Research Article MANOVA for Nested Designs with Unequal ...

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Research Article MANOVA for Nested Designs with Unequal Cell Sizes and Unequal Cell Covariance Matrices Li-Wen Xu College of Sciences, North China University of Technology, Beijing 100144, China Correspondence should be addressed to Li-Wen Xu; [email protected] Received 7 May 2014; Accepted 3 July 2014; Published 16 July 2014 Academic Editor: William Zhu Copyright © 2014 Li-Wen Xu. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We propose and study parametric bootstrap (PB) tests for heteroscedastic two-factor MANOVA with nested designs. For the problem of testing “main effects” of both factors, we develop a flexible test based on a parametric bootstrap approach. e PB test is shown to be invariant under affine-transformations. Moreover, the PB test does not depend on the chosen weights used to define the parameters uniquely. e proposed test is compared with the approximate Hotelling 2 (AHT) test by the simulations. Simulation results indicate that the PB test performs satisfactorily for various cell sizes and parameter configurations and generally outperforms the AHT test in terms of controlling the nominal size. For the heteroscedastic cases, the PB test outperforms the AHT test in terms of power. In addition, the PB test does not lose too much power when the homogeneity assumption is actually valid. 1. Introduction ere are many situations where we record more than one response variable from each sampling or experimental unit and where these units are allocated to or occur in treatment groups. Ecologists oſten record the abundances of many species from each sampling or experimental unit and physiol- ogists commonly measure more than one variable (e.g., blood pressure, heart rate, etc.) on experimental animals. With multiple response variables, we might be more interested in whether there are group differences on all the response variables considered simultaneously. is is the aim of mul- tivariate analysis of variance (MANOVA), the analogue of univariate ANOVA when we have multiple response variables for each experimental or sampling unit. e simplest design is single factor MANOVA (or one- way MANOVA) which tests the effect of a factor, having levels. If the cell covariance matrices are assumed to be equal, then there are some popular tests available to test the equality of the mean vectors. e tests that are commonly used are the Wilks likelihood ratio (WLR), Lawley-Hotelling trace (LHT), Bartlett-Nanda-Pillai (BNP), and Roys largest root tests ([1], chap. 8, sec. 6). When there are some departures from the standard assumption, that is, unequal cell covari- ance matrices, these solutions were proposed by James [2], Johansen [3], Gamage et al. [4], and Krishnamoorthy and Lu [5], among others. e more complex design is multifactor MANOVA especially when the homogeneity of the cell covariance matrices assumption is seriously violated. ere has been a continuous interest in the heteroscedastic two- factor MANOVA which tests in checking the significance of the effects of two factors and , each having and levels, respectively, in a multivariate factorial experiment with crossed designs. Recently, Harrar and Bathke [6] attacked this problem via modifying the WLR, LHT, and BNP tests. eir main ideas focus on modifying the degrees of freedom of the random matrices involved in the test statistics so that the heteroscedasticity of the cell covariance matrices is taken into account and the WLR, LHT, and BNP tests can still be used but with the degrees of freedom estimated from the data via matching the first two moments; see some details in Section 2.2 of Zhang and Xiao [7]. Zhang [8] proposed an approximate Hotelling 2 (AHT) test. A Wald-type test statistic is used. Its null distribution is approximated by a Hotelling 2 -distribution with one parameter estimated from the data. Some simulation studies conducted in Zhang [8] showed that the AHT test outperforms the modified LHT test of Harrar and Bathke [6]. Another useful two-factor design is the nested MANOVA [9]. Notice that each of the testing problems associated with Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 649202, 10 pages http://dx.doi.org/10.1155/2014/649202

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Research ArticleMANOVA for Nested Designs with Unequal Cell Sizes andUnequal Cell Covariance Matrices

Li-Wen Xu

College of Sciences North China University of Technology Beijing 100144 China

Correspondence should be addressed to Li-Wen Xu xulw163163com

Received 7 May 2014 Accepted 3 July 2014 Published 16 July 2014

Academic Editor William Zhu

Copyright copy 2014 Li-Wen Xu This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose and study parametric bootstrap (PB) tests for heteroscedastic two-factor MANOVA with nested designs For theproblem of testing ldquomain effectsrdquo of both factors we develop a flexible test based on a parametric bootstrap approach The PBtest is shown to be invariant under affine-transformations Moreover the PB test does not depend on the chosen weights used todefine the parameters uniquely The proposed test is compared with the approximate Hotelling 1198792 (AHT) test by the simulationsSimulation results indicate that the PB test performs satisfactorily for various cell sizes and parameter configurations and generallyoutperforms the AHT test in terms of controlling the nominal size For the heteroscedastic cases the PB test outperforms the AHTtest in terms of power In addition the PB test does not lose too much power when the homogeneity assumption is actually valid

1 Introduction

There are many situations where we record more than oneresponse variable from each sampling or experimental unitand where these units are allocated to or occur in treatmentgroups Ecologists often record the abundances of manyspecies from each sampling or experimental unit and physiol-ogists commonlymeasuremore than one variable (eg bloodpressure heart rate etc) on experimental animals Withmultiple response variables we might be more interestedin whether there are group differences on all the responsevariables considered simultaneously This is the aim of mul-tivariate analysis of variance (MANOVA) the analogue ofunivariate ANOVAwhenwe havemultiple response variablesfor each experimental or sampling unit

The simplest design is single factor MANOVA (or one-way MANOVA) which tests the effect of a factor having 119896levels If the cell covariance matrices are assumed to be equalthen there are some popular tests available to test the equalityof the mean vectors The tests that are commonly used arethe Wilks likelihood ratio (WLR) Lawley-Hotelling trace(LHT) Bartlett-Nanda-Pillai (BNP) and Roys largest roottests ([1] chap 8 sec 6) When there are some departuresfrom the standard assumption that is unequal cell covari-ance matrices these solutions were proposed by James [2]

Johansen [3] Gamage et al [4] and Krishnamoorthy and Lu[5] among others The more complex design is multifactorMANOVA especially when the homogeneity of the cellcovariance matrices assumption is seriously violated Therehas been a continuous interest in the heteroscedastic two-factor MANOVA which tests in checking the significanceof the effects of two factors 119860 and 119861 each having 119886 and 119887

levels respectively in amultivariate factorial experimentwithcrossed designs Recently Harrar and Bathke [6] attackedthis problem via modifying the WLR LHT and BNP testsTheir main ideas focus on modifying the degrees of freedomof the random matrices involved in the test statistics so thatthe heteroscedasticity of the cell covariance matrices is takeninto account and the WLR LHT and BNP tests can still beused but with the degrees of freedom estimated from thedata via matching the first two moments see some detailsin Section 22 of Zhang and Xiao [7] Zhang [8] proposedan approximate Hotelling 1198792 (AHT) test A Wald-type teststatistic is used Its null distribution is approximated by aHotelling1198792-distributionwith one parameter estimated fromthe data Some simulation studies conducted in Zhang [8]showed that the AHT test outperforms themodified LHT testof Harrar and Bathke [6]

Another useful two-factor design is the nestedMANOVA[9] Notice that each of the testing problems associated with

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 649202 10 pageshttpdxdoiorg1011552014649202

2 Journal of Applied Mathematics

the three null hypotheses in the heteroscedastic two-wayMANOVA can be equivalently expressed in the form ofthe general linear hypothesis testing (GLHT) problem asdescribed in Zhang [8] The GLHT problem is very generalIt includes not only the main and interaction effect tests butalso various post hoc and contrast tests as special cases Toconstruct the test for two-factor nested MANOVA we mayuse the sameWald-type test statistic as inZhang [8]Howeveraccording to our simulation studies the empirical sizes of theAHT test for the two-factor nested MANOVA model mayfar exceed the nominal level On the other hand the AHTtest needs to consider the the selected weights when the cellsizes are unequal Therefore it is important to develop a testprocedure for the nesting effects and the nested effects withsatisfactory size and power regardless of number of factorialeffects and the sample sizes

Accordingly the present paper will develop a parametricbootstrap (PB) test for heteroscedastic two-factor nestedMANOVA We use standardized effects sum of squares anda natural test statistic obtained by replacing cell covariancematrices by the corresponding sample cell covariance matri-ces The PB test admits several nice properties (1) it can besimply conducted by a routine Monte Carlo algorithm (2)it is shown to be invariant under affine-transformations (3)The PB test does not depend on choices of the weights usedto identify the parameters and (4) it works well Simulationresults reported in Section 4 indicate that the PB test per-forms satisfactorily for various cell sizes and parameter con-figurations when the homogeneity assumption is seriouslyviolated and generally outperforms the AHT test in terms ofpower and controlling size The main ideas of the proposedPB test are closely related to the work by Xu et al [10] Themethodologies for the PB test are presented in Sections 2and 3 Simulation results are presented in Section 4 Someconcluding remarks are given in Section 5 Finally sometechnical proofs of themain results are given in theAppendix

2 Tests for Nested Effects

Consider a two-factor nested designmodelwith factors119860 and119861 Suppose that the nesting effect corresponds to the factor119860 having 119886 factor levels and the nested effect correspondsto the factor 119861 having a total of 119887 = sum

119886

119894=1119887119894levels with 119887

119894

levels of 119861 nested within the 119894th level of the factor 119860 (119894 =

1 119886) Suppose a 119903-variate random sample of size 119899119894119895is

available under (119894 119895)th level 119894 = 1 119886 119895 = 1 119887119894 Let

Y119894119895119896 119894 = 1 119886 119895 = 1 119887

119894 119896 = 1 119899

119894119895represent these

random vectors and y119894119895119896

represent their observed (sample)values Assume that 119899

119894119895gt 119903 so that positive definite sample

covariance matrices can be computed for each cell of thedesign Suppose that Y

119894119895119896satisfy the following model

Y119894119895119896= 120583119894119895+ e119894119895119896 e119894119895119896sim 119873119903(0Σ119894119895) 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(1)

where 120583119894119895 119903 times 1 and Σ

119894119895 119903 times 119903 are the cell mean vector and

cell covariance matrix of the random sample at the (119894 119895)thcell and e

119894119895119896is an experimental error vector term All these

samples are assumed to be independent with each other Inthe two-factor nested model the cell mean vectors 120583

119894119895are

usually decomposed into the form

120583119894119895= 1205830+ 120572119894+ 120573119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 (2)

where 1205830is the grand mean vector 120572

119894is the effect vector of

the 119894th level of 119860 on each of the 119903 variables in Y119894119895119896 and 120573

119894119895is

the effect due to the 119895th level of the factor 119861 nested within the119894th level of the factor119860 so that (1) can be further written as thefollowing two-factor multivariate nested model with unequalerror covariance matrices

Y119894119895119896= 1205830+ 120572119894+ 120573119894119895+ e119894119895119896 e119894119895119896sim 119873119903(0Σ119894119895)

119894 = 1 119886 119895 = 1 119887119894 119896 = 1 119899

119894119895

(3)

In order to have 1205830 120572119894 and 120573

119894119895uniquely defined we need to

have additional constraints Let 1199061 119906

119886and V1198941 V

119894119887119894be

nonnegative weights such that sum119886119894=1

119906119894gt 0 and sum119887119894

119895=1V119894119895gt 0

We consider the following constraints

119886

sum

119894=1

119906119894120572119894= 0

119887119894

sum

119895=1

V119894119895120573119894119895= 0 (4)

where 1199061 119906

119886and V119894119895 V

119894119887119894are nonnegative weights such

that sum119886119894=1

119906119894gt 0 and sum119887119894

119895=1V119894119895gt 0 for each 119894

In this section we are interested in testing the nullhypothesis

1198670120573 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (5)

against its natural alternative hypothesis The null hypothesisin (5) aims to test if the nested effect corresponding to thefactor 119861 is statistically significant

The sample mean vector and the sample covariancematrix of the (119894 119895)th cell are denoted by Y

119894119895and S

119894119895 respec-

tively where

Y119894119895=

1

119899119894119895

119899119894119895

sum

119896=1

Y119894119895119896

S119894119895=

1

119899119894119895minus 1

119899119894119895

sum

119896=1

(Y119894119895119896minus Y119894119895) (Y119894119895119896minus Y119894119895)119879

(6)

119894 = 1 119886 119895 = 1 119887119894 andM119879 denotes the transpose of the

matrixM The observed values of these random variables aredenoted as y

119894119895and s119894119895 respectively 119894 = 1 119886 119895 = 1 119887

119894

We note that Y119894119895rsquos and S

119894119895rsquos are mutually independent with

(119899119894119895minus 1)S119894119895sim 119882119903(119899119894119895minus 1Σ

119894119895) and the model for Y

119894119895as follows

Y119894119895= 1205830+ 120572119894+ 120573119894119895+ e119894119895 e119894119895sim 119873119903(0

1

119899119894119895

Σ119894119895)

119894 = 1 119886 119895 = 1 119887119894

(7)

where e119894119895

= (1119899119894119895) sum119899119894119895

119896=1e119894119895119896

and 119882119903(119898 Γ) denotes the

119903-dimensional Wishart distribution with degrees of freedom

Journal of Applied Mathematics 3

(df) = 119898 and scale parameter matrix Γ Writing Y = (Y11987911

Y11987911198871 Y119879

1198861 Y119879

119886119887119886)119879 and e = (e119879

11 e119879

11198871 e119879

1198861

e119879119886119887119886)119879 the model (7) can be written as

Y = (1119887otimes I119903)1205830+ (diag (1

1198871 1

119887119886) otimes I119903)120572 + 120573 + e (8)

where 120572 = (120572119879

1 120572

119879

119886)119879 120573 = (120573

119879

11 120573

119879

11198871 120573

119879

1198861

120573119879

119886119887119886)119879 e sim 119873

119887119903(0Σ) Σ = diag((1119899

11)Σ11 (1119899

11198871)Σ11198871

(11198991198861)Σ1198861 (1119899

119886119887119886)Σ119886119887119886) 1119898denotes the 119898 times 1 vec-

tor of ones I119898

is an identity matrix with order 119898 V otimes

W denotes the Kronecker product of matrices V and Wand diag(M

1 M

119898) denotes a block-diagonal matrix with

M1 M

119898along the blocks

Define the standardized sum of squares due to the factor119861

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0minus 119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0minus 119894)

(9)

where 0and

119894are solutions of 120583

0and 120572

119894that minimize the

quadratic equation

119878 (Y11 Y

119886119887119886Σ11 Σ

119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 1205830minus 120572119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 1205830minus 120572119894)

(10)

subject to the constraints given in (4)In fact denoting 120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 and = (

119879

0 119879

1

119879

119886)119879 it follows from Theorem 525 in Wang and Chow

[11] that

= (119879

0 119879

1

119879

119886)119879

= (X119879Σminus1X + L119879L)minus1

X119879Σminus1Y (11)

where X = (1119887 diag(1

1198871 1

119887119886)) otimes I

119903and L = (0 119906

1

119906119886) otimes I119903 Then

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) = (Y minus X)119879

Σminus1(Y minus X)

(12)

Notice that indeed depends on L of chosen weightsHowever we can prove that the standardized sum of squaresin (9) does not depend on L This is the followingTheorem

Theorem 1 The standardized sum of squares 119878120573(Y11 Y119886119887119886 Σ11 Σ

119886119887119886) does not depend on L of chosen weights and

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(13)

whereMminus denotes any generalized inverse of matrixM

Note fromTheorem 1 that 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886)does not depend on L If Σ

119894119895are known then a natural

statistic for testing (5) is 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886) Infact Σminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ = (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12) (14)

is an idempotent matrix with rank 119903(119887 minus 119886) we have

Y119879Σminus12ΔΣminus12Y sim 1205942

119903(119887minus119886)(120583119879Σminus12ΔΣminus12120583) (15)

where 120583 = (120583119879

11 120583

119879

11198871 120583

119879

1198861 120583

119879

119886119887119886)119879 and 120594

2

119898(120575)

denotes a noncentral chi-square random variable withdegrees of freedom 119898 and noncentrality parameter 120575 Thenoncentrality parameter 120583119879Σminus12ΔΣminus12120583 is equal to zerowhen 120573

11= sdot sdot sdot = 120573

119886119887119886 Let y = (y119879

11 y119879

11198871 y119879

1198861

y119879119886119887119886)119879 be the observed value of Y Then the test that rejects

1198670120573 in (5) whenever

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus119886)120582(16)

is a size 120582 test where 1205942119898120582

is the upper 120582th quantile of a chi-square distribution with 119889119891 = 119898

In general the covariance matrices Σ119894119895are unknown in

this case a test statistic can be obtained by replacing Σ119894119895in

(13) by S119894119895 119894 = 1 119886 119895 = 1 119887

119894and is given by

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y(17)

where S = diag((111989911)S11 (1119899

11198871)S11198871 (1119899

1198861)S1198861

(1119899119886119887119886)S119886119887119886)

As shown in Zhang [8] the testing problem associatedwith the null hypothesis (5) can also be equivalently expressedin the form of the general linear hypothesis testing (GLHT)problem as119867

0120573 C120583 = 0 where C = H120573A120573 otimes I119903

H120573

= (

(I1198871minus1

minus11198871minus1

) 0 0 0

0 (I1198872minus1

minus11198872minus1

) 0 0

0 0 d 0

0 0 0 (I119887119886minus1

minus1119887119886minus1

)

)

A120573 = (

I1198871minus 111988711199061k1198791

0 0 0

0 I1198872minus 111988721199062k1198792

0 0

0 0 d 0

0 0 0 I119887119886minus 1119887119886119906119886k119879119886

)

(18)

and k119894= (V1198941 V

119894119887119894)119879 119894 = 1 2 119886 Then the associated

Wald-type test statistic is given as

119879 = (C)119879(CSC119879)minus1

(C) (19)

where = Y and S is defined as in (17)In the following we shall describe two tests based on (17)

and (19) respectively

4 Journal of Applied Mathematics

21 The Approximate Hotelling 1198792 (AHT) Test Similar to

Zhang [8] we proposed a test referred as the AHT test whichis based on the test statistic

(119889 minus 119902 + 1) 119879

119902119889

(20)

where 119902 = rank(C) = 119903(119887 minus 119886)

119889 =119902 (119902 + 1)

sum119886

119894=1sum119887119894

119895=1(119899119894119895minus 1)minus1

[tr2 (Ω119894119895) + tr (Ω2

119894119895)]

(21)

and Ω119894119895= 119899minus1

119894119895H119894119895S119894119895H119879119894119895and H

119894119895= (CSC119879)minus12C

119894119895with C =

[C11 C

11198871 C

1198861 C

119886119887119886] and C

119894119895 119902 times 119903 is the (119894119895)th

block matrix of C Zhang [8] showed that under 1198670120573 (119889 minus

119902 + 1)119879119902119889 is approximately distributed as 119865119902119889minus119902+1

randomvariable Thus the AHT test rejects the null hypothesis in (5)when the critical value (119902119889(119889 minus 119902 + 1))119865

119902119889minus119902+1(1 minus 120582) for

the nominal significance level 120582 is exceeded by the observedtest statistic 119879 in (20) The AHT test can also be conductedvia computing the 119875-value based on the approximate nulldistribution

22TheParametric Bootstrap (PB)Test Theparametric boot-strap involves sampling from the estimated models That issamples or sample statistics are generated from parametricmodels with the parameters replaced by their estimatesRecall that under119867

0120573 the vector Y have the mean X120579 where120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 As the test statistic in (17) is location

invariant under the group of location transformations G =

Y rarr Y + X120578 120578 isin R119903(119886+1) without loss of generality wecan take X120579 = 0 Using these facts the parametric boot-strap pivot variable can be developed as follows For a given(y11 y

119886119887119886 s11 s

119886119887119886) let Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and

S119861119894119895

sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Then the PB pivot variable based on the test statistic (17) isgiven by

119878120573B (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

(22)

where Y119861

= (Y11987911986111 Y119879

11986111198871 Y1198791198611198861

Y119879119861119886119887119886

)

119879

andS119861= diag((1119899

11)S11986111 (1119899

11198871)S1198611119887 (1119899

1198861)S1198611198861

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects 1198670120573 in

(5) when

119875 (119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573) lt 120582 (23)

where

119904120573 = 119878120573 (y11 y119886119887119886 s11 s119886119887119886) (24)

is an observed value of 119878120573(Y11 Y119886119887119886 S11 S119886119887119886) in (17)For fixed (y

11 y

119886119887119886s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so it canbe estimated using Monte Carlo simulation given in the fol-lowing Algorithm 2

Algorithm 2 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120573(y11 y119886119887119886 s11 s119886119887119886) in (17) and callit 119904120573For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Compute 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(22)If 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (23)

23 Some Desirable Properties of the PB Test The PB test (23)has several desirable invariance properties First of all thePB test is affine-invariant That is it is invariant under thefollowing affine-transformation

Ylowast119894119895119896= DY

119894119895119896+ 120585 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(25)

where D is any nonsingular matrix and 120585 is any given vectorThis property is desirable since in practice the cell responsesY119894119895119896

(1) are often recentered or rescaled before an inferenceis conducted The recentering and rescaling transformationsare special cases of (25)

Theorem 3 The PB test (23) is affine-invariant in the sensethat both the observed test statistic 119904120573 (24) and the distributionof PB pivot variable 119878120573B (22) are affine-invariant

Themodel (3) is identifiable when the constraints (4) areimposed In many situations there are no natural weightsto justify a particular test procedure For this we have thefollowing result

Theorem 4 The PB test (23) is invariant on the differentweight choices in constraints (4)

Notice that the test statistic (17) does not depend on thechosen weights The conclusion in Theorem 4 is obviousHowever we have the following property

Theorem 5 The test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(26)

Journal of Applied Mathematics 5

This property is desirable since in practice in order to avoidcomputation of generalized inverse of matrices and enhance theaccuracy of computation we have taken a particular L such asL = (0 1119886 1119886) otimes I

119903in the simulation study

3 Tests for the Nesting Effects

When the nested effects are present the nesting effect 120572119894can

not reflect the effect of 119860119894because it depends on which level

of factor 119861 it is in As pointed in Searle [12] Chap 7 in thecase of 119903 = 1 a popular solution to the problem is not quite atest for 120572

119894= 0 in the presence of nested effects but rather to

test the null hypothesis

1198670120572|120573 120572119894 + 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (27)

Wewill see that the problem of testing (27) is the same case ofthe problem of comparing 119887 = sum

119886

119894=1119887119894normal mean vectors

when the cell covariancematrices are unknown and arbitrary[5] In fact if Σ

119894119895are known then

0= (

119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895)

minus1119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895Y119894119895

(28)

is the best linear unbiased estimator of 1205830 and a natural

statistic for testing (27) is

119878120572 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0)

= Y119879Σminus12 (I119887119903minus Σminus12X1(X1198791Σminus1X1)minus

X1198791Σminus12

)Σminus12Y

(29)

whereX1= 1119887otimesI119903 Notice thatΣminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ1= (I119887119903minus Σminus12X1times (X1198791Σminus1X1)minusX1198791Σminus12

) is an idempotentmatrix with rank 119903(119887 minus 1) we have

Y119879Σminus12Δ1Σminus12Y sim 120594

2

119903(119887minus1)(120583119879Σminus12Δ1Σminus12120583) (30)

Thenoncentrality parameter1205831015840Σminus12Δ1Σminus12120583 is equal to zero

when 120572119894+120573119894119895= 0 119894 = 1 119886 119895 = 1 119887

119894Then the test that

rejects1198670120572|120573 in (27) whenever

119878120572 (y11 y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus1)120582(31)

is a size 120582 test In general the covariance matrices Σ119894119895are

unknown in this case a test statistic can be obtained byreplacing Σ

119894119895in (29) by S

119894119895 119894 = 1 119886 119895 = 1 119887

119894and is

given by

119878120572 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X

1(X1198791Sminus1X1)minus

X1198791Sminus12) Sminus12Y

(32)

In the following we describe the PB test for1198670120572|120573 in (27)

Recall that under 1198670120572|120573 the vector Y have the mean X

11205830

As the test statistic in (32) is location invariant under thegroup of location transformations G

1= Y + X

1120578 120578 isin R119903

without loss of generality we can take X11205830= 0 Using

these facts the parametric bootstrap pivot variable can bedeveloped as follows For a given (y

11 y

119886119887119886 s11 s

119886119887119886)

let Y119861119894119895

sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus

1))s119894119895) 119894 = 1 119886 119895 = 1 119887

119894 Then the PB pivot variable

based on the test statistic (32) is given by

119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X1(X1198791Sminus1119861X)minus

X1198791Sminus12119861

) Sminus12119861

Y119861

(33)

where Y119861= (Y11987911986111 Y119879

119861119886119887119886)119879 and S

119861= diag((1119899

11)S11986111

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects1198670120572|120573

in (27) when

119875 (119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572) lt 120582 (34)

where

119904120572 = 119878120572 (y11 y119886119887119886 s11 s119886119887119886) (35)

is an observed value of 119878120572(Y11 Y119886119887119886 S11 S119886119887119886) in (32)For fixed (y

11 y

119886119887119886 s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so itcan be estimated using Monte Carlo simulation given in thefollowing Algorithm 6

Algorithm 6 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120572(y11 y119886119887119886 s11 s119886119887119886) in (32) and callit 119904120572For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895)

119894 = 1 119886 119895 = 1 119887119894

Compute 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(33)If 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (34)

Note that the PB test provided in this section has the sameinvariance properties as in Section 2

4 Monte Carlo Studies

In this section we need to compare the PB test with theAHT test via comparing their empirical sizes (Type I error

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

Page 2: Research Article MANOVA for Nested Designs with Unequal ...

2 Journal of Applied Mathematics

the three null hypotheses in the heteroscedastic two-wayMANOVA can be equivalently expressed in the form ofthe general linear hypothesis testing (GLHT) problem asdescribed in Zhang [8] The GLHT problem is very generalIt includes not only the main and interaction effect tests butalso various post hoc and contrast tests as special cases Toconstruct the test for two-factor nested MANOVA we mayuse the sameWald-type test statistic as inZhang [8]Howeveraccording to our simulation studies the empirical sizes of theAHT test for the two-factor nested MANOVA model mayfar exceed the nominal level On the other hand the AHTtest needs to consider the the selected weights when the cellsizes are unequal Therefore it is important to develop a testprocedure for the nesting effects and the nested effects withsatisfactory size and power regardless of number of factorialeffects and the sample sizes

Accordingly the present paper will develop a parametricbootstrap (PB) test for heteroscedastic two-factor nestedMANOVA We use standardized effects sum of squares anda natural test statistic obtained by replacing cell covariancematrices by the corresponding sample cell covariance matri-ces The PB test admits several nice properties (1) it can besimply conducted by a routine Monte Carlo algorithm (2)it is shown to be invariant under affine-transformations (3)The PB test does not depend on choices of the weights usedto identify the parameters and (4) it works well Simulationresults reported in Section 4 indicate that the PB test per-forms satisfactorily for various cell sizes and parameter con-figurations when the homogeneity assumption is seriouslyviolated and generally outperforms the AHT test in terms ofpower and controlling size The main ideas of the proposedPB test are closely related to the work by Xu et al [10] Themethodologies for the PB test are presented in Sections 2and 3 Simulation results are presented in Section 4 Someconcluding remarks are given in Section 5 Finally sometechnical proofs of themain results are given in theAppendix

2 Tests for Nested Effects

Consider a two-factor nested designmodelwith factors119860 and119861 Suppose that the nesting effect corresponds to the factor119860 having 119886 factor levels and the nested effect correspondsto the factor 119861 having a total of 119887 = sum

119886

119894=1119887119894levels with 119887

119894

levels of 119861 nested within the 119894th level of the factor 119860 (119894 =

1 119886) Suppose a 119903-variate random sample of size 119899119894119895is

available under (119894 119895)th level 119894 = 1 119886 119895 = 1 119887119894 Let

Y119894119895119896 119894 = 1 119886 119895 = 1 119887

119894 119896 = 1 119899

119894119895represent these

random vectors and y119894119895119896

represent their observed (sample)values Assume that 119899

119894119895gt 119903 so that positive definite sample

covariance matrices can be computed for each cell of thedesign Suppose that Y

119894119895119896satisfy the following model

Y119894119895119896= 120583119894119895+ e119894119895119896 e119894119895119896sim 119873119903(0Σ119894119895) 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(1)

where 120583119894119895 119903 times 1 and Σ

119894119895 119903 times 119903 are the cell mean vector and

cell covariance matrix of the random sample at the (119894 119895)thcell and e

119894119895119896is an experimental error vector term All these

samples are assumed to be independent with each other Inthe two-factor nested model the cell mean vectors 120583

119894119895are

usually decomposed into the form

120583119894119895= 1205830+ 120572119894+ 120573119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 (2)

where 1205830is the grand mean vector 120572

119894is the effect vector of

the 119894th level of 119860 on each of the 119903 variables in Y119894119895119896 and 120573

119894119895is

the effect due to the 119895th level of the factor 119861 nested within the119894th level of the factor119860 so that (1) can be further written as thefollowing two-factor multivariate nested model with unequalerror covariance matrices

Y119894119895119896= 1205830+ 120572119894+ 120573119894119895+ e119894119895119896 e119894119895119896sim 119873119903(0Σ119894119895)

119894 = 1 119886 119895 = 1 119887119894 119896 = 1 119899

119894119895

(3)

In order to have 1205830 120572119894 and 120573

119894119895uniquely defined we need to

have additional constraints Let 1199061 119906

119886and V1198941 V

119894119887119894be

nonnegative weights such that sum119886119894=1

119906119894gt 0 and sum119887119894

119895=1V119894119895gt 0

We consider the following constraints

119886

sum

119894=1

119906119894120572119894= 0

119887119894

sum

119895=1

V119894119895120573119894119895= 0 (4)

where 1199061 119906

119886and V119894119895 V

119894119887119894are nonnegative weights such

that sum119886119894=1

119906119894gt 0 and sum119887119894

119895=1V119894119895gt 0 for each 119894

In this section we are interested in testing the nullhypothesis

1198670120573 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (5)

against its natural alternative hypothesis The null hypothesisin (5) aims to test if the nested effect corresponding to thefactor 119861 is statistically significant

The sample mean vector and the sample covariancematrix of the (119894 119895)th cell are denoted by Y

119894119895and S

119894119895 respec-

tively where

Y119894119895=

1

119899119894119895

119899119894119895

sum

119896=1

Y119894119895119896

S119894119895=

1

119899119894119895minus 1

119899119894119895

sum

119896=1

(Y119894119895119896minus Y119894119895) (Y119894119895119896minus Y119894119895)119879

(6)

119894 = 1 119886 119895 = 1 119887119894 andM119879 denotes the transpose of the

matrixM The observed values of these random variables aredenoted as y

119894119895and s119894119895 respectively 119894 = 1 119886 119895 = 1 119887

119894

We note that Y119894119895rsquos and S

119894119895rsquos are mutually independent with

(119899119894119895minus 1)S119894119895sim 119882119903(119899119894119895minus 1Σ

119894119895) and the model for Y

119894119895as follows

Y119894119895= 1205830+ 120572119894+ 120573119894119895+ e119894119895 e119894119895sim 119873119903(0

1

119899119894119895

Σ119894119895)

119894 = 1 119886 119895 = 1 119887119894

(7)

where e119894119895

= (1119899119894119895) sum119899119894119895

119896=1e119894119895119896

and 119882119903(119898 Γ) denotes the

119903-dimensional Wishart distribution with degrees of freedom

Journal of Applied Mathematics 3

(df) = 119898 and scale parameter matrix Γ Writing Y = (Y11987911

Y11987911198871 Y119879

1198861 Y119879

119886119887119886)119879 and e = (e119879

11 e119879

11198871 e119879

1198861

e119879119886119887119886)119879 the model (7) can be written as

Y = (1119887otimes I119903)1205830+ (diag (1

1198871 1

119887119886) otimes I119903)120572 + 120573 + e (8)

where 120572 = (120572119879

1 120572

119879

119886)119879 120573 = (120573

119879

11 120573

119879

11198871 120573

119879

1198861

120573119879

119886119887119886)119879 e sim 119873

119887119903(0Σ) Σ = diag((1119899

11)Σ11 (1119899

11198871)Σ11198871

(11198991198861)Σ1198861 (1119899

119886119887119886)Σ119886119887119886) 1119898denotes the 119898 times 1 vec-

tor of ones I119898

is an identity matrix with order 119898 V otimes

W denotes the Kronecker product of matrices V and Wand diag(M

1 M

119898) denotes a block-diagonal matrix with

M1 M

119898along the blocks

Define the standardized sum of squares due to the factor119861

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0minus 119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0minus 119894)

(9)

where 0and

119894are solutions of 120583

0and 120572

119894that minimize the

quadratic equation

119878 (Y11 Y

119886119887119886Σ11 Σ

119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 1205830minus 120572119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 1205830minus 120572119894)

(10)

subject to the constraints given in (4)In fact denoting 120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 and = (

119879

0 119879

1

119879

119886)119879 it follows from Theorem 525 in Wang and Chow

[11] that

= (119879

0 119879

1

119879

119886)119879

= (X119879Σminus1X + L119879L)minus1

X119879Σminus1Y (11)

where X = (1119887 diag(1

1198871 1

119887119886)) otimes I

119903and L = (0 119906

1

119906119886) otimes I119903 Then

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) = (Y minus X)119879

Σminus1(Y minus X)

(12)

Notice that indeed depends on L of chosen weightsHowever we can prove that the standardized sum of squaresin (9) does not depend on L This is the followingTheorem

Theorem 1 The standardized sum of squares 119878120573(Y11 Y119886119887119886 Σ11 Σ

119886119887119886) does not depend on L of chosen weights and

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(13)

whereMminus denotes any generalized inverse of matrixM

Note fromTheorem 1 that 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886)does not depend on L If Σ

119894119895are known then a natural

statistic for testing (5) is 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886) Infact Σminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ = (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12) (14)

is an idempotent matrix with rank 119903(119887 minus 119886) we have

Y119879Σminus12ΔΣminus12Y sim 1205942

119903(119887minus119886)(120583119879Σminus12ΔΣminus12120583) (15)

where 120583 = (120583119879

11 120583

119879

11198871 120583

119879

1198861 120583

119879

119886119887119886)119879 and 120594

2

119898(120575)

denotes a noncentral chi-square random variable withdegrees of freedom 119898 and noncentrality parameter 120575 Thenoncentrality parameter 120583119879Σminus12ΔΣminus12120583 is equal to zerowhen 120573

11= sdot sdot sdot = 120573

119886119887119886 Let y = (y119879

11 y119879

11198871 y119879

1198861

y119879119886119887119886)119879 be the observed value of Y Then the test that rejects

1198670120573 in (5) whenever

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus119886)120582(16)

is a size 120582 test where 1205942119898120582

is the upper 120582th quantile of a chi-square distribution with 119889119891 = 119898

In general the covariance matrices Σ119894119895are unknown in

this case a test statistic can be obtained by replacing Σ119894119895in

(13) by S119894119895 119894 = 1 119886 119895 = 1 119887

119894and is given by

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y(17)

where S = diag((111989911)S11 (1119899

11198871)S11198871 (1119899

1198861)S1198861

(1119899119886119887119886)S119886119887119886)

As shown in Zhang [8] the testing problem associatedwith the null hypothesis (5) can also be equivalently expressedin the form of the general linear hypothesis testing (GLHT)problem as119867

0120573 C120583 = 0 where C = H120573A120573 otimes I119903

H120573

= (

(I1198871minus1

minus11198871minus1

) 0 0 0

0 (I1198872minus1

minus11198872minus1

) 0 0

0 0 d 0

0 0 0 (I119887119886minus1

minus1119887119886minus1

)

)

A120573 = (

I1198871minus 111988711199061k1198791

0 0 0

0 I1198872minus 111988721199062k1198792

0 0

0 0 d 0

0 0 0 I119887119886minus 1119887119886119906119886k119879119886

)

(18)

and k119894= (V1198941 V

119894119887119894)119879 119894 = 1 2 119886 Then the associated

Wald-type test statistic is given as

119879 = (C)119879(CSC119879)minus1

(C) (19)

where = Y and S is defined as in (17)In the following we shall describe two tests based on (17)

and (19) respectively

4 Journal of Applied Mathematics

21 The Approximate Hotelling 1198792 (AHT) Test Similar to

Zhang [8] we proposed a test referred as the AHT test whichis based on the test statistic

(119889 minus 119902 + 1) 119879

119902119889

(20)

where 119902 = rank(C) = 119903(119887 minus 119886)

119889 =119902 (119902 + 1)

sum119886

119894=1sum119887119894

119895=1(119899119894119895minus 1)minus1

[tr2 (Ω119894119895) + tr (Ω2

119894119895)]

(21)

and Ω119894119895= 119899minus1

119894119895H119894119895S119894119895H119879119894119895and H

119894119895= (CSC119879)minus12C

119894119895with C =

[C11 C

11198871 C

1198861 C

119886119887119886] and C

119894119895 119902 times 119903 is the (119894119895)th

block matrix of C Zhang [8] showed that under 1198670120573 (119889 minus

119902 + 1)119879119902119889 is approximately distributed as 119865119902119889minus119902+1

randomvariable Thus the AHT test rejects the null hypothesis in (5)when the critical value (119902119889(119889 minus 119902 + 1))119865

119902119889minus119902+1(1 minus 120582) for

the nominal significance level 120582 is exceeded by the observedtest statistic 119879 in (20) The AHT test can also be conductedvia computing the 119875-value based on the approximate nulldistribution

22TheParametric Bootstrap (PB)Test Theparametric boot-strap involves sampling from the estimated models That issamples or sample statistics are generated from parametricmodels with the parameters replaced by their estimatesRecall that under119867

0120573 the vector Y have the mean X120579 where120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 As the test statistic in (17) is location

invariant under the group of location transformations G =

Y rarr Y + X120578 120578 isin R119903(119886+1) without loss of generality wecan take X120579 = 0 Using these facts the parametric boot-strap pivot variable can be developed as follows For a given(y11 y

119886119887119886 s11 s

119886119887119886) let Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and

S119861119894119895

sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Then the PB pivot variable based on the test statistic (17) isgiven by

119878120573B (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

(22)

where Y119861

= (Y11987911986111 Y119879

11986111198871 Y1198791198611198861

Y119879119861119886119887119886

)

119879

andS119861= diag((1119899

11)S11986111 (1119899

11198871)S1198611119887 (1119899

1198861)S1198611198861

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects 1198670120573 in

(5) when

119875 (119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573) lt 120582 (23)

where

119904120573 = 119878120573 (y11 y119886119887119886 s11 s119886119887119886) (24)

is an observed value of 119878120573(Y11 Y119886119887119886 S11 S119886119887119886) in (17)For fixed (y

11 y

119886119887119886s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so it canbe estimated using Monte Carlo simulation given in the fol-lowing Algorithm 2

Algorithm 2 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120573(y11 y119886119887119886 s11 s119886119887119886) in (17) and callit 119904120573For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Compute 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(22)If 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (23)

23 Some Desirable Properties of the PB Test The PB test (23)has several desirable invariance properties First of all thePB test is affine-invariant That is it is invariant under thefollowing affine-transformation

Ylowast119894119895119896= DY

119894119895119896+ 120585 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(25)

where D is any nonsingular matrix and 120585 is any given vectorThis property is desirable since in practice the cell responsesY119894119895119896

(1) are often recentered or rescaled before an inferenceis conducted The recentering and rescaling transformationsare special cases of (25)

Theorem 3 The PB test (23) is affine-invariant in the sensethat both the observed test statistic 119904120573 (24) and the distributionof PB pivot variable 119878120573B (22) are affine-invariant

Themodel (3) is identifiable when the constraints (4) areimposed In many situations there are no natural weightsto justify a particular test procedure For this we have thefollowing result

Theorem 4 The PB test (23) is invariant on the differentweight choices in constraints (4)

Notice that the test statistic (17) does not depend on thechosen weights The conclusion in Theorem 4 is obviousHowever we have the following property

Theorem 5 The test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(26)

Journal of Applied Mathematics 5

This property is desirable since in practice in order to avoidcomputation of generalized inverse of matrices and enhance theaccuracy of computation we have taken a particular L such asL = (0 1119886 1119886) otimes I

119903in the simulation study

3 Tests for the Nesting Effects

When the nested effects are present the nesting effect 120572119894can

not reflect the effect of 119860119894because it depends on which level

of factor 119861 it is in As pointed in Searle [12] Chap 7 in thecase of 119903 = 1 a popular solution to the problem is not quite atest for 120572

119894= 0 in the presence of nested effects but rather to

test the null hypothesis

1198670120572|120573 120572119894 + 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (27)

Wewill see that the problem of testing (27) is the same case ofthe problem of comparing 119887 = sum

119886

119894=1119887119894normal mean vectors

when the cell covariancematrices are unknown and arbitrary[5] In fact if Σ

119894119895are known then

0= (

119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895)

minus1119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895Y119894119895

(28)

is the best linear unbiased estimator of 1205830 and a natural

statistic for testing (27) is

119878120572 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0)

= Y119879Σminus12 (I119887119903minus Σminus12X1(X1198791Σminus1X1)minus

X1198791Σminus12

)Σminus12Y

(29)

whereX1= 1119887otimesI119903 Notice thatΣminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ1= (I119887119903minus Σminus12X1times (X1198791Σminus1X1)minusX1198791Σminus12

) is an idempotentmatrix with rank 119903(119887 minus 1) we have

Y119879Σminus12Δ1Σminus12Y sim 120594

2

119903(119887minus1)(120583119879Σminus12Δ1Σminus12120583) (30)

Thenoncentrality parameter1205831015840Σminus12Δ1Σminus12120583 is equal to zero

when 120572119894+120573119894119895= 0 119894 = 1 119886 119895 = 1 119887

119894Then the test that

rejects1198670120572|120573 in (27) whenever

119878120572 (y11 y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus1)120582(31)

is a size 120582 test In general the covariance matrices Σ119894119895are

unknown in this case a test statistic can be obtained byreplacing Σ

119894119895in (29) by S

119894119895 119894 = 1 119886 119895 = 1 119887

119894and is

given by

119878120572 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X

1(X1198791Sminus1X1)minus

X1198791Sminus12) Sminus12Y

(32)

In the following we describe the PB test for1198670120572|120573 in (27)

Recall that under 1198670120572|120573 the vector Y have the mean X

11205830

As the test statistic in (32) is location invariant under thegroup of location transformations G

1= Y + X

1120578 120578 isin R119903

without loss of generality we can take X11205830= 0 Using

these facts the parametric bootstrap pivot variable can bedeveloped as follows For a given (y

11 y

119886119887119886 s11 s

119886119887119886)

let Y119861119894119895

sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus

1))s119894119895) 119894 = 1 119886 119895 = 1 119887

119894 Then the PB pivot variable

based on the test statistic (32) is given by

119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X1(X1198791Sminus1119861X)minus

X1198791Sminus12119861

) Sminus12119861

Y119861

(33)

where Y119861= (Y11987911986111 Y119879

119861119886119887119886)119879 and S

119861= diag((1119899

11)S11986111

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects1198670120572|120573

in (27) when

119875 (119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572) lt 120582 (34)

where

119904120572 = 119878120572 (y11 y119886119887119886 s11 s119886119887119886) (35)

is an observed value of 119878120572(Y11 Y119886119887119886 S11 S119886119887119886) in (32)For fixed (y

11 y

119886119887119886 s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so itcan be estimated using Monte Carlo simulation given in thefollowing Algorithm 6

Algorithm 6 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120572(y11 y119886119887119886 s11 s119886119887119886) in (32) and callit 119904120572For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895)

119894 = 1 119886 119895 = 1 119887119894

Compute 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(33)If 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (34)

Note that the PB test provided in this section has the sameinvariance properties as in Section 2

4 Monte Carlo Studies

In this section we need to compare the PB test with theAHT test via comparing their empirical sizes (Type I error

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article MANOVA for Nested Designs with Unequal ...

Journal of Applied Mathematics 3

(df) = 119898 and scale parameter matrix Γ Writing Y = (Y11987911

Y11987911198871 Y119879

1198861 Y119879

119886119887119886)119879 and e = (e119879

11 e119879

11198871 e119879

1198861

e119879119886119887119886)119879 the model (7) can be written as

Y = (1119887otimes I119903)1205830+ (diag (1

1198871 1

119887119886) otimes I119903)120572 + 120573 + e (8)

where 120572 = (120572119879

1 120572

119879

119886)119879 120573 = (120573

119879

11 120573

119879

11198871 120573

119879

1198861

120573119879

119886119887119886)119879 e sim 119873

119887119903(0Σ) Σ = diag((1119899

11)Σ11 (1119899

11198871)Σ11198871

(11198991198861)Σ1198861 (1119899

119886119887119886)Σ119886119887119886) 1119898denotes the 119898 times 1 vec-

tor of ones I119898

is an identity matrix with order 119898 V otimes

W denotes the Kronecker product of matrices V and Wand diag(M

1 M

119898) denotes a block-diagonal matrix with

M1 M

119898along the blocks

Define the standardized sum of squares due to the factor119861

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0minus 119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0minus 119894)

(9)

where 0and

119894are solutions of 120583

0and 120572

119894that minimize the

quadratic equation

119878 (Y11 Y

119886119887119886Σ11 Σ

119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 1205830minus 120572119894)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 1205830minus 120572119894)

(10)

subject to the constraints given in (4)In fact denoting 120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 and = (

119879

0 119879

1

119879

119886)119879 it follows from Theorem 525 in Wang and Chow

[11] that

= (119879

0 119879

1

119879

119886)119879

= (X119879Σminus1X + L119879L)minus1

X119879Σminus1Y (11)

where X = (1119887 diag(1

1198871 1

119887119886)) otimes I

119903and L = (0 119906

1

119906119886) otimes I119903 Then

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) = (Y minus X)119879

Σminus1(Y minus X)

(12)

Notice that indeed depends on L of chosen weightsHowever we can prove that the standardized sum of squaresin (9) does not depend on L This is the followingTheorem

Theorem 1 The standardized sum of squares 119878120573(Y11 Y119886119887119886 Σ11 Σ

119886119887119886) does not depend on L of chosen weights and

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(13)

whereMminus denotes any generalized inverse of matrixM

Note fromTheorem 1 that 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886)does not depend on L If Σ

119894119895are known then a natural

statistic for testing (5) is 119878120573(Y11 Y119886119887119886 Σ11 Σ119886119887119886) Infact Σminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ = (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12) (14)

is an idempotent matrix with rank 119903(119887 minus 119886) we have

Y119879Σminus12ΔΣminus12Y sim 1205942

119903(119887minus119886)(120583119879Σminus12ΔΣminus12120583) (15)

where 120583 = (120583119879

11 120583

119879

11198871 120583

119879

1198861 120583

119879

119886119887119886)119879 and 120594

2

119898(120575)

denotes a noncentral chi-square random variable withdegrees of freedom 119898 and noncentrality parameter 120575 Thenoncentrality parameter 120583119879Σminus12ΔΣminus12120583 is equal to zerowhen 120573

11= sdot sdot sdot = 120573

119886119887119886 Let y = (y119879

11 y119879

11198871 y119879

1198861

y119879119886119887119886)119879 be the observed value of Y Then the test that rejects

1198670120573 in (5) whenever

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus119886)120582(16)

is a size 120582 test where 1205942119898120582

is the upper 120582th quantile of a chi-square distribution with 119889119891 = 119898

In general the covariance matrices Σ119894119895are unknown in

this case a test statistic can be obtained by replacing Σ119894119895in

(13) by S119894119895 119894 = 1 119886 119895 = 1 119887

119894and is given by

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y(17)

where S = diag((111989911)S11 (1119899

11198871)S11198871 (1119899

1198861)S1198861

(1119899119886119887119886)S119886119887119886)

As shown in Zhang [8] the testing problem associatedwith the null hypothesis (5) can also be equivalently expressedin the form of the general linear hypothesis testing (GLHT)problem as119867

0120573 C120583 = 0 where C = H120573A120573 otimes I119903

H120573

= (

(I1198871minus1

minus11198871minus1

) 0 0 0

0 (I1198872minus1

minus11198872minus1

) 0 0

0 0 d 0

0 0 0 (I119887119886minus1

minus1119887119886minus1

)

)

A120573 = (

I1198871minus 111988711199061k1198791

0 0 0

0 I1198872minus 111988721199062k1198792

0 0

0 0 d 0

0 0 0 I119887119886minus 1119887119886119906119886k119879119886

)

(18)

and k119894= (V1198941 V

119894119887119894)119879 119894 = 1 2 119886 Then the associated

Wald-type test statistic is given as

119879 = (C)119879(CSC119879)minus1

(C) (19)

where = Y and S is defined as in (17)In the following we shall describe two tests based on (17)

and (19) respectively

4 Journal of Applied Mathematics

21 The Approximate Hotelling 1198792 (AHT) Test Similar to

Zhang [8] we proposed a test referred as the AHT test whichis based on the test statistic

(119889 minus 119902 + 1) 119879

119902119889

(20)

where 119902 = rank(C) = 119903(119887 minus 119886)

119889 =119902 (119902 + 1)

sum119886

119894=1sum119887119894

119895=1(119899119894119895minus 1)minus1

[tr2 (Ω119894119895) + tr (Ω2

119894119895)]

(21)

and Ω119894119895= 119899minus1

119894119895H119894119895S119894119895H119879119894119895and H

119894119895= (CSC119879)minus12C

119894119895with C =

[C11 C

11198871 C

1198861 C

119886119887119886] and C

119894119895 119902 times 119903 is the (119894119895)th

block matrix of C Zhang [8] showed that under 1198670120573 (119889 minus

119902 + 1)119879119902119889 is approximately distributed as 119865119902119889minus119902+1

randomvariable Thus the AHT test rejects the null hypothesis in (5)when the critical value (119902119889(119889 minus 119902 + 1))119865

119902119889minus119902+1(1 minus 120582) for

the nominal significance level 120582 is exceeded by the observedtest statistic 119879 in (20) The AHT test can also be conductedvia computing the 119875-value based on the approximate nulldistribution

22TheParametric Bootstrap (PB)Test Theparametric boot-strap involves sampling from the estimated models That issamples or sample statistics are generated from parametricmodels with the parameters replaced by their estimatesRecall that under119867

0120573 the vector Y have the mean X120579 where120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 As the test statistic in (17) is location

invariant under the group of location transformations G =

Y rarr Y + X120578 120578 isin R119903(119886+1) without loss of generality wecan take X120579 = 0 Using these facts the parametric boot-strap pivot variable can be developed as follows For a given(y11 y

119886119887119886 s11 s

119886119887119886) let Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and

S119861119894119895

sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Then the PB pivot variable based on the test statistic (17) isgiven by

119878120573B (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

(22)

where Y119861

= (Y11987911986111 Y119879

11986111198871 Y1198791198611198861

Y119879119861119886119887119886

)

119879

andS119861= diag((1119899

11)S11986111 (1119899

11198871)S1198611119887 (1119899

1198861)S1198611198861

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects 1198670120573 in

(5) when

119875 (119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573) lt 120582 (23)

where

119904120573 = 119878120573 (y11 y119886119887119886 s11 s119886119887119886) (24)

is an observed value of 119878120573(Y11 Y119886119887119886 S11 S119886119887119886) in (17)For fixed (y

11 y

119886119887119886s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so it canbe estimated using Monte Carlo simulation given in the fol-lowing Algorithm 2

Algorithm 2 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120573(y11 y119886119887119886 s11 s119886119887119886) in (17) and callit 119904120573For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Compute 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(22)If 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (23)

23 Some Desirable Properties of the PB Test The PB test (23)has several desirable invariance properties First of all thePB test is affine-invariant That is it is invariant under thefollowing affine-transformation

Ylowast119894119895119896= DY

119894119895119896+ 120585 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(25)

where D is any nonsingular matrix and 120585 is any given vectorThis property is desirable since in practice the cell responsesY119894119895119896

(1) are often recentered or rescaled before an inferenceis conducted The recentering and rescaling transformationsare special cases of (25)

Theorem 3 The PB test (23) is affine-invariant in the sensethat both the observed test statistic 119904120573 (24) and the distributionof PB pivot variable 119878120573B (22) are affine-invariant

Themodel (3) is identifiable when the constraints (4) areimposed In many situations there are no natural weightsto justify a particular test procedure For this we have thefollowing result

Theorem 4 The PB test (23) is invariant on the differentweight choices in constraints (4)

Notice that the test statistic (17) does not depend on thechosen weights The conclusion in Theorem 4 is obviousHowever we have the following property

Theorem 5 The test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(26)

Journal of Applied Mathematics 5

This property is desirable since in practice in order to avoidcomputation of generalized inverse of matrices and enhance theaccuracy of computation we have taken a particular L such asL = (0 1119886 1119886) otimes I

119903in the simulation study

3 Tests for the Nesting Effects

When the nested effects are present the nesting effect 120572119894can

not reflect the effect of 119860119894because it depends on which level

of factor 119861 it is in As pointed in Searle [12] Chap 7 in thecase of 119903 = 1 a popular solution to the problem is not quite atest for 120572

119894= 0 in the presence of nested effects but rather to

test the null hypothesis

1198670120572|120573 120572119894 + 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (27)

Wewill see that the problem of testing (27) is the same case ofthe problem of comparing 119887 = sum

119886

119894=1119887119894normal mean vectors

when the cell covariancematrices are unknown and arbitrary[5] In fact if Σ

119894119895are known then

0= (

119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895)

minus1119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895Y119894119895

(28)

is the best linear unbiased estimator of 1205830 and a natural

statistic for testing (27) is

119878120572 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0)

= Y119879Σminus12 (I119887119903minus Σminus12X1(X1198791Σminus1X1)minus

X1198791Σminus12

)Σminus12Y

(29)

whereX1= 1119887otimesI119903 Notice thatΣminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ1= (I119887119903minus Σminus12X1times (X1198791Σminus1X1)minusX1198791Σminus12

) is an idempotentmatrix with rank 119903(119887 minus 1) we have

Y119879Σminus12Δ1Σminus12Y sim 120594

2

119903(119887minus1)(120583119879Σminus12Δ1Σminus12120583) (30)

Thenoncentrality parameter1205831015840Σminus12Δ1Σminus12120583 is equal to zero

when 120572119894+120573119894119895= 0 119894 = 1 119886 119895 = 1 119887

119894Then the test that

rejects1198670120572|120573 in (27) whenever

119878120572 (y11 y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus1)120582(31)

is a size 120582 test In general the covariance matrices Σ119894119895are

unknown in this case a test statistic can be obtained byreplacing Σ

119894119895in (29) by S

119894119895 119894 = 1 119886 119895 = 1 119887

119894and is

given by

119878120572 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X

1(X1198791Sminus1X1)minus

X1198791Sminus12) Sminus12Y

(32)

In the following we describe the PB test for1198670120572|120573 in (27)

Recall that under 1198670120572|120573 the vector Y have the mean X

11205830

As the test statistic in (32) is location invariant under thegroup of location transformations G

1= Y + X

1120578 120578 isin R119903

without loss of generality we can take X11205830= 0 Using

these facts the parametric bootstrap pivot variable can bedeveloped as follows For a given (y

11 y

119886119887119886 s11 s

119886119887119886)

let Y119861119894119895

sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus

1))s119894119895) 119894 = 1 119886 119895 = 1 119887

119894 Then the PB pivot variable

based on the test statistic (32) is given by

119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X1(X1198791Sminus1119861X)minus

X1198791Sminus12119861

) Sminus12119861

Y119861

(33)

where Y119861= (Y11987911986111 Y119879

119861119886119887119886)119879 and S

119861= diag((1119899

11)S11986111

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects1198670120572|120573

in (27) when

119875 (119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572) lt 120582 (34)

where

119904120572 = 119878120572 (y11 y119886119887119886 s11 s119886119887119886) (35)

is an observed value of 119878120572(Y11 Y119886119887119886 S11 S119886119887119886) in (32)For fixed (y

11 y

119886119887119886 s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so itcan be estimated using Monte Carlo simulation given in thefollowing Algorithm 6

Algorithm 6 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120572(y11 y119886119887119886 s11 s119886119887119886) in (32) and callit 119904120572For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895)

119894 = 1 119886 119895 = 1 119887119894

Compute 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(33)If 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (34)

Note that the PB test provided in this section has the sameinvariance properties as in Section 2

4 Monte Carlo Studies

In this section we need to compare the PB test with theAHT test via comparing their empirical sizes (Type I error

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Stochastic AnalysisInternational Journal of

Page 4: Research Article MANOVA for Nested Designs with Unequal ...

4 Journal of Applied Mathematics

21 The Approximate Hotelling 1198792 (AHT) Test Similar to

Zhang [8] we proposed a test referred as the AHT test whichis based on the test statistic

(119889 minus 119902 + 1) 119879

119902119889

(20)

where 119902 = rank(C) = 119903(119887 minus 119886)

119889 =119902 (119902 + 1)

sum119886

119894=1sum119887119894

119895=1(119899119894119895minus 1)minus1

[tr2 (Ω119894119895) + tr (Ω2

119894119895)]

(21)

and Ω119894119895= 119899minus1

119894119895H119894119895S119894119895H119879119894119895and H

119894119895= (CSC119879)minus12C

119894119895with C =

[C11 C

11198871 C

1198861 C

119886119887119886] and C

119894119895 119902 times 119903 is the (119894119895)th

block matrix of C Zhang [8] showed that under 1198670120573 (119889 minus

119902 + 1)119879119902119889 is approximately distributed as 119865119902119889minus119902+1

randomvariable Thus the AHT test rejects the null hypothesis in (5)when the critical value (119902119889(119889 minus 119902 + 1))119865

119902119889minus119902+1(1 minus 120582) for

the nominal significance level 120582 is exceeded by the observedtest statistic 119879 in (20) The AHT test can also be conductedvia computing the 119875-value based on the approximate nulldistribution

22TheParametric Bootstrap (PB)Test Theparametric boot-strap involves sampling from the estimated models That issamples or sample statistics are generated from parametricmodels with the parameters replaced by their estimatesRecall that under119867

0120573 the vector Y have the mean X120579 where120579 = (120583

119879

0120572119879

1 120572

119879

119886)119879 As the test statistic in (17) is location

invariant under the group of location transformations G =

Y rarr Y + X120578 120578 isin R119903(119886+1) without loss of generality wecan take X120579 = 0 Using these facts the parametric boot-strap pivot variable can be developed as follows For a given(y11 y

119886119887119886 s11 s

119886119887119886) let Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and

S119861119894119895

sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Then the PB pivot variable based on the test statistic (17) isgiven by

119878120573B (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

(22)

where Y119861

= (Y11987911986111 Y119879

11986111198871 Y1198791198611198861

Y119879119861119886119887119886

)

119879

andS119861= diag((1119899

11)S11986111 (1119899

11198871)S1198611119887 (1119899

1198861)S1198611198861

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects 1198670120573 in

(5) when

119875 (119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573) lt 120582 (23)

where

119904120573 = 119878120573 (y11 y119886119887119886 s11 s119886119887119886) (24)

is an observed value of 119878120573(Y11 Y119886119887119886 S11 S119886119887119886) in (17)For fixed (y

11 y

119886119887119886s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so it canbe estimated using Monte Carlo simulation given in the fol-lowing Algorithm 2

Algorithm 2 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120573(y11 y119886119887119886 s11 s119886119887119886) in (17) and callit 119904120573For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895) 119894 = 1 119886 119895 = 1 119887

119894

Compute 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(22)If 119878120573119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120573 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (23)

23 Some Desirable Properties of the PB Test The PB test (23)has several desirable invariance properties First of all thePB test is affine-invariant That is it is invariant under thefollowing affine-transformation

Ylowast119894119895119896= DY

119894119895119896+ 120585 119894 = 1 119886

119895 = 1 119887119894 119896 = 1 119899

119894119895

(25)

where D is any nonsingular matrix and 120585 is any given vectorThis property is desirable since in practice the cell responsesY119894119895119896

(1) are often recentered or rescaled before an inferenceis conducted The recentering and rescaling transformationsare special cases of (25)

Theorem 3 The PB test (23) is affine-invariant in the sensethat both the observed test statistic 119904120573 (24) and the distributionof PB pivot variable 119878120573B (22) are affine-invariant

Themodel (3) is identifiable when the constraints (4) areimposed In many situations there are no natural weightsto justify a particular test procedure For this we have thefollowing result

Theorem 4 The PB test (23) is invariant on the differentweight choices in constraints (4)

Notice that the test statistic (17) does not depend on thechosen weights The conclusion in Theorem 4 is obviousHowever we have the following property

Theorem 5 The test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(26)

Journal of Applied Mathematics 5

This property is desirable since in practice in order to avoidcomputation of generalized inverse of matrices and enhance theaccuracy of computation we have taken a particular L such asL = (0 1119886 1119886) otimes I

119903in the simulation study

3 Tests for the Nesting Effects

When the nested effects are present the nesting effect 120572119894can

not reflect the effect of 119860119894because it depends on which level

of factor 119861 it is in As pointed in Searle [12] Chap 7 in thecase of 119903 = 1 a popular solution to the problem is not quite atest for 120572

119894= 0 in the presence of nested effects but rather to

test the null hypothesis

1198670120572|120573 120572119894 + 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (27)

Wewill see that the problem of testing (27) is the same case ofthe problem of comparing 119887 = sum

119886

119894=1119887119894normal mean vectors

when the cell covariancematrices are unknown and arbitrary[5] In fact if Σ

119894119895are known then

0= (

119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895)

minus1119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895Y119894119895

(28)

is the best linear unbiased estimator of 1205830 and a natural

statistic for testing (27) is

119878120572 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0)

= Y119879Σminus12 (I119887119903minus Σminus12X1(X1198791Σminus1X1)minus

X1198791Σminus12

)Σminus12Y

(29)

whereX1= 1119887otimesI119903 Notice thatΣminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ1= (I119887119903minus Σminus12X1times (X1198791Σminus1X1)minusX1198791Σminus12

) is an idempotentmatrix with rank 119903(119887 minus 1) we have

Y119879Σminus12Δ1Σminus12Y sim 120594

2

119903(119887minus1)(120583119879Σminus12Δ1Σminus12120583) (30)

Thenoncentrality parameter1205831015840Σminus12Δ1Σminus12120583 is equal to zero

when 120572119894+120573119894119895= 0 119894 = 1 119886 119895 = 1 119887

119894Then the test that

rejects1198670120572|120573 in (27) whenever

119878120572 (y11 y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus1)120582(31)

is a size 120582 test In general the covariance matrices Σ119894119895are

unknown in this case a test statistic can be obtained byreplacing Σ

119894119895in (29) by S

119894119895 119894 = 1 119886 119895 = 1 119887

119894and is

given by

119878120572 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X

1(X1198791Sminus1X1)minus

X1198791Sminus12) Sminus12Y

(32)

In the following we describe the PB test for1198670120572|120573 in (27)

Recall that under 1198670120572|120573 the vector Y have the mean X

11205830

As the test statistic in (32) is location invariant under thegroup of location transformations G

1= Y + X

1120578 120578 isin R119903

without loss of generality we can take X11205830= 0 Using

these facts the parametric bootstrap pivot variable can bedeveloped as follows For a given (y

11 y

119886119887119886 s11 s

119886119887119886)

let Y119861119894119895

sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus

1))s119894119895) 119894 = 1 119886 119895 = 1 119887

119894 Then the PB pivot variable

based on the test statistic (32) is given by

119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X1(X1198791Sminus1119861X)minus

X1198791Sminus12119861

) Sminus12119861

Y119861

(33)

where Y119861= (Y11987911986111 Y119879

119861119886119887119886)119879 and S

119861= diag((1119899

11)S11986111

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects1198670120572|120573

in (27) when

119875 (119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572) lt 120582 (34)

where

119904120572 = 119878120572 (y11 y119886119887119886 s11 s119886119887119886) (35)

is an observed value of 119878120572(Y11 Y119886119887119886 S11 S119886119887119886) in (32)For fixed (y

11 y

119886119887119886 s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so itcan be estimated using Monte Carlo simulation given in thefollowing Algorithm 6

Algorithm 6 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120572(y11 y119886119887119886 s11 s119886119887119886) in (32) and callit 119904120572For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895)

119894 = 1 119886 119895 = 1 119887119894

Compute 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(33)If 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (34)

Note that the PB test provided in this section has the sameinvariance properties as in Section 2

4 Monte Carlo Studies

In this section we need to compare the PB test with theAHT test via comparing their empirical sizes (Type I error

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Differential EquationsInternational Journal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article MANOVA for Nested Designs with Unequal ...

Journal of Applied Mathematics 5

This property is desirable since in practice in order to avoidcomputation of generalized inverse of matrices and enhance theaccuracy of computation we have taken a particular L such asL = (0 1119886 1119886) otimes I

119903in the simulation study

3 Tests for the Nesting Effects

When the nested effects are present the nesting effect 120572119894can

not reflect the effect of 119860119894because it depends on which level

of factor 119861 it is in As pointed in Searle [12] Chap 7 in thecase of 119903 = 1 a popular solution to the problem is not quite atest for 120572

119894= 0 in the presence of nested effects but rather to

test the null hypothesis

1198670120572|120573 120572119894 + 120573119894119895 = 0 119894 = 1 119886 119895 = 1 119887

119894 (27)

Wewill see that the problem of testing (27) is the same case ofthe problem of comparing 119887 = sum

119886

119894=1119887119894normal mean vectors

when the cell covariancematrices are unknown and arbitrary[5] In fact if Σ

119894119895are known then

0= (

119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895)

minus1119886

sum

119894=1

119887119894

sum

119895=1

119899119894119895Σminus1

119894119895Y119894119895

(28)

is the best linear unbiased estimator of 1205830 and a natural

statistic for testing (27) is

119878120572 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

=

119886

sum

119894=1

119887119894

sum

119895=1

(Y119894119895minus 0)119879

(119899119894119895Σminus1

119894119895) (Y119894119895minus 0)

= Y119879Σminus12 (I119887119903minus Σminus12X1(X1198791Σminus1X1)minus

X1198791Σminus12

)Σminus12Y

(29)

whereX1= 1119887otimesI119903 Notice thatΣminus12Y sim 119873

119887119903(Σminus12120583 I119887119903) and

Δ1= (I119887119903minus Σminus12X1times (X1198791Σminus1X1)minusX1198791Σminus12

) is an idempotentmatrix with rank 119903(119887 minus 1) we have

Y119879Σminus12Δ1Σminus12Y sim 120594

2

119903(119887minus1)(120583119879Σminus12Δ1Σminus12120583) (30)

Thenoncentrality parameter1205831015840Σminus12Δ1Σminus12120583 is equal to zero

when 120572119894+120573119894119895= 0 119894 = 1 119886 119895 = 1 119887

119894Then the test that

rejects1198670120572|120573 in (27) whenever

119878120572 (y11 y119886119887119886 Σ11 Σ119886119887119886) gt 1205942

119903(119887minus1)120582(31)

is a size 120582 test In general the covariance matrices Σ119894119895are

unknown in this case a test statistic can be obtained byreplacing Σ

119894119895in (29) by S

119894119895 119894 = 1 119886 119895 = 1 119887

119894and is

given by

119878120572 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X

1(X1198791Sminus1X1)minus

X1198791Sminus12) Sminus12Y

(32)

In the following we describe the PB test for1198670120572|120573 in (27)

Recall that under 1198670120572|120573 the vector Y have the mean X

11205830

As the test statistic in (32) is location invariant under thegroup of location transformations G

1= Y + X

1120578 120578 isin R119903

without loss of generality we can take X11205830= 0 Using

these facts the parametric bootstrap pivot variable can bedeveloped as follows For a given (y

11 y

119886119887119886 s11 s

119886119887119886)

let Y119861119894119895

sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus 1 (1(119899

119894119895minus

1))s119894119895) 119894 = 1 119886 119895 = 1 119887

119894 Then the PB pivot variable

based on the test statistic (32) is given by

119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886)

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X1(X1198791Sminus1119861X)minus

X1198791Sminus12119861

) Sminus12119861

Y119861

(33)

where Y119861= (Y11987911986111 Y119879

119861119886119887119886)119879 and S

119861= diag((1119899

11)S11986111

(1119899119886119887119886)S119861119886119887119886

) For a given level 120582 the PB test rejects1198670120572|120573

in (27) when

119875 (119878120572119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572) lt 120582 (34)

where

119904120572 = 119878120572 (y11 y119886119887119886 s11 s119886119887119886) (35)

is an observed value of 119878120572(Y11 Y119886119887119886 S11 S119886119887119886) in (32)For fixed (y

11 y

119886119887119886 s11 s

119886119887119886) the above probability

does not depend on any unknown parameters and so itcan be estimated using Monte Carlo simulation given in thefollowing Algorithm 6

Algorithm 6 For a given (11989911 119899

119886119887119886) (y11 y

119886119887119886) and

(s11 s

119886119887119886)

Compute 119878120572(y11 y119886119887119886 s11 s119886119887119886) in (32) and callit 119904120572For ℎ = 1 119898Generate Y

119861119894119895sim 119873119903(0 (1119899

119894119895)s119894119895) and S

119861119894119895sim 119882119903(119899119894119895minus

1 (1(119899119894119895minus 1))s

119894119895)

119894 = 1 119886 119895 = 1 119887119894

Compute 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) using(33)If 119878120572119861(Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) gt 119904120572 set 119876ℎ =1(End loop)(1119898)sum

119898

ℎ=1119876ℎis a Monte Carlo estimate of the 119875

value in (34)

Note that the PB test provided in this section has the sameinvariance properties as in Section 2

4 Monte Carlo Studies

In this section we need to compare the PB test with theAHT test via comparing their empirical sizes (Type I error

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Page 6: Research Article MANOVA for Nested Designs with Unequal ...

6 Journal of Applied Mathematics

rates) and powers for two-factor nested MANOVA modelsvia simulations As pointed out in Section 3 the problem oftesting (27) is the same case of the problem of comparing119887 = sum

119886

119894=1119887119894normal mean vectors when the cell covariance

matrices are unknown and arbitrary Due to this reason wewill look only at the nested effects for comparisons

Let the two factors be 119860 and 119861 respectively Suppose thatthe nesting effect of the factor 119860 has 119886 factor levels and thenested effect of the factor119861has a total of 119887 = sum119886

119894=1119887119894levelswith

119887119894levels of B nested within the 119894th level of the factor119860 Let n =

[11989911 11989912 119899

119886119887119886] denote the vector of cell sizes For given n

and covariance matrices Σ119894119895 119894 = 1 2 119886 119895 = 1 2 119887

119894 we

first generate sum119886119894=1

sum119887119894

119895=1119899119894119895multivariate samples as

y119894119895119896= 120583119894119895+ Σ12

119894119895e119894119895119896 119896 = 1 2 119899

119894119895 (36)

where the cell mean vectors 120583119894119895= 12058311+ 119894119895120575h119887 with 120583

11being

the first cell mean vector h a constant unit vector specify-ing the direction of the cell mean differences and 120575 a tun-ing parameter controlling the amount of the cell mean dif-ferences We independently generate the 119903 entries of theerror terms e

119894119895119896from the 119873(0 1) distribution so that we

always have E(e119894119895119896) = 0 and Cov(e

119894119895119896) = I

119903 This means

that (36) will generate the (119894119895)th multivariate normal sampley119894119895119896 119896 = 1 2 119899

119894119895with the given mean vector 120583

119894119895and

covariance matrix Σ119894119895 Without loss of generality we specify

12058311

as 0 and h as h0h0 where h

0= [1 2 119903]

119879 forany given dimension 119903 and h

0 denotes the usual 1198712-norm

of h0 To estimate the sizes and powers of the AHT test

we used simulation consisting of 10000 runs and recordedcorresponding 119875 values Notice that two nested ldquodo loopsrdquoare required to estimate the sizes and powers of the PB testwe used 2500 runs for outer ldquodo loopsrdquo (for generating thedata) and 5000 runs for inner ldquodo loopsrdquo for estimating thebootstrap 119875 values The empirical sizes (when 120575 = 0) andpowers (when 120575 gt 0) of the two tests are the proportionsof rejecting the null hypothesis that is when their 119875 valuesare less than the nominal significance level 120582 In all thesimulations conducted we used 120582 = 5 for simplicity

Notice that the PB test does not depend on chosenweights Here we report only the comparative studies forthe equal-weight method to specify the weights of the AHTtest so that the AHT test in Zhang [8] may be used forshowing properties of the PB test The empirical sizes andpowers of the two tests for nested effect tests together withthe associated tuning parameters are presented in Tables 1ndash3 in the columns labeled with AHT and PB under ldquo120575 = 0rdquoand ldquo120575 gt 0rdquo respectively As seen from the three tables threesets of the tuning parameters for the cell covariance matricesare examined with the first set specifying the homogeneouscases four sets of the cell sizes are specified with the firsttwo sets specifying the balanced cell size cases To measurethe overall performance of a test in terms of maintaining thenominal size 120582 we use the average relative error defined inZhang [8] as ARE = 119872

minus1sum119872

119898=1|119898minus 120582|120582 times 100 where

119898

denotes the 119898th empirical size for 119898 = 1 2 119872 120582 =

005 and 119872 is the number of empirical sizes under consid-eration The smaller ARE value indicates the better overall

performance of the associated test Usually when ARE le 10the test performs very well when 10 lt ARE le 20 the testperforms reasonably well and when ARE gt 20 the test doesnot performwell since its empirical sizes are either too liberalor too conservative and hence may be unacceptable Noticethat for a good test the larger the cell sizes the smaller theARE values The ARE values of the two tests under the twoerror schemes are also presented in these three tables Noticethat for simplicity in the specification of the covariance andsize tuning parameters we often use (u

119905) to denote ldquou repeats

119905 timesrdquo Table 1 shows the empirical sizes and powers of thetwo tests for a bivariate case with 119886 = 2 and 119887

1= 16 119887

2= 20

With 119887 = 1198871+ 1198872= 36 one may be able to check how the

two tests behave when one of the factors has a large numberof levels Tables 2 and 3 show the empirical sizes and powersof the two tests for a three-variate case with 119886 = 3 and1198871= 8 119887

2= 10 119887

3= 12 and a 10-variate case with 119886 = 3

and 1198871= 3 119887

2= 5 119887

3= 7 respectively These two tables

allow us to compare the two tests for higher-dimensionaldata

In the following let us compare the AHT and PB testsvia examining their empirical sizes and powers It is seenfrom the three tables that for all the cases the PB testgenerally outperforms the AHT test for all the cases underconsideration as shown by their empirical sizes and theassociated ARE values presented in the three tables TheAHT test performs reasonably well only for 10-dimensionaldata In the homogeneous cases the AHT test appears tobe more powerful than the PB tests because of its inflatedempirical sizes exceeding the nominal level considerably Forthe heteroscedastic cases we once again observe from Tables2 and 3 that the PB test performs superior to the AHT test interms of controlling nominal level and powers

We conclude this simulation section via summarizingall the simulation studies conducted In terms of size theoverall conclusion is that the PB test is a flexible procedurethat performs satisfactorily regardless of the cell sizes valuesof the cell covariance matrices and the number of effectsbeing compared In terms of power the PB test generallyoutperforms the AHT test for most heteroscedastic casesunder consideration Thus one may recommend to use thePB test as a good alternative in practical applications becauseof its simplicity and accuracy

5 Concluding Remarks

The available classical tests for the two-factor MANOVAmodel with crossed designs and heteroscedastic cell covari-ance matrices have serious Type I error problems that havebeen overlooked To address this serious problem Zhang [8]proposed the AHT test In this paper we proposed and stud-ied the PB test and the AHT test for the two-factorMANOVAmodel with nested designs and heteroscedastic cell covari-ance matrices We showed that the PB test is invariant underaffine-transformations and different choices of weights usedto define the parameters uniquely We demonstrated viaintensive simulations that the PB test generally performs welland outperforms the AHT test in terms of size and power formost cell sizes and parameter configurations

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Stochastic AnalysisInternational Journal of

Page 7: Research Article MANOVA for Nested Designs with Unequal ...

Journal of Applied Mathematics 7

Table 1 Empirical sizes and powers of the two tests for nested effects for bivariate two-factor multivariate nested designs [119886 = 2 1198871= 16 119887

2=

20 (Σ11Σ12 Σ

119886119887119886) = 111987918otimes (I2 diag (120582))]

120582 n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

1205821

n1

0056 0046 0210 0162 0831 0793 0999 0999n2

0057 0050 0394 0372 0995 0994 1000 1000n3

0070 0052 0295 0232 0957 0930 1000 1000n4

0057 0054 0926 0923 1000 1000 1000 1000

1205822

n1

0058 0055 0108 0124 0379 0551 0826 0964n2

0068 0044 0173 0232 0718 0930 0995 1000n3

0071 0053 0152 0150 0549 0717 0958 0995n4

0059 0051 0475 0776 0999 1000 1000 1000

1205823

n1

0056 0045 0093 0114 0301 0507 0722 0943n2

0058 0046 0151 0204 0614 0878 0977 0999n3

0070 0048 0131 0145 0458 0654 0903 0990n4

0061 0055 0397 0755 0994 1000 1000 1000ARE 2403 683

1205821 = (1 1) 1205822 = (1 5) and 1205823 = (1 10) n1 = (7 7)20 n2 = (10 10)20 n3 = (7 10)20 and n4 = (30 15)20

Table 2 Empirical sizes and powers of the two tests for nested effects for 3-variate two-factor multivariate nested designs [119886 = 3 1198871= 8 1198872=

10 1198873= 12 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag (120582) (1 minus 120588)I

3+ 1205881311198793)]

(120582 120588) n 120575 = 0 120575 = 06 120575 = 12 120575 = 18

AHT PB AHT PB AHT PB AHT PB

(1205821 1205881)

n1

0052 0054 0099 0103 0369 0353 0840 0815n2

0056 0041 0164 0152 0705 0691 0994 0991n3

0092 0041 0302 0194 0944 0891 1000 1000n4

0096 0044 0288 0185 0927 0847 1000 1000

(1205822 1205882)

n1

0054 0049 0181 0213 0756 0860 0997 1000n2

0055 0048 0335 0422 0985 0997 1000 1000n3

0090 0050 0629 0519 1000 1000 1000 1000n4

0089 0061 0522 0545 0999 1000 1000 1000

(1205823 1205883)

n1

0049 0050 0271 0194 0727 0852 0995 1000n2

0052 0056 0308 0427 0978 0998 1000 1000n3

0094 0060 0593 0519 1000 0999 1000 1000n4

0098 0057 0486 0551 0999 1000 1000 1000ARE 4667 1083

(1205821 1205881) = (13 0) (1205822 1205882) = (1 15 01 01) and (1205823 1205883) = (1 10 01 05) n1 = (10 10 10)10 n2 = (15 15 15)10 n3 = (10 20 40)10 and n4 = (40 20 10)10

Appendix

Proof of Theorem 1 When 1198670120573 is true the model (8) can be

written as the following reduced model

Y = X120579 + e e sim 119873119887119903(0Σ) (A1)

Premultiplying Σminus12 in model (A1) we have the followingtransformed model

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I) (A2)

Under the constraint L120579 = 0 the model (A2) becomes

Σminus12Y = Σ

minus12X120579 + Σminus12e Σminus12e sim 119873119887119903(0 I)

L120579 = 0(A3)

In view of Theorem 525 in Wang and Chow [11] P 177 weshall prove that L120579 = 0 is a side condition on the matrixΣminus12X that is L satisfies conditions

R((Σminus12X)

119879

) capR (L119879) = 0

rank(Σminus12XL ) = (119886 + 1) 119903

(A4)

where R(M) denotes the subspace spanned by the columnof matrix M Denote X

0= [1119887 diag(1

1198871 1

119887119886)] and L

0=

(0 1199061 sdot sdot sdot 119906

119886) respectively Then X = X

0otimes I119903 and L =

L0otimesI119903 Note that 119906

1 119906

119886are nonnegative weights such that

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article MANOVA for Nested Designs with Unequal ...

8 Journal of Applied Mathematics

Table 3 Empirical sizes and powers of the two tests for nested effects for 10-variate two-factor multivariate nested designs [119886 = 3 1198871= 3 1198872=

5 1198873= 7 (Σ

11Σ12 Σ

119886119887119886) = 111987910otimes (I3 diag(120582) diag(120578))]

(120582 120578) n 120575 = 0 120575 = 1 120575 = 2 120575 = 3

AHT PB AHT PB AHT PB AHT PB

(1205821 1205781)

n1

0046 0042 0438 0415 0999 0998 1000 1000n2

0052 0050 0937 0932 1000 1000 1000 1000n3

0062 0046 0788 0762 1000 1000 1000 1000n4

0058 0056 0871 0846 1000 1000 1000 1000

(1205822 1205782)

n1

0052 0044 0842 0635 1000 1000 1000 1000n2

0052 0047 1000 0996 1000 1000 1000 1000n3

0067 0049 0995 0965 1000 1000 1000 1000n4

0059 0046 0999 0972 1000 1000 1000 1000

(1205823 1205783)

n1

0049 0042 0066 0101 0115 0401 0235 0877n2

0050 0054 0090 0236 0255 0930 0660 1000n3

0071 0045 0089 0145 0183 0622 0442 0982n4

0055 0047 0095 0236 0260 0961 0662 1000ARE 1378 867

1205821 = (110)5 1205781 = (110)5 1205822 = (123 13 243 1)5 1205782 = (13 013 22 24 21)5 and 1205823 = (13 33 93 20)5 1205783 = (53 153 453 100)5 n1 = (253)5 n2 = (503)5n3 = (25 35 50)5 and n4 = (70 40 35)5

sum119886

119894=1119906119894gt 0Then it can be easily verified thatX

0andL0satisfy

the following conditions

R (X1198790) capR (L119879

0) = 0

rank(X0L0

) = 119886 + 1

(A5)

Assume that d isin R(X119879) cap R(L119879) Then there exist twovectors g and h such that

d = X119879g = L119879h (A6)

Denote

X1198790= (

11990911

sdot sdot sdot 1199091119887

sdot sdot sdot

119909119886+11

sdot sdot sdot 119909119886+1119887

) L1198790= (

11989711

119897119886+11

) (A7)

and g = (g1198791 g1198792 g119879

119887)119879 where g

119894= (1198921198941 119892

119894119903)119879 and h =

(ℎ1 ℎ

119903)119879 are 119903 times 1 vectors respectively 119894 = 1 2 119887

Then (A6) may be rewritten as

X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h) = L119879h

(A8)

which is equivalent to X1198790glowast119896= L1198790ℎ119896 where glowast

119896= (1198921119896 1198922119896

119892119887119896)119879 119896 = 1 119903 It follows from (A5) that

X1198790glowast119896= L1198790ℎ119896 119896 = 1 119903 (A9)

and hence

d = X119879g = (11990911g1+ sdot sdot sdot + 119909

1119887g119887

119909119886+11

g1+ sdot sdot sdot + 119909

119886+1119887g119887

) = (

11989711h

119897119886+11

h)

= L119879h = 0(A10)

This means that

R (X119879) capR (L119879) = 0 (A11)

It follows fromR(X119879) = R(X119879Σminus12) that

R((Σminus12X)

119879

) capR (L119879) = 0 (A12)

On the other hand

rank(XL) = rank((X0L0

) otimes I119903)

= rank(X0L0

) times rank (I119903) = (119886 + 1) 119903

(A13)

The second equation of (A13) may be derived easily fromTheorem 221 in Wang and Chow [11] It follows from (A13)that

rank(Σminus12XL ) = rank((Σ

minus12 00 I)(

XL))

= rank(XL) = (119886 + 1) 119903

(A14)

Combining (A12) with (A14) we obtain that L120579 = 0 is a sidecondition on the matrix Σminus12X Thus it follows from Lemma522 in Wang and Chow [11] that

X119879Σminus1X(X119879Σminus1X + L119879L)minus1

X119879Σminus12 = X119879Σminus12 (A15)

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article MANOVA for Nested Designs with Unequal ...

Journal of Applied Mathematics 9

Thus we obtain that X119879Σminus1X(X119879Σminus1X + L119879L)minus1X119879Σminus1X =

X119879Σminus1X This imply that (X119879Σminus1X + L119879L)minus1 is a generalizedinverse of X119879Σminus1X Notice from Corollary 223 in Wangand Chow [11] that Σminus12X(X119879Σminus1X)minusX119879Σminus12 is invariantwith respect to the choice of the involved generalized(X119879Σminus1X)minus which leads to Σminus12X(X119879Σminus1X)minusX119879Σminus12 =

Σminus12X(X119879Σminus1X + L119879L)minus1X119879Σminus12 Therefore It follows that

X = X(X119879Σminus1X + L119879L)minus1

X119879Σminus1Y

= Σ12Σminus12X(X119879Σminus1X + L119879L)

minus1

X119879Σminus12Σminus12Y

= Σ12Σminus12X(X119879Σminus1X)

minus

X119879Σminus12Σminus12Y

= X(X119879Σminus1X)minus

X119879Σminus1Y

(A16)

Then we have

119878120573 (Y11 Y119886119887119886 Σ11 Σ119886119887119886)

= (Y minus X)119879

Σminus1(Y minus X)

= Y119879Σminus12 (I119887119903minus Σminus12X(X119879Σminus1X)

minus

X119879Σminus12)Σminus12Y(A17)

as desired Theorem 1 is proved

Proof of Theorem 3 Since the observed values of Y119894119895119896 119896 =

1 2 119899119894119895 are denoted as y

119894119895119896 we let ylowast

119894119895119896denote the observed

values of the affine-transformed cell responses Ylowast119894119895119896 119896 =

1 2 119899119894119895 given by (25) Then we have ylowast

119894119895119896= Dylowast119894119895119896+ 120585 On

the other handwe obtain that the samplemean vector and thesample covariance matrix of the (119894 119895)th affine-transformedcell are Ylowast

119894119895= DY

119894119895+ 120585 and Slowast

119894119895= DS

119894119895D119879 respectively The

observed values of these random variables are ylowast119894119895= Dy119894119895+ 120585

and slowast119894119895= Ds119894119895D119879 respectively 119894 = 1 119886 119895 = 1 119887

119894 Let

Ylowast = (Ylowast11987911 Ylowast119879

11198871 Ylowast119879

1198861 Ylowast119879

119886119887119886)119879 and ylowast = (ylowast119879

11

ylowast11987911198871 ylowast119879

1198861 ylowast119879

119886119887119886)119879 from the reduced model (A1) we

have the following affine-transformed reduced model

Ylowast minus 120585119887= D119887X120579 +D

119887e D

119887e sim 119873

119887119903(0D119887ΣD119879119887) (A18)

whereD119887= I119887otimesD 120585

119887= 1119887otimes 120585

Denote slowast = diag((111989911)slowast11 (1119899

11198871)slowast11198871 (1

1198991198861)slowast1198861 (1119899

119886119887119886)slowast119886119887119886) Let 119904

lowast

120573 be affine-transformedobserved test statistic of 119904120573 (24) Then by direct calculationswe have

119904lowast

120573 = 119878120573 (ylowast

11minus 120585 ylowast

119886119887119886minus 120585 slowast11 slowast

119886119887119886)

= (ylowast minus 120585119887)119879slowastminus12 (I

119887119903minus slowastminus12D

119887X(X119879D119879

119887slowastminus1D

119887X)minus

times X119879D119879119887slowastminus12) slowastminus12 (ylowast minus 120585

119887)

= y119879sminus12 (I119887119903minus sminus12X(X119879sminus1X)

minus

X119879sminus12) sminus12y = 119904120573(A19)

where s = diag((111989911)s11 (1119899

11198871)s11198871 (1119899

1198861)s1198861

(1119899119886119887119886)s119886119887119886) is the observed values of S The affine-invariance

of the observed test statistic 119904120573 (24) then followsLet

Ylowast119861119894119895

sim 119873119903(0

1

119899119894119895

slowast119894119895) Slowast

119861119894119895sim 119882119903(119899119894119895minus 1

1

119899119894119895minus 1

slowast119894119895)

(A20)

be mutually independent 119894 = 1 119886 119895 = 1 119887119894 Then

Ylowast119861119894119895

119889

= DY119861119894119895

and Slowast119861119894119895

119889

= DS119861119894119895D119879 where H 119889= G means that

H and G have the same distribution Using these facts theaffine-transformed parametric bootstrap pivot variable canbe given by

119878lowast

120573119861 (Ylowast

11986111 Ylowast

119861119886119887119886 Slowast11986111 Slowast

119861119886119887119886)

= Ylowast119879119861Slowastminus12119861

(I119887119903minus Slowastminus12119861

D119887X(X119879D119879

119887Slowastminus1119861

D119887X)minus

timesX119879D119879119887Slowastminus12119861

) Slowastminus12119861

Ylowast119861

119889

= Y119879119861D119879119887(D119887S119861D119879119887)minus12

(I119887119903minus (D119887S119861D119879119887)minus12

D119887X

times (X119879D119879119887(D119887S119861D119879119887)minus1

D119887X)minus

timesX119879D119879119887(D119887S119861D119879119887)minus12

)

times (D119887S119861D119879119887)minus12

D119887Y119861

= Y119879119861Sminus12119861

(I119887119903minus Sminus12119861

X(X119879Sminus1119861X)minus

X119879Sminus12119861

) Sminus12119861

Y119861

= 119878120573119861 (Y11986111 Y119861119886119887119886 S11986111 S119861119886119887119886) (A21)

where Ylowast119861

= (Ylowast11987911986111 Ylowast119879

11986111198871 Ylowast1198791198611198861

Ylowast119879119861119886119887119886

)119879 and

Slowast119861= diag((1119899

11)Slowast11986111 (1119899

11198871)Slowast11986111198871

(11198991198861)Slowast1198611198861

(1119899119886119887119886)Slowast119861119886119887119886

) The affine-invariance of the distribution of PBpivot variable 119878120573119861 (22) followsTheorem 3 is then proved

Proof of Theorem 4 The proof of Theorem 4 is obvious andhence omitted here

Proof of Theorem 5 Using the same proof as that ofTheorem1 it follows that L120579 = 0 is a side condition on the matrixSminus12X that is L satisfies conditions

R((Sminus12X)119879

) capR (L119879) = 0 ae

rank(Sminus12XL ) = (119886 + 1) 119903 ae

(A22)

where ae denotes ldquoalmost everywhererdquo Thus it follows fromthe similar proof of Theorem 1 that

X(X119879Sminus1X + L119879L)minus1

X119879Sminus12 = X(X119879Sminus1X)minus

X119879Sminus12 ae(A23)

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article MANOVA for Nested Designs with Unequal ...

10 Journal of Applied Mathematics

Then we obtain from (A23) that the test statistic (17)

119878120573 (Y11 Y119886119887119886 S11 S119886119887119886)

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X)

minus

X119879Sminus12) Sminus12Y

= Y119879Sminus12 (I119887119903minus Sminus12X(X119879Sminus1X + L119879L)

minus1

X119879Sminus12) Sminus12Y(A24)

as desired Theorem 5 is proved

Conflict of Interests

The author declares that there is not any conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11171002) and the Importation andDevelopment of High-Caliber Talents Project of BeijingMunicipal Institutions (CITampTCD201404002)

References

[1] TW AndersonAn Introduction toMultivariate Statistical Ana-lysis Wiley New York NY USA 3rd edition 2003

[2] G S James ldquoTests of linear hypotheses in univariate and multi-variate analysis when the ratios of the population variances areunknownrdquo Biometrika vol 41 pp 19ndash43 1954

[3] S Johansen ldquoThe Welch-James approximation to the distribu-tion of the residual sum of squares in a weighted linear regres-sionrdquo Biometrika vol 67 no 1 pp 85ndash92 1980

[4] J Gamage T Mathew and S Weerahandi ldquoGeneralized 119901-values and generalized confidence regions for the multivariateBehrens-Fisher problem and MANOVArdquo Journal of Multivari-ate Analysis vol 88 no 1 pp 177ndash189 2004

[5] K Krishnamoorthy and F Lu ldquoA parametric bootstrap solutionto theMANOVAunder heteroscedasticityrdquo Journal of StatisticalComputation and Simulation vol 80 no 7-8 pp 873ndash887 2010

[6] S W Harrar and A C Bathke ldquoA modified two-factor multiva-riate analysis of variance asymptotics and small sample approx-imationsrdquo Annals of the Institute of Statistical Mathematics vol64 no 1 pp 135ndash165 2012

[7] J Zhang and S Xiao ldquoA note on the modified two-wayMANOVA testsrdquo Statistics amp Probability Letters vol 82 no 3pp 519ndash527 2012

[8] J T Zhang ldquoTwo-way MANOVA with unequal cell sizes andunequal cell covariance matricesrdquo Technometrics vol 53 no 4pp 426ndash439 2011

[9] S Weerahandi Generalized Inference in Repeated MeasuresExact Methods in MANOVA and Mixed Models Wiley Seriesin Probability and Statistics JohnWiley amp Sons New York NYUSA 2004

[10] L-W Xu F-Q Yang A Abula and S Qin ldquoA parametric boot-strap approach for two-way ANOVA in presence of possibleinteractions with unequal variancesrdquo Journal of MultivariateAnalysis vol 115 pp 172ndash180 2013

[11] S G Wang and S C Chow Advanced Linear Models vol 141Marcel Dekker New York NY USA 1994

[12] S R Searle Linear Models John Wiley New York NY USA1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article MANOVA for Nested Designs with Unequal ...

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Al-Laymoun - MANOVA Article

Al-Laymoun - MANOVA Article

5 MANOVA Presentation Stats

5 MANOVA Presentation Stats

Multivariate Analysis of Variance (Manova) Description ...

Multivariate Analysis of Variance (Manova) Description ...

Aplicacion DOE Utilizando MANOVA

Aplicacion DOE Utilizando MANOVA

SPSS Tutorial Multiple Analysis of Variance (MANOVA)lc.gcumedia.com/psy380/spss-tutorials/v1.1/pdfs/MANOVA.pdf · SPSS Tutorial 01 Multiple Analysis of Variance (MANOVA) A MANOVA

SPSS Tutorial Multiple Analysis of Variance (MANOVA)lc.gcumedia.com/psy380/spss-tutorials/v1.1/pdfs/MANOVA.pdf · SPSS Tutorial 01 Multiple Analysis of Variance (MANOVA) A MANOVA

MANOVA - Analysis

MANOVA - Analysis

MANOVA - Equations Lecture 11 Psy524 Andrew Ainsworth.

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