•, iGf-j?. then
L#f i — I = I/q'.Wf. -I ~ N(L'r. -I, ífiL'Q'iQjL)
(3.279)
(iEf^). Because we are assuming that the above hypothesis is
testable for all i€f^, L/r. consists of a set of linearly estimable
functions of r• , ief^. Thus, there exists a full row rank matrix M
(rank(M)= r^) such that
L/ = M,Q,iQi, i G ff.
(3.280)
Then, taking r^ < m^, i Gí|, it is easily shown that M,Q#- has full row
rank r^, i€f^. ^r^ = rank(L') < rank(M'q^ ) , i G but M'q'^ is r^xm^
and r^ < m^ and so rank(M'Q^ ) < r^, i G ¡fO. Furthermore
consequently, for i6f^
L'Q,iqiL = M,q,iqiM, iejf,
(3.281)

-136-
rank(L/Q^Q jL) = rank(M,Q/£) = r^.
(3.282)
Thus, L/Q^Q^L is nonsingular for i6 íj. Therefore, if we define
SSHi = (L'Q^Wf i -|),(L'Q'iQiL)-1(L'Q/iyfi - i)
(3.283)
(i GÍ^), we obtain
~ XrL(^i ) > i
(3.284)
where
2ífiAi = (L/ri-|)(L'Q'iQiL)-1(L/ri-£), i€Jf. (3.285)
Now if Íq e is such that ífi =^fi’ then
SSH./r,
F“i-T%7~FrL>%(Ai)’
(3.286)
where, recall, MS^- = wip. u>_p• /m- . Under Hn, A- =0 and
tl0 ~il0~t;i0 10 U i
FH.~Fr m. • Thus, Fjj can be used to test Hg:L'r^=|, i G ip. We
i L’ i0 i
would reject Hq for large values of the test statistic.
An approximate test. When there does not exist a proper mean
square, with the property that E[MS^ ] = (i = 0,1, • • • ,r-p;
iz-p+1 < ig < i/), then we cannot apply the above methodology to construct
a test for Hg:L'r^=|, i=0,l ,•••,r-p. One common approximate
procedure is to develop a “synthetic” mean square that represents a
linear combination of the mean squares associated with the random

-137-
portion of model (3.243) such that its expected value is
(i = 0,1,•••,^-p). This is Satterthwaite’s procedure. Unfortunately,
the numerator and denominator of the resulting quasi F-ratio will not
be independent as required (see(3.250)). To overcome this difficulty
we first make a transformation (utilizing resampling) that will
result in a set of mutually independent, (scaled) chi-squared
distributed sums of squares. Then (rightly) apply Satterthwaite’s
procedure to construct the approximate test. Ve will illustrate the
idea by considering a three-way cross-classification design in which
factor A is fixed while factors B and C are random.
To begin with recall from (3.238) that
y'r
p+l:
:u Í,)'.
Let
= {0 < i (3.287)
(see (3.257) and (3.259)).
Where the vector v^ is defined as follows:
(i) Put u-(i = 0,1, • • • ,j/-p) in v^ if and only if there does not exist
an Íq G {i/-p+l, • • • ,v} for which 6- = <5^ (á^, k = 0,1, • • • ,!/, is
defined in (3.217)).
Thus, identifies the subscripts of the u. (i = 0,1, • • • ,i/-p)
are in v^. Note that every u^ (i = 0,1, • • • ,i/-p) is in either
that
Yf or
Define x as

-138-
(3.288)
In (3.288), is not independent of ur ^see (3.250)^.
Example 3.3
Consider a three-way cross-classification design in which factor
A is fixed while factors B and C are random. The model can be
expressed as in (3.260). In the notation of Section 3.2.2 we have
v— 1 and p = 6. The variance components are o\, 03»• • • ,07» and cr^.
Table 3.4 summarizes the pertinent information needed in the
determination of x.
From Table 3.4 we see that
Yf=[ugiu;] = uf, (3.289)
where was defined in (3.245). Thus, in this example, the vector x
is exactly the vector u ^see (3.243)^.
Now, let ‘T be the set
'T = í^ U {u-p+1, • ■ ■ ,u) .
Also, define
^ai —
c- £ -i
ieTc
(3.290)
na.2 = n"2c + £
i €
(3.291)

-139-
Table 3.4 Information needed to determine the vector x in (3.288)
for a three-way cross-classification model in which
Factor A is fixed while Factors B and C are random
i
*i
bi
V^see (3.
217)) «.(see (3.217))
0

abc
{2,3,4,5,6,7}
a.ca\ + C(T¿ + abo^ -f b i
{i}
be
{3,5,7}
c tr2 + b a2 + cr2
2
{j}
ac
{2,3,6,7}
2 2 2 2
ac<72 + C(T3 + a<7g -f (Tj
3
{i »j>
c
{3,7}
2 2
C<73 -f
4
W
ab
{4,5,6,7}
aber^ + ber2 + ao-g + cr2
5
{i>k}
b
{5,7}
b 6
{ J*k}
a
{6,7}
aa6 + °7
7
{i,J>k)
i
{7}
where l3’c is the complement of “T with respect to {0,1, • • •,r}. Note
that r;al + ija2 = n — c = rank(R) . Also, f?a2 > 0 by (3.195). The matrix R
is defined in (3.226). Ve now introduce a transformation that will
result in a set of independent, (scaled) chi-squared distributed sums
of squares. The development of this transformation is similar to the
development that led up to (3.265) and is thus only outlined.
Define the random vector u>a as
l
^a = £ + (^a,max^T7al — ^a)2^aiy
(3.292)

-140-
where Ga = QXKQX and Qx consists of those for which iG^T- (In
Example 3.3, Qx = Q. ) Aa,max Is ^e largest eigenvalue of Ga; and Cal
as an nxi¡al matrix constructed similarly to in (3.263), for
example.
Partition x as
x=(xi:ieg'), (3.293)
where each x^ is of dimension m^, i g Í. We note that {x• : i £ <3’}
represents a subset of {uq,u1, • • •,u^}. Partition wa just like x.
That is
ya=(yai:ieg')’ (3.294)
where u> • is of order m; xl, iGl”. We then have
~al 1
Theorem 3.9
f Q• t• , i£f.p
(a> E(tf«)= 0, i € {i'-p+l
(3.295)
(b) {yaj:i6 consists of independently distributed normal random
vectors with wa^ having variance-covariance matrix
Var(wa^ ) — (¿. + Aa^max«rf )Im^ , iGf.

-141-
Proof:
The proof of Theorem 3.9 is similar to that of Theorem 3.8 and is
thus omitted. â–¡
In order to construct an approximate test of the testable
hypothesis Hq:L/t.=£ vs. Ha:L'r^^£, iGf^, we apply Satterthwaite’s
procedure (in the usual manner) to the set of vectors {yaj:iG7}.
3.3.5 Reduction of the Tests When the Design is Balanced
If the design is balanced, that is, if nr=N for all rGT (see
(3.190) and (3.191)^, then the cxc matrix K given in (3.211) becomes
k = K-
Consequently, the matrices Gr, G^, and Ga become, respectively, j^I,j ,
^1^^ , and ^See (3.223), (3.266), and (3.292).^ Hence, each
matrix has ^ as its largest eigenvalue. Thus, when the design is
balanced, wr, and wa reduce to ur, v, and x, respectively, ^see
(3.237), (3.265), and (3.292), respectively.J Recall that ur= Qry
^see (3.220) and (3.221)^. Similarly, y = Qvy |see (3.271)j and x = Qxy
^see (3.288) and the discussion after (3.292)j.
Now, from (3.243), and (3.252), and the fact that the matrix Q
in (3.214) is orthogonal, we have
y'u = y'Q'Qy

-142-
= Ey'PiV-
i=0~
From Khuri’s (1990) Lemma 3.2 and (3.208) we have
y,piy = y'
= y
|0(giMj«)
(i = 0,1, • • • ,v) . Aj (j = 0,1, • • • ,i/) was defined in (3.207)
is a known constant whose possible values are -1, 0, and 1
(1982) for a more detailed definition of the A — ). Since,
is balanced
y = J(!c® lN)y-
Therefore (3.297) can also be expressed as
(ic®
(3.296)
(3.297)
Here A^j
^see Khuri
the design
y
(3.298)

-143-
(i=0,1,• • • ,u). Thus, when the design is balanced, Ny'PQy,
Ny'Pjy, • • • , Ny'PQy reduce to the usual balanced ANOVA sums of squares.
Recalling the expressions for ur, v, and x given above we see
that when the design is balanced
i>
UpUr = uj,ur = Z y'P^y,
i=i/-p+l~
Ul'rUr = vV = Z
ief
w^Wa = x'x = Z y'P;y-
ief
Consequently, since {v-p+1, • • • ,j/} , f, and if are all subsets of
{0,1, • • • ,i/}, the exact tests developed for the random and fixed
effects reduce to the usual ANOVA tests associated with a balanced
mixed model of the form given in (3.189). The approximate test
reduces to the approximate test (based on Satterthwaite’s procedure)
that would have been obtained if the design had been balanced to
begin with.
3.3.6 The Power of the Exact Tests
In this section we examine the power of the exact tests (when
such exist). To begin with we consider the power of the exact test
associated with the i^ variance component (i =v-p+1, • • • ,v) .
The power of the exact test associated with the hypothesis
Ho:o’|=0, i = i/—p+1, • • • ,v. For sake of convenience, let
SSr,i/+l =SSEr2> ^+1=°’ and Vr2 = "V+l/^r,max (see (3.232) and

-144-
(3.235)^. We wish to test Hq:o-|=0 vs. Ha:cr|^0, i =
Suppose that there exists a j( ^i)G{r-p+1,•••,u+l}
H0>
rss -i
SS -I
ri
L mi J
- ¿j + Ar ?maxt7^ _ E
r J
L mj.
as in (3.242). Then the appropriate test statistic
_SSri/"j T
SSrj/mj 0 mi’mj'
Under Ha:cr|^0, we have
SSri/mi
E bk°rk + , max'7?
k G Vj
E \a\< + •*
kevi
k k _r r, max“ e
SSrj/mj
This follows from (3.217) and Theorem 3.7. Hence ii
the power of the test for , i = r-p+1, • • • ,u, then
a,ri=P = 11 F™. „ >
mi>mj-l+bi0 • a5mi5mj
where
9 . =
ri E bkak , max^f
kGVj
:r-p+1,•••,u.
such that under
is given by
^ri rePresen^s
(3.299)
(3.300)
(i=r-p+l j( ^ i) G {r-p+1, • • • ,i/+l} such that <5^=6j). Since
increases monotonically with , i = i/-p+l, • • • ,v, it follows that

-145-
it is a monotone decreasing function of ^r>max ^for fixed ratios,
a\¡a\ ( i = v-p+1, ■ • • ,v) , of the variance components^. It is thus
desirable for Ar?max to be small.
The power of the exact test associated with the hypothesis
Ho:L,Tí=|, i € f-p. To test HqiL't;. =£, i the appropriate test
statistic was (see (3.286))
SSH./r, .
FHi“~MS ’ lG:f
fi
f *
0
Recall that r^ = rank(L). Also, SSII^ (iGf^) was defined in (3.283).
Furthermore, ÍqGfr is such that E[MS^j 3=^ ^see (3.259), (3.273),
and (3.276)^. Now, under Ha:L'ri/£, FHi has a noncentral F-
distribution with noncentrality parameter A^(i€f^) (see (3.285)^.
Thus, if is the power associated with this test, then
^fi “ P(FHi > FQ,rL,m. lHa)
l iQ
(Ai)>Fa>r
0
L>
m- )»
10
6J
f *
(3.301)
Now for i€f.p 4^ monotonically increases in A^ (see Graybill
(1976), p. 130^. However A^ monotonically decreases in (recall
(3.285)1. Since £*■ =6• + Xr a?, A- monotonically decreases in
x 7/ ti i t?max c i
A_f iGÍr. Therefore it is desirable to have Ax small.
x y niuX i x j ni3>x
The design parameters Armax and max* The following lemma
provides upper and lower bounds for Armax and A^ max-

-146-
Lemma 3.11
Let n^1^=mini nr and recall the definitions of rjrl and 77^ in (3.232)
and (3.264), respectively. Then,
c-1
Y' J__
C r£Tnr
c-nri-!
,(0
< A < 1
—,max — ^1y
c —
'jfi
< Xr <—r~s
— i,max — (x)
nv '
Proof:
(a) Recall that emax(A) denotes the largest eigenvalue of the matrix
A. Then, from (3.223),
â– V,max = emax(^r) = emax(^r^Qr)
— emax(^)emax(^r^r)
=emax(^)emax(^rQr)
=emax(^) (Qr has orthonormal rows)
.co¬
lt is also true that
emax(QrKQr) ^ f/rltr(Qr^Qr) *

-147-
The term on the right being the average of the eigenvalues of
QrKQ'r. However,
tr(QrKQ^) = tr(Kq,rqr)
= tr(K(Ic-Q'fqf))
(see (3.244)).
From Lemma 3.8(a) and for q^ = [q^: • • • :qj/_p]/ we have
M'
Vc~c
= jc + q£'q£.
Hence
tr(K( ic - q'f qf )) = tr(K( Ic - Jc) - Kq^)
= tr(K(Ic-Jc))-tr(Kq;'qp.
Now,
tr(h( Ic — Jc)) = tr(K) — tr(KJc)
= tr(K) -itr(Klcl'c)
= tr(K)-Il'cKlc
— V y' J-
~ /—'_nr c 2-» nr
r G T r r G T r

-148-
(b)
Also, tr(KQ^'Q^) = tr(Q^KQ^). But, since = Ic_^ _-p it
follows that / >, Q jQy ~ is p.s.d. Consequently,
n ' ^ '
tr(qjKqJ') <-iy(c-nrl-l).
Thus
tr(QrKQ'r) >
c-1
T —
â–  GT t
C-Dri~1
.0)
and
'r, max — r¡r j
c-1 v' 1 c ^ri ^
>_i_ Y' _L_
- c n,.
rlTA n(0
Af max is the largest eigenvalue of defined in (3.266). Thus
^f,max — emax(^f) — emax(^v^Qv)
— emax(^)emax(Qv^v)
.(0
(Qv has orthonormal rows).
On the other hand, it is true that Ar is greater than or
1 y llldA
equal to the average of the eigenvalues of QVKQ(,. That is,
emax(Qv^Qv) — T^brCQv^Qv) •

-149-
Now ,
tr(QvKQy)= tr(KQ(,Qv)
= tr(K( I - Qv'Qv))
where Qy consists of all those matrices, for which iGfc.
(Ve note that if Qv = Q then tr(QvKQy) = tr(K) = E w~‘ And thus
v r G T T
emax(QvKQ'v) >WT E it- Also, in this case r?f = c (see
'ti r g T T
(3.264))). Now,
tr(h(I — Q3'Qv)) = tr(K) “ tr(Q<-Kq$#)
= E ¿-tr^KQ").
r G T r
However, —trQvQv' “ QvKQv7 is P-s.d.
n' '
Consequently,
tr(Q3KQ") < —tr-tr(QyQy7)
n' '
(c-r?fl).
tr(qvKQ(,)> E WZ
rG T T
c-
’ifl
.(0
Thus,

-150-
Hence
f ,max _ emaxVHv
(qvKQv)>
Ifi
y JL_'
r G Tnr
^fl
.(0
â–¡
Some remarks are in order:
(i) In an unbalanced random effects model where the imbalance
affects the last stage only, such as described in Khuri (1990),
c — »7rl = 1. Thus the lower bound given for Ar?max in Lemma
3.11(a) reduces to the lower bound reported in Khuri’s (1990)
Inequality (4.1).
(ii) If Qq is not one of the matrices in the definition of C)v ^see
(3.271)), then similar to part (a) of this lemma, the lower bound
for Xn can be improved upon. The improved lower bound can be
x 9 max
shown to equal
1
’ifi
c-1
y -i__
r € t"7-
c
This follows similarly to the proof of Lemma 3.11(a).
(iii)Recall that Ar?max is the largest eigenvalue of QrKQ^ while
Af max is the largest eigenvalue of QVKQ(,. Similar to the
discussion concerning Amax presented after Lemma 3.7, Ar^max will
be invariant to the choice of Qr and A_p will be invariant to
the choice of Qv.

-151-
3.3.7 An Alternative Approach to the Analysis of the Unbalanced
Mixed Model with Imbalance Affecting the Last Stage Only
Introduction. In this section we reanalyze model (3.189).
Instead of making transformations on only portions of u ^as in
(3.237), (3.265), and (3.292)), we now transform all of u. We will
again utilize resampling in this transformation. The result will be
a set of sums of squares which are independent and distributed as
scaled chi-squared variates analogous to those in a balanced mixed
model.
This approach has advantages and disadvantages. One advantage
is that only a single transformation is needed to reduce the analysis
to that of a balanced mixed model. Thus exact tests (when they
exist) for the variance components and for estimable linear functions
of the fixed effects are easily constructed (as they are in the
balanced case). Also, when no exact tests exist, Satterthwaite’s
procedure is readily applied in the usual way.
A disadvantage to this approach is that the power of the
resulting exact tests cannot be any greater than the power of the
exact tests derived in the previous sections. In fact, we will show
that the power of the exact tests resulting from this new approach is
equivalent to the power of the tests we would have obtained if we had
just simply deleted observations until each cell contained
n(1)=minn_ observations.
r€T t
This disadvantage is the principal reason why we recommend
adopting this approach only when a few of the cell frequencies are
large relative to n^1^. In situations where only a few of the cells

-152-
contain n^-1^ observations and n^1^ is small relative to the other cell
counts, we recommend the use of the procedures developed in the
previous sections.
One final remark: Unfortunately some notations will need to be
reused. However, this should pose no difficulty since it will be
clear from the context when a symbol is uniquely defined for this
section or when it has taken on the same meaning it had in some
previous section.
Utilizing resampling. Recall the definition of u given in
(3.243), namely u = Qy is defined in (3.214)^. From (3.246) we have
u ~N|
’iViImi
+ where is given in (3.244) and G is defined in (3.247). The
(i =0,1,• • • ,u) are defined in (3.217).
The matrix R (see (3.226)1 can be written as
R = CAC'
(3.302)
where C is an orthogonal matrix and A is a diagonal matrix whose
diagonal entries consist of n-c ones and c zeros. Ve can partition C
and A as
C = [Ca:C2:C3]
A =diag(Ic,In_2c,0)
(3.303)

-153-
where C1,C2, and C3 are matrices of orders nxc, nx(n-2c), and nxc,
respectively. We must necessarily assume
n > 2c. (3.304)
This assumption is more stringent than the assumption in (3.195) in
the sense that if (3.304) holds then (3.195) will hold. But the
opposite may not be true. Even if (3.195) holds, (3.304) may not
hold.
Since C is orthogonal we have
C/iCi = I (i = 1,2,3)
(3.305)
C'jC-0 (i#j).
Consequently,
R — CjCj + C2C2 •
(3.306)
Also,
SSE = y'Ry = y'CjCjy + yC2C'y
where
= SSEj + SSE2,
(3.307)
SSEi=y'CiC'iy,
i = 1,2.
(3.308)

-154-
Furthermore y, SSEj, and SSE2 are mutually independent and
SSEj/cr* ~ x2
(3.309)
SSE2/^ ~ Xn_2c
Now, we utilize resampling to transform the vector u into a random
vector that has a diagonal variance-covariance matrix. We do this by
considering the cxl random vector w defined as
y = a + (Wic-G)5ciy,
(3.310)
where Amax is the largest eigenvalue of G. Because AmaxIc —G is
1
p.s.d., (AmaxIc —G)2 is well-defined with eigenvalues equal to the
square roots of the eigenvalues of AmaxIc —G. We note that in this
case
^max — emax(^) — emax(Q^Q )
=emax(K) (Q is orthogonal)
= - ) . (n^1^ = minnr)
.(0
r 6 T
(3.311)
Partition w just like u in (3.245). That is
«= (v'f-VrY = (y'o: • • • :^-p:4,-p+l: • • • (3.312)

-155-
where each is of order m- x 1, i=0,1,• • • ,v. The next theorem
provides some distribution properties for the , i =0,1,• • • ,i/.
Theorem 3.10
(a)
E(c-i)=|
Í QiHg,
i = 0,1, • • • ,v-p
[Q,
i = r-p+1, • • • ,i/.
(b)
Vq >-1’' '
•are
normally distributed random vectors with
having variance-covariance matrix
Var(w.) = («i+-^yor“)Im< , i =0,1, • • • ,v.
n' ' 1
(c) The vector w is independent of SSE2.
Proof:
The proof of parts (a) and (b) follows closely to that given in
Theorem 3.8. The proof of part (c) closely mimics that given for
Theorem 3.7(c). â–¡
Recall from (3.256) that Q|Hg = Q^r., i = 0,1, • • ■ ,r-p, where the
vector r-x is defined in Lemma 3.10(c).
The vector u. Recalling Lemma 3.9 and that
y = Xg + Zh + e (see (3.196)),
y = Hg + Fh + c
(see (3.206)),

-156-
and
u =Qy ^see (3.243)),
we see that the vector u) ( in (3.310)1 can be expressed as
y = Qy + (-rpj-íc _G)^cí(xg + zb + £)
n' '
= QHg + QFh + Qe + (—Jjylc - G^e .
n' '
The last equality follows since range(C1) Crange(R)
CiX = CiZ = 0. Now, let
and consequently
/Q_pH\ /Q^H\
H- = QH = (qrH) = ( o )
(3.313)
/ Qf F \
Fw = QF = (qrp)
(3.314)
£w = Ql+(-77yIc-G)5cif-
(3.315)
In (3.313) recall from Lemma 3.8(c) that Q^H = 0 for i = i/-p+l, • • • ,u.
Also was defined in (3.244) while Qr was given in (3.220).
Therefore we can express u as
y = Hwg + Fwh + ew. (3.316)
Clearly E(ew)=0. Furthermore, since 7 = Dc ^see (3.200)), it must be
that 7 and C^e are independent. This follows since DR = 0 ^Lemma
3.9(b)) and range(Cx)Crange(R). Consequently,

-157-
Var(íw) = QVar(l)Q' + (-Mc - G)^iVar(£)C1(-iyIc - G)¿
n'' ' n' '
^QKQ' + ^-^Ic-G)
n' '
_1 2 T
J¡VCc'
The last equality follows since QKQ'= G. Thus model (3.316) describes
a balanced mixed-model. Letting —T~^e^ i = 0,1, • • • , v,
or
we
obtain from Theorem 3.10 that
N(HWg> i®0^iImi)*
(3.317)
Let
SSi=w,iwi, i =0,1,
(3.318)
then, from Theorem 3.10, we have:
(i)
i =0,1, • • • ,v
where
2íiriqiqiri:
0,
i = 0,1, • • • ,i/-p
i = iz-p+1
(3.319)
(3.320)

-158-
(ii) For MS^=SS^/m^, i =0,1, •••,&'
E(MSi)
ii+iffr-r-q-Qiii,
i = 0,1, • • • ,I/-P
Í = V-p+1
(3.321)
Thus, SSqjSSj, • • • , SS„, and SSE2 act like sums of squares in an ANOVA
table for a balanced mixed model ^see Khuri (1982), pp. 2915-2917^.
That is, analyses concerning the variance components and estimable
linear functions of the fixed effects can proceed using these sums of
squares just like in a balanced data situation. In fact, if the
design was balanced, then similar to Section 3.3.5 we can show that
NSSqjNSSj,•••jNSS^ reduce to the usual balanced ANOVA sums of squares.
(Recall that N is the common last stage cell frequency.)
The variance components. Using SS1/_p+^,* • - jSSj,, and SSE2 we can
proceed to test hypotheses concerning the variance components
similarly to the methodology presented in Section 3.3.3. Suppose we
wish to test Hq: of the exact test (assuming such exists based on SSJ/_ , • • • ,SSV,
and SSE2) can be expressed as (take SS^.^ = SSE2, ¿^^ = 0, m^+^= —
n
/SS./m.
'p. — PI i-—I > F I
t\sSj/mj-*“»“i.“i0l
i = v-p+1,
(3.322)
rss-i
SS-
where SSj/mj is such that E
Lmi J
= E
J
mj
when Hq:(t|=0 is true.
Under
Ha>
SSi/mi
E
keWj
Vk + '
0Te
E
k€V.
bkak+-
(0
SSj/mj
x

-159-
Thus, for i=y-p+1,•••,v,
*i“ KFmi’mj - l+biíiFa’rai’mj)
(3.323)
where
key. >
(3.324)
^Compare (3.223) and (3.224) to their counterparts expressed in
(3.299) and (3.300), respectively.)
Now, let us suppose that the design is indeed balanced. That
is, suppose nr = N for all t£ T ^recall (3.191)). To test the above
null hypothesis we would form
NSSi/mi SSi/m{
FBi = NSSj/mj = SSj/mj’
rss-i
ss.
^Recall that SSj/mj is such that E
L-iJ
= E
1
3
^ c_
l
(3.325)
when
Hq: this test, Vl'g. , is
^Bi -KFmi>mj-l+bi0BiFa’mi’mj)
(3.326)
where
^Bi
keW.
(3.327)

-160-
( i = í/-p+l, • • • , t/) . Thus, the power of the above exact test for
Hq:o,|=0 in the unbalanced case ^see (3.323)^ is equivalent to the
power of the test based on the following procedure:
(i) Randomly delete observations, if necessary, until all r-
cells contain n^1^ observations.
(ii) Analyze the resulting data set as if it arose from a
balanced mixed-model with n^1^ observations per cell.
We would therefore expect the methodology developed above for testing
hypotheses concerning the variance components (for unbalanced models)
to be inefficient when most of the cell counts are much larger than
n^1). Under such situations, Armax will typically be less than
1/n^1^ and thus we would recommend using the procedures developed in
Section 3.3.3 ^see Lemma 3.11(a)j.
Table 3.5 compares the powers of the exact tests for different designs arising from an unbalanced mixed two-way cross¬
classification model such as described in Gallo and Khuri (1990).
(We assume the “a” effect to be fixed and the “/?” and “a/?” effects to
be random -- their variances being expressions used in the comparisons were derived in (3.299) and
(3.323). The values for and and Littell (1987).
Design I was chosen so that one cell would have a small cell
frequency relative to the rest; while Design II was chosen so that
n^1 ^ = 1. In both cases -V,max < l/11^ so that the testing procedures
developed in Section 3.3.3 are to be recommended. Design IV also had

-161-
n^1^ = l. However, unlike Design II, almost 25% of the cells had only
one observation in them. This is why Ar?max=l. In this case we
would recommend using the methodology developed in this subsection.
Finally, even though Ar>max < in Design III, the gain in power
is not worth the extra effort necessary to develop the test
statistics in Section 3.3.3. Thus in “nearly balanced” designs we
would again recommend using the procedures developed in this section.
We should point out that the unbalanced mixed two-way cross¬
classification model is a very special case of the general unbalanced
mixed-model (with imbalance affecting the last stage only). Thus, we
may need to modify our recommendations when more general models are
considered. Ve make two final remarks:
(i) In Section 3.3.4 we described a procedure for constructing
exact tests (when such exist) for testing a testable
hypotheses concerning estimable linear functions of the
fixed effects. The power of these tests was derived in
(3.301). Recall that the power is a monotonically
decreasing function of £_p- (see Section 3.3.6). Exact tests
for the fixed effects can also be constructed by utilizing
SS
, • • • ,SS„ ^see (3.318)^. As in Secti on 3.3.6, the power
of these tests will be a monotonically decreasing function
of ^=^4- l/n(l)or2. since if i = ^ + Af?max <1/n^1^ by Lemma 3.11(b), we can conclude, similar to the
recommendations made above, that when most of the cell
counts are large relative to n^1-^ the procedures developed

-162-
Table 3.5 A comparison of the powers of the exact tests for in an unbalanced mixed two-way cross-classification
model
Cell counts:
row=a effect
col=/3 effect
ff2t3
aal3
tfr2 (see (3.299))
(see (3.323))
.2
.2
.138
.121
2 10 9
.2
5.0
.059
.059
I 10 10 9
1.0
1.0
.229
.215
9 9 10
5.0
.2
.787
.750
5.0
5.0
.273
.268
Ar,max =-370; l/n^ = .5
.2
.2
.102
.091
14 4
.2
5.0
.058
.058
II 3 3 4
1.0
1.0
.193
.175
4 4 3
5.0
.2
.687
.633
5.0
5.0
.260
.252
^r,max= *750; 1/n
t1) = 1.0
5 5 5 5 6
.2
.2
.473
.470
6 5 6 5 5
.2
5.0
.082
.082
III 6 4 5 6 5
1.0
1.0
.668
.667
6 6 5 5 6
5.0
.2
.994
.994
6 5 5 5 6
5.0
5.0
.721
.721
^r,max — -2457; 1/n^ ^ —
.25

-163-
IV
Table 3.5 — continued
Cell counts:
row=a effect
col=/J effect
*r2 (see
(3.299))
(see
(3.323))
9 2
9
1 2
.2
.2
10 1
2
9 10
.2
5.0
In this
case the
powers are
identical
1 8
1
2 2
1.0
1.0
since Ar
,max = 1/
n(1) = 1.
9 10
1
9 3
5.0
.2
8 3
2
10 1
5.0
5.0
in Section 3.3.4 should be used. Otherwise construct exact
tests by utilizing the sums of squares developed in this
section.
(ii) Our methodology has an advantage over the methodology based
on deleting observations until all r-cells contain n^1-^
observations (as described earlier) in that it uses all the
data in the calculation of the test statistics. This is
clearly a desirable feature especially if n^^=l. For
example, if n^-1^ = 1 then the “deleting observations
methodology” could not produce a test for <7^ in a two-way
cross-classification model, whereas our procedure would
allow us to construct such a test.
Ve now turn our attention to the development of exact
simultaneous confidence intervals on estimable linear functions of
the fixed effects. Since model (3.316) describes a balanced mixed-

-164-
model, we will use the results of Khuri (1984) to obtain our
intervals.
The fixed effects parameters: Introduction. In Section 3.3.4
we developed exact procedures concerning estimable linear functions
of the fixed-effects parameters. The development involved first
constructing the vector v ^see (3.257)), which consisted of those u^ ,
i = 0,1, • • • ,i/-p (and the corresponding u^ , ig = i/-p+l, • • • ,i/) for which
¿i =6^, then making a transformation to arrive at ^see (3.265)).
The 5^, k = 0,1, • • • ,v were defined in (3.217). The , i £ If ^see
(3.259) and (3.268)) were then utilized in the construction of the
exact procedures. The motivation behind the construction of v came
from the procedures used in dealing with estimable linear functions
of the fixed-effects parameters in a balanced data situation. For
balanced data, exact methods can be derived if a proper mean square
can be found in the corresponding ANOVA table that can serve as an
“error” term in the F-ratio used for testing a hypothesis concerning
the parameters associated with the fixed-effects.
Since SSq, • • •,SS¿, and SSE2 ^see (3.318)) mimic the sums of squares
associated with a balanced mixed model, we can also develop exact
procedures concerning estimable linear functions of the fixed-effects
parameters (based on the methodology developed in this section) if we
can find a proper mean square that can serve as an “error” term in
the F-ratio similarly to the balanced data situation. If no mean
square exists that can serve as an “error” term in the F-ratio, then
we can apply Satterthwaite’s procedure in the usual way to develop
approximate methods. However, as pointed out in a recent article by
Ames and Webster (1991), in many instances the Satterthwaite

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estimator of the denominator (“error”) degrees of freedom will be
positively biased. This will cause the resulting inference to
overstate the actual Type I error rate. This same criticism also
applies to the approximate method developed in Section 3.3.4. ^In
certain layouts Ames and Webster (1991) develop an improved estimator
of the true denominator degrees of freedom. Details can be obtained
from this article.^
For balanced mixed models, Khuri (1984) developed a method for
the construction of exact simultaneous confidence intervals on
estimable linear functions of the model’s fixed-effects parameters in
situations when no proper mean square existed which could serve as an
“error” term. We now extend his results to the unbalanced mixed
model where the imbalance affects the last stage only. The idea is
to find a sequence of independent and identically distributed random
vectors each having a normal distribution. Furthermore their common
mean vector will consist of a vector of estimable linear functions of
the fixed-effects parameters. Exact simultaneous confidence intervals
on these estimable linear functions can then be obtained by utilizing
this sequence of estimators.
The fixed effects parameters: Notation and preliminaries.
Recall that r=(kx,k2, •••,kg_^)fsee (3.191)^. By rearranging, if
necessary, the order of the elements in r, we can consider the first sa
subscripts (1 (3.316); the next s2 subscripts (l subscripts, that is, subscripts associated with those random-effects
in (3.316) which do not nest each other and are not nested in any
other random effects; and the last s3 subscripts are associated with

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random effects nested in at least one of the preceding s2 effects. We
note that
si + s2 4- s3 = s-l. (3.328)
Define
s-l
d1= n a- (3.329)
l^Sj+1
d2 —
s-l
n a- ,
i=s1+s2+l
1,
if Sj + s2 + 1 > s-l
else
(3.330)
SJ + S2
d3= . n ai=d1/d2.
l=Sa+l 1
(3.331)
Recall that c =
of dimension s2
_II^a^ ^see (3.192)^.
of the form
Also let A be the set of vectors
A = {('■•!' = (i
Sj + 15 *s2+2’
51 + S2)'}
(3.332)
with
ig — 1)2)'* ' ?a^j £ — s^+1, • • • , Sj+s2.
We note that the number of vectors in A is d3. Now, from (3.203) and
(3.204) recall that the CXC| matrix was defined as
s-l
Hi = ® Lig» i =0,1, • • • ,u
1 1=1
where c^ was given in (3.194) and

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_ for
Li£ Ua£ for k£ e
(i=0,1,••• ,u; l=1,2,•••,s-l). Hence, for i=0,1,•••,v,
Hi=(«=lLi£) <8’^£=^>+lLi^' (3.333)
Thus, for i = 0,1, • • • ,i/-p (that is, H- is a matrix associated with a
fixed-effect) we have
H- — ( ® L • ® f <3> L • nj
1 V i= 1 ki=Sl+l
= (¿N
!d.
(3.334)
This follows since the subscripts associated with the random effects
in(3.316) cannot belong to , the set of subscripts associated with
the fixed i^ effect in the model, i = 0,1, • • • ,i/-p. Note that (3.334)
can also be expressed as ^see (3.331)^
Hi=(£=lLi£) ®id3®1d2’ i =0,1, ••• ,i/-p. (3.335)
Furthermore, from (3.313), let the cxcj matrix be
Hwi = QHi ’ i = (M> ' ' ’ >"-P>
(3.336)
then

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H«i = *
(&.«)
®ld3®Íd2
, i = 0,1,• • •,v-p.
For i = 0,1, • • • ,i/-p and v G A define the cxcj matrix as
Hj = ( ®Li£)®(S®S2la>ld
1 \l=i V£=Sl+l V d
where la^ is a vector of dimension ag (Í= Sj+1,•••,Sj+s2) whose i
entry is unity and the remaining entries zero (ig = 1,2,•• • ,ag).
Similar to (3.337), define cxcj the matrix as
H~i=QHj, i = 0,1, • • ■ ,i/-p; j/GA.
Finally, recall that = H-H'. , i = 0, • • • ,u ^see (3.207)^. Thus,
(3.208), we have for i=0,1,• • • ,u-p,
A
= ( ®M-gW V Ja )
W=1 xi> W=s1+1 V
= ( AMi>Jd3«Jd2-
Furthermore, define the cxc matrix A • as
’ UJl
Au,i = Hu,iHLi = QHiH'iQ' = QAiq'
(i =0,1,
a a _iiif uv>
wi wi wi
,j/-p). Also, for i=0,l,
. Then
• ,i/-p; VGA let k\ = H^Hj'
A
V
T
1
Sl + S2
£=Sj+l
(3.337)
(3.338)
th
i
(3.339)
using
(3.340)
(3.341)
and
(3.342)

-169-
and
A£i = QHjHi'Q' = QA?Q'
(3.343)
where Jaj| is an a^xa^ matrix whose (i£»ig) entry is unity and the
remaining entries are zero, ig = 1,2,•• • ,ag; £ = sa+l, • • • ,s1+s2. Both
and A^. are of order cxc, i = 0,1, • • • ,j/-p; t/gA.
The fixed-effects parameters: Initial development. From Khuri’s
(1990) Lemma 3.2 we have for i = 0,1, • • • ,v-p
j=ovuj
j=0'uj
(3.344)
The last equality follows since A-j=0 for i =0,1, • • • ,r-p and j=i/-
p+1, •••,!/ |see Khuri (1982)^. The cxc matrix is associated with
the sum of squares for the i^ effect in (3.213), i = 0,1, • • • ,i/-p.
Define the cxc matrix P • as
wi
b^A-=Q(MA>'’
r-pA-
(3.345)
(i = 0,1, • • > ,i/-p) . From (3.344), we can equivalently express (3.345)
as
Pwi=QPiQ', i = 0,1, • • ■ ,v-p.
(3.346)
Furthermore, define the cxc matrices P^ and P-. as
1 D\
i = 0,1,••• ,i/-p; y g A
(3.347)

-170-
pSi=0éfyzi=qp!q'’ i=o’1’
,v-p; ye A.
(3.348)
The last equality in (3.348) follows from (3.343).
Now, consider the Product for i , ir = 0,1,• • •,^-p. From
(3.337), (3.340), and (3.345) we obtain
PuáHuú'
QPjQ'QH.^QP.H.,
t/-PA. • Sj
-T-— ® M -^L
j=0 b j fci j£
(3.349)
Similarly, from (3.337), (3.342), and (3.347) we have
pEiHwi' = qpiQ/qHi' = qpiHi'
v-p X
j=0 bj ¿=1 J 1 U V
sl+s2 i
® la
i=Sl+l
5
(3.350)

-171-
(i , i7 = O, • • • ,i/-p; »^€A). Define the matrix N^./ as
v-p Ai
Nü'= T,
11 >0
bjJ Éf1l‘jíLi,í’ 1
i,i'= 0,1,
,í/-p.
Then, from (3.349) we see that for i = 0,1, ■ • • ,J/-p,
PwiHw = Pu,i(t3H0:‘iHi:---:^-p) (see (3-313))
= Pwi(HwO:Hwl: ‘ ' ' :Hw,i/-p) (^e (3.336))
= (d1Q[Ni0®ldi]:..-:diq[Nijl/_p®ldi])
= d1q(Ni®idi),
where
Similarly, from (3.350) we have for i = 0,1, • • • ,i/-p; vEA,
P£iHu = fcSj + 1
Thus, for i = 0,1, • • • ,u-p; vEA,
rank(pjf,iHu;) = rank(N- ) = rank(PwiHw)
(3.351)
(3.352)
(3.353)
(3.354)
= rank(qPiq,qH) (see (3.346) and (3.313))

-172-
=rank(P^H)
= rank(Q^Q^H) ^recall (3.254)j
= rank(Q-H) (Q^ has full column rank)
= m •
^by Lemma 3.10(a)j.
(3.355)
Now, there exists a matr
such that consists
i = 0,1, • • • Define
ix C| of order m^xdg (dg^II^a^) and rank
of m^ linearly independent rows of N^,
the m-xc matrix T» as
lL)i —
(3.356)
where
wi
s1+s2
®
£=S1+1
(3.357)
(i =0,1, • • • ,v-p; p€ A).
The fixed-effects parameters: The sequence of estimators. For
each i = 0,1, • • • jfc'-p; v 6 A define the cxl random vectors, zj, as
1/ rpV
-l = Twi^‘
(3.358)
Ve now establish some distributional properties for the z^. Before we
do so, however, we will need the following lemma.

-173-
Lemma 3.12
Recall the definitions of in (3.344) and Pj in (3.347),
i = 0,1, • • • ,i/-p; u£A. Then
(a)
pi'pi'
FiFi'
i = i' = 0,l,-
i ^ i'; v e A.
,p-p; v E A
(b)
K _ íár[Ido®Jdi]PI’ i = i' = 0,l,---,i'-P5 v£A
^i^i7 | 0, ijii'; v€A.
Proof:
(a) From the fact that P- is idempotent, i=l, •••,!/ ^see Khuri’s
(1990) Lemma 3.lj, we have
r-p A
\2 p-P A
A- -A. .,
iPi=( E -^a.) = £ -t¥a.a. + e ^bU'AjAi
v j=o bj j=o bj J J j ¿ y bjbj' J J
= d
v~ PA2i i
1 £ t¥
j=0 bj
g® Mj£Mj£
+ dj £
Viw
j^Wl
gfjpyt
&l
® M
«=1
jgMj£)®Jdi

-174-
v j + y bjby «=i J u dl
= dl{( Jd‘j
.iz-pA.-Sj v
( j?0"b“gf1Mj¿)® Jd1 = Pi’ 1 ~ °’1’' ' ' ’^"P-
Consequently, we must have
/i/-pA..s1 ,2 "_P A • • sa
^i( S "k— ® ^ itf) = S “ü ® M •£.
V >0 bj«=l j=0 bj«=l
(3.359)
Thus, for i = 0,1, • • • ,J/-p; ifGA, we obtain
/ i'-p A- • n2
p;p;=( P0TfAl)
(see (3.347))
u-p A, •
?-• bi
°1 / l/?\
® M -a®! da!)® ^d
t-1 V i=Sj+l a«' d
(see (3.342))
/ ^-P A• • sa . / Sj+s2 1N
( E-K-^®Mii)®( ® Ja!)®Jd
' j=o bj É=! J<7 V£=Sl+l V d2
= d,
,1/ P A* - Si \2 ✓ Si+So i/>\
( E 41® M.£) ®( ® Ja«)®Jd
V j=0 bj^l J ' ^fcSj+l V d2

-175-
Since
for i
After
/1 pA* • s, \ / í a / \
(Í j?oTfi®Uje^s%J‘e)®\ (from (3’359))
4-P;
d3 1
(see (3.331)).
P^P^, = 0 for i ^ i1 (see Khuri’s (1990) Lemma 3.1)) we have,
v-p A-
0 = PiP.,=
<-!• • v / V-P A././ \
"-P AijAi'j
AijAi'j'
= £ A -A • + £ K K A-A.,.
j=0 b2j J J j £ j' bjb j' J J
some simplification similar to the above we obtain
0 = d
11
"¿p x.iln %
h t
+
/ Ai iAi'P Sx \
= d.
u-p A- â–  sx v v-p A.,.,
E -r ® m J E -tH- m
'jio bj «=i J'Ajt0 by J"
«8» J.
i/-p i/-pAi iA.;., s
E E hK J ® M -»M ,. = 0.
3=0 bjbj' «=1 J£ J *
(3.360)
Thus, for i i'; i/€A, we have

-176-
v-p i/-p A- -A.,.,
P;Pf, = T. E a3aV
1 1 j=0 j'=o bjbj' J J
j/-p i/-pA- -A., ./(s, / si+s2 i,\ I /
= d2j?Oj5o bjbj' MJ£Mj'«®(«=s®+lJa^Jd2j (See (3,
342
= d„
v-p i/-p A- -A.,./ s, \ , s.+s., j.
= 0 (by (3.360)).
(b) For all i = i' = 0,1, • • • ,J/-p and u G A,
. i/-p A • • ., v-p A. ., .
J/~P A; â–  A- -A. .,
= -?n "ST AJAJ+. WVj'-
j=0
J 7
From (3.340) and (3.342) we can express this as
- u2
i/-p A|,
d2 £ -¿r
j=0 bi
® ( ® d-7)® Jj
g=l d3''g=s1+l d
+ d2 £
2 .j?., b-b.,
J # J J J
gflM j«M j'g ® Jd3(g=s®+1 Jag)® Jd
= d0
AijAij/ll
i/-p Aj • Sl
2 --j ® M • »M •»+ £
j=0 »j t=i J{ J{ j#y -j-j

-177-
sa+s2
¿=Sj+l
i/-pA. •
T -Ü
j?obj
u-pX. .
T —¿
j?0bj
S1
( sl+s2 i
n\ 1
® M
¿=1
® Jd
a3
UsV*
si
Sl+S2 in
i
(see (3.359))
® M 4 n
1=1
®Jd3(
tJ^4
)®Jd2|
[d0®Jd3® Jc
’1
^-pA- •
z-é2
j=0 bj ¿=1
s,+s,
6=Sj + l
When i ^ i' and v EA we have, reasoning similarly to the above and
using (3.360), that
P-P~, = 0.
i i
Now, recalling (3.358) and using Lemma 3.12, we have for
0,1, • • • ,J/-p; v G A,
(i) E(zj) =T^iE(w) =T~iHwg (see (3.317))
K¿iP~iHwg (see (3.356))
= K
MNi®(fc®s+id>i
V ~a2j
g (see (3.354))

-178-
= ® (1) ® d2j g (see (3.357))
= d2C?Nig. (3.361)
(ii) Var(z^) =T¿i[Var(y)}T
VI
CJÍ
v\Z
’Jy. ® , «’iK
' l=í/-p+l 7
+
_±__* T
Ff ‘ c
Ipü' v\L<
^wi wi '
The term in the brackets follows from (3.316) and the facts that
Var(£w) = —h)cre1c (see (3-315)) and k = (£i/-p+i: ‘ ’' :&y with
II ' '
Var(/?| ) = cr|Ic. , i = i/-p+l, • • • ,i/ ^see (3.196)J. Now from (3.314)
and (3.348) we have, since Pj is symmetric (see (3.347) and
(3.342)),
v»r(2«)=K¿iH[p5F(=i/®p+1d>'Pi+7rr?p;p; q'K
i'K-'.
wi 1
Now, the first term in the bracket can be written as
f( © = píj( £ a2.A.)pf (see (3.210))
1 —IS-p+1 7 V j=l/-p+l J J7 v 7
. £ /j*ijpI
j=i/-p+l
(see Khuri’s (1984) Lemma 5.2)
where
Í o, V’j
1 j lb j/d3, V’iCV’j,
• =
(3.362)
(i = 0,1, • • • ,t/-p; j = j/-p+l, • • • ,i/; v £ A) . Also

-179-
bj = iTTi¿ ’ ■i = "-p+1>
n
£=Sj+l
where
c _ í a£’ k£€^j
^ 1, G V1 j»
^see Khuri (1984)^. Therefore, recalling (3.217),
E ■7j',íjpÍ = Í( E b?j)pI
j=í^-p+l J J *^3' j G Wi
Hence, from Lemma 3.12(a) and (3.357),
Var(z^) = k£.Q
â– U E b*. ^ jeV- J 1 d3 ni1) f 1 "
vt
CJÍ
S=s1+1
Now, let
V-P i Sj
Bi = E -ü— i =0,1, • • • ,i/-p
1 j=0 bj 1=1 Jí
(3.363)
(3.364)
(3.365)
(3.366)
i =0,1, • • • ,i/-p,
(3.367)

-180-
1/ í S1^S2 i A /
then noting that P-r = B- ®( <8>
1 1 '£=Sj+l v 2 '
(3.347) and (3.342)), we have that (3.365) can be expressed as
í|[c?BiCÍ'®(l)0(d2)] = ^C?BiCÍ',
(3.368)
(i =0,1, • • • ,v-p; vG A).
(iii) For i ^ i7, v£A, and recalling P~- is symmetric
Cov(zf,z|/)=T^i[Var(y)]T^/./ (see (3.358))
^u;i* u>i
¡® {iIm¡ PL'Kui' (see <3-356) and <3-317))
(see (3.348) and (3.214))
%
%
1 CJ1
K5i«pí( |0f
K5iOpi(.£0ijpj>!'Q'K£'i'
(see (3.252))
= KiiQPI
ij' /
tp i(
q'K^7., = o
L>\
(3.369)
This follows from Lemma 3.12(b) and the fact that P^ is
idempotent for i = 0,1, • • • ,i/-p. We have therefore just proven the
following theorem.

-181-
Theorem 3.11
The random vectors zg, Zf,•••,zp_p are independent normally
distributed vectors with
(a) E(zj) = d2C?N^g, i =0,1, • • • ,i/-p; v G A
(b) Var(zJ)=^CÍB.CÍ/, i =0,1, • • • ,«/-p; v € A
where d2 was defined in (3.330); g is the vector of fixed effects
defined in (3.196); C| is such that consist of linearly
independent rows of (n^ was given in (3.353)); was gi^
(3.366); and £* was defined in (3.367).
Lven in
This theorem shows that the random vectors in the set {z^:^gA}
share the same mean vector, d2CjNjg, and the same variance-covariance
matrix, í^djC^B .C¡y, j=0,1,• • • ,r-p. The matrix C*-B-Cy ( j = 0,1, • • • ,u-
p) is nonsingular since if it were singular, the rows of T~ • (see
(3.356) and (3.358)) would be linearly dependent. But this implies
that the rows of T~jHw would be linearly dependent which is a
contradiction since T~ -Hw = d2C^N• by (3.361) and was determined so
that CjNj would have linearly independent rows.
Further, CjNjg is estimable for all j = 0,1, • • • ,//-p since
(by (3.361))
= T^jQHg
(by (3.313))

= T£jQE(y)
-182-
(by (3.210)).
Also, for each j = 0,1, • • • ,i/-p and r 6 A, the rows of P^jand those of
P -H,. span the same vector space spanned by the rows of N- and hence
w J J
by CjNj (see (3.354) and (3.355)). Since P^ = QP jQ'QH = QPjH (see
(3.346) and (3.313)), the space spanned by {QPgllg, • • • , QPj,_pHg) which is
the same as the space spanned by {QgQgHg, • • • ,Q^_pQj,_pHg} is the space
of estimable linear functions of the fixed-effects parameters in model
(3.316) (recall Lemma 3.10(c) and (3.352)). Thus {CgNgg, • ■ • ,CJ)_pNl/_pg}
must also span the space of estimable linear functions of the fixed
effects parameters. Consequently, we can restrict our attention to
the development of exact procedures concerning CgNgg,•••5C^_pNj,_pg.
The fixed-effects parameters: Exact procedures. Consider the
set
A0 = {v-v = qls2’ ! where q is constant and
Tq = mm a- ,
U s1+l < i < s1+s2 1
Also, define the subset of vectors Jj to be
*j = {?3:£ € A0}, j = 0,1, • • ■ ,i/-p.
(3.371)
(3.372)
Then as explained in Khuri (1984, p.21) the sequence of Tq random
vectors in fj (j = 0,1,•• • ,^-p) are statistically independent. Since
these random vectors are normally distributed with the same mean

-183-
vector and variance-covariance matrix (see Theorem 3.11), it follows
that an unbiased estimator of ^see (3.367) and Theorem 3.11(b)j is
given by
tr[S jídjC jB jC j/)-1/m j(r0 — 1) ] (3.373)
where
is the matrix
S
j
E (S3-Sj)(s5 — Sj)'
v e a
o
(3.374)
and
E (3.375)
3 0 if G AQ J
j = 0,1,
••,v-p (see Anderson (1984, p.
Further,
tr[S jid^C jBjC j')-1]/£ j
(3.376)
(j =0,1, • • • ,i/-p). Since
~j
N(d|CjNjg, Íjd^CjBjCjV^o)
* i2r.*i
(j = 0,1,• • • ,v-p), it follows that
C^Njg)'(d^BjCp-1(lj-d^Njg)/^
xly.
(3.377)
(j = 0,1, • • • ,i/-p) . Finally, because Zj and Sj are independent ^see
Anderson (1984, p. 71)) we conclude that

-184-
0 tr[SjCdjCjBjCj^-1]
d^Njg)
,j’mj(To-1)’
(3.378)
(j=0,1,••• ,r-p). This formula allows us to test a testable
hypothesis concerning CjNjg( j = 0,1, • • • ,i/-p) using an exact F-test.
Simultaneous confidence intervals on linear functions of CjNjg
(j=0,1,• • • ,r-p) can be obtained from the formula
Pjd^'CjNjg e «'fj ± Cotl'id^B jCp|]5} =
a,
(3.379)
( j = 0,1, • • • ,i>-p) , where i is a vector of constant coefficients and
c0 = {tr[Sj(d5C*jBjCp-]F(,,m.>m.(ro_1)/r0(r0-l)}5.
As a last remark, the matrix Cj ( j = 0,1, • • • ,i/-p) is not unique.
However, as pointed out in Khuri (1984, p. 22), the F-statistic given
in (3.378) and hence the confidence interval given in (3.379) is
invariant to the choice of the matrix Cj.
3.3.8. Numerical Examples
Dyeing of cotton-synthetic cloth. To begin with we use the data
in Table 3.6 to illustrate the methodology developed in Sections 3.3
and 3.4 of this chapter. The data in this table represent numerical
scores resulting from a comparison between a finished version of a
dyed cotton-synthetic cloth and a standard. Two temperatures (Factor
A), three cycle times (Factor B), and three operators (Factor C) were

-185-
selected and at most three small specimens of cloth were dyed under
each set of conditions ^see Montgomery (1984, p. 245)^. Ve assume that
Factors A and B are fixed while Factor C is random. The appropriate
model is given by
y i jk = A* + ai + 0j + (°/?)ij + 7k + («7)ik + (07)jk + («/?7)i jk +fijkp
i=l,2; j=l,2,3; k=l,2,3; 1=1,2,••• ,nijk-
Our first task is to analyze the variance components. To aid in
the understanding of the application of the testing procedures, the
reader is referred to Table 3.7 which lists some key quantities used
in the development of the exact tests. The expected mean square
values of MS j = SSrj/m^ , i=4,5,6,7, are provided in Table 3.8. The
results are summarized in Table 3.9.
Ve can see from Table 3.8 that no exact test (based on our
procedure) exists for testing H0:<72 =0. Thus we must apply
Satterthwaite’s procedure and construct an approximate test. The test
statistic in this case is given by
MSr4
4~MSr5 + MSr6-MSr7
1.547
which has an approximate F-distribution with 2 and rj* degrees of
freedom, where
V
*
(MSr5 -f MSr6 —
(MSr5)2 , (MSr6)2
2 + 4
MSr7)2
(MSr7)2
+ 4
2.728.
The observed level of significance is 0.3356.

-186-
Table 3.6. Numerical scores resulting from a comparison between
dyed cotton-synthetic cloth and a standard
Temperature
300“
350“
Cycle
Time
Operator
Operator
1
2
3
1
2
3
23
28
31
24
36
36
40
25
26
32
35
39
29
36
38
33
37
34
34
50
35
39
35
38
36
36
28
35
26
26
36
28
60
24
35
27
29
37
26
34
25
34
24
^source: Montgomery (1984, p. 245
Ve now turn our attention to the analysis of the fixed-effects.
The reader is referred to Table 3.10 which lists some quantities
needed in the construction of the exact tests (see also Table 3.7).
From Lemma 3.10(c) we may restrict our focus to testing hypothesis
concerning Zo’h’T2’ anc^ l3‘ In this case these quantities become:

-187-
lo = 0 + <* + /2 + 0/2))l18
Oi-a + (a/3)x> - (a/3))l9
1 (a2-a+ 0/2)2. - 0/2) )19
(/?1-/?+(a^).1-(a^))l6
(/?2-^+ (<*P).2~ 0/2)) *6
(/?3-^+(a/?)>3-(a/?))l6
(0/2)11-(0/5)1. “O0)2i +0/2)2. )13
((a/?)i2-(a/?)i. - (a/?)22+ (a/?)2. )13
((a/?)l3-(a/?)i. - (o/2)23 + 0/2)2. )13
( “ (a/2)n + 0/2)i. + 0/2 )2i - 0/2)2. )13
( “ 0/2) 12 + 0/2)i. + 0/2) 22 “ 0/2)2. )13
( “ 0/2) 13+ 0/2)i. + (Qr/2)23 0/2)2. )13
where
al + a2
/2 =
/2X + /22 + /3 3
2 3
0/2) =g £ £ 0/2)
i=lj=l
ij
0«i.=——3
a/?il+a/3j1+o^j3^ i _ j 2
a/? • + a/3 •
(a/3) j=—^2 —. j = 1,2,3.

-188-
Table 3.7 Some useful quantities used in the construction of the
exact tests for the variance components utilizing the
methodology developed in Section 3.3.3
Quantity
Formula cited
Value
V
(3.189)
7
p
(3.189)
4
r
(3.190)
(i,j,k)
c
(3.192)
18
n
(3.193)
40
b4
Table 3.3
6
b5
Table 3.3
3
^6
Table 3.3
2
b7
Table 3.3
1
P4
Khuri(1990)
J6 ® (I3 — J3)
P5
Khuri(1990)
(I2 — J2) <8> J3 ® (13-33)
P6
Khuri(1990)
J2®(I3 — J3)®(I3 — J3)
P7
Khuri(1990)
( 11 J2 ) ® ( ^3 — ^3) ® (I3 — J3)
m4
m4 = rank(P4)
2
m5
m5 = rank(P5)
2
m6
m6 = rank(P6)
4
Illy
m7 = rank(P7)
4
(3.217)(or Table 3.3) 60-4 + 305 + 2<7g + 07
6S (3.217)(or Table 3.3) 30^ + 07
56 (3.217)(or Table 3.3)
2 o\ + 07

-189-
Table 3.7 — continued
Quantity
Formula cited
Value
¿7
(3.217)(or Table 3.3)
°7
7rl
(3.232)
12
Vr2
(3.232)
10
,max
(3.237)
5/6
SSr4 = w4y4
(3.124)
87.816
SSr5 = ^5^5
(3.124)
2.885
SSr6 — ^7^7
(3.124)
134.599
SSr7 —- Ltj'jUj'j
(3.124)
26.825
SSEr2
(3.235)
31.5
Table 3.8
Expected mean square values
for the cloth example
Source
Mean Square
Expected Value
C
MSr4 = SSr4/2
6a4 + 3crl + 2 A*C
MSr5=SSr5/2
3 & l + 07 + §&£
B*C
MSr6 = SSr6/4
2 A*B*C
MSr7 = SSr7/4
J,5 2
“Error”
MSEr2 = SSEr2/10
5^.2
6^6

-190-
Table 3.9
Results from the analysis of
the variance components
for the cloth example
Source
F-value
P-value
C
(exact test does not exist)
A*C
MSr5/MSr7 = 0.215
0.8152
B*C
MSr6/MSr7 = 5.018
0.0737
A*B*C
§(MSr7/MSEr2) =2.555
0.1044
Choosing L7 =(l,0j7), | = 0, and recalling Table 3.3, we see that to
test H0: (/j+ a + (3 + (a/?)) = 0 we form ^see (3.286)^
_ SSH0/1
Ho~ MSf4 ’
where, recalling (3.265) and (3.268),
MSf 4 = y'f 4yf4/2 = 49.204.
In this case, then, we find Fjj =366.2 with p-value = 0.0027 (based on
1 and 2 degrees of freedom).
To test HqZOj + (a/?)j. = <*2 + (o/?)2. we would take L7 = (1,0g, — 1,0g)
and £=0. The appropriate test statistic would be
SSH./l
“i" "Sf5

-191-
Table 3.10 Some key quantities used in the construction of the
exact tests for the fixed-effects utilizing the
methodology developed in Section 3.3.4
Quantity
Formula cited
Value
bo
Table 3.3
18
W
Table 3.3
9
b2
Table 3.3
6
b3
Table 3.3
3
Po
Khuri(1990)
^18
Pi
Khuri(1990)
( ^2 ^2 ) ® Jg
P2
Khuri(1990)
J2 ® (I3 — J3) <8> J3
P3
Khuri(1990)
(I2 — J2) ® (I3 — J3) J3
m0
mQ = rank(P0)
1
ml
nij = rank(Pj)
1
m2
m2 = rank(P2)
2
m3
m3 = rank(P3)
2
¿0
(3.217)(or Table 3.3)
6 cr\ + 3og + 2 (3.217)(or Table 3.3)
3*1 + *?
^2
(3.217)(or Table 3.3)
2*1 + a\
^3
(3.217)(or Table 3.3)
ff7
*fl
(3.264)
18
^f,max
(3.265)
1
SSH0
(3.283)
18018.570
SSH1
(3.283)
5.701
ssh2
(3.283)
113.492
ssh3
(3.283)
39.666

-192-
Table 3.11 Summary of results for the cloth example
Source
Test Statistic
p-value
mean
SSH0/MSf4 = 49.204
0.0027
A
SSH1/MSf5 = 8.838
0.0970
B
SSH2/2MSf6 = 1.479
0.3305
A*C
SSH3/2MSf7= 2020
0.2476
C
f|=1.547
0.3556
A*C
MSr5/MSr7 = 0.215
0.8152
B*C
MSr6/MSr7 = 5.108
0.0737
A*B*C
§MSr7/MSEr2 = 2.555
0.1044
fBased on Satterthwaite’s procedure.
where
MSfs = w'f5yf,/2 = 0.645.
Thus Fjj =8.838 with a p-value of 0.0970. The degrees of freedom
associated with this test were 1 and 2.
Similarly to test + (<*/?) . j =/?2 + (a/?)-2 =/?34-(a/?) >3 we take
for our L matrix
(i,o',-i,o'u)
(1>9ii> -1>65)
(rank(L)= 2).

-193-
Also i =(0,0)'. In the present case the value of the test statistic
is
SSH,/2
H2- MSf6
1.479
with a significance level (based on 2 and 4 degrees of freedom) of
0.3305.
Finally, to test H0: (a/3) ^ j — (a/?) • -(a/?) - + (a/?) #=0 for all i
and j, we form
SSHo/2
H3_ MSf7
2.020
with a p-value of 0.2476 (based on 2 and 4 degrees of freedom).
Table 3.11 summarizes our results.
Performance of machines. The data in Table 3.12 will be used to
illustrate the methodology developed in Section 3.3.7. These data
represent performance values for three machines. Two power settings
(Factor A), three machines (Factor B), and three stations (Factor C)
were selected and either two or three runs were made under each
setting ^see Montgomery (1984, pp. 377-378)^. We assume that Factor A
is fixed while Factors B and C are random. The appropriate model is
given in (3.260).
The development of tests for the variance components follows
similarly to that given in the previous example and thus we only
summarize our results. This summary can be found in Table 3.13. Ve
note that to test for a significant B or C effect we must use
Satterthwaite’s procedure (see Table 3.4).

-194-
Table 3.12 Performance values for three machines
Machine
Station
1
2
3
1
2
3
1
2
3
1
2
3
34.1
33.7
36.2
32.1
33.1
32.8
32.9
33.8
33.6
Power
30.3
34.9
37.1
33.5
33.9
35.1
33.1
32.8
32.8
Setting 1
31.6
35.0
34.0
34.3
31.7
Power
24.3
28.1
25.7
24.1
26.0
27.1
24.2
27.4
24.7
Setting 2
26.3
28.6
26.1
25.1
27.1
23.9
26.1
28.0
22.0
24.9
27.9
23.9
25.3
24.8
^source: Montgomery (1984, pp. 377-378))
We next consider the fixed-effects. Recalling Table 3.4,
Theorem 3.10, (3.318), and (3.321), we see that no exact test exists
(based on our proposed procedures) for testing HQ:aj=a2. In this
example, the derived model is model (3.260) without the “f^term.
Thus, the fixed subscripts consist of just i, the nonnested random
subscripts consist of j and k, and there are no nested random
subscripts. Consequently, Sj = 1, s2 = 2, s3 = 0, s-l=3. We also have
aj=2, a2 = a3 = 3. Furthermore,
s-1
dj — II a • — a2a3 — 9
i=s1+l 1
(see (3.329))

-195-
Table 3.13
Results
from the analysis of the variance components
for the
machine example
Source
df
MS
F-value
D-value
B
2
14.023
2.2209*
0.6889
A*B
2
11.801
0.7582
0.5256
C
2
18.629
0.7552*
0.6195
A*C
2
30.155
1.9375
0.2580
B*C
4
10.077
0.6475
0.6580
A*B*C
4
15.564
8.9481
0.0024
“Error”
10
1.739
*Based on Satterthwaite’s Procedure
d2 = 1
tq = min a- = 3
Sj + 1 < i < si + s2 1
A0 = {v'-v = ql2> i ^0 = il8> ^1 = ^2<3)Í9
(see (3.330))
(see (3.371))
(see (3.370))
(see (3.334)).
The lj^ vectors (see (3.338)) are restricted to £ = 2,3 and to i^ = q
(q = 1,2,3). Therefore, these vectors are (since a2 = a3 = 3)
lá = (1,0,0)', 13= (0,1,0)', 1^ = (0,0,1)'.

-196-
Table 3.14 Some useful quantities needed in the development of the
exact confidence intervals for the machine example
Quantity
Formula cited
Value
H0~, K = (1,1)'
(3.338)
I2 ® I3 ® I3
Hff, v = (2,2)'
(3.338)
12® I3® I3
Hg, v =(3,3)'
(3.338)
12® I3® if
Hf j *=(1,1)'
(3.338)
I2® lf®lf
Hf, if = (2,2)'
(3.338)
I2® I3® I3
Hf, * = (3,3)'
(3.338)
I2® I3® I3
Pg, *eA0
(3.347)
HgH^/18
Pf, *GA0
(3.347)
Hf Hf'/9 — HgHg'/18
PwO’ ^ A0
(3.348)
QPgQ'
Pwi» ^ ^ ^0
(3.348)
QPf Q'
^00
(3.351)
i2/9
Noi
(3.351)
J2/i8
Nio
(3.351)
0
Nil
(3.351)
I2/9-J2/18
No
(3.353)
(212:J2)/18
Ni
(3.353)
(0:2I2 — J2)/18
C0*
(3.356)
(1,0)
cí
(3.356)
(1,0)
B0
(3.366)
J2/18
Bi
(3.366)
I2/9 — J2/18
^ICoNoS
Theorem 3.11
(// + o)/9
dlCfNjg
Theorem 3.11
(«1 ~q2)/18

-197-
Table 3.14 - continued.
Quantitv
Formula cited
Value
dlC*B0C¿'
Theorem 3.11
1/18
d&V*7
Theorem 3.11
1/18
?o
(3.375)
3.5281
?i
(3.375)
0.7491
So
(3.374)
0.5185
Si
(3.374)
0.6192
Some key quantities used in the development of the confidence
intervals given in (3.379) are listed in Table 3.14. In this case,
the space of estimable functions of the fixed-effects parameters is
spanned by CpNog and CjNjg, or equivalently by
p-fa, a1—a2 (see Table 3.14)
where a = (a1 + £*2)/2. In Table 3.15 we provide exact 95% confidence
intervals for /i + a and a1—a2 along with approximate ones based on
Satterthwaite’s procedure. To apply Satterthwaite’s procedure we
note that (see Lemma 3.10 and Theorem 3.10)
u0 ~
5

-198-
Table 3.15 Exact and approximate 95% confidence intervals for the
machine example
Exact
Estimable function confidence interval
Approximate
confidence interval
/i + a
(20.2696, 43.1367) (24.9762, 36.3594)
(-11.3968, 38.3637) (-31.8117, 12.9087)
wi ~N(^(ai_or2)> si+~é)
where 60 and S1 are given in Table 3.4. Thus, an unbiased estimator of
fi + a is w0/^18 and an unbiased estimator of a1—a2 is From
er2 cr2
Table 3.4 we see that unbiased estimates of (¿0 + _j')/18 and 2(¿j 4--t^-)/9
are, respectively,
yg(MS2 4- MS4 — MS6) = 1.2541
|( MS3 + MSs - MS7) = 5.8648.
As a final remark we note that since there is a large A*B*C
interaction, any conclusions which one reaches in regards to the
other effects in the model must be done so cautiously.

-199-
3.3.9 Concluding Remarks
In this section we analyzed the general unbalanced mixed model
with imbalance affecting the last stage only. We approached the
problem of obtaining exact tests for the variance components and for
estimable linear functions of the fixed-effects parameters in two
different ways. In Subsections 3.3.3 and 3.3.4 we utilized
resampling to first construct exact tests for the variance components
and then to develop exact and approximate tests for the fixed-effects
parameters. With the exception of the exact test for cr^,, our exact
tests for the variance components and the fixed-effects (when they
exist) are Wald’s tests. This is true since the degrees of freedom
associated with these tests match those of the corresponding tests
from the balanced data situation. A drawback to this approach was
the need to use two different transformations to arrive at our exact
procedures. An interesting side note to our methodology was the use
of resampling as a precursor to utilizing Satterthwaite’s procedure.
Instead of (wrongly) applying Satterthwaite’s methodology to sums of
squares which were not independent nor distributed as scaled chi-
squared variates, we first used our resampling scheme to transform
the model into one in which the required assumptions were met, and
then (correctly) applied Satterthwaite’s procedure.
In Subsection 3.3.7 only one transformation was needed to arrive
at a set of sums of squares which were independent and distributed as
scaled chi-squared variates. These sums of squares mimicked their
balanced counterparts. However, the resulting exact tests were seen

-200-
to be less powerful (in general) than the exact tests developed in
the two previously mentioned subsections. This led us to recommend
the use of those tests when most of the cell counts are much larger
than the minimum cell frequency.
Because the degrees of freedom associated with the exact tests
for the variance components >av-p+2’ ’ " ’ , fixed-effects parameters in which there existed a proper mean square
that could serve as an “error” term in an F-ratio, match the degrees
of freedom associated with the corresponding exact tests in a
balanced mixed model, we can conclude that the exact tests developed
in Subsection 3.3.7 are also Wald’s tests. In cases where no error
term existed, we utilized the results in Khuri (1984) and constructed
exact simultaneous confidence intervals on linear functions of the
fixed-effects parameters.
Finally, although not pursued in this dissertation, it would be
very simple to construct simultaneous confidence intervals for
continuous functions of the variance components by employing the
methodology presented in Khuri (1981). This may be preferable in
those situations where the only alternative is to use Satterthwaite’s
procedure (or some other approximate procedure) since the exact
converage probabilities associated with these (approximate) intervals
are typically unknown.

-201-
3.4 The Random Unbalanced Nested Model with Imbalance
Affecting the Last Two Stages Only
3.4.1 Introduction
In this section we extend the results of Khuri (1987) by
considering the random s-fold unbalanced nested model with imbalance
affecting the last two stages only. We again utilize resampling to
make transformations in a “step-down” manner to arrive at our exact
tests. This approach is similar to the approach employed in Section
3.2. We assume all cells contain at least one observation.
After some preliminary development we proceed to the
construction of the exact tests for the variance components in the
model. Furthermore, we show that our tests reduce to the usual ANOVA
tests when the design is balanced. Also, we develop expressions for
the power associated with the tests. A numerical example is given to
illustrate our proposed methodology.
3.4.2. Notations and Preliminary Development
In our development we will concentrate on the following model
(the notation being similar to that of the previous section),
y«.+l=" + 1,il+7*J+'"+T*s + ‘<’S+l
(3.380)

-202-
where the sets 9j are defined as
flj = (k1}k2, •••,kj), 1 < j (3.381)
and (since the design is balanced except for its last two stages)
( 1> 2,••• ,a j, j — 1j 2,• • • , s 1
kj={ J = s
1,2,••• ,n0 , j = s+1.
(3.382)
•th
In (3.381), 9j represents the subscripts associated with the j
effect in model (3.380) , j=0,l,---,s+l; (when j = 0, 9 and
7^ = n). Furthermore we will assume that Jq ,• • • , and Cq
independently distributed with jg ~N(0, o'2-), l j J
eg ~ N(0, cr2) .
s+1
In matrix notation (3.380) becomes
s+1
are
y-l4n + Xi/?i+ • • • +Xs/?s+£> (3.383)
where y is the vector of observations; e is the vector of errors;
is a random vector consisting of the elements of 7g (l Xj is a matrix of zeros and ones (l Xj=®.%.’ 1-j-S (3.384)
where
n =
(3.385)

-203-
(3.386)
where 9S — 9 j = (k j+j, k j+2» • ■ • , ks ) . Let us also define
(3.387)
(3.388)
where — 9j = (kj+^,kj+2*•• •,^s_i)• The matrix Xj is of order
J
(3.389)
The assumptions made earlier concerning the random effects can now be
restated as:
(i) ^,^2, • • • ,/3s, and e are independent and normally distributed
random vectors with zero mean vectors and variance-
covariance matrices given by
Var(/?i) = i = l,2,---,s
(3.390)
Var(c)=a\ln.
From (3.383) it is thus easy to verify that

-204-
E(y)=/4n
Q = Var(y) = £ HXjX'. + i = l
(3.391)
Besides the usual assumptions concerning the random effects we
will also assume:
(i) The model used is model (3.380) ^or model (3.383)^.
(ii) n>2m—1. (3.392)
(iii) m >2cg_i — 1. (3.393)
The last two assumptions are needed for technical reasons to insure
the validity of our methodology. Neither is overly restrictive. For
example, (3.391) will hold if n^ >2 for all k1,k2,'*’»ks while
(3.392) will hold if, for example, m^ >2 for all kx, k2, • • • , kg_ j .
s-1
3.4.3 An Exact Test for <7?.
Introduction. In this section we develop an exact test for the
hypothesis Hg:cr| = 0 vs Ha:cr|^é0. This test is similar to the
corresponding test in a balanced model in that it compares the sum of
squares associated with the yq effect to the error sum of squares in
model (3.380).
Preliminary development. To begin with let
vj = [ln:Xi: • • ' :Xjl> l (3.394)

-205-
Then, momentarily treating /?j ,/?2, • • • ,/?s as fixed, we define the sum
of squares Ry(/?s I @1 > ' ’ * »^s-l) as
Méslíi. • • • -és-i) = y'[in-vs.1(v's.1vs.1)-w's.1]y
-y'[In-ys(V's“s)X]y- (3.395)
Since
range(ln)C range(X1) C • • • C range(Xs), (3.396)
we can express (3.395) as
Ry(£sl£i> • • • ,^s_i) =y'[Ps-ps-l]y> (3.397)
where
pj = xj(xjxj)_lxj, j = 1 >2, • • • ,s. (3.398)
^From (3.384) it is clear that Xj has full column rank Cj,
j = 1,2, • • • , s-1; and Xs has full column rank m.) Ry(/?s I Pi > ' ' ' ’^s-l)
can be interpreted as the sum of squares for adjusted for
/?!,••• ,/9g_l |see Searle (1971) for a description of the R( • | • )
notation).
The error sum of squares associated with model (3.380) ^or model
(3.383)) i s given by
SSEy = y'(In-Vs(V'sVs)-V's)y

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= y'(In-ps)y-
(3.399)
The last equality follows from (3.397). The following theorem
provides the necessary tools to construct our exact test.
Theorem 3.12
(a)
Ry(/?s |/?j, • • • ,^s_i) is independent of SSEy and, under
is distributed as a\\^_c .
(b) SSEy/^~X2_m.
Proof:
(a) Ve first verify that Ry(/?S|/31S • - • »/?s_j) is independent of SSEy.
To verify this we must show that
(Ps-Ps_1)il(In-Ps)=0. (3.400)
In order to show (3.400) we need the following easily derived
results:
(i) XiX'iPj = XiX'i =PjXiX/i, 1 < i < j < s (3.401)
<“>. pj = pj’
j = 1,2, • • • ,s.
(3.402)

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(iii) PjPi=Pj = PiPj, 1 < i < j < s.
(3.403)
For example to show (3.401) note that range(X^X^) C range(Pj) for
i X^j such that
xix'i = pjxi j
= xijpj
(by symmetry),
Consequently,
XiXiP j =XijPj
^by (3.402) which clearly holdsj
= xiX'i
= p.x^x^ (by symmetry),
Formula (3.403) can be shown in a similar fashion.
Now, (3.401) implies ^see (3.391)^
«(In-Ps) = (In-Ps)ffc
(3.404)
while (3.402) and (3.403) imply

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(Ps-Ps_l)(ln-Ps)=0.
(3.405)
Thus (3.400) is verified and consequently SSEy is independent of
Ry(/?s I P\ > ’ ‘ ' > @s-1) ’
Ve now show that Ry(/?s | > • • • ,/?s_j )/~ X^-c ^ when <7g = 0.
To do so we need to verify that when <7g = 0 (Ps — P j)í2/ idempotent; and rank(Pg—Pg_^) = m—cg_^. Now, from (3.391) and
(3.401) we have
(Ps-Ps-l)Q=(Ps-Ps-l)
i = l
= (í/iXiX'i+^Ps)-(5ViXiX'i
+ °sPs-lXSXS + ^Ps-l)
= 4XsXs - 4Ps-lXsXs +^(ps - Ps_i)
= 4PsXsX,s-4Ps-lXsXs + ^(Ps-Ps-l)(see (3-401))
= (ps-ps-l)[4xsx's + ^l]-
(3.406)
Thus when Hg:cr| = 0 holds, (Ps — Pg_)£l/cr^ becomes
(Ps-ps-i)fi/^ = ps-ps-r

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Since Ps—Pg-i is idempotent ^from (3.402) and (3.403)) and
clearly has rank m — it follows that when HQ: Ry(/?s I @\ ’ ' ' ’ »£s-l)/ae ~ ^m-cs_i"
(b) Recalling from (3.404) that
n(in-ps)/^ = in-Ps
and using the fact that In—Ps is idempotent and has rank n—m,
it follows from (3.399) that
SSEy/ â–¡
The exact test. From Theorem 3.12, a test statistic for testing
Hq: /[Ps —Ps-i]y/Q-cs-i)
SSEy/(n-m)
(3.407)
Under Hq, Es~Em_c ^,n-m. Under Ha the distribution of Fs is
complicated. However, we still have
’y'CPg-Ps-i)?
= m^II[trC(Ps-P8-l)fi] + ^l,n(Ps-ps-l)ln]- (3-408)
Now, since range(ln) Crange(Pj) and Pj is a projection matrix (recall
(3.398)) we have that

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Pjln = ln> j = l,2,--.,s. (3.409)
Thus
ln(Ps-Ps-l)ln = 0-
Hence, from (3.406) and (3.409),
tr[(Ps-Ps_1)i2] = _ . m
= ^(m-cs-1)+crl E í'si(ps-ps-l)^si
j=l J
where
— Cxsi:xs2: ‘ ' ’:xsra]
Therefore,
y\ps-ps-i)y
m-c
s -1
_2 , ° S
— ae "P m-c
Y.EXsjiPs-Ps-l^sj^^-
1J=1 J
(3.410)
Consequently, we would reject Hg for large values of Fs in (3.407).
Note that resampling was not needed to construct this exact test.
We now develop an exact test for

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3.4.4 An Exact Test for
Introduction. In this section we develop an exact test for the
hypothesis Hg: we would simply compare, aside from a constant, RyI > • • • ,/?s_2)
to Ry(/?s | /?x , * - * ) (treating ,•••,/? s as fixed) where
®y(£s-l ^-1 ’ ’ ’ ’ ’-s-2) =¥ [Ps-l-Ps-2]y>
|see (3.396) and (3.398)J. Unfortunately, when the data are
unbalanced, we can not, in general, show that
(Ps-l-Ps-2)fi(Ps-Ps-l)=0
unless we assume
This follows since
(Ps-l-Ps-2)fi(Ps-ps-l)=4(Ps-l-Ps-2)XsX,s(Ps-Ps-l)’
^recall (3.391) and use (3.401)—(3.403)j. Thus these two sums of
squares will not, in general, be independent unless we (unreasonably)
assume this hypothesis. We might try comparing the sum of squares for ^
adjusted for all the other effects in the model (treating them as
fixed) to Ry(/?s | /Jj, • • • ,/?s_i) • Unfortunately, the former sum of
squares can be shown to zero (see (3.396)).

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The idea we develop is based on a simple observation that was
illustrated in Section 3.4.2:
For any random nested model (balanced or unbalanced) we can
construct an exact test for the variance component
associated with the penultimate stage by comparing the mean
square for the corresponding effect (adjusting for all the
effects preceding it in the model) to the error mean square
for the model.
Thus to test H0: (utilizing resampling) that will reduce model
one that mimics a random (s-l)-fold nested model. In the transformed
model, will be the variance component associated with the
penultimate stage.
Preliminary development. Define the cg x n matrix D as
= (X'SXS)-1X'S.
(3.411)
Also, let
y = Dy.
(3.412)

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Furthermore,
The matrix A
(j = 1,2,-- - ,
Consequently
where e = De.
This follows
define
Ai=®ln,. . j=l. 2. (3.413)
"j "j
j is or order mxcj and of full column rank Cj
s—1)• From (3.384) and (3.396) we can show that
1 — Y 1
±n—As±m
xj = xsAj,
j = l,2,
, s-1.
(3.414)
(3.412) can be expressed as
y - /ilm +Ai^i + A2^2 + • • • + As-l^s-l + Im^s + 1
(3.415)
Ve note that
range(lm)C range(Al) C • • • C range(Agl). (3.416)
since (writing Xq = ln and AQ = lm), for 0 range(A•)=range(XsA^) (Xs has full column rank)
= range(X^) ^see (3.414))
C range(Xj)
(see (3.396))

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=range(XsAj)
=range(Aj).
From (3.415) it is easily seen that
E(y) =/4m
£ = Var(y)
s-1
= £ <7iAiA,i+ i = l
' € >
(3.417)
where
K=(X,SXS)"1= ©n^1. (3.418)
s
Now, let H1 consist of the last m-1 rows of the mxm Helmert matrix
^see Searle (1982, p. 71)j. Then Ha has the following properties:
(i) HlHi = Im-l-
(ii) H'1H1 = Im-Jm. (3.419)
(iü) Hilm = -m-1 •
^Recall that Jn^ffilmlm') Define the (m-l)xl random vector p as
u = Hjy
(3.420)

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The purpose of this transformation is to cause u to have a zero mean
vector as illustrated below ^see (3.417)j.
E(u)=H1E(F)=/iH1lm = Om_1
Var(u)=E1H1AiHXi+4lm-l+HiKHi
i = l
(3.421)
where Á^=A^A^, i = 1,2, • • • ,s-1.
Utilizing resampling. Let R be the nxn matrix associated with
error sum of squares in model (3.380). That is, from (3.399),
R = In-Ps. (3.422)
The matrix R is symmetric, idempotent, and has rank n-m (see Lemma
B.4). Ve can write R as
R = CAC' (3.423)
where C is orthogonal and A is diagonal. The first n-m diagonal
entries of A are ones while the remaining m diagonal entries are
zeros. Partition C and A as
C — [Cx:C2: C3]
A = diag(I,7i,IJ?2,0)
(3.424)

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where
7jx = m-1
r¡2 — n-2m+l
(3.425)
and (^,02, and C3 are of orders nxi)¡, nxr/2, and nxm, respectively.
Note that t]1 + rj2 = n-m = rank(R) and r¡2>0 by (3.392).
Ve now introduce a transformation that will result in a random
vector which has a diagonal variance-covariance matrix. Let w be the
(m-l)xl random vector defined as
y = a + (Wif?1-L)^c;y (4.426)
where
L = H1KH,1 (4.427)
l
and Amax is its largest eigenvalue. The matrix ( AmaxI,^ — L)2 is well-
defined since AmaxIj^ —L is p.s.d. It’s eigenvalues are the square
roots of the eigenvalues of AmaxI,^ — L. The next lemma provides some
distributional properties for w.
Lemma 3.13
The random vector w is normally distributed with
(a) E(w) = 0
(3.428)
(b) Var(w) = ¿ H1AiHxo-|+<5Im l, where <5 = i = l

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Proof :
First note that u) is clearly normally distributed.
(a) From (3.421), E(u)=u. From (3.398), (3.414), and (3.422), it
follows that Rln = 0. Thus, since range(Cj) C range(R) , it must
be that Cjln = 0. Consequently the result follows after recalling
the definition of u in (4.426) and that E(y)=pln.
(b) Ve claim that u is independent of C^y in (4.426). To see this
first note that
DfiR = DR(Tj (see (3.404))
= 0. (see (3.398), (3.411), and (3.422))
Thus, since range(Ca) C range(R) it follows that DflCj = 0 => y is
independent of C^y. But this implies that u and C^y are
independent since u is a function of y. Consequently,
Var(w) = Var(u) + (AmaxI^ - L)^fiC1( AmaxI^ - L)Í
Now, from (3.391), (3.401), (3.422), and the fact that R is
idempotent, we have

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Thus, since range(C1)Crange(R), it follows that
C'ílCj^CÍCj^í^.
Hence, from (3.421),
Var(«) =S£ HjA.HXi +<41.-! +
^¿/.AjHXí +ilm_i
1 = 1
where ¿ = 4 + Amax From this lemma, (3.415), and (3.420), we can write w as
u = H1A1^1 + HjA2/?2 + • • • +H1As_1^s_1+e* (3.429)
l
where e* = Hj/?s + HjT + (-AmaxI- L)2C^€ ~ N(0 ,<5Im_ j ) independently of
/?!,■•• ,/?s_l. Model (3.429) resembles a random (s-l)-fold nested
model (with zero intercept) because of the distributional properties
of /?x, • ■ • ,/?s_i, and e*, and since for l H1A£=H1AjA*j ^for some matrix A*j, see (3.416)^
=>range(HjA^) C range(H1Aj).
(3.430)

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Before we proceed to the development of our exact test we need
the following result.
Lemma 3.14
(a) rank(AjHjHjAj) = c j — 1, j = 1,2, • • • ,s-l.
(b) A g-inverse of AjHjHjA- is given by
(AjHjH1Aj)-= (AjAj)-1, j = 1,2, ... ,s-l.
Proof:
(a) rank(AjHjHjA j) = rank(H1Aj)
< rank(Aj) = cj .
However, from (3.413) A-lc. = lm. Since Hilm:=Qin-l (see
J j '
(3.419)(iii)^, the columns of HjAj are linearly dependent. Thus
rank(AjHjHjAj) < cj — 1, j = 1,2, • • • ,s-l. Conversely, by the
Frobenius Inequality (see Lemma A.l), we have
rank(HjAj) >rank(H1)+ rank(Aj)— rank(Im)
= m — 1+Cj—m

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(b)
= cj - 1, j = 1,2, • • • ,s-l.
Thus, rank(AjHjH1A j) = cj — 1, j = 1,2, • • • ,s-l,
First note that (see (3.413)) for j = 1,2, • • • ,s-l,
(Aj(AjA.P lAj)J-n-C®4.S>
(3.431)
= Jm(Aj(AjAj)_lAj)- (3-432)
Hence, for j=1,2,•••,s-1
A7iH1Aj(A'jAj)-lA71H1Aj
= A'j(Im-Jm)Aj(A'jAj)-1A'j(Im-Jm)Aj (see (3.419))
= (Aj~Aj3mAj(AjAj)_1(Im - 3m)Aj
= (Vj-A'jJm)(Im-jm)Aj (see (3.431) and (3.432))
= A,j(Im-Jm)Aj
= AjHjHjAj.
â–¡
(Im —Jra is idempotent)

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Ve now develop the exact test for
The exact test. Let us define the (m-l)x(m-l) matrix Qj
(j = 1,2,•• • ,s—1) as
Q j= HiAj(AjHíHiA j)-ní
= HiAj(AjAj)_lAjHi' (4.433)
Furthermore let Rw(/?s_-^ | , • • • >/?s_2) be the sum of squares for the
effect adjusting for , • • • , P s_2 (treating them as fixed) for
the u model defined in (3.429). Also let SSEW be the error sum of
squares for model (3.429). Then
Mis-l I?.." - .5s-2> =»'[I»-1 -Vs-2(Vi-2V8.2)-V'8.2]tf
and
SSE„ = a'(In,_1-Vs_1(V's_1Vs_1)-V's_1)¥
where
Vj = [H1A1:H1A2: • • • :H1Aj], j = 1,2, • • • ,s-l. (3.434)
From (3.430), Lemma 3.14(b), and (3.433), Rw(/?s_^ | , • • • ,/?g_2) an<^
SSEW can be equivalently expressed as

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M£s-ll£i’---’£s-2)=*' Ws-l~%-2te (3-435)
SSEy = y'[Iin_1-Qs_1]y. (3.436)
The major result of this section is given in the next Theorem which
provides some distributional properties for Rw(/?s_i | /?i > • • • ,Ps-2^ an3
SSEy.
Theorem 3.13
(a) Rw(/?s-l I j ' ‘ ' >/?s_2) independent of SSEW and, under
HQ;(Tg_j=0, is distributed as 8\c _c 2'
(b) SSEW/6
*m-c
s —1
(ó was given in Lemma 3.13(b)j.
Proof:
(a) Ve first verify that hw(/?s_^ | /?a, ■ • • , /?s _2) an3 SSEW are
independent by showing
[Qs-i-Qs-2^Var(y)][im-i-Qs-il = °-
In order to show this we need the following results, which are
analogous to (3.401) - (3.403).
(i) HjA.A'.H'Qj = H1AiAjHj = Q jH^.A'.H',
l
-223-
(ii) Qj = Qj, j = 1,2, • • • ,s-l
(iii) QjQj = Qj = QiQj, l Formula (3.438) is obviously true. To show (3.437) note
for 1 < i < j < s-1
range(H1AiA,iHi) =range(H1Ai)
C range(H1Aj)
(see (3.430))
=range(Qj).
(see (3.433))
Thus there exists a matrix B?j such that
H1A.ViH; = QjBÍj
= BÍ,.q.. (by symmetry)
Hence, for l HiAiA/iHiQj = BÍ'jq2j
= BijQj (by (3.438))
(3.438)
(3.439)
that
= H1AiA/iH'1

-224-
= QjlljAjAjHj (by symmetry).
Formula (3.439) follows similarly. Now, from Lemma 3.13(b), we
have
1 = 1
= «(Im.l-»,-l) (see (3.437)). (3.440)
Thus
[Qs_l-qs_2][Var(y)][Im-l-QS-l]
= ¿(íís-l-t3s-2)(Im-l-«s-l)
= 0 (by (3.349)).
Consequently, Rw(/?s_^ | P-y, • • • ,/?s_2) an(^ SSEW are independent.
To show that Rw(/?s_^ | > ■ • • ,/?s_2)/^ ^as a chi-squared
distribution with cs_i~cs_2 degrees of freedom under Hq:ct^^=0,
we need to verify that (Qs_j—Qs_2)(Var(w))/¿ is idempotent (when
3.13(b) and (3.437), we have

-225-
Js-1
â– Us-2)
S¿1HiÁiHÍ i = 1
=8¿1H1AiHÍ i=1 i=l
s-2 _
= í7s-1^s-1H1^s-1HÍ - 'l-A-M-lH'K),.! - Qs_2) (“« (3.437))
Thus when Hq:(t2_^=0 holds, (Qs_^ — Qs_2)(Var(w))/<5 becomes
(^s-l-t3s-2)(Var(y))/¿ = «s-l-Qs-2
Since Qs_i~Qs_2 idempotent ^from (3.348) and (3.349)^ and
clearly has rank cs_i—cs_2 (see (3.433) and Lemma 3.14(a)^, it
follows that when Hq:c2_^=0 is true
Rw(^s-1l^i’ '' '’^s-2)/¿ ~ xcg_1— cg_2'
(b) From (3.440),
var(y)[im_1-qs_1] = ¿(im_1-qs_1)
Thus Var(w)[Im_j—qg_^]/0 is idempotent. Since rank(Im_^—qg _^)
= m-1 — (cg— 1)=m-cs_^ ^recall (4.333) and Lemma 3.14(a)^, it
follows that ^see (3.436)j
s — 1
SSEw/i5 ~ Xm — c
â–¡

-226-
To test Hgio-g_^=0 vs. Ha:<7gwe form
p _^(qs-l-qs-2)^/(cs-l-cs-2)
S-1 SSEy/im-Cg^)
(3.441)
Under Hq, Fs_^ ~Fc c 0 m—c • Under Ha, the distribution is
s — 1 s — s 1
complicated. However, we still have
y'(Qs-i-^s-2)y
cs-l _cs-2
= cs-l~cs-2 tr[(qs-l-C3s-2)(Var(y))]
cs-l~cs-2
tr
(Q,
1-Qs_2)(¿Ví"Xí+«b,-i) (
see Lemma 3
.13(b))
= c77r-cs.2tr[<-iqs-iHiAs-iHi-^s-iqs-2HiAs-iHi
+¿(qs-i-qs-2)]
^see (3.347))
= * + cs-lS-^s-2tr[(q-l ~ qs-2)HiAs-lHi]
= g+cr-is-^s-2 MWiu-Ws-J
= S + :
2 C -i
at i s-1 a
s-1 ^*/
_1~cS-2 as-l,j(Qs-l qs-2)as-l,j
where a*^,j = H1Ss_1? ■, j = 1,2,• ••,cs_1,
and
4s-l
:[*s-1,1!*b-1,2!‘’-:-s-l,csl]-
Thus we reject Hq for large values of Fg_j in (3.441),

-227-
3.4.5. Exact Tests Concerning Introduction. Recall that model (3.429) mimics a random un¬
balanced (s-1)—fold nested model where the imbalance affects the last
stage only. Thus we can adopt the procedures developed in Khuri (1990)
to aid us in the construction of our exact tests. In that article,
Khuri was able to develop his exact procedures by first averaging
over the last subscript and then making a transformation that
resulted in a random vector having a zero mean vector. However this
new vector had a variance-covariance structure that was not diagonal
^see Khuri (1990, p. 182)^. To obtain a diagonal variance-covariance
matrix, Khuri made another transformation, this time based on a
resampling scheme. Thus these transformations effectively reduced
the analysis of the unbalanced model to that of a balanced one.
We can employ the same ideas to our present situation. However,
some care needs to be taken since model (3.429) is not exactly (in a
statistical modeling sense) a random unbalanced nested model. This
is true since the HjAj (j=0,1,• •■ ,s-1) are not design matrices.
Preliminary development. Recall model (3.429). Namely,
w = HjAj/Ij + • • ■ +HiAs_i^s_i +Í*-
Define the cs.ix(m-l) matrix D* as
D* = (AB-1HiH1As_1rA;.lHi

-228-
= (As-lAs-l) 1Ag—iHj fsee Lemma 3.14(b)j. (3.442)
Compare the definition of D* to that of D given in (3.411). The
matrix D was used in (3.412) to average over the last subscript in
model (3.380) ^or model (3.383)^. Similarly to (3.412), define the
c i x1 random vector w as
w = D*w. (3.443)
Equation (3.443) can be viewed as “averaging over the last subscript”
in model (3.429).
Although E(y)=0 ^see Lemma 3.13(a)^, it is still convenient, for
mathematical reasons, to make a transformation that results in a
random vector having a “zero mean vector.” This transformation is
similar in spirit to that given in (3.420).
Let H* represent the last cs_i—1 rows of the cs_ixcs_i Helmert
matrix ^see Searle (1982, p.71)^. Then, similar to (3.419), H*
possesses the following properties:
(i) HiH*' = ICg — 1'
(ii) Hj'Hj = ICg j —*^Cg_1' (3.444)
(iii) H,*lr = 0_
V ’ 1~cs-l -cs-l
-1-

-229-
Let u* be the (cg_j —l)xl random vector
u* = H*w
(3.445)
The next lemma establishes some statistical properties for u*.
Lemma 3.15
The random vector u* is normally distributed with
(a) E(u*)=0 i
cs-l 1
(b)
s-2
“1*1
Var(u*) = EHÍBiH1+V|+^_1Ic . + ¿H1*(A'S_1AS_1)-1HÍ'
i=l s-i
where
®i — *Cj ® ^Cj »
s-1
c- =. n a-,
j=i+l J
i = 1,2, • • • ,s-2
(3.446)
^Note that cicj=cs_i> i = 1,2, • • • ,s-2, see (3.389).j
(c) u* is independent of SSEW, where SSEW was given in (3.436).

-230-
Proof :
First note that since w is normally distributed (see Lemma 3.13),
w is normally distributed ^see (3.443)^, and hence u* is normally
distributed.
(a) From (3.443), (3.445), and Lemma 3.13(a) we have,
E(u*) =H*E(w)
= H*D*E(w) = 0 ,
cs-l_i-
(b) From (3.443), (3.445), and Lemma 3.13(b) we have
Var(u*)= H*D*
1 = 1
D+/Hj;
s -1
= ¿2 H*D*H1A.H'1D*,H*V? +
i = l
(3.447)
Now,
(i) H1W'Hf' = Hf(A'8.1A8_1)-1A'8.1HiH1A8.1(A;.1As.1)-»Hr
^see (3.442)^
= H1*{(A's_1As.1)-Vs_1As_1(A's_1As_1)->
^see (3.419)j.

-231-
From (3.413),
(As-lAs-l) lAs-lJm-(. ® nV Amg )mlmlm
ys-l s 1 s-1
= -1 \
m¿c„ 1¿m-
s-1
Hence, from (3.444),
Hí(A's-lAs-l)"'A's-lJm = 4“f ics.,lL = 8C..J-1
Thus,
(3.448)
H*D*D+,H*^HÍ(A/S_1AS
(3.449)
(ii) In the first term of (3.447) consider H*D*H1A^, i = l,2,---,
s-1. We have, for i=1,2,•••,s-l,
H*D%Ai=HÍ(A's_1As_1)-1A's_1H'1H1Ai (see (3.442))
= H*(As_iAs_i)_lAs_iAi — Hi (As-lAs-l)~1As -l^niAi
(see (3.419))
= HÍ(A's-1As_1)-1A's_iaí. (see (3.448))
Now, since range(A^) Crange(Aj), 1 < i < j < s-1 (see (3.416)),
there exists matrices B- of order c iXc*, l J 1 J

-232-
Aj=Ag_1Bj, j = 1,2, • • • ,s-2.
(3.450)
Consequently, for i = 1,2, • • •,s-2,
Hr(A's-iAs-i)‘lA,s-iAi=Hr(A,s-iAs-i)"lA's-iAs-iBi
= HÍBi.
That is
HjBj, i=1,2,•••,s-2
i" “i"1 “I Hí, i = s-1.
HÍD*H,A.
(3.451)
Therefore, from (3.449), (3.451), and (3.447) we have
s-1
Var(u*) = ¿2 H*D*HjA• • + <5H*D*D*'H*'
i=l
s-2
= E HfBiB'iHfVl+HÍH*Vs.1 + ÍH*(A'8_1As.1)-1Hr
i = l
s-2
“EHfSjHfV.+I -1<4.1 + ®f(*í.1*s-1)'1HÍ
1=1 S_i
-III*/
where B.=B^Bj, i = 1,2, • • • ,s-l. From (3.413) and the fact that
As-1 Bas column rank, Bj can be (uniquely) expressed as
B- = Ic.l^ , j=1,2,•••,s-2
J J ~ J
(3.452)

-233-
where
s-1
c-= II a., j = 1,2, • • • ,s-2.
J k=j+1 K
Thus,
Bj = BjBj = Icj ®Jcj’ j = 1,2,••• ,s-2.
(c) Since u* = H*w, it suffices to show that w and SSEW are
independent. From (3.442), (3.443), and (3.436), we have
(Ai.iAs.1)-1A's.iH'I[Var(s)](Inl_1-Qs.1)
= S(A's-lAs.1)-,A^1H'1(Im_1-qs.1) (see (3
However, since
range(H1Asl)=range(Qsl) (see (4.333))
it follows that
A,s-iHi(im-i-qs-2) = °-
Thus u (and hence u*) is independent of SSEW.
.440

-234-
In the expression for Var(u*) given in Lemma 3.15(b), the
matrices i = 1,2,• • •,s-2, are analogous to their counterparts
in a balanced model in the sense that they can be shown to commute in
pairs. This is the focal point of the next subsection.
The matrices H*B^H*', i = 1,2,• • • ,s-2. In this subsection
we show that the matrices H*B^H*', i = 1,2, •••,s-2, commute in pairs
(see Lemma 3.16). To do so, we first note that for 1 BiBj = (Ic.®Jc.)(Icj®Jc.) (see (3.446))
= (Ia1® Xa2® • •
• ® Io ® Jo ® • • • ® Jfl • • • ® Jo )
ai ai + l aj as~r
® Ia ® Ia ® • • •
ai ai+l
® Io ® J„ ® • • • ® Jo )
aj aj+i as-r
(see (3.389))
= Ici®Jai+l®
‘ ' ’ ® Ja-j ® a j+i Ja ® • • ■ ®as-lJas_1
= c j(Ic ® Jc )
u 1 1
(3.453)
Consequently, B^Bj is symmetric which implies that B^ and Bj commute.
Lemma 3.16
The matrices H*B.H*7 commute in pairs for i = 1,2,• • •,s-2.

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Proof :
Without loss of generality take l = HÍBi(ICs i-JCs l)BjHr (see (3.444))
= HÍB; B -Hi' — HÍB • Jc B-H*'. (3.454)
1 1 J 1 1 1 s-1 J
Now, from (3.446),
BiJc
. — (Ic ® ^C- )^C 1
s-1 1 1 s-1
(IC. ® ^
c^r(Jc. ® ci JC. )
cs-l C1 1 S
recall that cici = cs_i> i = l,2,*
,s — 1 (see (3.389) and (3.446)).
Consequently,
HÍVc
s-1
CS-1 1 CS-1
= 0.
(see (3.444))

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Thus, the second terra in (3.454) vanishes and hence
(H1+BiHn(Ha*BjHÍ')=HÍBiBjH1*'.
Therefore, since B-Bj is symmetric ^see (3.452)^, it follows that
(H*B|Hj') (H* B jll*') is symmetric and consequently H*B^Hj' and H*BjH*/
commute. Ü
This lemma indicates that there exists a (c ^ — 1) x (cs_^ — 1)
orthogonal matrix S which simultaneously diagonalizes
HjBjH*',••• ,HjBs_2H*/. Before we apply this transformation let us state
some further properties concerning the H^B^H*', i = 1,2,•••,s-2.
(i) The matrices B^, i =1,2,•••,s-2, defined in (3.452),
satisfy
range(Bj) C • •• Crange(Bg_2)• (3.455)
Thus, it follows that
range(HjBj) C •• • Crange(H*Bs_2)• (3.456)
(ii) For i =1,2,•••,s-2,
rank(H*B^ ) = rank(H*B- H*') = c^ — 1.
(3.457)

-237-
To verify (3.455) let 1 < i < j < s-2 and note that (see (3.389) and
(3.446)),
c
Hence, from (3.452)
B- = Ic. ®le. = Ic. ® 1
1 ci ci ci
= Ic. ® lc • ® ir •
i _J J
= (Ici®Icj®lcj)(Ici®l
= (Ic.®lc.)(IC. ®lc.)
J J ^ ~
= B.(IC.®1C.).
J 1 J
ci
Therefore, range(B|) Crange(Bj), 1 < i < j < s-2. Since, for
1 < i < j < s-2,
H*Bi=H*Bj(Ic.®lcj),

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it follows that
range (H*B^ ) C range(H*Bj ), l To verify (3.457) first note that
B-lc.=lc „! i = 1,2, • • • ,s-2
1 i s-1
and
(see (3.444)).
Consequently, the columns of H*B^ are linearly independent =>â– 
rank(H*B|) Frobenius Inequality (see Lemma A.l), for i =1,2, • • • , s-2
rank(H*B^)> rank(H*)+ rank(B^)-rank(Ic ^)
= cs_1-1+ci-cs-i
= ci-l-
Thus, rank(H*B^ ) = rank(H*B^H*') = c — 1, i = 1,2, • • • ,s-2.
As indicated earlier, Lemma 3.16 implies that there exists
(cg_j-l)x(cg_^xl) orthogonal matrix S which simultaneously
a

-239-
diagonalizes HjB1H*/,•••>H*Bs_2Hl,• The construction of S is our focus
in the next subsection.
The construction of S. Ve now describe how to construct the
(cg_i — 1) x(cs_i — 1) orthogonal matrix S. To begin with consider the
matrix B'^Hj'HjBj, i = 1,2, • • • ,s-2. From (3.446) and (3.452) and
recalling that c-c- = c , i = 1,2, • • • ,s-2, we have
11 o 1
3/-Hj'HjB- = B'- (Ic -Jr )B•
i i i i i'- cg_1 cg_1/ i
(Ir )(Ir -Jr )(Ir l? )
^ -C^V Cg_1 Cg.jA Ci ~Ci ^
(Ic. ® 1^.) -^(Ic. ® Ife. )(JCi ® Jcj) (ICi ® Icj)
(Ic-®lc-) c ^ (^c-® lc-
1 1 0—1 1 1
) (
^Cj ®ic- )
—((^C. ^C^)®-C^)(®Icj)
— ci(Ic^ Jc^)’ i —l,2,*--,s 2.
Hence, a generalized inverse of B'^Hj'HjB^ is
(B/iHÍ,H1*Bi)- = Xlc., i = 1,2, • • • ,s-2.
Consequently, the matrix H*B^(B;-IIj'HjB^ )-B/^H*/= ¿-H*B^Hj' is symmetric,
idempotent, and has rank c. — 1, i = 1,2,• • •,s-2. Now, consider the
identity

-240-
T _ U*|J*'
cs-l-1_ 1 1
^see (3.444)j
-u1 + u2 + • • • +us_1,
where
Uj =ÍH*BiHÍ'-=^_H*Bi_1HÍ/, i = 1,2,- • -,s-l.
(We take CQ = cg_^; Bq = 0; cg_i = l; Bg_j = I.) From (3.456) and the
fact that ^-HjB^Hj' is a projection matrix, i = 1,2,• • •,s-2, it follows
that U- is a projection matrix and furthermore range(LL) is the
orthogonal complement of range^=-^—H*B^ i = 1,2, • • • ,s-l (s
Christensen (1987, p. 339)J. Also, for Cq = 1,
see
rank(Ui) = cj-ci;l, i = 1,2, • • • ,s-l,
Let S- be a (c^ — c-_^) x (cg_^ — 1) matrix of rank c-— whose
rows of orthonormal and span the row space of U-, i = 1,2,•••,s-1.
Then similar to Khuri’s (1990) Lemma 3.4 (ii), it is easy to verify
that
S-S'.
i J
i = j = 1,2,
i ¥= J-
, s—1
Now, define the (cg_j — 1)x(cg_j — 1) matrix S as
S = [S'1:S':-..:S'g_1]/.

-241-
Then S is orthogonal and diagonalizes H*B1H*/,• • •
simultaneously.
In the following subsection we consider the transformation
x = Su*.
The transformation x = Su*. Define the (cg_^-l) xl vector x as
x = Su*
where S is the orthogonal matrix mentioned above and u* was defined
in (3.445). From Lemma 3.15 we have
E<*)=Vi-i
(3.468)
s-2
Var(x) = £ ¿A + n-Jc .-l+^rCA's-lAs-l) HÍ'S
i=l s-l
where = SH*B^H*/S/ is a diagonal matrix consisting of the
eigenvalues of HjB^H*'. The matrix Aj is of order (cg— 1)
x(cs-i-l)»
i = 1,2,••• ,s-2. Lemma 3.17 provides the form for A - , i = 1,2,• ••,s-2.
Lemma 3.17
The eigenvalues of H*1LH*' are 0 with multiplicity cg — c- and
c- with multiplicity c^ — 1, i = 1,2,• • • ,s-2, ^c^ is given in (3.446)^
Furthermore, we can express A^ as

-242-
Ai = ‘i
ci“l
> i = 1,2,
,s-2.
(3.459)
Proof:
Since is of order (cg— 1) x (cg— 1) and has rank c* — 1
(see (3.457)), 0 is an eigenvalue with multiplicity cg_^ — 1 — (c^ — 1) =
cg_-^ — c^, i = 1,2, • • • ,s-2. Also, the nonzero eigenvalues of HiB^H*'
are equivalent to the nonzero eigenvalues of B^H*'H*, i =1,2,■••,s-2.
However, for i = 1,2, • • • ,s-2
ft.Hj'H^CIc^JcJCIc 1 -Jc J (see (3.444) and (3.446)).
Recalling that c-cj=cg_i for i = 1,2, • • • ,s-2 (see (3.446) and (3.389)),
we have
—J = tí ——[J„ cs-l cs-l Cs-1L ci ciJ
Thus, for i=1,2,•••,s-2,
iHi Hi-(IC.®JC.) c . (^c• ® ^c• )(*^c• ® ^c• )
1 1 Si 1 1 1 1
— (Ic.^Jc.) C .(^C.^^C.)
1 1 S— i. 1 1
= (Ici ® Jcj) - (JCi ® Jc-) (recall, cici=cg_1)
— (^c* JCi)®JCi'

-243-
Now, the nonzero eigenvalues of (Ic. ~3c.)®Jc. are the products of
the nonzero eigenvalues of (Ic. —Jc.) with those of . Thus, since
Ic. —Jc. is idempotent with rank c^ — 1 and the only nonzero eigenvalue
of J^. is Cj, the nonzero eigenvalues of are c^ with
multiplicity c^ — 1, i =1,2,•••,s-2.
Since the diagonal elements of are the eigenvalues of HjB^H*7
and since (3.456) implies that
range(Aj) C range(A2) C • • • Crange(Ag_2) which in turn implies that
rank(Aj + A2 + • • • + Aj) = rank(Aj) = cj — 1, j = 1,2, • • • ,s-2, we can,
without loss of generality express A^ as
Ai=5i
V1
, i=1,2,
, s-2.
â–¡
In (3.458) the variance-covariance matrix of x is not diagonal.
To diagonal this matrix we make a transformation utilizing
resampling. This is done in the next subsection.
Utilizing resampling. Consider once again SSEW defined in
(3.346). Let R* be the ( m-l)x(m-l) symmetric, idempotent matrix
m-l— 4s-l'
(3.460)
j = 1,2,• • ■,s-1, is defined in (3.433)\ Then
SSEW = wR*w.
s-1 ‘
Clearly R* has rank m —c
Now, we can express R* as

-244-
R* = C*A*C*'
(3.461)
where
C* = [C*:C*:C*]
A* = diag( I * , I * ,0)
Vl 72
(3.462)
r?í==cs-l-1
772 - m_2cs-l + 1
(3.463)
and C*, C2, and C| are of orders m X 77*, mX 772 , and mxcs_-p
respectively. Ve note that 77* + ?72 = m — cg_-^ = rank(R*) . Also, rj^ > 0 by
(3.393).
In order to diagonal the variance-covariance matrix of x ^see
(3.458)) we introduce the following transformation
I = *+(4axI (3.464)
”l
where L* = SH*(Ag_■^Ag_2)-1HJ,S, and A¡Jjax is its largest eigenvalue. The
vector x was defined in (3.457) and C* in (3.462). The random vector
l
w was given in (3.429). The matrix (A^axI * — L*)2 is well-defined
Vi
since A*axI * — L* is p.s.d. It’s eigenvalues are the square roots of
Vl
the eigenvalues of A*axI * — L* .
'll

-245-
Theorem 3.14.
The random vector r is normally distributed with
(a) E(t)=0 1
cs-l 1
(3.465)
(b) Var(r) = ¿ +5*1 _ 1
i=l s-1
where ¿*=(Tg_^+A*ax^ and ^ was defined in Lemma 3.13(b).
Proof:
The proof of Theorem 3.14 follows similarly to that of Lemma
3.13 and is therefore omitted. â–¡
The exact tests. Let us partition r as
r = (ri:ri:---:r'B_1)
(3.466)
where rj is of order (c--c._1)xl, i = 1,2, • • • ,s-l (cg = l). Note that
s-1
¿2 (cj—)=cs_i—1• Then the following facts are apparent:
i —1
(i) r¡ ~n(o, ('J¿i»2jcj+S*)Ic. _C¡1), i = 1,2, • • • ,s-2
rs_i ~N(0,
(3.467)

-246-
This follows since we can use Lemma 3.17 to express Var(r) in Theorem
3.14(b) as
Var(r) = diag((£ *jcj + <5*)Ic1 - 1 * (.E a]cj + O1^-Cl» • ' * »
( IT +^*)IC. — c- ( j — j J J 1 1 -1 S — ¿ S — o S— J- S — jL'
(ii) Ii>f2»■’ ’’-s-1 are mutually independent.
(Hi) r'sri/iif^cj + s^-x^-cj.j. i =
=1,2,•••,s-2
(3.468)
Thus, if we wish to test Hg:cr|=0 vs. Ha:a|^0, i = 1,2, • • • ,s-2, an
appropriate test statistic would be
p _ ?jij/(ci-ci-l) • , 9
1 — / / / \ 5 1 1 J ^ J
1 ii+lli+l/(ci+l-ci)
, s-2.
(3.469)
Under Hn, F ■ ~ Fr _r r _ r , i = 1,2, • • • , s-2. Under H~,
0 1 ci ci-l»ci+l ci a
Fi ~ (1 + c-C: )FC. _c. c, _c. where
1 V lSl' Ci C._1,C.+1 Cj
s-2
E
j=i+l
'j5j+¿*
(3.470)
i = 1,2,• • • ,s-2. Since c• >0 for all i=1,2,••■,s-2, we would reject
Hq for large values of our test statistic, F^ (i = 1,2, • • • ,s-2).

-247-
3.4.6. Reduction of the Exact Tests when the Design is Balanced
Introduction. In this section we show that the exact tests
reduce to the usual ANOVA tests when the design is balanced. For
testing HQ:<7g = 0 vs. Ha: clearly reduces to the usual ANOVA test since it is based on the same
mean squares as are used in a balanced data situation. In the
remainder of this section then, we concentrate on showing that the
exact tests developed for testing Hq: i =1,2,• •• ,s-1, reduce to their ANOVA counterparts when the design is
balanced.
Reduction of the exact test for Hgtffg_^=0. The test statistic
derived for testing Hg: It was based on the random vector w defined in (3.426). Now, when
the design is balanced, that is, when
s-1
and
(3.471)
the matrix K (see (3.418)) becomes

-248-
Hence L (see (4.427)) reduces to
as 1 1 as cs - 1
(in
this case cs = m = Il^a^ (see (3.389))).
Thus Amax, the largest eigenvalue of L, is Consequently u (in
(3.426)) simplifies to
u> = u = Hay (see (3.420)).
From (3.441) we can express the test statistic as
F _^(Qs-1-Qs-2)^/(cs-1-cs-2)
s-1 SSEy/(m-cs_1)
..rH/i(qs-1-qs-2)Hiy/(s-i-cs-2)
fHÍCím-l-ls-lWÍ—cs-l)
(see (3.426)).
(3.472)
Now, from (3.433) and (3.413), when the design is balanced we have
for j = 1,2,•• ■ ,s-1,
^ j = HlA j ( A jA j ) _1A jHi
= c-^a Hl[Ic®Jg.as3H/i- (recall (3.446) and (3.471
J ^ J J

-249-
Also, from (3.419),
H1H1 — Im Jm — ICg Jcs
(cs = m when the design is balanced)
— (Ic.®Ic.as) cs(^c • ® -as) •
J J s s J J
Consequently for j =1,2,•••,s-1,
HiQ jHi =
(Ic <8>Ic.a ) (Jc ®JC a )
J J “ J J
Ic . ® Jq -a
J Cjas
(Ic ®Ic.a„) (Jc • ®^c-as)
J J s J J s
(Icj®Jcjasy ^cj^"cjas'
,) (JC.®JC -a„ )
(Ic ®Ic.a ) (^c-®^c-as)
J J b J J “
((icj JCj)JCj= o)*
(3.473)
Thus, when the design is balanced, HÍ(QS_i - Qs_2)^i t^le numerator
of (3.472) becomes
[(:
's-1
- J
â– laJ [(
Ics-2 Jc
s-2
)®Jc
(0
's-2
® I,
)-(Jc
-V K cs-
„ ® Ja
2 as
-l))®3cs
-las

-250-
— (^c (J
JaJ
s-2 as-l “s' ' ^s-2 as-l as
[(Ics-2 Jcs-2)® Jas-1® Jas]
(recall that cs_2 = as_l an<^ c i=l). Hence the numerator of
reduces to, finally
4Ics-2 ® (‘Vl - J»s-1) ® J“s^/(cs-l - cs-2>
Similarly, we can show that Hj(I ^ — Qg_^)H1 reduces to
»;(I„_l-«s-l)». = Ics.1®(Ias-Jas)
when the design is balanced (recall that cg = m) .
In a balanced data situati on, y |see (3.411) and (3.412)^
y = —(Ir la )y.
- as+lv CS ~as+l -
Therefore the numerator and denominator of (3.472) become,
respectively, aside from a constant,
jr^yflr _®(Ia — J-, ) ® J;, ® Ja Ty
as¿[ cs-2 v as-l as-ly as as+lJ£
J_vf
as~
Ir ®(Ia — Ja )®«Ja ly-
L cs-l v as asy as+lJi
(3.474)
(3.475)
becomes
(3.476)
(3.477)
(3.478)

-251-
Hence asw,(Qg_^ — Qs_2)y and asSSEw reduce to the usual ANOVA sums of
squares from a balanced model ^see Khuri (1982)). Thus Fg_^ ^in
(3.472)) reduces to the usual ANOVA F-test associated with the
hypothesis _^=0.
Reduction of the exact tests for Hg: testing Hq: vector r defined in (3.464). When the design is balanced, the matrix
L* (given in the expression for r) reduces to
L* = SH*(Ag_^As_1)_1H*,S'
^see (3.413) and (3.471))
= §^SS' (see (3.444))
= g^I _-i . (S is orthogonal)
s cs-l 1
Consequently, its largest eigenvalue, A*ax, is Therefore, when
the design is balanced, r becomes
r = x = Su* (see (3.457))
=SR*w (see (3.445))
=SH*D*w
(see (3.446)).

-252-
Hence,
t't = w'D^'Hj'S'SHj D*w
= w'D*'(Ir -Jr )D*w
K cs_l cs-l ~
^see (3.444))
= w'D*,D*w-w'D*'Jf. D*w
" ~ cs-l “
= y/[iHi(ics_1®Jas)H;]y
-y'[ÍH1(JCsi®Jas)H'1]y
^see (3.442) and
(3.413))
l'w.
(3.479)
As mentioned previously, when the design is balanced, w reduces to
u) — H.y = —H, (Ir ®li )y.
x~ as+l 1V cs ~as+l -
Consequently, after some simplification, we can express (3.479) as
T'T = a—^r yf(Ir - Jr ®J~ V •
-- asas+l-[v cs-l cs-ly aS^ as+l/
Now, recalling (3.398), when the design is balanced we have,
j = 1,2,•••,s,
(3.480)
for
Cj(X'jXj)
-â– x',

-253-
= 1,
(see (3.384) and (3.471)),
where
s+1
n . = n/c • = . II a-, j = 1,2, • • • ,s
J J i=j+l
s+1
n = II a- .
i = l 1
Therefore (3.480) can be written as (letting Pq=Jn)
asas+l-
(!<
s — 1
L ) — ( Jr
as+r v CS-1
lS+l
)>
1
asas+l-
y'[Ps-i-po]+
(3.481)
However,
SE(Pj-Pj-i) = Ps-i-P0>
j=i
therefore
asas+l
/ s-1
y'( E (Pj
j = l J
1
s -1
^rrS^Pj-Pj-Or
From Khuri (1982), the sums of squares y'(Pj — Pj_^)y, j = 1,2, • • • ,s-1,
are the usual ANOVA sums of squares from a balanced model associated

-254-
with the effects /3X, • • ■ ,/?s_j (note that our “Pj — F>j_i” is Khuri’s
“P.”). Thus asa t'-t • reduces to the usual ANOVA sum of squares,
Pj”). Thus asas+1 t'-t-
j = 1,2,• • • ,s-l. Consequently, the test statistic developed to test
reduces to the ANOVA F-test
when the design is balanced.
Expressions for the power of the exact tests are developed in
the next section.
3.4.7. The Power of the Exact Tests
Introduction. We now investigate the power of the exact tests.
We first concentrate on developing an expression for the power of the
test associated with , i = 1,2, • • • ,s-2. Since the imbalance affects
the last two stages only, our expression for this power function will
be exact.
The approximate power of the test associated with is
derived using Hirotsu’s (1979) procedure (see Appendix D). A similar
expression can also be developed for the power of the test associated
with The power of the exact test associated with Hq:(t|=0,
i = 1,2, • • • , s-2. In testing Hq:ct|=0 vs. Ha: test statistic was given in (3.469). If 4^ denotes the power
associated with this test, then, for i =1,2,•••,s-2,
(3.482)

-255-
^i + lW(ci+l-Ci) “’ci~ci-l’ci+l
Ha .
— i a
Under H~ , F- ~ (1 + c • C-)F~ ~ ~ where c- was defined in
a’ 1 V lSl' ci ci l,ci + 1 Ci 1
(3.446) and Cj in (3.470), i = 1,2, • • •,s-2. Consequently, (3.482)
becomes
*i -P(Fci-ci-i’ci+i~ci >i+^iCiFa’ci-ci-i’ci+i-ci)
(3.482)
(i = 1,2, • • • ,s-2). Since '5'. is a monotone increasing function of ,
it is a monotone decreasing function of 6* (everything else being
fixed) where 6* was defined in Theorem 3.14(b). (Recall that
Ci - s_! 1 »
£ ^. + 6*
j=i+l J J
, s-2,
(see (3.470)).^ From the definition of <5* in Theorem 3.14(b) we have
S*=al_1+X*&xS
~ ^s-l ^max(ffs ^max^f ) (see Lemma 3.13(b)^.
It is thus desirable for A*ax and A*axAmax small. A sufficient
condition for this to occur is when both A*nv and Amav are small. The
next lemma provides bounds for A*ax and Amax.

-256-
Lemma 3.18
A^ax and Amax satisfy the double inequalities:
O) c~T E BñT— 1 ' HI/]
S-^S-1 ®8-l
(1)
(b) m En^ — Amax — (i) ’ w^ere
es »s
(0
nv ' = min na
*s 6s
(l)
mv y = min nu
«S-1 s-i
Proof:
(a) -A^ax the largest eigenvalue of the (cg— 1) x (cg_j — 1) matrix
L^SHÍCA'g^Ag^)-^' (see (3.464)). Thus
Amax = emax(sdi (Ag_^As_i) Hj S )
= emax(Hí(As-lAs-l)”lHr) (S is orthogonal)
— emax[(As-lAs-l) ]emax(^l )
= emax[(As-lAs-l) 1] (H* has orthonormal rows)
= en
<(e ® mdl_^) (see (3-413))

-257-
(l)’
1 ' '
where m
(1)_
nun mn
fi8-l
7s-l
It is also true that
Amax > c~~l itr[SH*(A7s_1As_1)-1HrS7],
the term on the right being the average of the eigenvalues of L*.
Consequently,
Amax — c * _ itr[(As-lAs-l) ^l]
s — 1
= d7itl(A,s-lAs-l)"L(A®-lAS-l)"1JcS-1]
^see (3.444)^
= cs_J-l|tr[(As-lAs-l) 1]-C¿iics_1(As-lAs-l)lcs.1}
cs-l-1
“0* 1 ct
E
-1
»s-l "s"1 s_1«s-l "s"1
isee (3.413)^
* i;t.
Hence, the result follows. The proof of (b) is similar.
â–¡
It is more important that A^ax be small than for Amax to be
small. Even if Amax =1 ^it’s maximum possible value regardless of
(nested) design), the product A^axAmax will be small if A^ax is small.

-258-
The power of the exact test associated with IlgitTg_^=0. We now
obtain an expression for the power of the test for Hg:^_^ = 0. In
this case the appropriate test statistic is (see (3.441)1
P _^/(qs-l-tJs-2)^/(cs-l-cS-2)
s-1- SSE^m-c^)
where Qj was defined in (3.433), j = 0,1, ■ • •,s-1, and the denominator
in (3.436). The random vector w was given in (3.429). Under Ha,
w,(Qg_j—Qg_2)y is distributed like a random variable of the form ^see
Box (1954a)^
T =
cs-l-cs-2
E
j=i
(3.484)
where the x’s are independent chi-squared variates with one degree of
freedom and the 7r’s are the nonzero eigenvalues of the matrix
•& = (Qs-l-Qs-2)(Var(y)) (3.485)
Similarly to the latter part of the proof to Theorem 3.13(a), we can
simplify (3.485) to
^ = (Qs-i-Qs.2)[4-iMs-iH¡ + ¿iln_1].
The exact distribution of T is complicated. To obtain an
approximate expression for the power we use the approach described in
Hirotsu (1979) (see Appendix D). Let

-259-
MSEy = SSEy/(m - cs_1)
y/(Qs-i-qs-2)^’/c*f
MSE js
h =
¿(c£
-1 ~cs-2
)
c*f
a.c.
-1 “ cs-2’m “ cs-
where in this case
tr(Jt2)
tr( Jk)
f _[tr(^)f
tr( Jt2)
Then the approximate power is given by
*s-l = P(H > h |Ha) = P(Ff )f2 > h) +[p/{3(f+2)(f+4)B(If ,If2)}]
x
x
(f+2)(f+4)-
2(f+f2)(f+4)
l+f2/(fh)
+
(f+f2+2)(f+f2)
(l+f2/(fh)}2
where
*2
m — c
s-
1
(3.486)
(3.478)
(3.488)
(3.489)
(3.490)
(3.491)
(3.492)

(3.493)
[tr(-^t)][tr( Jt3 ) ]
P= [tr(Jt2)]2
and B(a,b) is the beta function.
Similar to the derivation of E[w7(Qg_^ — Qs_2)y/(cs_ j — cs_2) ]
given after (3.441), we can find the trace of Jk as follows:
tr(-^)-tr[(Qs_1 —Qs_2)(a2s_1H1Ás_1H;+5Im_1)]
= ¿(cs_1-cs_2)+ = '5(cs_1-cs_2)+ir2_1tr[A/s_1Hi(Qs_1-qs_2)H1As_1]
cs-l
,2 v- „*/
^(cs_i — cs-2^ ^s-l -s-l :(QS-1 — ^s-2^-s-l i
j=l
(3.492)
where, recall
-s-l, j-Hl-s-l, j’ J ~ 1 >2> • ' • >c 1
and
A -i —- [” a -i -\ • • • • i a -i "1.
s —1 *—s — 1 ? 1 ~s — 1 j Cg_|
In order to evaluate (3.491) we also need tr(Jt2) and tr(Jt3). These
quantities can be expressed in much the same way as tr(-Jb) in (3.494).
This method of evaluating these traces is similar to the idea of
synthesis as described in Milliken and Johnson (1984, p. 219-231).

-261-
3.4.8. A Numerical Example
Data were generated according to the following 4-fold random
nested model:
yi jk£t - ? + “i +13i j + T'i jk + ¿i jk£ + ei jk£t ’
i = l,2,3; j = 1,2,3; k = l,2,3; l = 1,2, • • • jk; t = 1,2, • • • ,ni jk£;
where p is an unknown constant parameter; and 7jjk, ^ijk£’
and are independent normally distributed random variables with
0 0 0 0 o
zero means and variances aa, a^, aty, aand ae, respectively. The
mijk values were randomly generated from the integers {2,3,4} and the
nijk£ values were randomly selected from {1,2,3,4}. Furthermore, we
took p = 10, a& = 3, = 0.2, aiy — 1, a^ = 5, and ae — 1. These values
were only used to generate this (artificial) data set.
Using the methodology developed in Subsections 3.4.3—3.4.5, we
tested the following hypotheses: H0:tr^ = 0, H0: H0: important quantities used in the development of the exact tests.
Table 3.17 summarizes our findings.
3.4.9. Concluding Remarks
In this section we developed exact tests for testing the
variance components in a general unbalanced s-fold random nested
model where the imbalance affects the last two stages only. In

-262-
Table 3.16 Some useful quantities used in the construction of the
exact tests for the four variance components in the
example
Quantity
Formula cited
Value
s
(3.380)
4
h
(3.381)
(i)
92
(3.381)
{i »J}
^3
(3.381)
{i,J»k}
*4
(3.381)
{i > j
n
(3.385)
216
m
(3.387)
78
ci
(3.389)
3
c2
(3.389)
9
C3
(3.389)
27
c4 = m
(3.389)
78
(3.407)
9.6053
^max
(3.422)
1.0
6
(3.428)
as + ^max^e
f3
(3.441)
2.5540
Cl
(3.446)
9
c2
(3.446)
3
li
(3.446)
I3 Jg
b2
(3.446)
Ig <8> Jg
A*
"'max
(3.464)
0.5
6*
(3.465)
°y + ^max5

-263-
Table 3.16 - continued
Quantity
Formula cited
Value
(3.469)
0.7592
Fi
(3.469)
1.9929
Subsection 3.4.3 we derived an exact test for testing H0:(7g=0. This
test is the usual ANOVA test. Next we considered testing H0:(7g_j=0.
In this case, we employed resampling and made a transformation that
resulted in a model which resembled an unbalanced (s-l)-fold nested
model. Using this fact, we were able to consturct an exact test by
deriving a test statistic that was similar to the one used in testing
the equivalent null hypothesis in an unbalanced (s-l)-fold nested
model in which all the effects are considered fixed.
Finally, in Subsection 3.4.5, we constructed exact tests for
testing crj, 0^2, • • • , crg_2 • The motivation behind our procedure came
from noting that after making the above transformation, the resulting
model resembled an unbalanced random nested model with imbalance
affecting the last stage only. Consequently, we could apply the
results in Khuri (1990) to aid in the development of our methodology.
From Seely and El-Bassiouni (1983), it is easily seen that the
exact test for testing H0:<7| = 0 is a Wald’s test associated with model
(3.383). Similarly, the exact test for testing H0:c-g_^=0 is a Wald’s
test associated with model (3.429). Recalling (3.464) and (3.466), let
r(i) = (rí:T2:
••I'i)', i = 1,2, • • • ,s-l.

-264-
Table 3.17 Results from our testing procedures
Test for
factor
Numerator
degrees of freedom
Denominator
degrees of freedom
F-value
p-value
A
2
6
Fj=l.9929
0.2169
B(A)
6
18
F2=0.7592
0.6109
C(AB)
18
51
F3=2.5540
0.0045
D(ABC)
51
138
F4=9.6053
<0.0001
Then, the exact test for testing H0:cr|=0, i = 1,2, • • • ,s-2, is a
Wald’s test based on the model involving only If>t2, • • •,r-+^ (that is,
based on * + ^ ^ ) .
As a final remark, the extension to more general nested models,
including models in which some of the effects are fixed, should offer
no resistance. We merely invoke our resampling scheme (when
necessary) and make transformations in a “step-down” manner similar
to the methodology presented in Section 3.2 and in this section to
derive the exact tests.

CHAPTER FOUR
CONCLUSIONS AND DIRECTIONS FOR FUTURE RESEARCH
In the teaching of linear models there seems to be an
underlining hierarchy that is followed. The first linear model most
students and/or researchers are exposed to is one in which the
effects are fixed and the data balanced. For such models, the idea
of an “analysis of variance” is used to construct tests of hypotheses
concerning the fixed-effects in the model. These tests are easily
derived from the corresponding ANOVA table. Furthermore, they enjoy
many optimal properties. The next step up this hierarchy brings us to
unbalanced fixed-effects models. Now the picture is much more
muddled. For example, it is quickly learned that there is no unique
partitioning of the total sums of squares (as there is in the
balanced data situation) and thus no unique way to construct the
ANOVA table. Hence, the sums of squares used in a test of a
particular hypothesis must be chosen carefully by the researcher.
Otherwise, very spurious results may be obtained.
The concept of a balanced random or mixed model is usually not
introduced until after thoroughly analyzing the fixed-effects model.
After discussing the differences between these types of models, the
focus typically shifts to the problem of constructing point
estimators for the variance components in the model. Exact tests of
hypotheses concerning the variance components and/or estimable linear
-265-

-266-
combinations of the fixed-effects parameters rest upon finding mean
squares that can serve as “error” terms in appropriate F-ratios.
However, unlike the balanced fixed-effects case, exact tests (based
on the ANOVA mean squares) do not always exist. This phenomenom is
even more apparent when the data are unbalanced. Consequently, for
unbalanced random or mixed models, tests of hypotheses are usually
only developed in the context of an exercise in the application of
Satterthwaite’s procedure. Therefore, we are left with the idea that
in the unbalanced data situation exact tests either do not exist at
all or only exist for very simple models. Although Satterthwaite’s
methodology may provide adequate (approximate) tests when the design
is balanced, there is no theoretical justification for its use when
the data are unbalanced since in these situations the necessary
assumptions will not, in general, be satisfied. The need for exact
testing methods is therefore clear. The derivations of some such
methods are described in this thesis.
In the development of our procedures, we made the “usual”
assumptions about the random-effects. That is we assumed they were
independent of each other and had normal distributions with zero
means and “a2I” variance-covariance structure. This led to a
particular structure for the Var(y). But, are these assumptions
correct? Some diagnostic tools would be of help here and perhaps the
results of Hocking (1988), Hocking, Green, and Bremer (1989), and
Beckman, Nachtsheim, and Cook (1987) could be of use. Also, it would
be of interest to develop exact tests under more laxed assumptions.
Except for a few special cases, this has not been pursued.

-267-
The problem of constructing optimal tests in unbalanced random
or mixed models seems to be a difficult one. Besides the results of
Spj0tvoll (1967), Mostafa (1967), and Mathew and Sinha (1988), little
is known about this area. Even these papers deal only with very
simple models. Therefore it is not known if our tests (or tests
constructed similarly to ours) possess any optimal properties similar
to the ones mentioned in the above articles. A full solution to this
problem may be unattainable. Partial solutions may possibly be
obtained through intensive computer simulation.
Ofversten (1989) also constructed exact tests for the variance
components in unbalanced random and mixed models. His methods
differed from ours in the way he initially attacked the problem —
orthogonally transforming the model matrices into layered-triangular
form ^see Allen (1974)^. However, this transformation did not always
lead to an exact test. When it did not, Ofversten relied on a
resampling scheme similar to ours to aide in the construction of the
test. Although his procedure is adaptable to many unbalanced
models, it is not clear how to utilize it in models not discussed in
his dissertation. Still it would be of interest to compare his
results to ours.
A drawback to our procedures, and indeed, to all procedures
utilizing a transformation based on a resampling scheme, is that the
values of the test statistics are not unique. This is true since
only a portion of the error vector is utilized in these
transformations. However, the distributional properties of the
resulting vector remain unchanged no matter what portion of the error
vector is used in the transformation. Consequently, the power of the

-268-
exact tests will not depend on what portion is used in the
transformation. The power will, however, depend on certain design
parameters (eigenvalues of certain matrices). Thus, it is of
interest to know exactly how much influence these design parameters
have on the power. This has yet to be resolved.
In this dissertation we have developed exact procedures to
handle the variance components and estimable linear functions of the
fixed-effects in some unbalanced random and mixed models. As noted
above, there are still many unresolved problems left to tackle. It
is our hope that the techniques put forth in this work and the ones
sure to be developed elsewhere will help in the construction of
solutions to these challenging problems.

APPENDIX A
MATHEMATICAL RESULTS
We present some useful mathematical facts in this appendix.
When necessary, proofs will be supplied. Otherwise, the relevant
source will be cited. We begin with a result dealing with the rank
of a product of matrices
Lemma A.l (Frobenius Inequality)
Let A, B, and C be matrices of dimensions nxp, pxq and qxt,
respectively. Then
rank(ABC) > rank(AB)+ rank(BC)— rank(B)
Proof:
First note that if either A, B, C, AB, or BC is a zero matrix
then the result trivially follows. So, suppose that none of these
matrices have rank zero. Then, there exists nonsingular matrices P
and Q, both of order qxq, such that
P(AB)'(AB)Q =
0
Or r
rB"rAB
0
0
0 0 0n_r
q-rB
(A.l)
-269-

-270-
where r^g=rank(AB)>0 and 0 < rg = rank(B) < q.
Define the qxq matrix X as
X = P
-3
AB
0 I
0
0
rB_rAB
0 0
0
0
q-rB
(A.2)
From (A.l) and (A.2) we have
(AB)'(AB) +X = P'1
q-rB
i-i
(A.3)
Write Q 1 as Q 1=
Qi
Q2
where Q1 is of order rgXq while Q2 is of
order (q —rg)xq. Then from (A.3),
_
p-i
Ir 0
rB
«1
= p_1
Qi
0 0n r
q-rB
Qi
0
(A.4)
Also, write B as
B =
Bi
b2
(A.5)

-271-
where Bj is of order rgXq and B2 is of order (p — rg)xq (rg < P since B
is pxq). The rg rows of Bj are chosen to be linearly independent.
Since Q-1 is nonsingular it follows that the rg rows of are also
linearly independent. Thus there exists a nonsingular matrix M of
dimension rgxrg such that
Qj = MBj (A.6)
Now
rank(BC) = rank(BjC) = rank(M ^C) = rank(Q1C) .
But
rank(QjC) = rank]
= rank
= ran
0
= rank((AB)'ABC + Xc)
^see (A.3)^
< ran k((AB)'ABCl) + rank(XC)

-272-
< rank(ABC) + rank(X)
— rank(ABC)+ rank(B)— rank(AB)
^from (A.2)^
Thus,
rank(ABC)>rank(AB)+ rank(BC)— rank(B)
Remark A.1.1
Suppose q = p and let B be the pxp identity matrix. Then, by
this lemma we have
rank(AC)>rank(A)+ rank(C)— p
(A.7)
Furthermore, if n= p then we get, as a special case to this lemma,
(1982)
Remark A.1.2
Most often we will use Lemma A.l in the following context: A is
(p-l)xp of rank p-1; B = Ip; C is pxt of rank r will be able to find a vector x^O such that Cx ^ 0 and ACx = 0.
Consequently it is clear that

-273-
rank(AC) By the Frobenius inequality we will also have
rank(AC)=rank(A)+rank(C)— p
= p-1 + r-p = r-1
Thus, rank( AC) = r-1.
The next three results deal with the simultaneous
diagonalization of two symmetric matrices. The proofs to all three
of these lemmas can be found in Searle (1982, pp. 312-314). However,
we will present a slightly different proof to Lemma A.4 then what is
found in Searle.
Lemma A.2
For symmetric A and B of the same order there exists an
orthogonal matrix P such that P'AP and P'BP are both diagonal if and
only if AB = BA.
Remark A.2.1
Since A and B are both symmetric, the condition AB=BA is
equivalent to showing that AB is symmetric.

-274-
Lemma A. 3
For symmetric matrices A and B of the same order, with A being
positive definite (p.d.), there exists a nonsingular matrix M (not
necessarily orthogonal), such that M'AM = I and M'BM is a diagonal
matrix whose diagonal elements are solutions A to |B — AA| =0.
Remark A.3.1
The diagonal elements of M'BM are called eigenvalues of B with
respect to A. Since A is p.d. it is clear that these diagonal
elements are also the eigenvalues of A_1B which are equivalent to the
eigenvalues of BA-1 which are equivalent to the eigenvalues of
1 1
A 2BA 2.
Lemma A.4
For real symmetric matrices A and B of the same order and non¬
negative (n.n.d.) there exists a nonsingular matrix P such that P'AP
and P'BP are both diagonal.
Proof:
Let the order of A and B be n, with ranks a and b, respectively,
and without loss of generality suppose a
-275-
and n.n.d., there exists a nonsingular matrix T such that
T'AT =
I
a
0 0
n-a
Denote T'BT by
T'BT =
E
F'
F
G
(A.8)
(A.9)
where E is axa and G is (n-a) x (n-a). Now, let C be any (n-a)xa
matrix which satisfies
F + C'G = 0.
(A.10)
One such choice for C is
C = -G~F'
(A.11)
where G is a generalized inverse of G. That is, G satisfies
GG—G = G. Consider the nonsingular matrix S defined as
S =
la
C
0
M
(A.12)
where M is an (n-a) x (n-a) nonsingular matrix satisfying
Ig 0
0 0n_a_g
(A.13)

-276-
where g = rank(G). Now S'T'ATS = T'AT by the construction of S. While,
by (A.10) and (A.12), S/T,BTS becomes
S'T'BTS =
H
0
0
where H is the axa matrix
0
h
0
0
0
®n-a-g
(A.14)
H = E + C'F'.
(A.15)
Note that the rank of H is b-g. Since C'F' can also be expressed
as— C/GC ^see (A.10)j, we have that H is symmetric. Thus there exists
an orthogonal matrix Q such that
q,Hq = Da = diag(A1,..-,Ab_g, 0, • • • ,0) (A.16)
where Aj, A2, • ■ • , are the nonzero eigenvalues of H. Therefore,
upon defining R as
R =
q
o
o
o
0 I
0
0
n-a-g
(A.17)
we have from (A.14) and (A.16),

-277-
R'S'T'BTSR =
O
O
O
I
g
O O
o
o
n-a-g
(A.18)
Also,
R'S'T'ASTR =
q'Q o
o o
n-a
(A.19)
Hence, upon letting
Aa
0 0r
(A.20)
P = TSR
(A.21)
we have P'AP and P'BP both diagonal.
â–¡
Corollary A.4.1
Suppose the conditions of lemma A.4 hold. Further, if
a = rank(A) and b = rank(B), suppose that rank(A+B) = a+b < n. Then
there exists a nonsingular matrix L such that
L'AL =
'â– a
0
0
0 0
0
0
n-a-b
(A.22)

-278-
and
L'BL =
0
0
0 0
Tb 0
(A.23)
Proof:
From lemma A.4, there exists a nonsingular matrix P such that
P'AP and P'BP are both diagonal. Since the rank(A+B) = a+b = rank(A)
+ rank(B), it follows that the nonzero diagonal elements of P'AP
correspond to zero diagonal elements of P7BP and vice-versa. Clearly
then, there exists nonsingular matrices Mj and M2 for which NfP'APM and
M'P'BMP have the forms indicated in (A.22) and (A.23), respectively,
where M = M1M2. Upon defining L = PM, the result follows. â–¡
Remark A.4.1
Although not explicitly stated in the proof to lemma A.4 or
corollary A.4.1, it should be noted that H, defined in (A.15), is
n.n.d. This follows from the fact that C'F' can be written as
C'F' = -FG~F' (A. 24)
^see (A. 11) and use the facts that FG~F/ is invariant to choice of G
^Corollary 12.2.20.2 of Graybill (1983, p. 416)^ and that E—FG—F' is

-279-
n.n.d. ^Theorem 12.2.21(3) of Graybill (1983, p. 417)^. Thus, the
eigenvalues of H are nonnegative.
We present one last result concerning the diagonalization of
symmetric matrices. This will extend Lemma A.2. The reader is
referred to Graybill (1983, p. 408).
Lemma A.5
Let Aj ,A2, • • • ,A^ be symmetric nxn matrices. A necessary and
sufficient condition that there exists an orthogonal matrix P such
that P'A^P is diagonal for all i=l,2,---,k is A^Aj=AjA^ (or A^Aj
symmetric) for all i and j.
is

APPENDIX B
STATISTICAL RESULTS
In this appendix we give some statistical results. Most of the
following concepts are well known and are only presented for the sake
of completeness. Unless otherwise stated, these can be found in
Graybill (1976, pp. 134-140).
Lemma B.1
Let the nxl random vector y be distributed as N(p,£) where £ has
rank n. If B£A = 0, the quadratic form y'Ay is independent of the
linear form By where B is a qxn matrix.
Lemma B.2
Let the nxl random vector y be distributed as N(/¿,£) where £ has
rank n. If A£B = 0, then the two quadratic forms y'Ay and y'By are
independent.
Lemma B.3
Let y be an nxl random vector and let E(y)=¿í, Var(y)=£. Then
-280-

-281-
E(y'Ay) = tr(AE) + //A/i
The next two results deal with the distribution of a quadratic form.
Lemma B.4
Let the nxl random vector y be distributed as N(^,£) where E has
rank n. Then the quadratic form U = y,Ay~Xp(^) where p is the rank(A)
and X = if and only if one of the following three conditions are
satisfied: (1) AE is an idempotent matrix of rank p; (2) EA is an
idempotent matrix of rank p; (3) E is a generalized inverse of A
(i.e., AEA = A).
Lemma B.5
Let the nxl random vector y be distributed as N(/i,E) where E has
rank n. The random variable y'Xy has the same distribution as the
n
random variable U, where U= where the d— are the
i = l
eigenvalues of the matrix AE, and where U1,U2, • ••,Un are n independent
noncentral chi-square random variables, each with one degree of
freedom.
Our last lemma applies some of the above results to a general
mixed-model. This result is particularly useful since it is used
throughout this dissertation.

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Lemma B.6
Consider the general mixed-model
y = Xg + Zh + e (B.l)
where X = (XQ: • • • :X^) ; Z = (X„_p+1: • • • :X„) ; g = (ffoi ■ • • :£i,_p)' is the
vector of fixed effects; and h = : • • • ‘-P'p)' is the vector of
random effects. We will assume that the nxl vector of errors, e, is
distributed as N(Q, y~N(Xg,£) (B.2)
where
E = Z«Z, + (B • 3)
For ease of notation let V=[X:Z] and let r = rank(W). Also, define
the nxn matrix R to be
R=In-V(W)~V
(B.4)
Then
(i) R is symmetric, idempotent of rank n-r.

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(ii) Clearly RW = 0. Since range(X^) C range(V) for all
1 = 0,1, • • • ,r, it follows that RX^=0, i = 0,1, • • • ,v.
(iii) The error sum of squares can be written as
SSE = y'Ry • (B. 5)
Since from (i) and (ii), R£/<7j = R (which is idempotent)
and RXg = 0, we see, using lemma B.4, that
SSE/o-2 ~ Xn-r •
(iv) From (i), there exists an orthogonal matrix C such that
R = CAC' (B. 6)
where
A = diag(I^,I^,0)
(B.7)
and
>?i + »72 = n-r.
(B.8)
Suppose r?1>0, then for r;2 to be positive we must
necessarily assume

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n > r?j + r. (B.9)
Partition C as
C=[C1:C2:C3] (B.10)
where Cj is n xj^; C2 is n xt¡2; and C3 is nxr. Then
C'iCj = I (i = 1,2,3)
(B.11)
C'iC^O i#j.
Furthermore
SSE = y'Ry = y'CAC'y
= y'CjCjy + y,C2C2y
= SSE1 + SSE2 (B. 12)
and SSEj/o-j ~ independently of SSE2/ This forms the basis behind the idea used in
resampling from the error vector.
(v) Suppose range(X^) C range(XJ/) for all i = 0,1, • • • ,v-l.
Define the matrix D as

-285-
D=(XX)"lxí,-
(B.13)
Further, let the random vector x be
x = Dy.
(B-14)
Then, by lemma B.l, since
DER = (B-15)
we have that x = Dy and SSE = y'Ry are independent. ^The
fact that DR = 0 follows after noting that range(XJ/)
C range(V) and RW^O.j

APPENDIX C
AN ILLUSTRATION OF THE UTILITY OF RESAMPLING
In this appendix we consider the general unbalanced mixed cross¬
classification model to illustrate the idea of resampling from the
error vector. Ve assume the imbalance affects the last stage only.
Thus, this model is a special case of model (3.189) in Section three
of Chapter three.
As it applies to models having imbalance in the last stage only,
resampling serves to reduce the analysis of the unbalanced model to
that of a balanced one. For designs having more general forms of
imbalance, resampling will typically need to be employed more than
once to construct exact tests. This is the case for the designs
studied in Sections two and four of Chapter three.
Ve start by considering the following model:
y = Xh + Zg + f (C.l)
where X = (XQ: • • • :XJ/_p) ; Z = (XJ/_p+1: • • • :X„) ; h = (#,: • • • is the
vector of unknown parameters; g = (,0't/_p+i: • • • :P'u)' is the vector of
random effects, y is the nxl vector of observations; and e is the
nxl vector of errors. We assume that 0u_p+i > • • • and e are
independent, normally distributed random vectors with zero mean
vectors and variance-covariance matrices given by
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-287-
Var(^i) = Var(e) - a\In
(C.2)
where c- is the number of elements in /?^ (i = 0,1, • • • .
The vector of cell means, y, can be expressed as
y = Dy
(C.3)
where
D=(XX)"lxr-
(C.4)
we can also express (C.3) as
y = Hh + Gg+7
(C.5)
where H = (Ag: • • • : Aj,_p) and G = (Ay : • • • : Aj, = Ic^) . Also, 7 = Dt.
The A^ (i = 0,1, • • • ,j/-1) are described in Khuri (1990). In this same
article, Khuri mentions that the matrices B- = A^A'^ (i = 0,1, • • • ,v)
commute in pairs. Consequently, by lemma A.5, there exists an
orthogonal matrix Q which simultaneously diagonalizes them. Write Q
as
(C.6)

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where Q2 is constructed so that
Q2H = 0.
(C.7)
From (C.5) and the distributional properties of g and £, it can be
easily shown that
E(y) = Hh
Var(y) = £ i=i/-p+l
(C.8)
= £ ^iBi +CTl1cl/ + Ka¡
i=I/-p+l "
where
k = (x'„xl/yK
(C.9)
Now, define the qxl random vector u as
u = q2y
(C-10)
where q is the number of rows of Q2. Then
E(u) = q2Hh = 0
Var(u)
= E1 ^q.B-q'+^i +q2Kq'^.
i=i/-p+l
(C.ll)

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If not for the Q2KQ2 could analyze model (C.10) as a balanced model. This is where the
utility of resampling appears. Coupled with the results of Lemma
B.6, resampling will allow us to reduce the analysis of model (C.10)
to that of a balanced model.
From (C.l), write
V= [X: Z]
(C•12)
and
R= In-V(V'V)“V'
(C.13)
Assuming
n > q + cv
(C.14)
and using lemma B.6(iv), we can write R as
R = CAC'
(C-15)
where C is orthogonal and A is diagonal. Furthermore, we can
partition C and A as

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C — [Ci:C2:C3]
A — diag(Iq,In_c^_q,0)
(C•16)
where Cj is nxq; C2 is nx(n-c„-q); and C3 is nxc„. Define the qxl
random vector uj as
y = u + (Wiq-L)^iy (c-17)
where L = Q2KQ2 and Amax is its largest eigenvalue. Cj is defined in
(C.16). Notice that AmaxIq - L is positive semidefinite and hence
l
(AmaxIq—L)2 is well-defined. From lemma B.6 the following facts are
apparent:
(i) u is independent of C^y.
(ii) CjECj = CfIq.
Consequently, we obtain from (C.ll):
(C-18)
E(w) =0
Var(y) = ^Q2B.Qi + «q
i=j/-p+l
(C-19)

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where 6 = unbalanced model (C.10) to that of a balanced one (C.17). ^See, for
example, Searle (1971), Graybill (1976), or Khuri (1982) for a
discussion on the analysis of balanced models.
In (C.17) note that C'jy is a portion of the error vector, Ry.
The actual value of C^y depends on which q columns are chosen from the
first n—cu columns of C to form Cj ^see (C.16)^. However, the
distribution of uj will not depend on the choice of Cj. Thus, this
approach is similar in spirit to methodologies employing resampling,
such as the jackknife and bootstrap.

APPENDIX D
HIROTSU’S APPROXIMATION
In this Appendix we describe the F approximation (used in some
of our power calculations) as developed in Hirotsu (1979). We use
his notation.
Let
x = N(/?>E)
and let Xf be a x2 variable distributed independently of x. Further,
take n = 0 (as in the case in our approximations). Then we are
concerned with finding P(Z>z) where
Z =
(x'Ax)/cf
(D.l)
and A is a nonnegative definite matrix. The values of c and f are
given by
c = ( x7Ax ) / /Cj ( x'\x )
f = 2k\ ( x'Ax) /«2 ()
(D.2)
where /c^(x'Ax) represents the i^ cumulant of x'kx. In this situation,
these quantities become:
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-293-
c = tr[(AE)2]/tr(AE)
f =[tr(AE)]2/tr[(AE)2].
Hirotsu’s approximation is then given by
(D • 3)
P(Z > z) =P(F, , >z)
Ti > J2
+ [«/{ 3(f+2)(f+4)B(|f,if2)}]
x(l + fz/f2)_^(f+f2)(fz/f2)5f
x
(f+2)(i+4)—
2(f+f2)(f+4) (f+f2+2)(f+f2)
l+f2/(fz) + {l+f2/(fz)}2
(D.4)
where B(•, • ) is the beta function and
_ |«i(x/Ax)/c3(x/Ax)
k2(x'Ax)
_{tr(AE)}{tr[(A£)3]} B
{tr[(AE)2]}2
The approximation is based on a generalized Laguerre polynomial
expansion of the true distribution of (x7Ax)/2c and has been shown to
provide satisfactory results.

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BIOGRAPHICAL SKETCH
I was born in Ft. Wayne, Indiana, on November 4, 1960. When I
was six years of age we moved to California. I resided there for 20
years. In 1986 I came to Gainesville to continue my education at the
University of Florida. I am engaged to be married to Debra Thurman
in August of 1991.
My B.S. degree was in applied mathematics with an emphasis in
astronomy and was received from San Diego State University in
December of 1983. My M.S. degree, in statistics, was also awarded
from San Diego State. I received this degree in August of 1986.
From August of 1986 until August of 1988 I worked in the
Biostatistical Unit located in Shands hospital. While there I worked
under Dr. Randy Carter and Dr. Jane Pendergast as a statistical
consultant. From August 1988 to the present I have been employed as
a teaching assistant in the Department of Statistics and have taught
many of the undergraduate courses.
-298-

I certify that I
conforms to acceptable
adequate, in scope and
of Philosophy.
have read this study and that in my opinion it
standards of scholarly presentation and is fully
quality, as a dissertation for the degree of Doctor
VUMit
André I. Khuri, Chairman
Professor of Statistics
I certify that 1
conforms to acceptable
adequate, in scope and
of Philosophy.
have read this study and that in my opinion it
standards of scholarly presentation and is fully
quality, as a dissertation for the degree of Doctor
I certify that 1 have read this study and that in my opinion it
conforms to acceptable standards of scholarly presentation and is fully
adequate, in scope and quality, as a dissertation for the degree of Doctor
of Philosophy.
Associate Professor of Forest
Resources and Conservation
This dissertation was submitted to the Graduate Faculty of the
Department of Statistics in the College of Liberal Arts and Sciences and
to the Graduate School and was accepted as partial fulfillment of the
requirements for the degree of Doctor of Philosophy.
August 1991
Dean, Graduate School

UNIVERSITY OF FLORIDA
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