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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by Robert Calvin Capen.
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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


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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


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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.


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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))