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...... ...! .~ ·. )

f

¡ ..

.

AN INTRODUCTION TO LINEAR STATISTICAL MODELS Volun1e I

FRANKLIN A. GRAYBILL L-'rojl'.~sor

of 1lfoth e111ulical S tatistics Colorad o S tutc University F ort Colli11.$, Colorado

:rifoGRA \\'- H I LL BOOK COl\IPA N Y, INC.

Nc w York

'l'oron to l !Hi 1

Londo n

AN INTRODUCTION TO LINEAR STATISTICAL MODELS VOLUME I Copyright © 1961 by the McGraw-Hill Book Company, Inc. Printed in the United States of America. All rights reserved. This book, or parts thereof, may not be reproduced in any forro without permission of the publishers. Library of OongretJs Oatalog Oard Number 60-9841 ISBN 07-024331-x

12 13 14 15 16-MAMM-7 6 5 4 3 2

FAC. CIENCIAS

TO JEANNE

Preface

This book was \vritten with the intention of fulfilling three needs: ( l) for a theory textbook in experimental statistics, for undergraduates

or first-year graduate students; (2) for a reference book in the area of regression, correlation, least squares, experimental design, etc., for consulting statisticians with limited mathematical training; and (3) for a reference book for experimenters with limited mathematical training who use statistfos in their research. This is not a book on mathematics, neither is it an advanced book on statistical theory. It is intended to be an introductory mathematical treatment of topics which are important to experimenters, statistical consultants, and those who are training to be statistical consultants. The only mathe- matics that is required is generally obtained before the senior year in college. The mathematical leve] of this book is about the same as "An Introduction to the Theory of Statistics" by A.M. Mood, except for the fact that a great 0, ... ,

1 ª21

5

MATHE.MATICAL CONCEPTS 1i

+ Theorem 1.31 For every symmetric matrix A tl . ie:c ex.1sts an orthogonal matrix C such that C' AC - D. ·h .· . w h ose d'1agonal elements are th- 1 '·'\ ere. D• is a d1auonal 0 mat 11x e e 1aracter1strn roots of A. + Theorem 1.32 Let A A A b · • 1, 2, • · • , t e a collect1on of sym metric n .?< n matrices. A necessary and sufficient condition that there ex1sts ~n orthogonal ~ransformatiOñ C such that C' A C C' A C 1 2 ' · · · , C AtC are all diagonal is that A A b .' . d . s· i i e symmetr1c for all i an J.t . ~fnce dall the A; are symmetric, it follows that A A is symme r1c l an only if A; and A, commute. ¡ i

>0

ª22

+ Theorem 1.24 If A is an n x m matrix of rank m < n, then A' A is positive definite and AA' is positive semidefinite. + Theorem 1.25 If A is an n x m matrix of rank k < m and k < n, then A' A and AA' are each positive semidefinite. + Theorem 1.26 If C is an orthogonal matrix, and if the transformation Y = CZ is made on Y'Y, we get Y'Y = Y'IY = Z'C'ICZ = Z'C'CZ = Z'Z. In order to develop the theory of quadratic forms, it is necessary to define a characteristic root of a matrix. A characteristic root of a p x p matrix A is a scalar A such that AX = .A.X for sorne vector X =I= O. The vect.or X is called the characteristic vector of the matrix A. It follows that, if A is a characteristic root of A, then AX - .A.X = O· and (A - .A.l)X =O. Thus, A is a scalar such that the above homogeneous set of equations has a nontrivial solution, i.e., a solution other than X =O. It is known from elementary matrix theory that this implies IA - .HI =O. Thus the characteristic root of a matrix A could be defined as a scalar A such that IA - .A.11 =O. It is easily seen that IA - .UI is a pth degree polynomial in .A.. This polynomial is called the characteristic polynomial, and its roots are the characteristic roots of the matrix A. We shall noi.v give a few theorems concerning characteristic roots, characteristic vectors, and characteristic polynomials. In this book the elements of a matrix will be real.

+ Theorem 1.27 The number of nonzero characteristic roots of a

+ Theorem 1.33 Let an n X n matrix C be written

is the ith row of c. Thus e 1-8 th t f · i e ranspose o an n x 1 coffiu~n vt~ctoCr. The following two conditions are necessary and su cien .ior to be orthogonal: ( l) where l

C;

for ali i -=!= .i for ali i

(2)

That is t~ s~y, any two rows of any orthogonal matrix are ortho onal (the1r mner product is zero) and th . g row with itself is unity. ' e mner product of any

•

Theore~

1.34 Let C - ( •

·

matrix A is equal to the rank of A.

+ Theorem 1.28 The characteristic roots of A are identical with the characteristic roots of CAC-1 • lf C is an orthogonal matrix, it follows that A and CAC' have identical characteristic roots. + Theorem 1.29 The characteristic roots of a symmetric matrix are real; i.e., if A =A', the characteristic polynomial of IA - .U¡ = O has all real roots.

+ Theorem 1.30 The characteristic roots of a positive definite matrix A are ¡Jositive; the characteristic roots of a positivo semidefinite matrix are nonnegative.

P

::¡

x! - . . .

be

f P rows o an n

X

n orthogonal

Cv

matrix. That is to sav let e vectors suc'Ii that e¡¿ ~ O

(i' ;

2 '. ·~. ' e,, be the transpose~ of p (i - 1 ? ) i J - l, 2, · · · , p) and e.e. = 1 - , -, ... , ? . ~hen there exist n - p vectors L su~h ti1 t C; f J = O for all i andJ and f~f. = 1 (i = 1 i , ª 2' • • • , n - P) ·f · -1- • 'I,h '· 1 ' f f - O I i ,_J. us the theorem states that if we are i ' i i -. such as C1, there exists a matrix C2 of di:Uension (n g ve) .a matr1~ C 1) - p X nsuc that ( C = C (C1 forms the first p rows of C and C the last n _ p

2

2

rows of C), where

9 is orthogonal.

G

1.3

LINEAR STATISTICAL l\WDELS

MATHEMATICAL CONCEPTS

7

The trace of a matrix A, l'i'l1ich will be \vritten tr(A), is equal to the

Determinants

ta

In this section a few of the important theorems on determinants will be given. It will be assumed that the student knows the definition of a determinant and knows how to evaluate small ones. In linearhypothesis applications it is often necessary to..§.olve systems involving a great ma~y equations. It might at times be necessary to evaluate large determinants. There m·e many methods of doing these tlúngs that are adaptable to automatic and semiautomatic computing machines. These methods will be discussed in detail later. It will be assumed here that the student knows how to evaluate determinants by the method of minors or by sorne other simple method.

+ Theorem 1.35 The determinant of a diagonal rnatrix is equal to the product of the diagonal elements. IBA! = IAI IBI. + Theorem 1.37 If A is singular, IAI = O.

• Theorem 1.38 If C is an orthogonal matrix, then ICI = +1 or ICI = -1. + Th~orem 1.39 If C is an orthogonal matrix, then IC' ACI = l~I­

+ Theorem 1.40 The determinant of a positive definite

~atrix

is

positive.

¡.,. ¡

• Theorem 1.45 tr(AB) = tr(BA). Proof: By definition, tr(AB) is equal to 2,a¡;bii.

By definition,

ij

tr(BA) is equal to¡ b,kaki· ~

.

But it is clear that 2, a;;bii = 2, bikaki; ij

~

therefore, tr(AB) = tr(BA).

• Theorem 1.46 tr(~BC) = tr(CAB) = tr(BCA); that is, the trace of the product of matrices is invariant under any cyclic permutation of the matrices. Proof: By Theorem 1.45, tr[(AB)C] = tr[C(AB)].

• Theorem 1.48 If C is an orthogonal matrix, tr(C' AC) = tr(A). Proof: ByTheorem 1.46, tr(C'AC) = tr(CC'A) = tr(IA) = tr(A). It is sometimes advantageous to break a matrix up into submatrices. 'fhis is callcd partitioning a matrix into submatrices, and a matrix can be partitioned in many ways. For example, A might be partitioned into submatrices as follows:

A= (Au Aj2 A21

+ Theorem 1.41 The determinant of a triangular matrix is equal to the product of the diagonal elements.

+ Theorem 1.42 1n-11 = I/IDI, if !DI -:F o. + Theorem 1.43 If A is a square matrix such that

=

Oor A 21 =O,

+ Theorem 1.44 If A 1 and A 2 are symmetric and A 2 is positive definite and if A 1 - A 2 is positive semidefinite (or positive definite), then IA 1 1 ~ IA2I·

1.4

= 2, aii·

• Theorem 1.47 tr(I) = n, where I is an n x n identity matrix.

+ Theorem 1.36 If A and B are n x n matrices, then IABI =

where Au and A 22 are square matrices, and if A 12 th_en IAI = IA11l IA22I·

snm of the diagonal elements of A; that is, tr(A)

Miscellaneous Theorems on Matrices

In this section we shall discuss sorne miscellaneous theorems concerning matrices, which we shall use in later chapters.

~9

where A is m X n, A 11 is m 1 x n 1 , A 12 is m 1 x n 2 , A 21 is m 2 x n 1 , and A 22 is m 2 x n 2 , and where m 1 + m 2 = m and n 1 + n 2 = n. The product AB of two matrices can be made symbolically even if A and B are broken ~nto SQbmatrices. The multiplication proceeds as if the submatrices were single elements of the matrix. However, the. dimensions of the matrices and of the submatrices must be such that they will multiply. For example, if Bisan~ x p matrix such that

B=

(Bn Bj2 B 21

B

wherc Bii is an ni X P; matrix, then the product AB exists; and the corresponding submatrices will multiply, since A¡1 is of dimension m; X n; and B;k is of dimension n 1 x P1:· The. resulting matrix is as follows:

9

l\:IATHEMATICAL CONCEPTS

8

LINEAR STATISTICAL MODELS

+ Theorem 1.49 If A is a positive definitc symmetric matrix such that

A=(:: j

This is clear since, by Theorem 1.43,

IBI = IIA;21I = IA2ll IAI

so

=

IA22I IAI IA2il

Proof:

=

B12) B21 B22

The corresponding sub matrices are such that they multiply; so

Thus,

A11 A12\ (Bn B12)

~1 ~J

=

¡

IAI =

B21 B22

+ A12B21 = 1

A11B12

and

An - A12A;;l A21 A12Aii 1 _1 o 1 = IA22l IA11 -· A12A22A21I

l~I 1

and we get the following two matrix equations: A11B11

IAI

A~)I

mi, etc.,

X

B11 - B12B;iB21

Since A= B-1 , then AB =l. (

=

~I

= (Bn

and if Bu and Au are each of dimension mi

A;l

IAI IA2il IA22l

Replacing A and B by their submatrices, we get

and if B is the inverse of A such that B

=

as was to be shown.

+ A12B22 =O

1 Solving ~he second equation for A 12, we get A12 = --:'-11B12B:?2 · Substituting this value for A 12 into the first equat10n, we get

· Consider the system of equations AX =Y, where A is an n x m matrix, X is an ni x 1 .vector, and Y is an n X 1 vector. Writing this linear system in detail, we get

An Bn - AnB12B2f B21 = 1

Multiplying by Ali 1 gives the desired result. . . It is known that Bti1 and Añ 1 exist, since A and B are pos1t1ve definite matrices and since A 11 and B 22 are principal minors of A and B, respectively. By Theorerr_i 1.23, t.he. dete~"?inant ?f ~he principal minor of a positive defimte matr1x is pos1tive. S1m1lar 1 equations can be derived for At:i.1, Bñ1, and B; •

+ Theorem 1.50

Le.t the square matrix A be such that

A= (Au A12) ~l

A.¿2 1

lf A 22 is nonsingular, then IAI = IA 22 l IA11 - A12A~ A21I· Proof: The determinant of A can be written as follows:

where

B=

1 (

-A;;iA21

o)

A;f

For a given set of aii and Y; (that is to say, for a given matrix A and vector Y), does there exist a set of elements xi (that is, a vector X) such that the equations AX = Y are satisfied ~ Three cases must be considered: 1. The equations have no solution. In this case there exists no vector X such that the system of equations is satisfied, and the system is said to be inconsistent. 2. There is just one set of X¡ that satisfies the system. In this case, there is said to exista unique solution. 3. There is more than one vector X that satisfies the system. If more than one such vector X exists, then an infinite number of vectors exist that satisfy ~he system of equations. We shall consider two matrices: the coefficient matrix A and the

10

LINEAR STATISTICAL l\IODELS

augmented matrix B = (A, Y), which is the matrix A with the vector Y joined to itas the (ni + l)st column; that is to say,

r :1 ~

t

(~

;

"

B=

We shall now state sorne important theorems concerning solutions to the system of equations AX = Y. ·

MATHEl\IATICÁL CONCEPTS

11

~t

t.5

The Derivatives of Matrices and Vectors

We shall now...--discuss sorne theorems relating to the differentiation of quadratic forros and bilinear forms. lt will sometimes be advantageous in taking derivatives of quadratic and bilinear forms to be able to take derivatives of matrices and vectors. Let X be a p X l vectOr with elements x,, let A be a p x 1 vector with elemen~s ai, and let Z = X' A = A'X (Z is a scalar). The deriva.ti ve of Z w1th respect to the vector X, which will be written oZ/oX, will mean the vector

az

+ Theorem 1.51 . A necessary and sufficient condition that the system of equations AX = Y be consistent (have at least one vector X satisfying it) is that the rank of the coefficient matrix A be equal to the rank of the augmented matrix B = (A, Y). = p(B) = p, then m - p of the unknowns x, can be assigned any desired value and the remaining p of the x, will be uniquely determined. It is essential that the m - p of the unknown xi that are assigned given values be chosen such that the matrix of the coefficients of the remaining p unknowns have rank p.

+ Theorem 1.52 If p(A)

+ Theorem 1.53 lf p(A)

= p(B)

= m < n, there is a unique vector X

+ Theorem 1.54 If X, A, and Z are as defined above, then ·

az¡ax

P1·00J:

=A. To find the ith element of the vector

that satisfies AX =Y.

f 1 a.x.) _a_z = a(;-:: , '.

Asan example, consider the system of equations x1

-

X2

ax,

=6

2x1 -2x2 =3

This can be put into matrix formas

oZ/oX, we find

which equals a,. az¡ax =A.

Thus the ith element of

oZ/oX

is a,; so

+ Theorem 1.55 Let A be a p x 1 vector, -B be a q x 1 vector, and X be a p X q matrix whose ijth element equals x... ,, Let f/

It can easily be verified that the rank of the augmented matrix (

1 -1

ª)

2 -2 3

is 2. Therefore, the system of equations is not consistent, and there exist no values x 1 and x 2 that satisfy it. This fact is also easily seen if we multiply the first equation by 2 and subtract it from the second equation. We get O = -9, an impossible result.

11

~ k, 111==1 n=l

Z = A'XB = k, ""

a n'x nmbm

Then az¡ax = AB'. Ptooj: éJZ/oX will be a p X q matrix whose ijth element is o Z/ oxiJ· Assuming that X is not symmetric and that the elements of X are independent,

az

12

LINEAR STATISTICAL l\IODELS

Thus the ijth element ?f oZ/oX is aib;.

oZ =

ax

AB' 1~

ij

+ Theorem 1.56 Let X be a p x

á then az¡ oA = " tJ

1 vector, let A be a p X p symp

p

= X' AX = 2 L xixp¡¡;

metric matrix, and let z

MATHEl\IATICAL CONCEPTS

Therefore, itfollows that '

i=lj=-1

¡;

2XX' - D(XX'), where D(XX') is a diagonal matrix whose fl ~~ diagonal elements are the diagonal elements of XX'. Proof: By oZ/oA we shall mean a matrix whose ijth element is ~·~ ~ oZ/oa¡;. Thus,

iJZ =

{t J, "'m"'nªm•)

ºª'iJ

lfi =j, oZ/oaii = ing that a¡¡ =a;¡}.

ºªiJ

x¡.

~ ~-

(:

.,

Ifi =Fj, then oZ/oail = 2x¡X; (rememberThus oZ/oA = 2XX' - D(XX').

+ Theorem 1.57 Let X be a p x 1 vector and let A be a p x p symmetric mat.rix such that Z = X' AX; then oZ/ oX = 2AX. Proof: The derivative of the scalar Z with respect to the vector X : will mean a p X 1 vector whose ith element is oZ/ox¡.

az

acx'AX) oxi

OX¡

aCt J 1

13 use of idempotent matrices in our ensuing work. A square matrix A ¡8 a synunetric idempotem matrix if the following two conditions hold: ( 1)

A=A'

(2)

A=A2

For brevity we shall omit the word "symmetric." That is to say when we say a matrix · is idempotent, we shall mean symmetri~ idempotent. We shall make no use whatsoever of idempotent matrices that are not symmetric.

+ Theorem 1.58 The characteristÍc roots of an idempotent matrix . are either zero or unity. Proof: If A is idempotent and if A is a characteristic root of A there exists a .vector X =F Osuch that AX = .!.X. If we multipl; both sides by A, we get . A2X

OX¡

).2X

But A 2X =AX= .A.X; so we have A.X= A. 2X (..\

But X =F O; so A. xmxnamn)

= Á.AX =

2

-:- ).

2

-

Á)X =O

must .be zero.

Thus J.

= o or ). =

l.

+ Theorem 1.59· If A is idempotent and nonsingular, then A = I. Proof: AA __: A. l\foltiply both sides by A-1.

+ Theorem 1.~0 If A is idempotent ofrank r, there exists an orthog-

p

11

=

2xiati

+ 2 11=1 L xna'in = 2 n::::l L xnain n,i!:i

but

2AX = 2

,, L

x 11a¡ 11

n=l

1.6

ldempotent Matrices

W e shall now pro ve sorne theorems concerning a special type of matrix, the idernpotent matrix. Since many elementary textbooks on matrix algebra include few theorems on idempotent matrices and since these theorems will play so important a part in the theory to follow, we shall supply the proofs to the theorems. We shall make extensive

onal .ma~r1x P. su ch· that P' AP = E,, where E, is a diagonal matr1x w1th r diagonal elements equal to unity and the remaining diagonal elements equal to zero. Proof: This follows immediately from Theorem 1.31.

+ Theorem 1.61 Aff idempotent matrices not of full rank are positive semidefinite. Proof: Since A = A'A, the result follows from Theorem 1.25. This theorem ~ermits us to state that no idempotent matrix _can have negat1ve elements on its diagonal. • Theorem 1.62 If A is idempotent with elements a .. and if the ith diagonal element of A is zero, then the element~ of the ith row and the ith column of A are all identically zero. Proof: Since A = A 2 , we get for the ith diagonal element of A n

ªii

= i=l 2 ª'ltu

14

LINEAR STATISTICAL l\IODELS l\IATHEl\IATICAL CONCEPTS n

Suppose (1) and (2) are given.

Proof:

ª" = i=l I a7;

P'BP

By Theorem 1.31, there exists an orthogonal matrix P such that P' AP = E,.. But tr(P' AP) = tr(A); thus tr(A) = tr(P'AP) = tr(E,) = r.

+ Theorem 1.64 If A is an idempotent matrix and B is an idempotent matrix, then AB is idempotent if AB = BA. Proof: If AB = BA, then (AB)(AB) = (AA)(BB) = AB.

+ Theorem 1.65 If A is idempotent and P is orthogonal, P' AP is idempotent. (P'AP)(P'AP) = (P'A)(AP) = P'AP.

Proof:

=

P 'BP =

An are p X p idempotent matrices, a necessary and sufficient condition that there exist an orthogonal matrix P such that P'A 1 P, P'A 2 P, ... , P'AnP are each diagonal is that AiA; = A;Ai for all i and j. Proof: This theorem is a very special case of Theorem 1.32. Because of its importance, we ha ve stated itas a separate theorem. ••• ,

Am are symmetric p x p matrices, any two of the following conditions imply the third: (1) A 1, A~, ... , Amare each idempotent. ••• ,

1>&

(~

=

:)

º)

(1,o

Lm P'A,P o . = i=l

But P' A,P are each idempotent . , by Theorem 1. 65 . B y Theorem 1 ~~t p - r diagonal elements of each P'A,P must be 1.61, tTiel zero. us 10 ows since the diagonal elements of an 1·dcmpo t en t t . ~na r~x arn nonnegative; so, 1f their sum is zero, they must all be 1dentrnally zero. Also bv Theorem I • 62 , the Jast p - r rows and J p - r columns of each P' AiP must be zero. Thus we may write r

•

11

•

•

•

º)

P'A,P = ( B.' O 0

O.

(I - A)(I - A) = 1 - IA - Al + A 2 = 1 - A. Thus B 2 = B. We must now show that AB = BA =O. We have A+ B = l. Multiply on the right by B and obtain AB + ··B 2 = B or AB = O. If we multiply the quantity A + B = 1 on the - right by A, the result BA = O follows.

+ Theorem 1.68 If A 1 , A 2,

.

1, B is idem-

Proof: We shall first show that B is idempotent. Now B = 1 - A, and we must show that B 2 = B. We get (1 - A) 2 =

+ Theorem 1.67 If A 1 , A 2 ,

I A¡ is idem-

wher?. we suppose that B is of rank r and I, is the r x r identit matnx. Thus we have y

Proof:

=

=

1

+ Theorem 1.63 If A is idempotent of rank r, then tr(A) = r.

potent and AB = BA

Then B

potent, and there exists an orthogonal matrix p su~h that

But if aii = O, then a¡; = O (for j = 1, 2, .•. , n); that is, the elements of the ith r~w are all zero. But A = A'; so the elements of the ith column are also ali zero.

+ Theorem 1.66 If A is idempotent and A + B

15

So~ using.only the first r rows and first r columns of m

P'BP =}: P'A¡P i=l

m

we have

l=í: B¡ i=I

wh~re the B¡ are idempotent. Let us assume that the rank of B, is r¡. Then there exists an r x r orthogonal matrix e such that C'B,C = Then

C'IC

=

I =

i

~;!

(~

C'B¡C

:)

+(

1 '

o

º)

0

Since C'B,C is idempotent, by Theorems 1.61 and 1.62 we haYe

(2) The sum B = LA¡ is idempotent. i=l

(3) AiA; =O for all i =F j.

i = l, 2, ... , m; i =fo t ·

16

LINEAR STATISTICAL MODELS

where K; is an {r - t)

X (r -

t) matrix.

l\IATHEMATICAL CONCEPTS

17 Proof: If any .two conditions of Theorem 1.68 hold, this implica that there ex1sts an orthogonal matrix P such that the following are true:

Thus we see that

C'B;CC'B,C = O

which implies :~

and

i

=

1, 2, ... , ·m; i =F l

Since t was arbitrary, thc proof of condition (3) is complete. Now suppose (1) and (3) are given. Then we have 2

B =

)2 L A; = m

(

i=l

m

L i=l

A¡

+2

A;A;

=

i>Fi

m

2

A¡

...

(a)

(b)

P'BP =(~' :) P'A P=(~· :);PV={:

where the rank of B is r

1

=B

i=l

We have shown that the sum is idempotent, and condition (2) is satisfied. Finally, suppose (2) and (3) are given. By Theorem 1.67, there exists an orthogonal matrix P such that P' A 1 P, P' A 2 P, . . . , P' AmP are each diagonal (since A;A; = A;Ai = O). Since the sum of diagonal matrices is a diagonal matrix, it also follows that P'BP is diagonal. By condition 3, we know that P' A;PP' A;P = Ofor ali i =F j. But the product of two diagonal matrices P' A¡P and P' A;P is zero if and only if the corresponding nonzero diagonal elements of P' A;P are zero in P' A;P. Thus, if the tth diagonal element of P' A;P is nonzero, the tth diagonal element of P' A 1P (for allj =F i) must be zero. Since P'BP = Er, the tth diagonal elemcnt must be O or 1 for each P' A¡P. For, if the tth element of P' A;P is equal to k =F O, then the tth diagonal element of the remaining P' A¡P (i =F j) is zero. But the tth diagonal element of B is O or l and is the sum of the tth diagonal elements of P' A;P (i = 1, 2, ... , m). Thus k = O or k = l. Since P' A;P is diagonal, the characteristic roots of A; are displayed dowll' the diagonal, and, since these roots are either O or 1, A¡ is idempotent, and the proof is complete.

o 1,::

o

:}

... ;

o

P'V=(: where the rank of A, =

1·;·

Ir,,,

o

:)

Thus the result follows .

1.7 Maxima, Mínima, and Jacobians We shall now state some theorems concerning the maxima and minima of functions.

+ Theorem 1.70 If Y = f(x1,X 2, .•• ,xn) is a function of n variables ~nd if a~l partial d~ri:atives oy/axi are continuous, then y attains 1ts max1ma and mnuma only at the points whei:e

Theorem 1.71 If f(x1,x 2 , ••• ,x11) is such that all the first and second partial derivatives are continuous, then at the point where

It is of special interest to note that, if m

2A;=1 i=l

then condition (2) of Theorem 1.68 is satisfied. condition ( l) implies condition (3) and vice versa.

In this situation,

+ Theorem 1.69 If any two of the three conditions of Theorem

the function ·has ( 1) a minimum, if the matrix K, where the ijth element of K is 02.f/ ox;ox1, is positive definite. . · (2) a maximum, if the matrix -K is positive definite.

tn

1.68 hold, the rank of

2 A; equals the sum of the ranks ofthe A¡. i=l

In the above two theorems on maxima and minima, it must be remembered that the xi are independent variables. Many times it is

LINEAR STATISTICAL l\IODELS

18

r ,.

desired to maximize or minimize a function /(x1 ,x 2 , ••• ,xn) where the xi are not independent but are subject to constraints. For example, suppose it is necessary to minimize the function /(x 1 ,x 2, ••• ,xn) subject to the condition h(x 1 ,x 2 , ••• ,xn) = O. Since the x, are not independent, Theorems l. 70 and l. 71 will not necessarily give the aesired resu}t. If the equation k(x 1,x 2 , ••• ,xn) = 0 could be so}ved for x t such that

1

._

re want to minimize f(x 1,x2, ... ,xn) subject to the constraint li(x1,X2, · · · ,xn) = O, we form the equation

i'J l~

f~ {~

í1

The x 1 ,

fJ;=;

~ j;i

OX¡

oF oF

oF

oF

OX¡

oxn -

o.A. -

-=-=···=----0

a¡

if oh/ox, =P

-=2x2 -6=Ü

OX2

ofor ali i at the point.

Thus we now have n + 1 equations and n + i unknowns and we need not worry about which variables are independent for w~ treat all n + l as ií they were independent variables. ' This will be generalized in the following.

OX2

The solutions yield x 1 = 1, x 2 The matrix K is given by

= 3.

• Theorem 1.73 To find the maximum or minimum of the function f~x1,X2, · .. ,xn) subject to the k constraints h¡(x 1 ,x 2 , ••• ,xn) = O (i form.. the function F =f(x1> x 2> • • • 'x) 1: = 1, 2, ... , k), n

K=

.I l,h,(x1,x2, · · · ,xn)·

=

(1 - x2) 2

-

2(1 - ~)

+~ -

6x2

then/, subject to the constraints, can have its ma~ima and minima only at ~he points where the following equations are satisfied (if a Jacob1an loh,/ox;I =PO at the point):

or

'

oF

+ 16

oj = -2(1 - x2 ) + 2 + 2xi - 6 = Ü OX2 The solution gives x 2 = .¡.. In this case the matrix K consista oí a single term o 2f/ox~. So K = 4 and is positive definite. Thus /, subject to the constraint x 1 + x 2 = 1, attains a minimum at the point x 1 = -!, x 2 = !. If the constraint is a complicated function if there are many constraining equations, this method may become cumbersome. An alternative is the method of Lagrange m:uUipliers. For example, if

If oF/ox, (i = 1, 2, ... , n) are continuous

i= 1

K is positive definite; so f has a minimum at the point x 1 = 1, x 2 = 3. If we want to find the minimum of f subject to the condition x + x = 1, we proceed as follows. Substitute x 1 = 1 - x 2 into 2 1 the f function, and proceed as before. This gives j

1 independent vari-

••• ,xn ) and the constraint h(x l> x 2' • • • ' xn ) = O are such that all first partial derivatives are continuous then the m~ximum or mínimum oí /(x 1 ,x 2 , ••• ,xn) subject t~ the constramt h(x 1,x 2 , ••• ,xn) = O can Qccur only at a point where of F = f(x ¡, x 2> ... , x) . hthe . derivatives h n - .lh(xl' x 2' • • • ,Xn) vams ; 1.e., w ere

~

=o

+

,xn, l can now be considered n

+ Theorem 1.72 Iff(x 1 ,x 2 ,

i

= 2x¡ - 2

X2, • • •

3:z ables. We now state the theorem:

then this value of xt could be substituted into /(x 1 ,x 2 , ••• ,xn) and ~ Theorems 1.70 and 1.71 could be applied. ~ Asan example, suppose we want to find the minimum of .f =xi - ~ 2x 1 + ~ - 6x 2 + 16. Using Theorem 1.70, we get

of

19

l\IATHEMATICAL CONCEPTS

OX¡

=

al,

oF

OX2 = ... =

oP"'

aF

oF

a-x-n = -OA-1 = -º~- = ... = -OA-k = o

Let g(x1,X2, · • · ,xn)? where - oo < x, < oo (i = 1, 2, ... , n), representa fre~~ency-dens1ty function. This is equivalent to the following two cond1t10ns on the function g(x l' x 2' • · • ' x n ) ·• (1)

(2)

g(x11x 2 ,

Í

f f

ao 00 -oo -oo

• • •

00

__ 00

•••

,xn) ~ O

g(x1,X2, • · • ,xn) dx¡ dx., • • • dx = 1 -

n

20

LINEAR STATISTICAL MODELS

If we make the transformation X¡ = hi(y 1 ,y 2 , ••• ,yn), where i = 1, 2, ... , n, the frequency-density function in terms of the new variables y 1 , y 2 , ••• , Yn is given by

rl

MATHEMATICAL CONCEPTS

In the example abo ve, if we solve the equations for y 1 and y 2 , wc get

,, ~'

r.¡;

Y1

!.' i'

X¡+ X,. =---()

(il -5·!)

~

(we shall assume ccrtain regularity conditions on the transformation equations). The symbol IJI denotes the absolute value oftheJacobian of the transformation. 'l'he Jaco bian is the determinant of a matrix K whose ijth element is ox;f oy¡. For example, if -

00


(Y - X f3)'(Y - Xf3)

=

(Y - X~)'(Y - X~ ) + ( ~ - f3 )' S ( ~ - (3)

can be rcadily establis hed . joint freq ue ncy fun ction is f(c)

=

1

=

l

(2 77rocedure to obta m 1.e max1mum r J r¡ d · · one t ra nsformation of t hc '/J. 'I'h - i rn.1t 1oot· est 1ma tor of any one-to,. e s1 ua 10n for least sq . t· . m a.tors is not so simp le T I . . . - uares es 1. . . . rnre is one important case h o . . w 111ch t he m van a nce property hold e . l , . wevei, ll1 tl . . 1 s .ior east-squares est1mator d us I S t le case of linear fonctions of t he fJ TI . . ~, a n following theorem. i· us is stated m t lie

+ Theorem 6.4 If the covarian cc matrix of the vector e in t he

+ Theorem 6 3

would play a n impo rtant part in estim ation and testing h ypoth escs about t he parameters in m odel l. In t he previo us ar t iclcs of t his chapter the normal cquations occnr ma n y ti 111es; th is is so me indication of their imporLance in point estinrnLi on. T he student s honld beco me fam iliar wit h equations of t hi s type n,nd with Yarious methocls of solving them , since in most cases of esLimation and tcsting hypothescs their solu t ion will be needed. \Ve s hall not d isc uss compu ting p rocedures in t his chapter but dcfer t his disc ussion unt il Chap. 7. The ijth e le ment of X ' X is t he quant,ity 2: X;kX;k , a ncl the ith elem ent

=?

U d . ti . Theorem B2 ti ~ ~L t ~ ? general -~mear-hJ'.'pothesis model given in bi na . . , w. es mear unb1ased estimate of a ny linear corn.t10n of tl~e {J; JS t li e same li near combination of thc be t r un b1ased est1mates of t he (J .· t hat . th b . s ~ near estima te of t ' . . · " e JS, e est Jrn ear 1111 biased whe· ~ (\\ hei~ tis ap X lknownvectorofconstants)is t'~ te ~is t he bcs t 1111ear unbiased estímate of r.>. t l1at . t ' fi. , t ' S - lX' Y. < f"" • IS, t" = A.

Th~ proof follows t he Ji ne of Theorem 6.2 a nd will b J f excr c1se for t he read er. e e t as a n l 1·r t h e 1east-squa res estimators of {J fJ . F01· exam · Pe, d (3 . tively are 3 6 el v 2, an 3 , rcspecesti m~tor of 'say' ~{J1 - iR th+cn5/Jth~ val ue of the best linear ~mbiasccl , < ' l 1' 2 3 l S (3)(3) - (2)(6) + (5) ( - 4) = -23.

general-lincar-h ypothesis m odel of foil rank is equal to Va 2 , wherc V is a known positive dcfinite m atri x , t he maximum -likcli hood estimator of í3 is ~ = (X' V- 1X ) 1 X ' V -1 Y a nd t his has propcrt ies simil ar to t hosc listed in Thcorem 6.1. T he proof of t his t heore m will be left for t he reader. The Normal Equa tion s . Early in th is chapter we s tated t hat thc norma l equations 6.2.4

~/X 'X ~ = X'Y

k

of X ' Y is

2: X;k'!Jk ·

T hcse quant it;ics cn.n 1·cadily be fo und by us ing a

"

desk calculator. 6.2.5 Choosing the X matrix .

The

X;;

values rnust be kn own

IIS

i\IODEL

LIN EAR ST A'.l.'I S'l'TCAL l\IODELS

b efore the Y; valnes are sclected at ra nclom. In some cases X;J Yalues rnay h e p ickccl or chosen by t he expcrimc nter in a ny way he wishcs· in other cases t hey ca.n not be so con trnlled. lf t hc experim enter ca1~ ch oose t hc x i; valu cs, t he question of h ow to select t he m arises. In general, it wo nld secm t hat the best way to pick th em is so t hat thc vari ance of certain estim ators will be as s mall as possible . F or exarnple, the varia nce of the estimator of A.' 13 is a 2 A.' S- 1 A., a nd wc rnigh t wa nt to ehoose t he X;; t hat minimizo A.' S- 1 A.. \ Ve cannot do t his in general for a ll vectors A., but we mig ht be a ble to for so me. \~Te shall discuss t his fur t her when we discuss sp eci fie examples. r:.. •. 6.2.6 Exa mples . A simple linear mod el is

X1)

1

X

l

x,.

I t is easy to see t hat S

=

X 'X

=

s-1=

( L X;

l

LX~ - L X;)

(

- - - - -2 nL(x; - .¡;¡ - LX;

n

and

Thus,

. (Pi) = s- x 'Y = - -1- - (I,x7I,Y; -

13 =

1

µ2

n L (x; - :f:)

2

J _ L(x; - x)(y¡ - ?i)

0 1'

f'9 -

-

P

· =-

' - ( 11P2)

and

s uch t hat L(x; - x) 2 is as la rge as ºssible. To minimizo var (fj 1 ), we choose t he X; such t hat L.xlf po 11 as poss1. ble. • ,.incc ,¿,,, ,~( . ,~ 2 )'(°¡; . _ x}- 1s as sma x, - x-)2 ~ kJX¡, t hc .., · ·('p· ) is mi n imum if t he X; a.re chosen s uch t hat x = O. Th is ~Liso va1 1 • • . niakcs t,hc cov(f3 1 ,f3 2 } = O. Note that we ha ve assumed t hat n is 9

X;

•

fi xc_l)t 1.o - .6_') X'Y

18

e = s-1 =

(

=

- .67 Furthc r,

and ¡3'X'Y = 49.5.

~? l ' XoX-

_

[l + -1+

l:(x, _ x)2

n

Therefore ii/a,, is distributed 1V(O, 1).

á2 ~

J.

_.,J

-

(6. 15)

The quantity

2) a ncl is inclcpendent of it. V2 2 G

is distribu ted a,s 7., 2 (k -

is distributed as the quantity

x2 (n

= 11+"'"'k L.,;

i =n+l

(

-

Thcrefore, a l. X" -

Also, the quantity

)"

Y; - Yo ? v·

1) and is ind ependent of u and v 1 .

+ le - 3) ancl is indepcnden t of u

Hence,

Thcrefore, .

u.Jn + k -

3

2

ªu·./vja is distributcd as t(n If we dcsire a 1 -

+ C1.

.

~.o)

(x -

J ~~ ta/2 2

2

(6.16)

L(x; - i)2

n

Yo) ± ta¡2Ó" J.

J1.(~ + ~) + l.:

n

2

(Yo - 9) 2

L(x; - i)

2

~

x2 (n

-;

2

/ /Jl[J -

~i is dis tributed as

+~+

k

Xo = X + /J2(i¡ -

(6. 14)

whcre

:i:) L:(x; - :c)(x0 -

~

fJ1 - fJ2~;0)

(Yo -

Using the equ a lity, we h ave a quadratic equ ation in x 0 , the only unknown in (6.16). Solving for x 0 gives

F rom t his we obtain v,.2 -_ a 2 k

+

\Ve sha ll let á2 = v/(n k - 3). Therefore, the prob a bili ty is 1 t.hat t he fo llowing inequality h olds:

P2x0

Clearly, ~¿ i s distributed n orm a lly wi t h mean O and vari a n ce a.,2

127

GEN ERAL LINEAR HYPO'l'HESIS OF FULL RANK

[

f31-

Yo -

1:

k - 3). confidence interval abo ut x 0 , we can write

A-

-

Yo) -

C1.

confiden ce inte rva l on x 0 is given by

la12Ó"J;. (~ + ~) , k n

11.

f32W - Yo) + },

-1-

"' "-'(X; -

ta12Ó"JA(~ + ~) A

n

k

Wo -

m-2· ~~ Xo x)-

+ Wo - '[/)

2

L(x, - i)2

if A. > 0

(G. 17)

If }. < O, it can be seen by exarninin g (6.16) t h at confidence lirnits on x rnay not exist. Even if co nfidc n ce limits do exist when J. < O, t hcy 0 will in general b e too wide to be of a.ny p ractica! use. 6.3.6 Width of Confidence Intervals. When an experimenter decides to u se a con fidence in terval, h e s h ould decide wh at widt h is necessary in order to make a decision . Y.,Te must, t herefore, investigate the width to see wh a t is in volved. W'e shall not d iscuss this h ere, but shall de vote a ch apter in Vol. II to this i m portant s ubject. 6.3.7 Simultaneous Confidence Intervals. The frequency interpretation of the foregoing resu lts on confide nce inter vals is that, if man y samples are taken a ndan inter val su ch as that give n b y (6.9) is constructed , then, on the average, 100( 1 - a) per cent of these intervals will cover t he t rue (unkn own ) value {3¡. This is t rue for one par t icular value of i only; i.e ., a sarnple can be selected anda confiden ce interval set on {3 1 , and the above frequency interpretation will h old. B u t if t he same data are used to set confiden ce intervals on {3 1 and {3 2 , the peobability is not equal to 1 - C1. t hat t h e result ing confidence intervals will include both {3 1 and {3 2 • For each set of ob.servations only one confidence statem.enl can be made using the above form:ulas. T h e same is true for Eqs. (6. 8), (6.1 0), (6.11), a nd (6.12) . This procedure does n ot seem to be realistic-setting a confidence

128

LINEAR STA'l'ISTlCAL J\lODELS

)JOPEL ) : C:E1\E.RAL LINEAR HYPOTJIESTS 01'' Jo' ULL RANK

intcrval on only one {l¡. It seems t hat a n experimente r will want to set a confidence in terva l on cach (Ji and to .know the p robability that al! these interYals contain t hcir respective {l¡. If the inteiTals were independent, t his p roba bility would be the product of t he individua! probabilities; i.e., in simp le linear models we should calculate two intervals similar to (G.ü), one corrcsponding to {3 1 and one for {3 2 , and t he probabili ty t hat both intervals wcre correct, would be (1 - 0:)2. Unfortunately , the intervals are not independent, and (1 - cx) 2 is not the correct probabi lity. The problem is more complicated than this and will be dealt with in sorne detail in Vol. II. \

6.4

6.4.1 Testing the Hypothesis (3 = (3*. To test the hypothcsis (3 = (3 * ( (3* is a known vector) in the general-linear-hypothesis moclei of foil rank, we s ha ll assume that the vector e is distributed N( O,a2I) · un less it is specifically stated otherwise. Testing (3 = (3* ( (3 * known) is equivalent to testing simultaneously that each cocffi cicnt {3 1 equals a given co11stant fJi. It is quite important fo r an experimentcr to have · such a test available. In t he model Y = X (3 + e, for examplc, if (3 = O, a knowledge of t he factors that correspond to t.he X; does not aid in the prediction of E(y). However, if (3 -:p. O, t he x 1 factors wi ll be val uable in predicting E(y). To test the hy pothcsis H 0 : (3 = (3 * wc must devise a test fun ction, say, u = g(y 11y 2 , •• • ,y,.,x 11 ,x12 , ••• ,x,.v), t hat is a function of the observations Y; a nd x 1; such t hat t he distribution of u is known when (3 = (3 * . In orcler to cvaluate the powcr of the test, the distribution must also be known whcn the a ltcrnative h ypothcsis H 1 : (3 -:/=- (3 * is true. \Ne shall use the likelihoocl ratio Las t he test function. The likelihood equation is

= j(e1 ,e9, ... ,e,.; (3,a-) = _-2

-

. ilar set of values of t he paramcters a 2 , {J 1 , (J 2 ,

• • • , {J'P) in Q t h at . ) . t he likclihood fun ction , and let /_,(L) be t he max1mum value. • . • • J,(cíi) has a simila1·.defi111t1on 111 the para.meter space r cstncted by 1:fo· 'fhat is to say, w 1s t he space of a 2 , {3 1 , {J 2 , ••• , (J,, defined by the 111~quafüy o < a 2 < oo a.ncl by /3 1 = f3i, {3 2 = {3i, ... , {J,, = p;. Actually, · one-dim ensiona. l spacc. To fincl L .(w) and L(ñ) we shall work (o) IS a . . with the logarithm of t he hkcl1hood funct1on. To find L(w} we procccd as follows: ~ n ., (Y - Xf3*)'(Y - X(3*) ¡ f(e· a2 (3*) = - - log 27T - - log a- - .;._---'-----og ' ' 2 2 2a2

nn.r t JCt

r .· 1 ·zes 111nx1111

•

Since the {l[ are fi xed, this eqnation is a function of a 2 only, and the value of a2 t hat maximizes it is the solntion to

T ests of Hypotheses

f (e; (3,a2)

129

1 e(27T) n/2an

e'e/20•

1 exp - (Y - X (3)'(Y - X(3) (27Ta2)"'2 2a2

-

d (l f ( . 2 A*)) = (Y - X(3 *)'(Y - X(3 *) - _!:..._=O og e,a , l'"' -2 •4

dcr

Thc solut ion is

á2

The test criterion is L = L(w)/L(Ó.), where L(w) a nd L(Ó.) are as defined below. The likelihood function is regarded as a function of the p + 1 parameters {J 1 , {32 , • • • , f3v and a 2 • 'Í'he pa.ramcter space Q is the p + 1-dimcnsional space clcfined by the incqua lit ies O < a 2 < co; - oo < (Ji < oo (i = l, 2, ... , p) . Let ñ be the point (that is, the

= (Y - .X(3*)'(Y - X(3 *)

n n"'2e- 11f2

L( ') - - - - - - - - -- ---::: w - (27T)"' 2 ((Y - X (3 *)'(Y - X (3 *)]" 12

and

ProceeclinCY in simila r fashion to obtain L(~)), we see t hat t he point 0 where Jogf(e ; a 2 , (3) attains its maximum is t he solu Lion to t he equations

o X'Y X'X(3 - [lo • ' ' á á -

()

ºª2

~

(logf(e· a- (3))

' '

(Y - X{3)'(Y - X(3)

=

n

- -

2á4

2á2

= O

n"'2e-,,¡2

Wc get (G.18)

2a

2 'a

whcre (3

L(Ó.)

=

s - 1X'Y.

=

(277

)"' 2 [(Y

- X{3)'(Y - X {3 )]"12

T hcrcfore,

L = [ (Y - X{3)'(Y - X{3) ]"' (Y - X (3 *)'(Y - X(3 *)

2

If g(L; (3 *) is the clistributio n of L nnder H 0 : (3 region is O ~ T, ~ A, where A is su ch t hat

f

1

g(L; (3 *) clL =

ex

=

(3 *, the critica!

(6.19)

130

LINEAR STA'l'lS'l'JCAL llfODELS

where o: is the probability of typc I e rror. To determine A from Eq. (G. l!)) we must find g(L; (3 *), 'the distribution of L when (3 = (3 *. Then Lis de ter mi ned fro m collected data, and H 0 is rejcctcd if L ::;;: A. Before determining the distribution of L, we shall study the quant itjes that a_re involrnd, t hat is, (Y - X(3 *)'(Y - X(3*) a nd (Y _ X(3 )' (Y - X(3). From (6.18) we obtain the identity (Y - X(3*)'(Y - X (3 *) =(Y - X~)'(Y - X~) +( ~ - (3 *)'X'X(~ - (3 *) (6.20)

If we s nbstitute S-1 X'Y for ~ ' Eq. (ü. 20) becomes (Y - X(3*)'(Y - X(3*) = (Y - X(3 *)'(1 -

xs-1X')(Y -

131

l\fODEL ] : GENERAL LINEAR HYPO'l.'HESIS 01< l!'ULL RANJC

and j,

=

((3 - (3 *)'X'(XS- 1 X')X((3 - (3*) _ ((3 - (3 *)'S( (3 - (3*) "~ 2a2 ~u-

definite, Z' A 2 Z/a 2 has a central chi-square TI1Cfefo re ' since Sis positive . . . . * distribution if and only 1f (3 - (3 * = O; t h at is, 1f and only 1f H 0 : (3 = (3 is t rue. . . Sta.tement (G .22c) follo'ws 1mmed1ately, by Theorem 4.16. Beca.use of t he distributional propcrties of Z ' A 1 Z/a2 and Z ' A 2 Z /a2 "iven in (G.22) i t follows t h at

"'

X(3*)

+ (Y - X(3 *)'XS- 1 X'(Y - X(3 *)

To see how the theor~ implies (6.22b) wc continuc as follows : Z' AzZfa2 is distribu ted as x' 2 (7c 2 ,}.), wherc 7c 2 = rank of A 2 = tr( A 2 ) = V

Z'A Z n - p

2 i¿= - --

(6.21)

Z'A¡Z

(6.23)

p

If we let Z = Y - X(3*, we get isdist ributcd asF'(p, n - p , ?.) a nd reduces to Sncdecor 'sF d istribution ¡f and onl y if H 0 is true. It al so follows t hat

Z'Z =(Y - X(3*)'(Y - X(3 ~ ) = Z 'A 1 Z + Z' A 2 Z

where A 1 = I - XS - 1 X ' and A 2 = XS- 1 X' are idempotent m atrices. Since Y is distributed N(X(3 ,a2 1), it follows tha:t Z is distributed N( X(3 - X(3 *, a 2 l). And, s ince A 1 and A 2 are idempotent m atrices, we note (see Theorem 4. IG) that (a )

Z' A 1 Z. - - . - 1s

ct·1stn.b utcd as x·(n •

- p).

(j"

(l) >

Z'A. . where - ; Z. - 1s a·1s t·'ri uuted as X'º· (p, A),

a·

(G.22) ). = ((3 -

(e)

Z'A1 Z

- -9

a-

(3 *)'X'A2X((3 - (3 *) = ((3 - (3 *)'S((3 - (3 *) 2a2 2a2

Z 'A 2 Z

and - -0 - are indepc11dent. a·

In a rd er to examine state me nts (G.22a) , (6.22b), (ü. 22c) m ore closely a nd to see how they follow from Theorem 4.. 16, we note that Z'A 1 Z/a2 is distributed as x' 2 (/c 1 ,}. 1 ), wherc lc1 is t hc rank of A 1 ancl A _ ((3 - (3 *)'X'A1 X((3 - (3 *) t -

2a2

But, since A 1 is idempoLen L, thc rank equals the trace, a nd so lc 1 = n - p. Also, it is evident t hat ?. 1 = O. Thus x' 2 (n - p, }, 1 = O) is equivalent to x2 (n - p); hence we have (6.22a) .

v = _ _ -=.p_u_ _ (n - p) + pu

=

Z ' A 2Z Z 'Z

(6.24)

is distributed as E 2 (1J, n - p, A.). Return ing now to t he cqu ation for the likelihood mtio L , we h ave found that _ L -

- r(Y - X[3) (Y - X(3)

Cv -

J -= n/?

(l _ v )"12 =

X(3*)'(Y - X(3*)

1

{

} n/ 2

1 + [p/(n - p)]u (6.25)

We observed from (6.25) that Lis a monoto ni c function of v, an d L is also a m o notonic fonction of i¿. From thcsc facts we see that either v or u can be used as a test function fo r t hc hy pothesis H 0 : (3 = 13 * · For t he distri bu t ion of ii (Snedecor's F) the crit ica,l region that corresponds to O ~ L ~ A is F"' ~ u < oo, wh cre.F"' is a constant such that

f

00 F (i¿;1J, n - p , },

=

O)

rli ¿

= (/.

F,,,

Similarly, for t hc distribution. of v (Tang's E 2 ) the critical region is E2ex -.....;.:;::: ~ v '..;;:;;:: .,:::: 1 where E ex2 is sn ch that _

f

1

E (1, '

E2(v; p, n - 1J, A. = O) dv

=

o:

,·

132

i\! OD'EL ] : Cl~l' 'ERAL LINE Alt HYl'O'l'HBSI S OF FULL RANK

L1NEA R s·r A'l'lS'r ICAL i\lODELS

thc rninimum \\'it h respcct to í3 whcn t hcre is no rcstriction on f3 ,

The power of t he LcsL is gi \'en by

fJ(J,) =

J:,,/

and 2

.?.

(v; p , n - p, },) dv

and m ay be eYaluated for differcnt Yalucs of p, n - p , and ). by usin Tang's tables. g TAUL"E

sv Total Duo to !3 E LTO I'

Ü. J

ANA l.Y:-iTS OF VAlUANCE FOB. '!'ES'J'l NG

n JJ

n - 1J

{3

s~

Dl[l .): whic h is R(P 1 ) + R(p 2 ) . 6.4.6 Example. T he distance a part icle Lra vels fro m a givcn refere nce p oint is g iven t heoretically by t he curve

I

SoJdng the e- l

Y; =

2: {J;z;; + e1 i= O

Í: (Y;

- 'fi) 2

j - 1

(7.3)

p- 1

n - 11

-

Í:

rfoRfo

i - l

T he valucs r;'b a nd RJi a re valucs tak en fro m a table s imila r to Table 7.2except that t he matrixen tered in the table is (p - 1) x (p - 1) a nd the elemen ts a re t he corrected (d eviations from t he means) sums of squa res and cross products. T he row and colu mn correspondi ng to µ are om itted . T hc ele men ts of t he 0 0 colu mn a re t he corrected cross products of t he x a nd y values. 7.3

Computin g the Inverse of a S y mme tric Matrix

7 .3.1 The In ve rse. If a confiden ce in tcr val is dcsired on a ny elemen t of (3 or on a ny linear com bi nation of fJ ;, then the elements of t he

15G

LINEAll. S'l'A'l ' CS'J'ICAL i\IOJ> t> LS

invorse of X ' X are nccd cd [seo Eqs. (G. ü) a nd (H. l O}). T ho in VC' rsc car¡ be fo und by using t he abbreviated Dooli ttle techni q uc. The t heory can be expla ined by supposing t,hat, A is a symmctric matrix whose inverso is desircd . L ct B = A - 1; t hen AB = l. L et B = (B 1 , B 2 , . . . , B 1,), whcrc Bi is t he 1ºth col umn of A.-•. Then A (B 1 , B 2 , . . • , B p) = (E v E 2, .• • , E p), whcre E ; is the ith column of the ident ity matri x l. \Vecan then o btain systems of peq uatio ns AB ; = E ; (i = 1, 2, . .. , p). \ Ve must solve t hese p systems and t hen obtain tbe elements of A - 1 . By using a table s uch as Table 7.2, wc can soh-e ali JJ of the systems by reducing A to t ri a ng ula r for m. We s hall illus tratc by finding t he inverse of t he matrix in (7.1) . X 'X

2 4 2) = ( 4

10

2

2

2

12

vVe can find B 1 by solving

TAUL E

7.4.

T11E !NVJ

Thc F valne found in Table 7.8 is less t ha n t he tabulated Ji' val ue ¡1,t t hc 5 per cent lcvel. A test of t he bypothesis µ eq~ al to zern (or to so me other constant) can be made by t be t test, us111g the elements from (X'X)-1giyei1in lines B 1 , B 2 , and B 3 ofTable 7. 7. Similarl y for tcsting

o

O'l

M'-'l

lQ

ANALYSJS 01•' VARTANCE FOR DATA OF TABLE

Total R(¡L,{11 ,pz) R(p) R(PpP2 11)

CN P2 Due to {3 0, {J 1 (unacl j )

n :J 2

B(f30,fl1.f32) R({J 0 ,{31 ) = R(cx 0 ,cx 1 )

Dueto {3 2 (ad j ) Err o1·

MS

F

R(fJ2 1 flo,f31)

n - 3

Y 'Y - R(fJo,fl1,f32)

E2

These are obtained fro m (8.2) by striking out the Jast row and column from t he 3 x 3 matrix a nd t he last ele me nt from each of the vectors. Clear ly, Po a nd /i 1 in (8.3) are t he same as &0 and &1 in (8. 1) a nd, hence, do not nec 11 P 2 · · · L.J>"!.JI So we mus t choose the coefficients a11 in (8. l 7) so a s to rn ake X 'X 2%

Quarlic Erro1· for quar Lic

l

34.43 28. 70 1

:!4.43

8.40

> 2%

7 1

i

4-.

lo

If we desire t o fi 11 d t hc clegree polynomia l that represents a give11 set of data a ncl to estímate Lhc p a ramet crs in the poly nomial, t ite p rocedure is (1) to use t he res ults of this section to find the desired d egrec polynomia l, a nd (2) to use Lhe results of Sec. 8.2 (the results of Chap. 6) to estímate t he parameters in the poly no mial orto test hypotheses about them.

8.5

Repeated Observations for Each X

Suppose t hat t he observecl random variab les y havo t he fo llowing structure: Y ;; = f(x,.)

+ e;;

j

=

i

= 1, 2, ... , k

1, 2, . .. ' ni; 1n

> 1

wh cre we postu latc t hat f(x) is a fonction "·hich ca n be expanded into a Taylor series, a ncl e; 1 ;"\,re 11n correlatcd norm a l ra ndom v ariables with mean s zero a nd varitLnces a 2 . The prob lem is to find a polynomial that adequately represe nts f( x) . This model may be realistic in a situa tion

184

Ll N l ~A l t

S 'l'A'l' JS'l'lCAL J\'lODELS

POLYKOMTAL OR ClJH.VI LlNEAlt l\10DELS

wherc t he exporimcnter can contro l !,he poi nt,s X; " ·here t he obsorvations Y;; are takon. 'l'he mod el assum es t hat at each X; point, rn (m > l) valu es of Y;; are o btain cd. Thus wo can gct a n unbiased estímate of a 2 by using t he guant,it,y á

A lso

2

=

1 k (rn -

/,·(m -

1)6 2

a2

k

1)

l

111

2.: 2.: {?/¡; - y,y =

i=J ;=1

k(m -

k

"'

2.: I

1)

(e;; - é¡f·

i= J ; = 1

185

(b) E sLim ate t he cooffi cionts in Lhis poly n om ia l by us ing tho abbrov iated olitt lo m eLhod . Do(c) F ind the standard e rro rs of Lhe cooffi c ie nts. s.S U so Lhe r csul Ls of this ch a.p ter Lo íintl t ho firs t throe orthogonal poly n.ornia ls for n = 5. . . .2 S.6 In a quadmt1c p oly nom1al. m odol . !J = {1 0 + (J 1x + {J2x + e, fmd t he use Lho m odol Y = Yo + y 1z + coeffi c1·en ts y·' in tcrms o[ tho cooffic1e nls {J,. 1( we . . 2 + e whero z is a coded vo.lue of x; Lhat 1s, z = :>; - h, wh e1·0 h is known i'tz ' constan t.

is distri buted as ;¡: 2 [k(m - l )] regardless of what

fun ctionf(x) is. The co mputations procced as o ut,li ned in (,bis chapt.er cxcept t hat in each table t he error sum of squares can now be broken into two parts: one term called t he lack of fit, t he other t he error su m of s quares k(rn - 1)62 • The e rro r sum of squares is easily computed, and so it is su btracted from the re ma inder to o btain t he lack of fi t tei·m. If the modcl is act ua lly linear , then the mean s qua re lack of fit t erm, which is t he er ror sum of s qu arcs for linear minus k(ni - l )á2 ali dividecl by t he d.f. , is a n unbiasecl estimate of a2 . If the model is not lin e~tr, t he n this tc nn is o n t he ave rage la rger tha n a 2 . The lack of fit mean squa re di\"id cd by á 2 is d istribu tcd as F if the model is linear . If t hc hypoLhcsis is rejected , t hen the laok of fit term for quadratic can be used in t ite numorator and 6 2 in t ho denorninator of the F to test for qua O; that is, t his is a onesidccl test. Calcul a.te t he power if p 1 = .2, .3. 2. p3 = .5 ; a lternative hypothesis is p3 =I= .5; that is, t his is a t wosided tes t. Calc ulate t he power if p 3 = .2, .3. 3. p L = p 2 = O; a lternative hypothesis is t hat one of t he equalities docs not hold. 4 . p 1 = p 2 = p 3 = p 4 ; a lternative hypothesis is tha t at lcast one eguali ty does not h old . l. We can use Thcore111 10.15 and have an exaet t est.

\;Ve get

10.19 a nd l =

1 ""A p * =-¡ L..P; m

211

REGRESSlON J\IODELS

• Theor e m 10.21 T osetan approximate l - 1:1. confidenee limito n p. wc use t he fact tha t (n - 3)11 2 (a.rctanh p - arctan h p) is approximately distributccl N(O, l ). A 1 - 1:1. confidence in ter val on pis

+ Theorem 10. 18

U=

3:

MODEL

Lf(n; - 3)/(n; -

0'.)8

-./i 1)1

-'-;......;;.~_;..'-'-"'-~~

Z(n; - 3)

T he improvcd es t i mate givc n iu Theo re rn l O. l Gis used to obt,ain t hc pooled estimatc.

/'>->

. ~ -.~-

-

. 13 1

2

(.02 )

which is not signi íiean t at the 5 per cent level. H 0 • The power functio n for p = p* is {J(p*)

=

f

P:z.

f(p ;p*) dp

Hence \\"C eannot reject

2 12

L I NE AR STA'rlST lCA L i\IODELS

whe re p" is such t hat p(f.

.95

=

J f (p;O) dp - 1

From D a ,-jd's tables fo r n = 24 a nd p = O we get Pa = .3447. l?t-oin David 's tables (using t hi s va l11e of Pa) we obtain {J (p* = .2) = .2·11 ; {3( p* = .3) = .42 1. The power using the Z transform ation givcs /3(.2) = . 239; {J(.3) = .413. . 2. \Ve can use D aYi n 32• n 33• a.nd n3 4' 13 .2 In P ro b. 13 . l , fln d N 1., Nv N a.• N. 11 N _2 , N _3 , a nd NA. 13.3 ln P rob. 13. 1, find Y 1 ... Y 2 __, Y3 .. , Y_ 1_, Y_ 2 _, Y_3 _. a ncl Y. 4 . • 13.4 In Prob. 13. 1, flnd t.he normal equat,io ns. 13 .5 In P rob. 13. l , fin

Y •• (1.'ss

2

bt'l.'ssBss

+ Bss + 1Jfss)]

76

+

(1)(3)

(146)(14)

+ (0)(4) + (4)(8) = 63

+ (76)(8) + (63)(7) =

3,093

2

Also,

11-fss

(29) =-=

12

2

i

iJ

(2)(14)

.f: (t y,,l' J) l',, =

2

t)

""' k i

+ (1)(8) =

the third element in column B is

-------------------==--

0 ..::::

(4)(14) + (0)(3)-+ (3)(4)

i

70.08

B 88 = (l 4 )

+ (8 )2 + (7 )2 -

Tss = (14)2

+ (3)2: (4)2 + (8)2 -

2

Mss = 7.17

4 j fss =

24.92

332

LINEAR STATISTICAL MODELS

Then

[~Y;;Y;.Y., -

Y .. (Tes

+ B, + 1lfss)] = 8

801\IE ADDITIONAL TOPICS ABOUT MODEL

[3,093

2

Nss = (130.07) = (130.07)2 = 7 _89 btTssBss (12)(178.68)

TheAOVisgivenin Table 15.4. TheFvalueis2.64;so thereisnoindication of interaction in these data. 15.4

ANALYSIS OF VARIANCE FOR NONADDITIVITY

sv

DF

SS

MS

JP

Total Mean Treatments Blocks Nonadditivit.y Balance

12

125.00 70.08 24.92 7.17 7.89 14.94

7.89 2.99

2.64

l

3 2 l 5

333

and the conditions may be satisfied if xii = log Yii is used instead of ~here may be many reasons for using transformations, but we shall d1scuss only one. Let y be a random va~iable, and suppose that we want it to satisfy conditions A to E. Suppose it is known that the mean of y is related to the variance of y by the function/(t). That is to say, if E(y) = µ and var(y) = a 2 , a 2 = /(µ). For example, if y is a Poisson variable with parameter m, µ ='In and a 2 = m; soµ = a 2• If y is a binomial variable '~ith p~r~meter p, µ = p and a2 = p(l - p); so a2 = µ(1 - µ). Now, 1f cond1t10ns A to E are to be satisfied, y must he a normal variable, and in general we cxpect the mean and variance to be unrelated. Therefore, if it is known that a 2 = /(µ), we shall try to find a transformation h(y) su ch that var[h(y)] is not related to E[h(y) ]. Suppose that x = h(y) is such a transformation. The problem is to find a function h(y) such that the variance of x is a constant unrelated to E(x). Suppose that h(y) can be expanded about the point µ by a Taylor series. This gives

Yw

- (29)(102.17)] = 130.07

TABLE

4

X=

}t(y) = h(µ) +(y - µ)h'(µ)

+ ··•

If we ignore all but the first two terms, we get X=

15.5

Transformation

If conditions A to E on model (15.1), listed in Sec. 15.2, are not mct, then, as explained in previous sections, the conventional tests of hypotheses and methods of defining confidence intervals may not be strictly valid. In this case there are various things that might be done: l. We might use nonparametric procedures that are valid for very general assumptions (along this line the reader should investigate "randomization" procedures). 2. vVe might ignore the fact that conditions A to E are not met and proceed as if they were. 3. We might transform the observed random variables Yi; in ( 15. l) so as to meet conditions A to E. Suppose we are interested in testing a certain hypothesis H 0 in model 4. If the test function is insensitive to the conditions A to E, it will be said to be robust. If a test is robust, then procedure 2 above will be usc-

ful. If enough information is available about the random variables y,1, it may be possible to transfor!Jl them so as to meet conditions A to E (procedure 3). For example, suppose y¡ 1 = µT;eii, where log e¡; is a normal variable. Then log Yii = log µ + log Ti+ log e¡¡

h(µ)

+

(y -

µ)h'(µ)

where h'(µ) is the derivative of k(y) evaluated at y = µ. E(x) = E[h(µ)

since E(y) = µ.

+ (y -

µ)h'(µ)] = h(µ)

Also,

var(x) = E[x - E(x)] 2 = E[x - h(µ)] 2 = E[(y -

since var(y)

=

Now

E(y - µ)

f(µ) is known; hence,

2

= a 2•

µ)lt'(µ)]2 = a2[h'(µ)]2 .

We have assumed that a2 = /(µ), where

var(x) = f(µ)[h'(µ)]2

Since var(x) is to be independent of µ, we set var(x) equal to a constant cz. 'Ve have c2 = f(µ)[lt'(µ)]2

h'(µ)

or

which gives

h(µ) =

=

cf .Jdtf(t)

e .JJ(µ) = cG(µ)

+k

where G(µ) is the indefinite integral of the function 1

.JJ(t)

and k is the constant of integration.

Actually, the constants c2 and k

334

SOME ADDITIONAL TOPICS ABOUT MODEL

LINEAR STATISTICAL MODELS

are immaterial, since thcy do not depend on µ. They can be given any convenient value (so long as they are not related to µ). For example, suppose/(t) = t; that is, the mean and variance of y are equal; then h(µ)

e-A¡:i:

P(x)

~t

Problems 15.1 Suppose the data of Table 15.5 satisfy the model given in Eq. (15.2) and Test the hypothesis

Ti

=

Tz =Ta

with a type I error

· TABLE 15.5 Block

Treatment

l

2 3

1

2

3

4

5

6

7

8

52.3 53.2 30.6

54.l 53.0 54.7

54.3 56.3 55.2

55.l 36.9 32.4

56.4 54.4 55.7

59.2 58.5 58.3

60.2 60.1 59.7

53.l 54.l 54.7

15.2 Prove Theotem 15.3. 15.3 Prove Theorem 15.4. 15.4 Prove Theorem 15.5. 15.5 Show how the abbreviated Doolittle method can be used to find u in Theorem 15.1. 15.6 Use the data in Prob. 15.l to test the hypothesis T1 - 2T2 + Ta = O. Use Theorem 15.5. 15.7 Test the data of Table 15.6 to see whether there is interaction (use a type I error probability of 5 per cent). TABLE 15.6 Treatment 2

3

4

5

4 6 1 8 1

3

10 19 7 14 9

5 12 2 9 6

1

2

1 7 4

7

1 4 2

x = O, 1, •••

X

=

What transformation makes

0, 1, ..• , n

Find the mean and variance of this distribution. What transformation ~akes the mean and variance independent? 15.14 Find the mean and variance of a chi-square distribution with n degrees of freedom. What transformation leaves the mean and variance independent? 15.15 Prove that the (t - 1) x (t - 1) matrix V with elements vN in Eq. (15.4) is positiva dafinite if t.he t x t matrix with elements ai¡' in Eq. (15.3) is positiva definite. 15.16 In Theorem 15.1, prove that u is the same if y 2 ; instead of Yu is subtracted from the other Yii to form Y;.

Further Reading

t 2

3 ·4:

5

7

1

= -x!- --

Find the mean and variance of this distribution. the mean and variance independent? 15.13 The binomial distribution is

6

Block

l 2 3 4 5

335

Prove Theorem 15.7. 15.8 Prove Theorem 15.8. 15.9 15.10 Prove Theorem 15.10. 15.11 Prove Theorem 15.11. 15.12 The Poisson distribution is

= cf ~- = 2c.[µ. + k

So we could use the transformation x = vy. This chapter is far from complete. For more complete information on the consequences that follow when the assumptions underlying model 4 are not satisfied, the reader is referred to the material in the bibliography.

the assumptions in (15.3). of 5 per cent.

4

8 9

10

F. N. David and N. L. Johnson: A Method of Investigating the Effect of Non N ormality and Heterogeneity of Variance on Tests of the General Linear Hypothesis, Ann. Math. Statist., vol. 22, pp. 382-392, 1951. · G. Horsnell: The Effect of Unequal Group Variances on the F-Test for the Homogeneity of Group Means, Biometrika, vol. 40, pp. 128-136, 1953. G. E. P. Box: Sorne Theorems on Quadratic Forms Applied in the Study of Analysis of Variance Problems, I, II, Ann. Math. Statist., vol. 25, pp. 290-302, pp. 448-498, 1954. P. G. Moore: Transformations to Normality Using Fractional Powers of the Variable, J. Am. Statist. Aasoc., vol. 52, pp. 237-246, 1957. M. S. Bartlett: The Use of Transformations, Biometrics, vol. 3, pp. 39-52, 1947. F. J. Anscombe: The Transformation of Poisson, Binomial and NegativeBinomial Data, Biometrika, vol. 35, pp. 246-254., 1948. J. H. Curtiss: Transfonnations Used in the A.O.V., Ann. .JJ1ath. Statiat., vol. 14, pp. 107-122, 1943. G. Beall: The Transformation of Data from Entomological Field Experiments so that the Analysis ofVariance Becomes Applicable, Biometrika, vol. 32, pp. 243-262, 1941-1942. H. Scheffé: Alternativa Models for the Analysis of Variance, Ann. Math. Statist., vol. 27, pp. 251-271, 1956. C. Eisenhart: The Assumptions Underlying the A.O.V., Biometrics, vol. 3, pp. 1-21, 1947.

336

11 12

13 14

LINEAR STATISTICAL MODELS

\V. G. Cochran: Sorne Consequences whcn the Assumptions far the A.O.V. Are Not Satisfied, Biomet'rics, vol. 3, pp. 22-38, 1947. G. \V. Snedecor: "Statistical Methods," Iowa State College Press, Ames Iowa, 1958. ' O. Kempthorne: "Design and Analysis of Experiments," John Wiley & Sons, Inc., New York, 1952. J. \V. Tukey: One Degree of Freedom for Nonadditivity, B-iometrica, vol. 5, pp. 232-242, 1949. -

16 l\fodel 5: Variance Components; Point Estimation

In this chapter we shaU discuss the subject of variance components. Suppose there are two populations designated by 7T 1 and 77' 2 • Let the mean and variance of 77' 1 be Oando!, respectively. Similarly, let O and a¡ represent the mean and variance of 1T 2 • Further, let a 1 , a 2 , • • • represent elements ofpopulation 7T 1 and bw b12 , ••• represent elements of 7T 2 • The elements from population 7T 2 ha ve two subscripts; this is solely for identification. Now let an observable random variable y be such that YH = µ

l

+ ai + b¡¡

where µ is an unknown constant. The thing to notice is the structure ofthe observed random variable Yii· The object in this model is to observe the y¡ 1 and, on the basis ~f this observation, to estimate orto test hypotheses about µ, o;, and a¡. -Another way to look at this is as follows: The variance of an observable random variable y is var(y) = a~+

a!

Now sup¡)ose that we can stratify the random variables represented by y so ~hat the variance of the random variables in a given stratum is a&. Then let YiJ be thejth observed random variable in the ith stratum, and Yu can be written Yu = µ

+ a; + b;;

For example, suppose a horticulturist wants to determine how a certain treatment has affected the nitrogen in the foliage of the trees in an 337

338

LINEAR STATJSTICAL MODELS

MODEL

orchard. He cannot examine every leaf on every tree; so he selects at random a group of trees forstudy. He cannot examine every leaf on the trees he has selected; so he chooses a set of leaves at random from each selected tree. Let Yii be the observed nitrogen content of the jth leaf from the ith tree, and the structure is assumed to be of the form

The assumptions are:

YiJ

= µ + a¡ + bii

where a¡ is a random variable with variance able with variance ut. Other models could be written Yii

a;, and bu is a random vari-

+ ai + b¡ + e,¡ µ + ªi + b¡ + ciJ + diik

5: VARJANCE COMPO~ENTS; POINT ESTIM:ATION

Case A. The a¡ are distributed N(O,a!); the b,1 are distributed N(O,ul); all the random variables a¡, b¡; are independent. Case B. The ai have mean O and variance a!; the b¡; have mean O and variance u:; ali the random variables ai, b¡; are uncorrelated. 16.1.1 Case A. In this case Yi; and Y;;· are correlated even if j =f= j'. This is given by a! if j =¡6 j' cov(y¡;1Yw) = ( (16.3) a¡ + o! if j = j' Also notice that ifi:¡!:p

= µ

Yiik =

(16.4)

The likelihood function for the random variables Yi; is

etc.

y(Y) = (J(Y11·Y12' ... 1Y111·Y21• ... ·Y2s• ... , Yr11 ... 'Yrs>

Notice that the structure of the observation is similar to model 4, the only change being that the a,, b¡;, etc., are random variables. Now we shall define model 5.

+ Definition 16.1 Let an observable random variable YH ... m be sucb that Yu ... m = µ +a,+ b,¡

+ · · · + eii ...m

= f(YwY12·

· · · ·Y1s)f(Y21·Y22, • · · ·Y2s) · • ·.f

~

expected mean squares. follows.

s"'

~

n - A

Cl1

c-1

¡:e

rn

00

¡¡: ~ e-a=:' ~

..

:¡~

~·To..

vl

~

~

1

1

~

i4'

rx. A

~

~ ~

~

~

.......

~

í4'

~;;; 1

........:.i

~

í4'

.:.i

::::

i

~

1

1

e ~

~

í4'

r:

1

.......

;::¡-

~

1

~

~

1

~

~ ~

= n

E

~

;;)

l

~

- --1

. n

~

::::

~

1

~

rn

> rn

¡:::: d

d+>

Q)

~

~

o

r/l r/l

UJ

Q)

UJ

al

~

tS

..o

o

'()

UJ

UJ

Q)

Q)

!11

rn

rn

m

o

f.)

UJ

d

o ~

a: + k 0 u!

1

Bms

u2b

i

y2) L __:h

-

ii

i

ni

1A E[¿ (y., - y,.)2] 1A E[.fr (µ+a,+ b;¡ -

µ - a, -

6..i•]

- E[2: (bi; - btJ 2] = -n -A ii 1

Using Lemma 16.3, we get -~

~

~

~

Nextwe find E(Ams) = -

Q)

Áms

The expected mean squares are obtained as

E(Bms) = -l- E ( L Y~i Cl1

y• - -·· n

n,

~>r, - ¡ ( ~r-)

n-A

b classes

r)

2: ¿

A - 1

'I)

b

EMS

n

i

Ol

MS

y:

+

Cl1'1)

5

- A -

UJ

Q)

!

o

Q)

1

where

-

E(2: l1. 1 i

n¡

Y~.) n

= -

1

- E[2: (y,_ - y_.)2]

A - 1

ii

354

This gives 1

E(Ams) = -

-

A-1

E[.Lu (µ + ai + bi. - µ- .!.n112 n.,,a11 -

.!: L bii)]

2

~

TABLE

BLOCK PATTERN FOR UNBALANCED TWOFOLD

2

])

Cin

Cuu

C112 C122 C113 C¡23

_I -[I (o! _~n n¡a! + L~! o!) + L (Din¡ _2n¡oi + ª~)] A - 1 ii n nn¡ n

...

~

ª1

bu b12

=

16.7

NESTED CLASSIFICATION

1 {E[¡ (a¡ - .!. 2 nf)a.,,\ +E[¡ (6¡. - .!. 2 b.,,q\ A -1 ., n 11 J " n.,,q J ]

355

units of the bi; (or the a, unit contains ni units of C¡;k), according to the pattern shown in Table 16. 7.

nu

2

=

5: VARIANCE COI\IPONENTS; POINT ESTIMATION

MODEL

LINEAR STATISTICAL l\IODELS

...

ªA bAlbA2···bABA

b21b22 • ' • b2Bi

b1B1

...

...

H

-20 = -1- [""' ¿,, ( n,u A - 1

¡

-2 -2n7 o;

n

l:n~) = A -1 1 o; n - --;;--

[-2 (

=

O'

2

+ n2 -

ri

l:n~

n(A - 1)

O'

+ n, -:I:n! a_-2) ¿,, (--2 u¡¡ 0 + ""' n2 i

+

2 A ab( -

C11n11C12n 12 • • •

-n¡ ab) ] 2

TABLE

n

ANALYSIS OF VARIANCE FOR THE UNBALANCED

SS

DF

5

MS

El\fS

1

Total

2

n

LYiik

ijk

ª Mean

ko =

n2

l:n~ '

-

a classes

n(A - 1)

b classes

k = 1, 2, ... , n,1 {

znij =

j = 1, 2, ... , B,

i

=

(16.15)

L (Yi••

- Y .. )

2

Áms

a; + q1u& + q2a:

L (Yi;.

-yi_.) 2

Bms

u; + q0u¡

Oms

a2e

A - I

B-A

ijk

n-B

n,

Y;;) 2

An AOV is given in Table 16.8, where zn¡j n-z-i' n;; 2f qo = B i Ani = ~ ii -

iJ

L (n¡; n7

1)

q1 = ii

2n¡ = n

n, n A - 1

=

L n¡¡f¡ ii

zn~ n--i-

i

'LB,= B This implies that there are A groups of the a,.

L (Y¡;k -

ijl.;

1,2, ... ,A

i

.Lnu = ii

n

e classes

Twofold Nested Classification in Model 5 with Unequal Numbers in the Subclasses

The model is

y_:_

1

ijk

The cstimates obtained from Table 16.6 do not satisfy the optimum properties of Theorems 16.4 and 16.6.

where

--

sv

1)]

So, in Table 16.6,

16.6

16.8

TWOFOLD NESTED CLASSIFICATION IN MODEL

n

The a, unit contains B,

q2 =

A - 1

=

~ n¡j,

(16.16)

l\IODEJ...

LINEAR STATISTICAL l\IODELS

356

1

1 n.

where

1

357

5: VARIANCE 001\IPONENTS; POINT ESTil\lATION

Next consider

1

n f; = ~ -1

E(Ams)

= -1- E [ L (Yi .. A - 1

Y.. J2 ]

Uk

= -1- E A - 1

(y¡ 2-·· n, i

Y2)

~ n

The thing to notice is the EMS column and especially the coefficients q0, q 1 , and q 2 • Next we shall show how to obtain them. First consider E(Bms) =_!__A E[_¡ (Yii. - Yi.J

.,k

B -

2 ]

=-

-[¡ n~(_!_ - !)u!+ L n¡;(_!_- !)~+(A -

1

A - 1

i

ni

n

i;

ni

u:

=

E(ciik) = O

n¡/i,2+ n¡a! + ~ n~t~ + ni~) -2i t ni =

1 (nµ2 B - A

+ n~ + n~ +Bu~ - nµ 2- no!- Ln~t ~ -Aa~) a

=

~ e

+

it

and

it ni

are easily computed.

2

""nit n-,¿_-

The results are

n¡ ~

B-A

Y ... = 1,423

b

n= 52

y2

So

as given in {16.16).

l)a!]

'fhe coefficients of anda! correspond to q 1 and q 2 in (16.16). · If expected mean squares are equated to mean squares in Table 16.8 and if the resulting equations are solved for uf, ando;, these u2 's are the analysis-of-variance estimates and are unbiased. The general unbalanced n-fold classification is an extension of the twofold. The analysis-of-variance estimators for the unbalanced classification do not lead to the same estimates as maximum-likelihood estimators. The maximum-likelihood equations are very difficult to solve in unbalanced classifications; so we shall rely completely on analysis-of-variance estimators. 16.6.1 Example: Twofold Nested Model. Suppose the data in Table 16. 9 are assumed to satisfy a twofold nested model. The data are artificial, but we assume that it is a breeding experiment. The number of sires is A = 4, the number of dams is B = 12, and n = 52. All the quantities needed for an AOV table can easily b.e computed on an automatic desk calculator. Quantities like

a;,

Taking expectations [remembering that E(ai) = _E(bH) and that all random variables are uncorrelated] g1ves

n

.2 _E.:_= 40,861.201 ij

'n¡¡

L Y~Jk = 41,811

ijk

358

l\IODEL 5: VARIANCE CO~IPONENTS; POINT ESTJl\IATION

LINEAR STATISTICAL MODELS TABLE

Si res

16.9

1

i

The AOV is given in Table 16.10.

16.6.1

DATA FOR EXAMPLE OF ART.

3

2

a;= 23.744

-

4

'rABLE

Dams

j

Total of jth dam in ith sire

1

2

32 31 23 26

30 26 29 28 18

4

yii·

n¡;

6

5

7

9

8

10

11

30 20 31 26 18 20 34 21 32 31 26

21 21 24 30 26 18

112 131 213 64 94 112

5

41

43 40 35 29

25 29 40

37

72

68

65

146 189 157

ANALYSIS OF VARIANCE FOR ARTIFICIAL

sv

SS --~~-i.--~-fs __,

DF

Total l\{ean a classes b classes e classes

52 1

41,811.000 38,940.942 1,669.943 250.316 949.799

3 8

40

4

5

3

3

4

4

5

16.7

Number in ith si re

Yi ..

456

ni

16

2 11.. =

342

133

492

15

7

14

a: + 4.410a: + 12.679a! a: + 4.270a:

1

40,610.885

{

5

.2

;

n¡.

.2 n¡¡ =

n.;

=

17.844

2

¿nis= 4.615 ii n 2

y2 -··· =·38,940.942

n

- 52 - 17.844 - 4 270 8 - .

- 17.844 - 4.615 -- 4 .410 q¡-

3

q2 = 52 - 13.961 = 12.679 3

n¡¡

J = 1, 2, ... , B i = 1, 2, ... , A

.2 n¡; =

n

ii

i

The ai, b¡, ciJk are random variables with zero means and variances a;, a¡, anda:, respectively. We find that the maximum-likelihood equations are very difficult to salve. Hence we shall rely on analysis-ofvariance estimators. This, however, presents a difficulty. In the unbalanced cross classification discussed in Chap. 13 we saw that

+ R(µ,{J)

y2

and

¿ni= 13.961 i n

qo -

~ = O, 1, ... ,

and R(µ,-r,{J) = R(íJ 1 µ,-r) + R(µ,-r) So we can partition the total_ sum of squares in two ways: 1,.2 :Ey¡;k =---:.:.:. + R(T 1 µ,{J) + R(µ,{J) + (error) n

ii n¡

So

2 n¡; =

where

R(µ, -r,p) = R( -r 1 µ,{J)

2 n;;

e

The Unbalanced Two-way Classification in Model 5

ni

i

EMS

556.648 1 31.289 23.744 1 a2

1

Total of ith si re

a;= 41.416

The model for this case is

- - - - -- - - -- - - -

3

7

34 41 40

a;= 1.767

-- ------ -- - - -

-

-- -- -

18 16 17

20 32

The estimators are

BREEDING DATA

12

- - - - -- ---- ---- - -34 26 22 23 21 16 14 31 42 26

-- - - -

Number injth dam in ith si re

3

16.10

359

:Ey¡Jk =--:.:.:. + R(fJ I µ, T) + R(µ, T) + (error) n If this is done for the model in this section and if mean squares are set equal to expected mean squares, we shall obtain two sets of estimators. Each· set is unbiased. Another method that is sometimes used is to partition the total sum - of sq uares as follows:

};!fi;. =

( y2 ~·· )

J [t (..:;· - ~-·J+

+ [ ~ (y; ..::·) - y2~·· +

y2. )

y2

(remainder)

360

LINEAR STATIS'l'ICAL l\'IODELS

l\IODEL

This can be put in an AOV table such as Table 16.11. TABLE

ANALYSIS OF VARIANCE FOR UNBALANCED Two-WAY CLASSIFICATION MoDEL

SS

DF

'l'otal

n

Mean

1

MS

2Y'f;k n y2

a classes

A -1

2--k ---= i ni. n

b classes

B - l

2---± i n.;

Remainder

n-A-B+l

Ams

a; + r ac + r a! 5

1

yz ( 2 __.!.:.:.

Bms

n

Rms

a: + r ac + r a: a; + r ac + r2a! 3

y2 . 'Ly¡ik =-···+(A - l)Ams + (B - l)Bms n

4

E(Rms) =

1

y2 )

-

l

n-A-B+I

+ (n -

B

+ l)Rms

(A -

l)E(Ams) - (B -

l)E(Bms)]

The quantities on the right have been computed, except for EC~Y~·1c)·

~~

u

Hence,

_ ...

n

l E(Rms) = - - - - - n - A - B + I

n(µ2

- a~ -

+ a! + a: + o!> -

2

" -n7; ªIJ -r- A u;; -2) + na; + ~ -2

9

n

n a!- 4--il n2 - 2--=1 n2) a~ (n - ¡-..!:. n ,, ni. n )

(

1 1i -

2

n7 ) a! + -l- ( 2 -n7; - 2 ~ n 1) ªb2 + 30, the confidence interval given in Theorem 17 .2 is probably adequate. If n < 30, the

and

:E g¡x¡ = (0) 2 n;

11

+ (.40180) 2 + (-.20375) 2 = 48

120

_

003709

'l'ABLE

i

MODEL

LINEAR STATISTICAL MODELS

372

\

TABLE OF QUANTITIES FOR lNTERVAL EXAMPLE OF ART. 17.1.l

11 48 120

2

3

.00000 .40180 -.20375

3.5629 1.2055 .6113

.0000

1

VARIANCE COl\IPONENTS; INTERVAL ESTIMATION

.!l333

-.3333

.1981

1 1 1

p(2.179 ~ ª2 & 2.179) 25.41 ~ b ~ 7.31

iM/ni

g¡xf/ni

¡

=g

1

·ººººººººº

·ºººººººº .0033630

.000028150 -.000000587

.0003459

=s

.003709

.0000276

=t

.003709

and we shall use n

=

ll.

ªª + 2

2

ab

+ ); so k(w) is the central F distribution. To evaluate the limits on the integral, we note that, when 'lt =Fa., w = AF'a, and when u = ,ex:>, w = ex:>; so

+ Theorem 17.4 Let the model satisfy the conditions of Theorem

=

+ Baa2

Bms

s:= Wms We shall test H 0 : ui =a~; that is, u! =O, and HA is a; > O. The test criterion is to reject H 0 at the ex level of significance if u > Fª, where

00

JFa

/i(u) du

where h(u) is the F(n;,ni) distribution. depends on the quantity

Bms

U=-Wms

The power of the test Also

A=~< 1

u¡

A=

uf

a;+s~

As another example, suppose we wish to test H 0 : in Table 17.1. We get

f lt(u) du = {J(A.) );.¡.'~ Proof: Let g(u;J.) be the frequency function of u for various values of )~. If ,l = 1, u is distributed as F(n1 ,n¡); so g(u;I) is Snedecor's F distribution with n¡ and ni degrees of freedom. Of course, if ,l ~ 1, then g(1t;J.) is not the F distribution. The test procedure is to reject H 0 if u >Fa., where Fa. is the upper « point of the F distribution; that is, where Fa. is given by and is

375

a!

= O using

the data

00

«=

f

u= 3.5629

=

2.956

1.2055

The tabulated F value for 11 and 48 degrees of freedom for « = .05 is F. 05 = 2.00. Hence we reject H 0 • 17.2.2 ·Test of a Linear Combination of In the previous article a t_est criterion was de,rised for testing the hypothesis that aj = O. This is equivalent to testing the hypothesis that af/aj = l. This test will satisfy many experimenter's requirements. However, a more general test may be desired. That is to say, we may need

ar.

ar -

00

g(u;l) du

JFa

1

MODEL

LINEAR STATISTICAL MODELS

376

to test H 0 : í:,gia¡ = O, where the gi are known constants that are either plus or minus unity. Of course, a more general test would be one in which gi were any known constants, but the case in which they are plus or minus unity covers many important situations. In most cases an exact test for this hypothesis has not been found, but an approximate test will be the subject of the next theorem.

+ Theorem 17.5 Let the model satisfy the conditions of Theorem 16.5, and let the AOV be represented by Table 16.2. To test the hypothesis a; + · · · +a: =a; + · · · + df against the ~lternative hypothesis + · · ·· + > a; + · · · + Uf, the quant1ty

a;

a;

u=

82

+ ... + 82

P

P

VARI.ANCE COMPONENTS; INTERVAL ESTIMATION

q

+ · · · + s!/fq (s; + ... + s~)2 s:ffr + · · · + s1fft s!/f

377

further approximation. Also, m and n will not be integers, but their values can be approximated in any F table. It should be pointed out that sorne care should be exercised in using any of these approximate methods where no bounds on the error of approximation are known. 17.2.3 Example: Testing a Linear Combination of Variance Components. The details will not be given, but suppose a three-way classification model with equal numbers and all interaction is given by i = l, 2, ... ' 10 Yiik =

µ

+ ai + b; + (ab)i:i + ck + (ac}ik + (bc);k + eiik

j = l, 2, ... , 8 k = l, 2, ... '4

"

8; + · · · + 8~

can be considered to be approximately distributed as F(n,m,), where (s2 + ... + 82)2 n=

5:

Let the distributional properties satisfy the requirements for case A in Chap. 16. Let the variances be u;, u:, u;b, a;,

- µ -

7

fJ

i -

The fJ equation can be written

¡

]

µ

are a sufficient complete set. Also, the quantities Ó'2 , p, LA¡(Yi. - Pxi.) for LAi = O are unbiased estimators of a 2 , p, LA¡T¡, respectively; hence, they are the best (minimumvariance) unbiased estimators. Next we state an important theorem on the distributional properties of the above estimators. • Theorem 18.2 Let the covariance model be given by Eq. ( 18.1) with the distributional properties noted there. Then (1) /j is distributed N({J, a 2 /E=:)· (2) LA¡f i is distributed N(LA;T,, ai), where

\J

a2 =

1\lIXED

,'V E:i:x

~ {J ~

/3 + ta./2

- a is

Ó'

A

/.

'\ Ex:x:

A confidence interval on LA;T¡, where LA¡ = O, with confidence .coefficient 1 - a is µ - Ti - ¡Jx¡_)2

..

LA¡T¡ -

Therefore, thet + 2 quantities &2,p, y 1 _, y 2 ., ••• ,yt. areasetof sufficient statistics. They are also complete. The foregoing can be summed up in the following theorem.

,. [LÁ¡ + (LA¡X; .)2]

ta.¡2Fa, where Fª is the upper ex percentage level of the F distribution with 1 and t(r - 1) - 1 degrees of freedom. The power of the test is

f,

E2 E2 2 -E!.= E - .......=!! = [t(r - 1) - l]Ó' E :cz uu E :i::i: T2

= ···

=

Ti

=

T)

can be written

+ {Jx¡¡ + e;;

where we ha ve written µ * in place of µ + -r. The normal equations can be obtained directly from this model and from the normal equations (18.3). They are

00

g(w) dw

µ*:.

Fa

+ Px..

= Y ..

¡3:

where w is distributed as F'(l, t(r - 1) - 1, l], where ;. =

rtP,!

p2E:c:i:

The solution is

2a2

~ .. píJ = t't~yu..x .,

N ext we shall prove a theorem that can be used to test the very important hypothesis H 0 : T 1 = T 2 = · · · = Tt· This hypothesis is estimable, sin ce LAi'T i is estimable if Lili = O. We can use Theorem 11.15. Using this theorem and the normal equations (18.3), we get

"" y i.1x u.. - X .. Y .. /rt - X .. y .. = 4 .,;:;11_ _ _ _ _ __

rt I tj; - X~. ii

I x¡, -

x:.Jrt

ii

If the quantities in Table 18.1 are used, we obtain

Further,

1

fi* =Y.. - iJx ..

So, referring to Theorem 11.15,

-

- {J ~Yii X ii

-

~

So H 0 is rejected if v > Fa.. in Tables 18.1 and 18.2.

1

y2.. ,.,iJy .... X +rt

393

l\IIXED l\IODELS

LINEAR STATISTICAL MODELS

392

rt

TABLE

18.2

SV

A computational procedure is given

AuxILIARY TABLE FOR TESTING (SS:xy) 2

DF

SS:yz -

SS:x2

T1

=

T2

= ··· =

Te

MS

F

B

A - B t(r - 1) -1 V=-t- 1 B

1

y2 rt

= -··

X y ) -··-·-· rt

+ {J-(LYuX;; -

Trcatmcnt 1 plus error

2

rt rt - t -

Error

s!v

S 1111

--=A szz

E;,, En-E=B

l

t(r -

zz

A-B

A-B

t- 1

Difforcnce

e::-T

1

The proof has been given above with the aid of Theorem

Proof: 11.15.

So noting that s of Theorem l l.15 is t - 1, that k = t + 1, and that n ~ k is the degrees of freedom for error t(r - 1) - l, we find that the quantity v in Theorem 11.15 here is

L y¡ + E2 [

1 i • v = (t - l)a2 - r -

y2

(T :r.11 +E XII

__!!! _ _ •• _

Exz

Txx

rt

+

>2]

1) -1

The noncentrality parameter A. can be found by using Theorem 11.16. '1Ve need the expected value of the treatment mean square, i.e., the treatment mean square Tms = (A - B)/(t - 1). This gives f

L

Exx

E(T ) = E-1-[T t _ l

ms

- (T:r;y + Ez,,)2 T +E xz :r:i:

+E 1111

1111

E + E~j E :r 1111

(18.9)

¡ 1'7.

y2 - ' - - - · · = T 1111 r rt

Sin ce

But, by the de_finition of a2 , we get

we can write vas

_ 1 [T

V -

(t -

l)a"2

+E tlll

1111

_ (Tx 11 + Ex 11 ) '1'x:c + E x:i:

2

-'--(E _E

[t(r -

1) - l]E(G2) =

E(E

1111

1111

xx

(18.S)

_1_ E[T

+ Theorem 18.5 Let the covariance model be given by Eq. (18.1) with thc distributional properties noted there. To test the hypothesis Ho: Ti = T 2 = · · · = Tt, the function v in ( 18.8) can be used. vis distributed as F'[t - 1, t(r - 1) - l, J.], where

A. =

~ (..-, -

T.)2

__:Í;..,____

2a2

[

~

(..-, -

•

2

2a ( T xz

T,)X,.]' + E :i::i:)

E~)

= [t{r -

l) - l]a2

(18.10)

E:r:r

So we nced only evaluate

This is the text of the following theorem.

r

-

E!.11)]

t- 1

+E 1111

1111

- (T:r11 + E:i:11)2] T :i::i: +E:i::i:

(18.11)

Using the notation of (18.4) and Table 18.1, this can be written

1 E{ --=-1 t

2

¡(Y;; - Y.J u

[fr

(y¡¡ - y.J(x., - x ..

T

+ E~ ~

)T}

394

Substituting by model (18.1), we get t

l l E{~(µ+ .,., + px., +e,, -

µ - T. -

l

Px.. - '

18.1.4 Example. The model for the following artificial data is Eq. (18.1). The problem is to estimate {3, estimate contrasts of the treatment effects T¡, and test the hypothesis Ti = T 2 = T 3 = T 4 • TABLE

_

[~ (µ + .,., + px., +e., -

µ - T. -

Px .. - O.J(x,1 -

x ..

>J"}

-{2 (T' -

1 t - 1

18.3

DATA FOR EXAMPLE OF ÜOVARIANCE MODEL

-

1

T:r::r:+E:r::r: = -

395

l\fIXED MODELS

LINEAR STATISTICAL MODELS

y

2 1

y

X

4

3 y

X

y

X

X

1

+ f3xii - fJx.J 2 + (rt -

f.

¡;

5.0 5.3 5.2 7.2 8.5

2

l)a

11

6.2 6.7 7.0 7.1 6.9

2.0 3.0 3.5 6.1 8.3

9.4 11.5 10.9 13.0 13.1

2.1 3.3 3.5 4.1 4.2

22.9

33.9

17.2

57.9

1

:Ex¡¡

+ 2{J 2 ('Ti -

7.)(xii -

ii

P'[.fr

(x,1

-

x.

.>'J"

2{J z[( 'Ti

-

x.J + (rt

f.}X¡¡]

ii

_¿(x110 - x.J2 flO

¡1

Tx:r:+E:r::x:

Txx+E:r::r:

94.5

+ '91 =

T:r::r:+E:i::r:

!

(x¡; - x..) 2

T :u

=

+E=,

TABLE

sv

this can be written

ij

18.4 1 1

Total Mean Treatments Error

Putting (18.10) and (18.11) together, we get for (18.9) l

{

¡

E(Tms) = - - r ('T¡ - f.)2 t- 1 ,

[

¡

'

:Ey;¡

=

1

33.9

70.4

20.5

193.4

¡1 + -92 = 4.92 ¡1 + '94 = 11.87

3.77

The average variance is .46la2•

If we use the fact that

2.3 3.1 4.0 5.1 6.0

From the row of totals in Table 18.3 and from Pas calculated in Table 18.4, the quantities ¡1 + -T, = y i. - Px,. can be computed. The results are as follows:

l)a2

-

=

13.5 13.4 14.0 14.5 15.0

1

1

31.2

2.7 6.3 6.4 8.3 10.2

Treatment plus error

¡1+-93 =7.93

From these values any contrast ofthe

ANALYSIS OF VARIANCE FOR THE DATA IN TABLE

SS:x2

DF

SS:X1J

T:r::r:

+ E:r::r:

+ (t -

l)a2

20 1 3 16

547.13 446.51 31.43 69.19

= T:r::r: = E:r::r:

977.94 913.82 26.89 = T:r:v 37.23 = EZfJ

19

100.62

= S:r:x:

64.12 = SZfJ

}

;;

.,. i

can be estimated.

E:x: 11 E:e:r:

=

37.23 69.19

= ·53 S

The auxiliary table for testing Ho:

Ti

= 'T2 ='Ta= 'T4

corresponding to Table 18.2 is Table 18.5.

1

2,107.66 1,870.18 216.06 21.42 237.48 1

P =

By using Theorem 11.16, the value of A can be evaluated. A covariance model for a one-way classification has been presented in sorne detail. Other covariance models can be analyzed in similar fashion.

SS:y 2

1

1

(-r, - T.)X,J

1

18.3

= =

Tvv Ev 11

= Bvv

396

LINEAR STATISTICAL MODELS TABLE

sv

18.5

AUXILIARY TABLE FOR TESTING 'T1

DF

SS:y 2

1

-

(SS:xy)2 SS:x 2

l\IIXED l\IODELS

=

'T2

=

T3

397 discussed. If an experimenter desires distributional properties other than those given in Eq. (18.12), tests and confidence intervals must be altered accordingly. The two-way classification mixed model \Vith interaction and with equal numbers in the subclasses will be defined by

= T4

MS

F

.087 65.077

748+

1

Treatment plus error Error Difference

-

19

16 3

1

196.62 1.39 195.23

ri =l. 2 •...• t j = l, 2, ... 'b

t

(18.12)

k = 1, 2, ... , 1n

The result indicates that the treatment effects are certainly different; i.e., we reject H 0 • 18.2

where µ and -r i are assumed to be fixed unknown parameters su ch that t

.2

Ti=

¡::::::¡

Two-way Classification Model with lnteraction and with Fixed and Random Effects

and where

In Chap. 12 the two-way classification model with interaction was discussed. In that chapter model 4 was assumed; that is to say, µ, -ri, f.J ;' and (-r{J) i; were assumed to be fixed unknown constants. In Chap. 1 7 various types of model 5 were discussed. In these models ali effects were random variables except the O\Ter-all mean µ. In man y experiments the model that fits the situation is neither 4 nor 5 but a combination of the two. In a mixed model such as this, there could be many specific types; that is to say, in an n-way classification model, any of the main effects could be fixed and any could be random. However, in a realistic model, the interaction terms are fixed or random depending on the main effects that occur in them. Since so many different types are possible, all cannot be described. Only one will be discussed: the two-way classification model. From this model the general procedure can be inferred. The distributional properties for model 4 and model 5 were fairly straightforward, but this is not true of the mixed model. If any element in an interaction term is random, it may be realistic to assume that the interaction term is random also. However, in that case we· must postulate the distributional properties of the interaction terms. The proper error term to use to test certain hypotheses and set certain confidence intervals depends on what distributional properties are assumed to hold. 1 The model and distributional properties given in Eq. (18.12) will be assumed, andan AOV for this situation will be

0

p1, ( -rP),.;, and e,.;k are random variables such that E(f.J;) = E( -r{1)¡1 = E(ew.:) = O

the eiJk are independent and distributed N (O,a;) ; the eiik are independent of the {.J1 and ~he (-r{.J)i;; the P; are independent and distributed N(O,o¡); the {.J1 are mdependent of the (-r{J)ii; the {-r{J);; are distributed t - 1 N(O, - t- tr;p); E[( -r{.J),.1( -rf.J)i·;] = -

E[ ( -r{.J)¡¡( -rf.J)w]

and

.2i (T{.J)¡;

=O

~ a;p

if i =I= i, if.i =I= j'

= O for allj.

· If all the terms were fixed except em.:• the AOV would be as given in Chap. 12. This AOV is also given in Table 18.6, except for the EMS column, which is different for the mixed model ( 18.12). The object in this model is to test H 0 : a~ = O, to test H 0 : a;tJ = O, to test H 0 : Ti = -r 2 = · · · = -rt, and to set confidence intervals on contrasta of the Ti. To accomplish this, we shall find the distribution of the sums of squares in Table 18. 6. The following theorem will be proved concerning the distributional proper~ies of the sums of squares in Table 18.6.

• Theorem 18.6 Let the model and distributional properties be as given by Eq. (18.12), i.e., a mixed two-'''ªY classification model. Then

l There are many sets of distribut.ional properties that have been advancecl as realistic for various situations [3, 4, 5, 6]. A discussion of these various assumptions is beyond the scope of this book. The concem hero is to explain methods of analysis when various assumptions hold.

(1)

1

: 1 '

398

LINEAR STATISTICAL MODELS

is distributed as x'2[(t - 1), A.], where

2 (r, - f.} 2 2(a! + ma~p) Í

SS

For a complete discussion of this combined test the reader may consult Zelen [7] and Weeks [l]. 18.3.4 Computing Instructions for the Balanced Incomplete Block Model with Recovery of Interblock Information. Computing instructions for f{ in Theorem 18.13 will be given. A format such as that ofTable 18.9 will be useful. From this table anAOV can

416

1

be computed as shown in Table 18.8. be needed: ~T~

B2 = __i

SS for treatments-(adj):

k Á2=-~~ At 'li

SS for blocks (unadj):

-

-

l

p2 ---=bk

r

A3 -

Y2.

~-± 7 k

(18.32)

y2 ble

A2

= 3.2908

Á3 =

2.0347

Total Mean Treatments (unadj) Blocks (adj) Intrablock error

1

y

1

1 y

X

4.0

LO

8.0

2.1 3.0 4.2 5.0

7.1 9.0 13.2 12.8 15.5

1.3 .2.0 3.3

7.0 8.3 9.2 10.4 11.5

3.4 4.1

X

.3 .8 1.1 1.5

1

1.8

18.8 In Prob. 18.7, find and set a 90" per cent confidence interval on a2. 18.9 In Prob. 18.7, find -91 - '92 • 18.10 In Prob. 18. 7, set a 95 per cent confidence interval on T¡ - T2 • 18.ll In Prob. 18.7, .find the average variance of the estimates of the difference-of-treatment means. 18.12 In Prob. 18. 7, test the hypothesis T¡ = T2 = Ta with a type I error probability of 5 per cent. 18.13 Suppose the data of Table 18.12 satisfy the assumptions for a balanced incomplete block with blocks random. Find the intrablock estimates of T. - -r • The numbers in parentheses are treatment numbers. ' ·

B 3 = 1.5375

ANALYSIS OF VARIANCE FOR EXAMPLE OF ART.

SV

X

3

82

The AOV is given in Table 18.10. 18.10

2

------------

9.4 12.8 15.7

--:..:..!

The example given in Art. 14.4.1 will be used to illustrate the above procedure. A 2 and A 3 can be obtained from Table 14.6; the other quantities are easily computed from (18.32).

TABLE

1 y

SS for blocks (adj):

B 2 = 3.7880

18.11 Treatment

TABLE

The following quantities will

SS for treatments (unadj):

419

l\:CIXED .l\fODELS

I ...INEAR STATISTICAL MODELS

418

DF

SS

21 1 6 6 8

169.1162 161.8519 3.7880 1.5375 1.9388

14.4.l

TABLE

MS

Block

1

.2563 .2424

1 2 3 4 5 6

= Eb = E6

18.12

Treatments (1)

1.2

(1)

.8

(1) (2) (2)

1.1 2.2 1.6 4.5

(3)

(2) (3) (4) (3) (4) (4)

1 1

2.2 1.8 7.1 4.2 6.7 6.3

1

-r, -

18.14 In Prob. 18~13, find the interblock estimates of -r • 18.15 Find the variance of the estimates in Probs. 18.13 and 18.14. 18.16 Run anAOVon thedata in Prob. 18.13 such as that given in Table 18.7. 18.17 Find combined estimators of Ti - 7- for the data in Prob. 18.13. 18.18 Use Theorem 18.15 and the data in P roblem 18.13 to test the hypothesis T1 = Tz = T 3 = T 4 with a 5 per cent type I error. 18.19 Prove parta (3) and (4) of Theorem 18.10. 18.20 If b > t, show that Ras and Esa in Theorem 18.10 can be used to form a combined estimator of T 1 - -r.•

Problems Prove that E(}.:i ..r,) = l:A¡T¡ in Eq. (18.7). Show that the r~nk of the normal equations (18.3) is t Show that E(P) = {J in Eq. (18.5); 18.4 Show that var(P) = a 2/E32 • 18.5 Show that

18.l 18.2 18.3

18.6

+ l.

0

Show that E(ii2) = t(r -

1) -

Further Readinf!

1 a2

rt

l

18.7 The covariance model given in Eq. (18.1) is assumed to fit the data in Table 18.l l. Find fi.

1

D. L. Weeks: An Exact Test of Significance in the Balanced IncompJete Block Design with Recovery of Inter-block Information, unpublished M.S. thesis, Oklahoma State University, Stillwater, Okla., 1957.

420

LINEAR STATISTICAL l\IODELS

2

C. R. Rao: General Methods of Analysis for Incompleta Block Dcsigns, J. Am. Stati8t. Assoc., vol. 42, pp. 541-561, 1947. 3 M. B. 'Vilk: Linear Models and Randomized Experiments, Iowa Ph.D thesis, Iowa State College, Ames, Iowa, 1955. 4 M. B. Wilk and O. Kempthome: Sorne Aspects of the Analysis of Factorial Experiments in a Completely Randomized Design, Ann. Math. Statist., vol. 27, pp. 950-984, 1956. 5 H. Scheffé: A "Mixed Model" for the Analysis of Variance, Ann. Math. Statist., vol. 27, pp. 23-36, 1956. 6 H. Scheffé: Altemi:i.tive Modela for Analysis ofVariance, Ann. Math. Statist., vol. 27, pp. 251-271, 1956. 7 M. Zelen: The Analysis of Incomplote Block Designs, J. Am. StatÜJt. Assoc., vol. 52, pp. 204-217, 1957. 8 D. A. Sprott: A Note on Combined Interblock and Intrablock Estimation in Incomplete Block Designs, Ann. Math. Statist., vol. 27, pp. 633-641, 1956. 9 E. S. Pearson: A Note on Further Properties of Statistical Tests, Biometrika, vol. 32, pp. 59-61, 1941-42. 10 F. Yates: Incomplete Randomized Blocks, Ann. Eugenics, vol. 7, pp. 121-140, 1936. 11 F. Yates: The Recovery ofinter-block Information in Balanced Incompleta Block Designs, Ann. EugenW8, vol. 10, pp. 317-325, 1940.

1

Appendix

Four tables are presented: central chi-square, Student's t, central F, noncentral beta. (The cumulative normal distribution can be obtained from Table A.2 with 11 = co.)

1

Table A.. l

Central Chi-square

The entries in this table are x ~ { 11), where co

P=

J

g(u)du=P[u

~ x~(11)]

x~(v)

for probability values P = .0001, .001, .005, .O 1, .025, .05, .1, .25, .5, .75, .9, .95, .975, .99, .995, .999, .9999, and for degrees of freedom values v = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120. Table A. 2 Student' s t The entries in this table are tr (11 ), where co

P/2 =

f g(t) dt = P[t ~

tp( 11)]

tp(v)

~

for probability values p = .0001, .001, .005, .O 1, .025, .05, .1, .25, .5, . 75, .9, .95, .975, .99, .995, .999, .9999, and for degrees of 421

freedom values v = 1, 1.2, 1.5, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120, oo.

~ ~ ~ 0.

Table A.3 Central F The entries in this table are Fp( v 1 , v.J, where

p

f

=

g(F) dF= P(F ~ Fp(vu v:J)

f~(v1 , v2)

CO

;~ N

•

~ ~ H

! : ~ ! s ~ 55! ; ~

~ ~ ~ ~ ~ ~ ~

i;:~

~

~ i ~: ~

;: :!: ~

H---'

C,

a.!"l __

o

~

:

:

;

:

! : ! ª ! 5: 5! ~ ª : ~ ~ +,:t.++

~ ~ ~ ; ~ i ~ ~ l

where v1 is the numerator degrees of freedom and 112 is the denominator degrees of freedom. The values of p are .0001, .001, .005, .O 1, .025, .05, .1, .25, .5, . 75, .9, .95, .975, .99, .995, .999, .9999; numerator degrees of freedom v 1 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120, 00 ; and denominator degrees of freedom 11 2 = 1, 1.2, 1.5, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120, oo.

~ ~

~

.

!~

~

~++.t.•

Table A.4 Noncentral Beta The entries in this table, except for the E~ column, are {31,(/1 , f 2 ,

0

s. 8715

4. 3513

2. 9747

1.4037 (-"l) 4.7192 1-11 l. 0437 (-Z) 1.6197 1-31 4. 032. 1 (-3) l. 0069 1·411.6106 (-5) 4.0264 (-611.6105 (-8) l. 6105

Z4

(+11 2.1663 (+I) l. 4028

9. 5513

7. 8229

s. 7167

4. 2597

z. 9271

1. 3898 1·11 ... 6902 1-1) l. 0389 (·2) l. 61l.9 1-Jl 4. 0153 (-31 1. 0028 HI 1.6040 (·SI 4.0096 (·61 1. 6039 (-8) l.603'!

24

30

1+11 z. 0092 (+11 l. 3Z93

9.1797

7. 56l.5

5. 5675

4. 1709

2. 8807

l. 3761 (-114.6616 1-111.0341 (-21 l. 6060 (·31 J. 9986 1-4) 9. 9860 1-41 l. ~973 (· 51 l. 9930 (-611.5972 (•81 l. 597Z

30

40

+I) l. 8668 1+11 l.Z.609

8. 8278

7. 3141

s. 4239

4.0848

2. 8354

l. 3626 (-1) 4.6330 1-11 l. 0294 1·21 l. 5993 (·31 3. 9818 (·419.9443 1·41 l. 5906 (·SI J. 9765 (·61 l. 5906 (-8) l. 5906

40

60 1+111.nn 1+11 1.1913

8.4946

7. 0771

s. 2857

4. 0012

2. 7914

1.3493 (-11 4. 6053 1-111.0247 1-21 1. 5925 1-3) J. 9651 1-419.9030 (·41 l. 5840 (- 51 3. 9599 (·61 l. 5839 (-811. 5839

60

120

+I) 1.6?04 (+I) l.1380

8.1790

6. 8510

S.1524

3. 9201

z.. 7478

l.3362 01 (-1) 1.7742 1-111.4551 1-11 9. 5413 •-21 5. 5187

5

6 ~+1) 4. 1071 (+11 1.8411 (+ll l. 0250

7. 8741

5.4613

4,060C

2. 9369

l. 7708

1.0418 (-11 6. 3432 1-ll 4.0640 (-1) 3.1083 1-11 2. 4557 1-11 1.8567 (·t) l. 5280 1-11 l. 0075 1-21 5. 8546

&

7

+l}

8. 3803

6, 6201

4. 7611

3.636!

2. 702!

1,6898

l.0304 1-116.3743 1-1) 4, 1425 1-11 3.1893 (-11 2. 5318 l·ll 1.9UO 1-11 l. 5867 1-11 l. 0507 (·2) 6.127'>

8

+11 2. 1683 ~+11 1.1540

7. 2107

5.8143

4.2951

3. 3472

z. 5380

l.6310

1.017! 1-11 6.4016 1-114.2066 l· 1) 3. ZS55 1-11 z. 5941 1-11 1. 9776 1-11

9

+11 l. 759C

9. 8943

6.4171

5. 2S6!

3,963!

3.1373

2.4163

l. S86l

1.0077 1-116.4255 (-114.2602 (-11l.3108 3

z. 0548

1.8337

1.6012

l. 2621 1·11 9. S03Z 1-ll 6. 9881 1-ll 5. 1752 1-114.2717 1-1) 3, 5876 l·ll z. 8991 1-11 2. 4907 -11 l. 7879 (•I) 1.1490

11.0

m

l.2612

Z. 7425

z. 3583

z. 1848

1. 9447

l. 7SZl

1. 5458

1. ll7l 1-119.4503 1-1) 7. 0319 l-11 s. 2531. 1-1) 4, 3550 1-ll 3. 6699 HI 2.9755 1-1) z. 5615 (·ll l.845Z -11l.1896

a>

IJ1

"z

p•0.001

p•0.0001

p•O.OOS

p•0.01

p•O,ZS

p110, 5

p• 0.15

p•0.9

P"º· 975

p•O, 995

= 15 p•O. 9999

p=0.999

6. 6939

1,8112 1-11 6. 7903 1-1) 3. 3603 (-1) z. 3Z78 1-11 l. 7328 l-1) 1.2546 (-1) 1.0166 l·ZI 6.6968 1-Zl 4. 0868 1.2

l. 5 15) l. 1787 131 8.US7 121 9. 6518 12) 3. BZ.75 IZI 1.1247 (1) 4.4349 lll l. 1313

4, 7594

l. 575G

1·214.6800 l. s HI 6.6473 1-11 3. 5055 1-112.4922. 1-111. 8878 1-111. 3887 1-11 1.1350 1-217.5809 z.

2 1+31 9. 9994 1+21 9. 9943 1+21 l. 9943 1+11 9.902 C+ll 3.901 1+11 l. 9429

9.4247

l.4098

l. 3771 1-11 6. 5673 1- ll 3. 1103 (-l) 2.7157 (-11 2.0986 l·ll l. 5726 1-111.2986 1·21 8.8190 1-2) 5. 5221

8. 7029

s. 2003

Z.4552

l. Zll l 1-ll 6. 5781

1+21 5, '1384 (+2) l. 2737 t+ll 4, 3085 (+112.6872 1+111. 4253

e-ti 4. 0164 1-1) 3.0419 1-11 2. 4080 ·ll l. 8460 l·ll 1.544.t (-11 l. 0712 1·21 6. 8329 ~-Z)

7.8228

3 4

IUI 1.4971 (+I) 4. 6761 (+I) 2.0438 (+11 l. 4198

8. 6565

5. 8576

3, 8689

Z.G829

1.1386 1· ll 6. 6353 1-114.2348 1-ll 3. Z.127 1-1) 2.6286 l·ll z. 0437 (-1) 1.7233 k-ll 1.U17

9. 7ZZZ.

6,4277

4.6198

3. 2380

1.8851

1.0980 (-11 6.6943 (-11 4. 3995 1-1) 3,4467 1-ll 2. 7961 HI Z.1951 1-111.8615 (•l) 1.3215 ~-2) 8. 6050

5

l.0722 1-116,7476 1-11 4, 5288 1-ll 3. 5836 1-112.9285 1-112.3157 1-11 l. 9721 1-ll 1.4101 1-21 9.2427

6

1+11 3. 9061 (+11 l. 7559:

6

7 1+1)

Z.681~

C+ll l. 3314

8 1+11 z. OZ7~ 1+11 1.08"1 9

1+1) 1.6331

9.2381

9.8140

7. 5590

5. 2687

3. 9381

Z.8712

l. 76ZI

6,310

4. 5678

3. 5108

Z. 632Z

1.6781

1.0543 1-116.1944 1-11 4. 633! 1-11 3. 6947 l-1) 3.0364 -ll 2. 4146 1-11 2.0630 1-1) 1.4835 C-219.7743

7

7. 9678

8

6. 8143 6. 0325

5. 5151 4. 9621

4.1012 3, 7694

l.21H

2. 4642

1.6170

1.0412 1-11 6.1346 (-1) 4. 7203 l·l) 3. 7867 1-1) l. 1Z63 -11 2.4972 (-2) Z, 1394 1-11l.5454 1-1) 1. 02Z5

3,0061

Z. 3396

1. 570!

1.0311 1-1)

1) 3,2024 1-11 2. 567! 1-l) z.zou 1-1) l. 5985 1-111.0614

9

l. 5H8

1.0l.32 1-1) 6,9008 (-114.8560 (-11 3. 9311 l•ll 3. Z678 1-ll 2. 6282 l·ll z. 2606 c-11 1. 6445 1-11 l. 0952

10 12

6,870~

1-1) 4. 7934 1-11 3.8646

( 0

10 1+1) l. 374i

8.1288

5.4707

4, 5582

3, 5217

2. 8450

z. 2435

12

l+I) l,0631

6. 7092

4.1214

4.0096

3, 1772

Z.6169

Z.1049

l.4796

1.011! 1-11 6, 9527 1-114.9576 1-ll 4,0399 1-11 l. 3746 1-ll 2, 7276 1-11 2.3531 1-111.1206 1·1) l.1513

8.U91

5,5351

4. 0698

3, szzz

z. 8621

2,403!

l. 9722

1.4263

1.0000 C·ll 7.0111 1-11 5. 070: 1-ll 4.1606 l-11 l.4939 1•11 Z.8391 1•1) Z.4571 (-1) l.8067 1·11 1. Zl52

15

15

zo

zo 24

6. 3741

4. 5618

3. 5020

3,0880

2, 57ll

z. Z033

1.8449

l. 3Jl6 1-1) 9. 8870 1-11 7,0786 1-1) 5.1967 l-1) 4. 296! C·ll 3.6286 1-11 z. 9657 1·112.5756 1-11 l. 9053 c-11 1. ze90

5. 6112

4, 1387

3. 2.456

z. 8887

z. 4374

Z.1077

l. 7831

l. 3474 1-11 9. 8312 1-117,1164 l·I) 5.2659 (-1) 4. 3710 1-11 3. 702.9 1-1) 3. 0358 1-1> 2.6414

30

4.938!

3, 7527

3. 0057

z. 7002

2, 307Z

2. 0148

l. 7223

40

4. 345'

3, 4003

z. 7811

z. 5216

Z.1819

l. 924!

l. 6624

60

3. 8ZZI

3.0781

2. 5705

2. JSU

2. 0613

l. 8364

1.6034

IZO

3.36~

2. 7833

2. 3727

2.191S

l. 9450

l. 750!

l. 5450

CD

Z.950'

z. 5132.

Z.1868

z. 0385

1.8326

1,666~

l. 4871

1. 9604 (-JI J. 3304

24

1. 3213 1-11 9. 7759 1-11 7, 1567 1-1) 5. 3396 1-11 4. 4508 1-11 3. 7826 1-1)3.llll 1-11 z. 7lZ5 l·I) 2.0201 (-11 l. 3754 1-1) J. 4246 1.2952. 1-119.7211 1-111. zoo• 1-11 5. 4189 1-114.5366 1·11 3.8685 1-11 3. l9Z9 1-112.7894 l·l) z. 0851

30

(-11 l. 4787 l. 2691 1-11 9. 6667 (-1) 7. 248 1-11 5. 504Z -11 4,6294 1-11 3.9617 l-11 3, Z818 1-11 z. 8733 l-llZ.1562 1-1) l. 5386 l. 2428 1-119.61Z8 1-117,3003 1- ll s. 5')64S

2. IZ42

l. 7938

l. 3580

1.0000 1-1)7.3638 1-115.57-48 l·ll ... 7077 1-11 4. 0576 1-113.4040 1·113.0140 1-1) 2. 3310 (-1) l. 6802

zo

Z4

S.2084

3.8'13Z

3.0624

2. 7380

2. 3273

z. 0267

l. 7302

l. 3307 1-119.9436 1·11 7.4107 l· 11 s. 6603 1-114.8019 (-1)4.1535 1-11 ). 4972 1-113.1037 1-11z.4100 1-111. 7441

z.t

JO

4. 5540

3. 4928

z. 8230

z. 5487

l. 19SZ

l. 9317

1.6673

1.3033 1-11 9. 8877 1-11 7.4f>Zl l· ll 5. 7531 1-114.9041 1-11 4. Z579 1-11 3. 5991 1-11 3. ZOl6 l•ll Z.4969 1-1) 1.8147

30

40

l.977Z

3. 1450

l. 5984

l. 3689

z. 0677

hU89

1,605Z

l. Z758 1-119.83Zl 1·117. H8Z 1-11 5. 8S38 l· l) S.0155 l• 11 4, l7ZO 1-ll 3. 7110 1-11 3, 30'16 l·I) z. 5931 1-1) l.8933

40

10

"º

).4681

z. 8266

2. 3872

Z.1978

l. 9445

l. 7480

l. 5435

l. 2481 1-119.7773 (-117.5798 1-11 5. 9637 l· 11 5. 1377 1-114.4976 1-11 3. 9346 1-11 3, 4Z95 1·11 2. 7005 1-111.9817

60

IZO

), 0180

2, 5344

z. 1881

z. 0346

1, 8Z49

1.6587

1.4821

l. zzoo 1·119.7ZZ8 1-11 7,6488 1-11 "· 085J 1-1) S.Z734 1-ll 4. 6378 1-11 3. 9733 C-1) ], 5640 (•!) 2,8Zl8 1-1) 2. 0820

IZO

Q)

z. 6193

Z, 2657

l. 9998

l. 8783

l. 7085

l. S705

l. 420(¡

1.1914 1-119.6687 l· 11 7. 7l62 l· 116.221Z 1- ll s. 4253 1-114.7955 1-11 ... 1302 1-11 3. 7169 1-112.9605 {-11 2. 1976

co

111 p•0,01

p•0.02.5

p•O.OS

p•0.001 p•0.005 p•0,0001 ltll 6, 2002 1+7) 6. 235~ (+SI 6. USO (+t) 2.4940 (+3) 6. Z346 (ti) 9. 97ZS 1+z1 z.4905

"z l

p•O, 5

p• o.zs '1.6255

2.1321

p:oO. 7'• 1-1) 7.

p•O, 9

I°'~ (-1)

p•O. 95

paO, 975

p•0.99

l. 4164 1-112.3476 1-11 l. 7493 1-1) 1.2783

p.o. 995

1-ll

= 24 p• o. 9999

p• o. 999

l. 0470 1-Z) 7. 1286 1-2) 4. 6161

"z 1

(1) 3.2649 1.2 16) 3. 3044 (4) 7.1192 (3) 4.8691 131 l. 5334 IZ) 3.U67 (2) 1.0452

6. 7759

1.8444 1-1) 7. 006Z 1-11 3. 5378 (-1) 2. 4916 (-1) l. 8872 (-1) l. 3996 (-1) l.1556 l-2) 7.9704 1·2) 5, 2180 l,Z

(1) l. 7438 l. 5 (5) l. 7897 Ul a. 3065 121 9.7113 (Z) 3, 8512 12) l, 1318 (1) 4,4641

1. 8075

l. 6034 (-1) 6. 8707 1-1) 3. 7039 (-1) 2. 6799 (-1) z. 0673 (•I) l. 5592 (•I) 1.2.994 1-2) '1. 0987 (·2) 6. 0349 l. 5

2

(+3) 9. 999' (+2) 9. 9946 l+Z) 1. 9946 l+ll 9, 9458 l+l) 3. 9456 l+ll l. 9454

9. 4496

3.4345

1.4014

(-1) l. 5013 (-1) l. 0707 l·Z) 7, 2185 1-ll 6. 8050 l· 11 3. 9396 Hl 2.ua8 (-1) 2. 3155 1-1) 1. 7814

2

8,6385

s. 1764

z. 462&

1, 2322

(-1) l. 3237 l·Z) 9, 12.08 1-1) 6.8423 (-1) 4.2966 1-ll 3. 323& (-1) 2.6874 1-ll z. 1195 1-ll l. 8119

3

3 1+2) s. 870~ 1+2) 1. 2.593 l+l) 4. Z.622 l+ll 2.65911 l+l) 1.4124 4 K+2) l.4642. (tl) ... 5766 (+l) Z.0030 1+11 l. 3929

8. 5109

s. 7744

3. 8310

Z,0827

l. 1583

1·1) 1. 5116 (·l) l.0&11 1-1) 6.9219 1-11 4. 5560 1-ll 3,6019 1-1) 2. 9591 1-1) 2.3706 1-11 z.001

4

9. 4665

6. Z780

4. 5Z7Z

l. 1905

1.8802

1.1170 1-1)7.0004 1-114.7551 (-l} 3.8158 (-1) 3. 1698 1-1) 2. 5673 (•l) 2.2293 (•1) 1. 6732. (•1) 1.1823

5

7. 3127

5. 1172

3. 841S

Z.8183

l. 7540

l.0907 (-l)7.0706 l·ll 4. 9138 1-ll l. 9869 l•ll 3. )393 1-ll z. 7Z7l. l·l) 2. 3799 l·I) l. 8017 (·1) 1.2838

6

l.667S

1.0724 (-1) 7. 1317 1-1) 5.0439 1-11 4.1278 1-l) l.4797 1-11 2. 860! (-1) 2. 5060 1-111.9103 (-111. 3702

7

1.0591 1-1) 7.1849 1-11 5. 1528 (-11 4.2461 1-11 3. 5983 (·l) 2. 9736 (•11 Z.6134 l·l) 2. 00)5 1-11 l. •451 1.0489 (.117. 2317 (· 1) 5. 2458 1·114.3474 1·11 3. 7000 (·ll 3.0713 1-11z.7064 l·l) z. 0847 1-11 l. 5106

8

1.0(08 1-11 7.2727 (-11 5. 3262 1-11 4.4352 1·113.7885 1-11 l. 1565 (-1) 2. 7878 l•l) Z.1561 1-11 l. 5686 l. 0289 l·ll 7, 3416 1-11 5.4588 1-11 4. 5800 l·ll l. 9351 1·113.2986 1·11z.9241 (•l) 2. 2.764 l·l) l. 6669

10 IZ

20

l+ll 6.445! l+ll 2. 5133 l+I) l. 2780

s

6 l+I) l. 7512 (+ll l,6891 7 ~+l) 2. 555( l+l) 1.2.732

t 00

p•O, l

9. •741 7. 6450

6,0743

3,4105

4.4150

2. 5753

8 l+l) 1.917 (+I) 1. 0295

6. 5029

5. 2793

3. 9472

3.1152

2,4041

1.6043

9 (+l) l. 5349

5. ?il.92

4. 7290

l. 6142

z. 9005

Z. l768

l. 5560

8. 7239

10 t+I) 1.284'

7.6376

s. 1732

4. 3269

3. 3654

Z. 737Z

z. 1784

l. 5179

lZ

6.2488

4. 4315

l. 7805

3.0187

z. 5055

z. 0360

1.4613

3. 2940

2. 7006

2. 2878

l. 8990

l. 4052

9. 8314

15 20

5. 7336

Z4

5. oooz

30 40

4.1493

3. zzzo

2. 8594

2. 4076

z.o8Z5

l. 7667

1, 3494

z.6591

2, 2693

1,9838

1. 7019

l. 3214

l. 0000 1-ll 7. 5(>71 l·ll 5,8158 l·ll 5,0408 1·1) 4.4066

(·11 3. 7607 (· ll 3. 3707 1·1) z. 6771 1-11 1.9999

24

3. 7354

2. 9667

l. 6377

l. Z.933 1·119.9438 (•I) 7.6260 (-11 5. 9805 (·I) S. 1Sl3 l·ll 4.5i69 l· 113,8800 (-1) 3.4869 1-1) 2. 7828 (-1) 2. 0891

JO

3. 3572

4. 3545

3. 0111

3. 785Z

15

1.0112 l·I) 7.4Zl7 1-11 5,6082 l·ll 4. 7445 (·114.1027 1-11 3.4618 l·ll 3. 0811 1·1) Z.4162 1-11 l. 7821 l. 0057 1·117. 5148 1-1) s. 7797 1-114.9341 1-114. 2968 (· ll 3. 6523 l· ll 3. 2654 (-l) 2. 5918 l·l) 1.9200

5, 1009

7. 5168

9

3. 7859

z. 7272 z. 5020

2.4689 2. 2880

1.8874

2, 1359

1. 7929

2. 0069

l. 5741

60

3.ZBZ5

Z.6938

z. 2898

z. 1154

l. 8817

l. 7001

l. 5107

120

2.8373

2.4019

z. 0890

1. 9500

l, 7597

1.6084

l.447Z

(1)

Z.4422

2. 1324

l. 8983

1. 7908

1.6402

l. 5173

l. 3832

----------~~~--~---------------------------

-----

1. 2649 1· ll 9. 8880 1-117.6899 1-11 6, 0950 1-ll 5,2854 1-11 4,6598 l•l) 4.0lZ4 l-IJ l.6161 1-11 2.9012 l·l) 2.' 897 l. 2361 l·l) 9.8U8 1-1)7.7610 l· ll 6, Z216 (.l) 5.4277 1-1) 4, 8079 1-n 4.1606 (-1) 3. 7615 1-113.0352 1-11 Z. 30of3

40 60

UD l. 2068 1·119.7780 1-ll 7, 8407 l· 116.3633 1-11 s. 5875 l·ll 4, 9754 (·ll 4. 3292 1-ll 3. 9273 (·11 l. 1890 (·I) Z.4368 (1) l·l) 2. 5929 1-113.3687 1-U 4.1193 (-11 4. SU! l.1767 1-ll 9. 7236 1-117. 93Zl (.11 6. 5244 1-11 5. 7700 (.1) 5. 1672

....,...

...____ --

-~-----"'·----.....-------~·"

rtJ

v.= 30 "z

p•0.001

p•0,005

p•O. 01

p•0.025

p•0.05

p•O, I

p.o. Z5

p•0.9~

p• o. 9999

"z

9. 6698

Z.145Z (-1} 7. 2669 l-lll.4714 1·11Z.3976 1·11 l. 7961 (-11 l. 3ZZ3 1-11 1.0894 1-211. 52ze (·Z) 4. 9771

1

l.Z 16) 3. 3154 141 7. 14Z8 131 4. 8852 (l) l. 5385 IZ) 3. 3378 121 l. 0487 111 3.2763

6.80)'.

1.8555 1-11 7. 0790 (-1) 3. 5981 1-112.5476 1-11 l. 9405 l-ll l.'450Z 1-11 l.2isson distribution, 76 functional relationship, 99, 104, Point estimation, 34, 35 186ff. components-of-variance (see Comgeneral-linear-hypothesis, 106ff. ponents of variance) incomplete blocks, 306ff. covariance (see Covariance) mixed (see Mixed models) experimental design modeis, 240, N-way cross, no interaction, 261, 255, 289 262 functional relationships, 187 one-way, 255, 338 model l. 110, 117 polynomial, l 65ff. regression models, 197, 217, 226 regression, 101, 104, 195ff. three-way with interaction, 272, 27 4 Power of test, 41, 42 on correlation, 212 twofold nested, 349 'lOdel 1, 132, 140 two-way, balanced, with interáct'. lOncentral F, 80, 81 265 :º• no interaction, 258, 321 unequal numbers, 287ff., 292 Moment, generating function, 32 , multivariate normal, 54 noncentral z2 , 75, 76 Móments, 32 Multivariate normal distribution, 49, 197 conditional, 62 linear functions, 56 marginal, 51, 52

463

INDEX