Chapter 3 Analysis of Variance (ANOVA; ”””†††óóó555ŠŠŠ...Relation between...

Chapter 3 Analysis of Variance

(ANOVA;變變變異異異數數數分分分析析析)

許湘伶

Applied Linear Regression Models(Kutner, Nachtsheim, Neter, Li)

Chapter 16Design and Analysis of Experiments

(Douglas C. Montgomery)

hsuhl (NUK) DAE Chap. 3 1 / 84

Part I

Supplement


Relation between Regression and Analysis of Variance

Regression model:

yi = β0 + β1X1i + · · ·+ βkXki + εi, i = 1, . . . , n

ANOVA Model or One-way model:

yij = µi + εij = µ+ τi + εij,

{i = 1, . . . , aj = 1, . . . , n


Relation between Regression and Analysis of Variance (cont.)

Analysis of variance models differ from ordinary regression models intwo key respects:

1 The explanatory or predictor variables in ANOVA models may bequalitative.

2 If the predictor variables are quantitative, no assumption is madein ANOVA models about the nature of the statistical relationbetween Xs and Y .



When indicator variables are so used with regression models, theregression results will be identical to those obtained with ANOVAmodels.

ANOVA models and regression models with indicator variableswill lead to identical results.



Figure : Regression model

Figure : Figure 16.4 Illustration of Partitioning of Total Deviations Yij − Y··hsuhl (NUK) DAE Chap. 3 7 / 84

Part II

Chapter 3 The Analysis of Variance


Outline

1 Example2 The analysis of variance3 Analysis of the fixed effects model4 Model adequacy checking5 Practical interpretation of results6 Sample computer output7 Determining sample size8 Other example of single-factor experiments9 The random effect model10 The regression approach to the ANOVA11 Nonparametric methods in the ANOVA


Example

methods for the design and analysis of single-factor experimentswith a levels of the factor (or a treatments)

Assume: completely randomized

wafer(晶片)Relationship: RF powersetting vs. the etch rate(蝕刻速率)

I RF power: 4 levels:160, 180, 200, 220 W

I 蝕刻速率: 測量物質從晶圓表面被移除的的速率有多快

n = 5 replicates- 20 runs inrandom order


Example (cont.)


Example (cont.)

no strong evidence to suggest that the variability in the etch ratearound the average depends on the power setting

Test: differences between the mean etch rates at a = 4 levels ofRF power

1 t-test for all six possible pairs of means: inflates the type I error2 the analysis of variance


ANOVA

a treatments of a single factor

yij: the jth observation taken under treatment i

means model:

yij = µi + εij

{i = 1, 2, . . . , aj = 1, 2, . . . , n


ANOVA (cont.)

Model:

yij = µi + εij

{i = 1, 2, . . . , aj = 1, 2, . . . , n

mean model

= µ+ τi + εij effect model

yij: the ij observationµi: the mean of the ith factor levelµ: overall meanτi: the ith treatment effectεij: the random error component; sources of variability

I measurementI variability from uncontrolled factorsI differences between the experimental unitI noise in the process


ANOVA (cont.)

yij = µi + εij

{i = 1, 2, . . . , aj = 1, 2, . . . , n

mean model

= µ+ τi + εij effect model

linear statistical models

one-way or single-factor analysis of variance model (單因子變異數分析)

the effect model is more widely encountered in the experimentaldesign literatureobject:

I test hypotheses about the treatment meansI estimate model parameters: (µ, τi, σ

2)

εij ∼ NID(0, σ2)⇒ yij ∼ N(µ+ τi, σ2)


ANOVA (cont.)

fixed effects model (固定效應模型): chosen by experimenter

random effects model (隨機效應模型; components of variancemodel變異數成分模型): (Chap. 3.9; Chap. 13)a treatment could be a random sample from a larger population oftreatments


Notation

yi·: the average of the observations under the ith treatment

y··: the grand total of all the observations

y··: the grand average of all the observations

yi· =n∑

j=1

yij yi· = yi·/n i = 1, 2, . . . , a

y·· =a∑

i=1

n∑j=1

yij y·· = y··/N, N = an


Testing

Testing the equality of the a treatment means E(yij) = µ+ τi = µi:

Hypothesis: {H0: µ1 = µ2 = · · · = µa

Ha: µi 6= µj for at least one pair (i, j)

⇔{

H0: τ1 = τ2 = · · · = τa= 0Ha: τi 6= 0 for at least one i

∵

∑ai=1 µi

a= µ ⇔

a∑i=1

τi = 0

The appropriate procedure for testing the equality of a treatmentmeans is the analysis of variance.


Testing (cont.)


Decomposition of the Total Sum of Squares

ANOVA: derived from a partitioning of total variability into itscomponent parts

1 SST : the total corrected sum of squares2 SSTreatment: the sum of squares due to treatments (between

treatment)3 SSE: the sum of squares due to error (within treatments)

SST(N−1)

=a∑

i=1

n∑j=1

(yij − y··)2

= na∑

i=1

(yi· − y··)2 +a∑

i=1

n∑j=1

(yij − yi·)2

= SSTreatment(a−1)

+ SSE(N−a)


Decomposition of the Total Sum of Squares (cont.)

SST(N−1)

=a∑

i=1

n∑j=1

(yij − y··)2 = na∑

i=1

(yi· − y··)2 +a∑

i=1

n∑j=1

(yij − yi·)2

= SSTreatment(a−1)

+ SSE(N−a)



Total variability: can be partitioned into1 the total corrected sum of squares

SST =

a∑i=1

n∑j=1

(yij − y··)2 =

a∑i=1

n∑j=1

y2ij −

y2··

N

2 a sum of squares of the differences between the treatment averageand the grand average

SSTreatment = na∑

i=1

(yi· − y··)2 =1n

a∑i=1

y2i· −

y2··

N

3 a sum of squares of the differences of observation withintreatments from the treatment average

SSE =

a∑i=1

n∑j=1

(yij − yi·)2 = SST − SSTreatment



1 a pooled estimate of the common variance within each of the atreatments

SSE

N − a

2 an estimate of σ2 if µis are all equal

SSTreatment

a− 1

3 ANOVA identity: provide two estimated of σ2



Error mean square (MSE;誤差均方):

1 MSE =SSE

N − a2 E(MSE) = σ2

Treatment mean square (處理均方):

1 MSTreatment =SSTreatment

a− 1

2 E(MSTreatment) = σ2 +n∑a

i=1 τ2i

a− 13 if there are no differences in treatment means (i.e. τi = 0),

MSTreatment also estimate σ2

A test of hypothesis of no difference in treatment means can beperformed by comparing METreatment and MSE


Statistical Analysis

Assumptions

εij ∼ NID(0, σ2)⇒ yij ∼ NID(µ+ τi, σ2)

Cochran’s Theorem

SST : a sum of squares in normally distributed r.v.1 SST/σ

2 ∼ χ2N−1

2 SSTreatment/σ2 ∼ χ2

a−1 if H0 : τi = 0 is true3 SSE/σ

2 ∼ χ2N−a

4 SSTreatment/σ2 and SSE/σ

2 are independent χ2 r.v.

⇒ test statistic: F0 =SSTreatment/(a− 1)

SSE/(N − a)=

MSTreatment

MSE

H0∼ Fa−1,N−a


Statistical Analysis (cont.)

Cochran’s Theorem

Let Zi ∼ NID(0, 1), i = 1, . . . , ν, and

ν∑i=1

Z2i =

s∑i=1

Qi,

where s ≤ ν and Qi has νi d.f. (i = 1, . . . , s). Then Qi, ı =

1, . . . , s are independent χ2νi

r.v., if and only if

ν =s∑

i=1

νi



If H0 is false, MSTreatment > MSE

⇒ reject H0 if F0 is too large, i.e., F0 > Fα,a−1,N−a

ANOVA table:



The Plasma Etching Experiment

H0 : µ1 = µ2 = µ3 = µ4 vs. H1 : some means are different



## ANOVA tableetch$FRF <- as.factor(etch$RF)etch.aov <- aov(rate˜FRF,data=etch)summary(etch.aov)

Df Sum Sq Mean Sq F value Pr(>F)FRF 3 66870.55 22290.18 66.80 0.0000Residuals 16 5339.20 333.70

F0 > F(0.99, 3, 16) = 5.29


Estimation of the Model Parameters

Model:

yij = µ+ τi + εij

{i = 1, . . . , aj = 1, . . . , n

Parameter: µ, τi, σ2

Estimates: (Least Squares Estimation)I overall mean: µ = y··I treatment effect: τi = yi· − y··, i = 1, . . . , aI µi: µi = µ+ τi = yi·I σ2: σ2 = MSE


Estimation of the Model Parameters (cont.)

εij ∼ NID(0, σ2)⇒ yi· ∼ N(µi, σ2/n)

100(1− α)% Confidence interval:

yi· − tα/2,N−a

√MSE

n≤µi ≤ yi· + tα/2,N−a

√MSE

n

yi· − yj· − tα/2,N−a

√2MSE

n≤ µi−µj ≤ yi· − yj· + tα/2,N−a

√2MSE

n


Estimation of the Model Parameters (cont.)

Ex 3.3

overall mean: µ = 617.75

treatment effect:

i 1 2 3 4RF power 160 180 200 220

τi -66.55 -30.35 7.65 89.25

95% confidence interval for µ4: (one-at-a-time)

689.6815 ≤ µ4 ≤ 724.3185

Bonferroni method: correct level α/2r


Unbalanced data

ni observations under treatment i (i = 1, . . . , a)

N =∑a

i=1 ni: total sample size

SST =a∑

i=1

ni∑j=1

y2ij −

y2··

N

SSTreatment =a∑

i=1

y2i·

ni− y2

··N


Model Adequacy Checking

yij: estimate of yij

yij = µ+ τi = yi·

residual eij: investigating violations of the basic assumptions andmodel adequacy

eij = yij − yij

I The checking should be automaticI Model is adequate⇒ eijs should be structurelessI graphical analysisI how to deal with commonly occurring abnormalities

standardized residual: dij =eij√MSE


Model Adequacy Checking (cont.)

Residual plot## Residual plotopar <- par(mfrow=c(2,2),cex=.8)plot(etch.aov)par(opar)



eij vs. time: independence assumption

eij vs. yij: nonconstant variance-variance-stabilizingtransformation


Statistical Tests for Equality of Variance

Bartlett’s test:

H0 :σ21 = σ2

2 = · · · = σ2a

Ha : above not true for at least on σ2i

a modification of the corresponding likelihood ratio test designedto make the approximation to the χ2 distribution better (Bartlett,1937)

very sensitive to the normality assumption

log10(e) = 2.3026


Statistical Tests for Equality of Variance (cont.)

Test statistic:

χ20= 2.3026

qc

H0∼ χ2a−1

q = (N − a) log10 S2p −

a∑i=1

(ni − 1) log10 S2i

c = 1 +1

3(a− 1)

(a∑

i=1

(ni − 1)−a − (N − a)−1

)

S2p =

∑ai=1(ni − 1)S2

i

N − a

Reject H0: χ20 > χ2

α,a−1



> bartlett.test(rate˜RF,data=etch)

Bartlett test of homogeneity of variances

data: rate by RFBartlett’s K-squared = 0.4335, df = 3, p-value = 0.9332

> qchisq(0.95,3)[1] 7.814728



Modified Levene test:

robust to departures from normality

considering the absolute deviation of yij from the treatmentmedian yi·:

dij = |yij − yi·|{

i = 1, 2, . . . , aj = 1, 2, . . . , n

The test statistic for Levene’s test is simply the usual ANOVA Fstatistic for testing equality of means applied to the absolutedeviations



Peak Discharge Data



> library(lawstat)> peak.aov<-aov(Observ˜as.factor(Method),data=peak)> summary(peak.aov)

Df Sum Sq Mean Sq F value Pr(>F)as.factor(Method) 3 708.3 236.1 76.07 4.11e-11 ***Residuals 20 62.1 3.1---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1> leveneTest(peak$Observ,as.factor(peak$Method))Levene’s Test for Homogeneity of Variance (center = median)

Df F value Pr(>F)group 3 4.5684 0.01357 *

20---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Transformation: y∗ij =√yij



Formal method: Box-Cox Method## Box-Cox Methodlibrary(MASS)boxcox(Observ ˜ Method, data = peak,lambda = seq(-1, 1, length = 10))


Comparing Among Treatment Means

ANOVA:

reject H0 ⇒ differences between the treatment means

which means differ is not specifiedmultiple comparison methods

yi· ∼ N(µi, σ2/n), σ2 = MSE

⇒µ1 6= µ2 6= µ3 6= µ4


Contrasts(對對對比比比)

200W and 220W produce the same etch rate:

H0 : µ3 = µ4

H1 : µ3 6= µ4⇔ H0 : µ3 − µ4 = 0

H1 : µ3 − µ4 6= 0

the average of the lowest levels of power did not differ from theaverage of the highest levels of power:

H0 : µ1 + µ2 = µ3 + µ4

H1 : µ1 + µ2 6= µ3 + µ4⇔ H0 : µ1 + µ2 − µ3 − µ4 = 0

H1 : µ1 + µ2 − µ3 − µ4 6= 0


Contrasts(對對對比比比) (cont.)

contrast: a linear combination of parameters

Γ =a∑

i=1

ciµi

c1, . . . , ca: contrast constants sum to zero⇒∑a

i=1 ci = 0

H0 :∑a

i=1 ciµi = 0H1 :

∑ai=1 ciµi 6= 0

1 c1 = 0 = c2, c3 = 1, c4 = −12 c1 = 1 = c2, c3 = −1 = c4



Testing Hypotheses involving contrast: t-test and F test

Sample sizes are different:

a∑i=1

nici = 0

treatment average:

C =a∑

i=1

ciyi·

V(C) = σ2a∑

i=1

c2i

ni



Under H0:

t statistic: t0 =

∑ai=1 ciyi·√

MSE∑a

i=1c2

ini

∼ tN−a

⇒Reject H0 if |t0| ≥ tα/2,N−a

F statistic: F0 = t20 =

(∑a

i=1 ciyi·)2

MSE∑a

i=1c2

ini

=MSC

MSE∼ F1,N−a

⇒Reject H0 if |F0| ≥ Fα,1,N−a

contrast sum of squares:

SSC =(∑a

i=1 ciyi·)2∑a

i=1c2

ini

(d.f.=1)



C.I. for a contrast:

a∑i=1

ciyi· − tα/2,N−a

√√√√MSE

a∑i=1

c2i

ni≤

a∑i=1

ciµi ≤a∑

i=1

ciyi· + tα/2,N−a

√√√√MSE

a∑i=1

c2i

ni

Orthogonal Contrasts:

Two contrasts with coefficients: {ci}, {di},

orthogonal ifa∑

i=1

nicidi = 0

Coefficients forTreatment Orthogonal Contrasts1(control) -2 02(level 1) 1 -13(level 2) 1 1



For a treatments, the set of a− 1 orthogonal contrasts partitionthe sum of squares due to treatments into a− 1 independentsingle-degree-of-freedom components.

Tests performed on orthogonal contrasts are independent.

many ways to choose the orthogonal contrast coefficients



Example 3.6: plasma etching experiment

> cont.matrix<-matrix(c(1,-1,0,0,1,1,-1,-1,0,0,1,-1),byrow=T,ncol =4)> effect<-aggregate(etch$rate, list(etch$FRF), mean)> Ci<-effect$x%*%t(cont.matrix)> Sci2<-apply(cont.matrixˆ2,1,sum)> Ciˆ2/(Sci2/5)

[,1] [,2] [,3][1,] 3276.1 46948.05 16646.4


Sheffe Method for Comparing all Contrasts

a method for comparing any and all possible contrasts betweentreatment means

Type I error is at most α

a set of m contrasts:

Γu = c1uµ1 + c2uµ2 + · · ·+ cauµa, u = 1, 2, . . . ,m

⇒Cu = c1uy1· + c2uy2· + · · ·+ cauya·

Standard error:

SCu =

√√√√MSE

a∑i=1

(c2iu/ni)


Sheffe Method for Comparing all Contrasts (cont.)

Critical value against which Cu:

Sα,u = SCu

√(a− 1)Fα,a−1,N−a

Reject Γu = 0 if |Cu| > Sα,u

Simultaneous C.I. for all contrasts among treatment means:

Cu − Sα,u ≤ Γu ≤ Cu + Sα,u


Sheffe Method for Comparing all Contrasts (cont.)

plasma etching experiment

Γ1 = µ1 + µ2 − µ3 − µ4; Γ2 = µ1 − µ4

i 1 2Ci -193.80 -155.80SCi 16.34 11.55S0.01,i 65.10 46.03

⇒ conclude Γi 6= 0

MSE<-summary(etch.aov)[[1]][2,3]cont.matrix2<-matrix(c(1,1,-1,-1,1,0,0,-1),byrow=T,ncol =4)Ci2<-effect$x%*%t(cont.matrix2)SCi2<-sqrt(MSE*apply(cont.matrixˆ2/5,1,sum))CVSi<-SCi2*sqrt((4-1)*qf(1-0.01,3,16))xtable(rbind(Ci2,SCi2,CVSi))


Comparing Pairs of Treatment Means

Γ = µi − µj, i ≤ j

Scheffe method is not the most sensitive procedure

Tukey’s Test:

H0 : µi = µj vs. H1 : µi 6= µj

the distribution of the studentized range statistic

q =ymax − ymin√

MSE/n

qα(p, f ): appendix VII;

f the number of d.f. with MSE; p: a group of p sample means


Comparing Pairs of Treatment Means (cont.)

Tukey’s Test: a procedure for testing hypotheses for which the overallsignificance level is exactly α when nis are equal; at most α when nisare unequal.

two means significantly different if

|yi· − yj·| > Tα = qα(a, f )

√MSE

n

C.I.: i 6= j

yi· − yj· −qα(a, f )√

2

√MSE

(1ni

+1nj

)≤ µi − µj

≤ yi· − yj· +qα(a, f )√

2

√MSE

(1ni

+1nj

)hsuhl (NUK) DAE Chap. 3 59 / 84


The Tukey procedure indicates that All pairs of means differ.

> TukeyHSD(etch.aov)Tukey multiple comparisons of means

95% family-wise confidence level

Fit: aov(formula = rate ˜ FRF, data = etch)

$FRFdiff lwr upr p adj

180-160 36.2 3.145624 69.25438 0.0294279200-160 74.2 41.145624 107.25438 0.0000455220-160 155.8 122.745624 188.85438 0.0000000200-180 38.0 4.945624 71.05438 0.0215995220-180 119.6 86.545624 152.65438 0.0000001220-200 81.6 48.545624 114.65438 0.0000146> plot(TukeyHSD(etch.aov),las=1)


The Fisher Least Significant Difference (LSD) Method

LSD:費雪爾最小顯著差異法

H0 : µi = µj

t statistic: t0 =yi· − yj·√

MSE

(1ni

+ 1nj

)significant differ if

|yi· − yj·| > LSD = tα/2,N−a

√MSE

(1ni

+1nj

)

LSD= tα/2,N−a

√MSE

(1ni

+ 1nj

): least significant difference

the overall α risk may be considerably inflatedhsuhl (NUK) DAE Chap. 3 62 / 84

The Fisher Least Significant Difference (LSD) Method (cont.)

> library(agricolae)> lsd2<-LSD.test(etch.aov,"FRF",group=F) # without grouping> lsd2$statistics

Mean CV MSerror617.75 2.957095 333.7

$parametersDf ntr t.value16 4 2.119905

$meansrate std r LCL UCL Min Max

160 551.2 20.01749 5 533.8815 568.5185 530 575180 587.4 16.74216 5 570.0815 604.7185 565 610200 625.4 20.52559 5 608.0815 642.7185 600 651220 707.0 15.24795 5 689.6815 724.3185 685 725

$comparison


The Fisher Least Significant Difference (LSD) Method (cont.)

Difference pvalue sig. LCL UCL160 - 180 -36.2 6.416224e-03 ** -60.69202 -11.70798160 - 200 -74.2 8.438627e-06 *** -98.69202 -49.70798160 - 220 -155.8 3.728560e-10 *** -180.29202 -131.30798180 - 200 -38.0 4.624381e-03 ** -62.49202 -13.50798180 - 220 -119.6 1.693894e-08 *** -144.09202 -95.10798200 - 220 -81.6 2.683834e-06 *** -106.09202 -57.10798

$groupsNULL


Which pairwise comparison method do I use?

No clear cut(明確的) answer to this question

Carmer & Swanson (1973): Monte Carlo simulation studiesI LSD is a very effective test for detecting true differences in means

if it applied only after the F test in the ANOVA is significant at 5percent

I But LSD does not contain the experimentwise error rate(實驗誤差率)

I Tukey method does control the overall error rate

Other multiple comparison procedures are recommended inliteratures.


Comparing Treatment Means with a Control

Comparing each of the other a− 1 treatment means with thecontrol

H0 : µi = µa H1 : µi 6= µa

Dunnett’s procedure(鄧奈特): a modification of the t-test

reject H0 if |yi· − ya·| > dα(a− 1, f )

√MSE

(1ni

+1na

)dα(a− 1, f ) in Appendix Table VIII

α: the joint significance level


Comparing Treatment Means with a Control (cont.)

##Dunnett’s Procedure> K <- rbind( "160 - 220" = c( 1, 0, 0, -1),+ "180 - 220" = c( 0, 1, 0, -1),+ "200 - 220" = c( 0, 0, 1, -1))>> dunnett2 <- glht(etch.aov,linfct=mcp(FRF=K))> summary(dunnett2)

Simultaneous Tests for General Linear Hypotheses

Multiple Comparisons of Means: User-defined Contrasts

Fit: aov(formula = rate ˜ FRF, data = etch)

Linear Hypotheses:Estimate Std. Error t value Pr(>|t|)

160 - 220 == 0 -155.80 11.55 -13.485 < 1e-06 ***180 - 220 == 0 -119.60 11.55 -10.352 < 1e-06 ***


Comparing Treatment Means with a Control (cont.)

200 - 220 == 0 -81.60 11.55 -7.063 1.93e-06 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Adjusted p values reported -- single-step method)

more observations for the control treatment than other treatment

Choose na/n =√

a


Determing Sample Size

Operating Characteristic Curves (OC curve;操作特徵曲線): a plotof the type II error probability

Appendix, Table V: β vs. Φ (Φ2 =n∑a

i=1 τ2i

aσ2 )


Determing Sample Size (cont.)

Used to guide the experimenter in selecting the number ofreplicates so that the design will be sensitive to importantpotential differences in the treatments.

Type II error of the fixed effects model:

β = 1−P{Reject H0|H0 is false} = 1−P{F0 > Fα,a−1,N−a|H0 is false}

the test statistic F0 if H0 is false:

F0 =MSTreatment

MSE

H0 is false∼ noncentral F with a− 1,N − a, δ

Φ2 is related to δ



A very common way to use these charts is to define a differencein two means D of interest, then the minimum value of Φ2 is

Φ2 =nD2

2aσ2

Typically work in term of the ratio of D/σ and try values of nuntil the desired power is achieved

Plasma etching experiment

Reject H0 with probability at least 0.9 (α = 0.01)

any two treatment means differed by as much as 75A/mm

σ = 75 psi⇒ Φ2 = n(75)2

2(4)(252)= 1.125n

n = 4 (d.f . : 3, 12)⇒ Φ = 2.12⇒ Power = 1− β ≈ 0.65



Confidence Interval Estimation Method

95% C.I.:

±tα/2,N−a

√2MSE

n

n = 5; σ2 = 252 ⇒ ±tα/2,N−a

√2MSE

n = ±33.52

n = 7; σ2 = 252 ⇒ ±tα/2,N−a

√2MSE

n = ±27.58

±30A/min


The Random Effects Model

A Single Random Factor

linear statistical model:

yij = µ+ τi + εij

{i = 1, 2, . . . , aj = 1, 2, . . . , n

1 εij ∼ NID(0, σ2)2 τi ∼ NID(0, σ2

τ )3 τi⊥εij

V(yij) = σ2τ + σ2

Cov(yij, yij′) = σ2τ j 6= j′; Cov(yij, yi′j′) = σ2

τ i 6= i′


The Random Effects Model (cont.)

y =

y11

y12

y21

y22

y31

y32

⇒Cov(y) =

σ2τ + σ2 σ2

τ 0 0 0 0σ2τ σ2

τ + σ2 0 0 00 0 σ2

τ + σ2 σ2τ 0 0

0 0 σ2τ σ2

τ + σ2 0 00 0 0 0 σ2

τ + σ2 σ2τ

0 0 0 0 σ2τ σ2

τ + σ2


ANOVA for the Random Model

SS:SST = SSTreatment + SSE

Testing Hypotheses:

H0 : σ2τ = 0 vs. H1 : σ2

τ > 0

(If σ2τ = 0⇒ all treatments are identical, but variability exists

between treatments)

SSE/σ2 ∼ χ2

N−a; SSTreatments/σ2 H0∼ χ2

a−1; Both are indipendent

⇒F0 =SSTreatments

a−1SSE

N−1

=MSTreatments

MSE

H0∼ Fa−1,N−a


ANOVA for the Random Model (cont.)

The expected mean squares: Under H0, these two components areunbiased estimators of σ2

E(MSTreatments) = σ2 + nσ2τ

E(MSE) = σ2

Decision:Reject H0 if F0 > Fα,a−1,N−a

The computational procedure and ANOVA for the random effectsmodel are identical to those for the fixed effects case.



Estimating the Model Parameters{E(MSTreatments) = σ2 + nσ2

τ

E(MSE) = σ2 ⇒{σ2 = MSE

σ2τ = MSTreatments−MSE

n

method of moments procedure: not require the normalityassumption

σ2, σ2τ : best quadratic unbiased (minimum variance)



εiji.i.d.∼ N(0, σ2)⇒ (N−a)MSE

σ2 ∼ χ2N−a

100(1− α)% C.I. for σ2:

(N − a)MSE

χ2α/2,N−a

≤ σ2 ≤ (N − a)MSE

χ21−α/2,N−a

σ2τ = MSTreatments−MSE

n(a−1)MSTreatments

σ2+nσ2τ

∼ χ2a−1;

(N−a)MSEσ2 ∼ χ2

N−a

σ2 ⇒ linear combination of χ2a−1 and χ2

N−a, i.e.,

⇒ u1χ2a−1 − u2χ

2N−a, u1 =

σ2 + nσ2τ

n(a− 1), u2 =

σ2

n(N − a)



Intraclass correlation coefficient: σ2τ

σ2τ+σ

2 (MSTreatments ⊥ MSE)

⇒MSTreatments/(nσ2

τ + σ2)

MSE/σ2∼ Fa−1,N−a

⇒P(

F1−α/2,a−1,N−a ≤MSTreatments

MSE

σ2

nσ2τ + σ2

≤ Fα/2,a−1,N−a

)= 1− α

⇒P

1n

(MSTreatmens

MSE

1Fα/2,a−1,N−a

− 1

)(L)

≤σ2τ

σ2≤

1n

(MSTreatmens

MSE

1F1−α/2,a−1,N−a

− 1

)(U)

= 1− α

100(1− α)% C.I. for σ2τ

σ2τ+σ

2 :

L1 + L

≤ σ2τ

σ2τ + σ2 ≤

U1 + U



Estimation of the Overall Mean µ:

µ = y··

V(y··) = nσ2τ+σ

2

an ⇒ V(y··) = MSTreatmentsan

100(1− α)% C.I. on µ:

y·· − tα/2,a(n−1)

√MSTreatmens

an≤ µ ≤ y·· + tα/2,a(n−1)

√MSTreatmens

an


Nonparametric Methods in the Analysis of Variance

Kruskal-Wallis Test (1952)

The normality assumption is unjustified(未被證明為正當的).1 rank yij in ascending(上升的) order2 replace yij by its rank Rij; (ties: average rank to each of the tied

observations)3 Ri·: the sum of the ranks in the ith treatment4 It ni are reasonably large⇒ H

H0∼ χ2a−1:

Reject H0: Test statistic H > χ2a−1


Nonparametric Methods in the Analysis of Variance (cont.)

5 Test statistic:

H =1S2

[a∑

i=1

R2i·

ni− N(N + 1)2

4

]

S2 =1

N − 1

a∑i=1

ni∑j=1

R2ij −

N(N + 1)2

4

(Variance of the ranks)

no ties=

N(N + 1)

12

Using ANOVA ranks F0 = H/(a−1)(N−aH)/(N−a) is equivalent to H


Nonparametric Methods in the Analysis of Variance (cont.)

> kruskal.test(rate ˜ FRF, data = etch)

Kruskal-Wallis rank sum test

data: rate by FRFKruskal-Wallis chi-squared = 16.907, df = 3, p-value = 0.0007386


Chapter 3 Analysis of Variance (ANOVA; ”””†††óóó555ŠŠŠ...Relation between...

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