Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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Source of Variation df Sum of Squares Mean Square f

Treatments 2 7.93 3.965 12.56

Error 19 6.00 .3158

Total 21 13.93

■

Multiple Comparisons When Sample Sizes Are Unequal

There is more controversy among statisticians regarding which multiple compar-

isons procedure to use when sample sizes are unequal than there is in the case of

equal sample sizes. The procedure that we present here is recommended in the

excellent book Beyond ANOVA: Basics of Applied Statistics (see the chapter

bibliography) for use when the I sample sizes J

, J

, ..., J

are reasonably close

to each other (“mild imbalance”). It modifies Tukey’s method by using averages of

pairs of 1/J

’s in place of 1/J.

Let

¼ Q

a;I;nI



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

MSE



Then the probability is approximately 1–a that

i

 X

j

 w

 m

 m

 X

i

 X

j

þ w

for every i and j (i ¼ 1, ..., I and j ¼ 1, ..., I) with i 6¼ j.

The simultaneous confidence level 100(1 – a)% is only approximate rather than

exact as it is with equal sample sizes. The underscoring method can still be used,

but now the w

factor used to decide whether



i

and



can be connected will

depend on J

and J

Example 11.9

(Example 11.8

continued)

The sample sizes for the elastic modulus data were J

¼ 8, J

¼ 6, and

I ¼ 3, n – I ¼ 19, MSE ¼ .316. A simultaneous confidence level of approximately

95% requires Q

.05,3,19

¼ 3.59, from which

¼ 3:59

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

:316



¼ :713 w

¼ :771 w

¼ :771

Since



1





2

¼ 44:71  44:06 ¼ :65 < w

, m

and m

are judged not signifi-

cantly different. The accompanying underscoring scheme shows that m

and m

differ significantly, as do m

and m

2. Die 1. Permanent 3. Plaster

44.06 44.71 45.58

■

578 CHAPTER 11 The Analysis of Variance

Data Transformation

The use of ANOVA methods can be invalidated by substantial differences in the

variances s

; ...; s

(which until now have been assumed equal with common

value s

). It sometimes happens that VðX

Þ¼s

¼ gðm

Þ, a known function of m

(so that when H

is false, the variances are not equal). For example, if X

has a

Poisson distribution with parameter l

(approximately normal if l

 10), then

¼ l

and s

¼ l

,sog(m

) ¼ m

is the known function. In such cases, one can

often transform the X

’s to h(X

)’s so that they will have approximately equal

variances (while hopef ully leaving the transformed variables approximately nor-

mal), and then the F test can be used on the transf ormed observations. The basic

idea is that, if h(·) is a smooth function, then we can express it approximately using

the first terms of a Taylor series, h(X

)  h(m

)+h

)(X

– m

). Then V[h(X

)] 

V(X

)·[h

)]

¼ g(m

)  [h

)]

. We now wish to find the function h(·) for which

g(m

)·[h

)]

¼ c (a constant) for every i. Solving this for h

) and integr ating

gives the following result:

PROPOSITION

If V(X

) ¼ g(m

), a known functi on of m

, then a transformation h(X

) that

“stabilizes the variance” so that V[h(X

)] is approximately the same for each i

is given by h(x) /

[g(x)]

1/2

dx.

In the Poisson case, g(x) ¼ x,soh(x) should b e proportional to

1/2

dx ¼

2 x

1/2

. Thus Poisson data should be transformed to hðx

Þ¼

ﬃﬃﬃﬃﬃ

before the analysis.

A Random Effects Model

The single-factor problems considered so far have all been assumed to be exam ples

of a fixed effects ANOVA model. By this we mean that the chosen levels of the

factor under study are the only ones considered relevant by the experimenter. The

single-factor fixed effects model is

¼ m

þ a

þ e

¼ 0 ð11:6Þ

where the e

’s are random and both m and the a

’s are fixed parameters whose

values are unknown.

In some single-factor problems, the particular levels studied by the experi-

menter are chosen, either by design or through sampling, from a large population of

levels. For example, to study the effects on tas k performance time of using different

operators on a particular machine, a sample of five operators might be chosen from

a large pool of operators. Similarly, the effect of soil pH on the yield of maize plants

might be studied by using soils with four specific pH values chosen from among the

many possible pH levels. When the levels used are selected at random from a larger

population of possible levels, the factor is said to be random rather than fixed, and

the fixed effects model (11.6) is no longer appropriate. An analogous random

effects model is obtained by replacing the fixed a

’s in (11.6) by random variables.

The resulting model description is

11.3 More on Single-Factor ANOVA 579

¼ m þ A

þ e

with EðA

Þ¼Eðe

Þ¼0

Vðe

Þ¼s

VðA

Þ¼s

ð11:7Þ

with all A

’s and e

’s normally distributed and independent of each other.

The condition E(A

) ¼ 0in(11.7) is similar to the condition Sa

¼ 0in(11.6); it

states that the expected or average effect of the ith level measured as a departure

from m is zero.

For the random effects model (11.7), the hypothesis of no effects due to

different levels is H

: s

¼ 0 which says that different levels of the factor contrib-

ute nothing to variability of the response. Although the hypotheses in the single-

factor fixed and random effects models are different, they are tested in exactly the

same way, by forming F ¼ MSTr/MSE and rejecting H

if f  F

a,I–1,n–I

. This can

be justified intuitively by noting that E(MSE) ¼ s

(as for fixed effects), whereas

EðMSTr) ¼ s

I  1

n 



ð11:8Þ

where J

, J

,..., J

are the sample sizes and n ¼ SJ

. The factor in parentheses on

the right side of (11.8) is nonnegative, so once again E(MSTr) ¼ s

if H

is true

and E(MSTr) > s

if H

is false.

Example 11.10 The study of nondestructive forces and stresses in materials furnishes important

information for efficient design. The article “Zero-Force Travel-Time Parameters

for Ultrasonic Head-Waves in Railroad Rail” (Mater. Eval., 1985: 854–858)

reports on a study of travel time for a type of wave that results from longitudinal

stress of rails used for railroad track. Three measurements were made on each of six

rails randomly selected from a population of rails. The investigators used random

effects ANOVA to decide whether some variation in travel time could be attributed

to “between-rail variability.” The data is given in the accompanying table (each

value, in nanoseconds, resulted from subtracting 36.1 ms from the original obser-

vation) along with the derived ANOVA table. The value of the F ratio is highly

significant, so H

: s

¼ 0 is rejected in favor of the conclusion that differences

between rails are a source of travel-time variability.

Source of

Variation

Sum of

Squares

Mean

Square

Treatments 5 9310.5 1862.1 115.2

Error 12 194.0 16.17

Total 17 9504.5

■

Rail Travel time x

i·

1 555354162

226373295

3 789185254

4 92 100 96 288

5 495150150

6808583

248

··

¼ 1197

580

CHAPTER 11 The Analysis of Variance

Exercises Section 11.3 (22–34)

22. The following data refers to yield of tomatoes (kg/

plot) for four different levels of salinity; salinity

level here refers to electrical conductivity (EC),

where the chosen levels were EC ¼ 1.6, 3.8, 6.0,

and 10.2 nmhos/cm:

1.6: 59.5 53.3 56.8 63.1 58.7

3.8: 55.2 59.1 52.8 54.5

6.0: 51.7 48.8 53.9 49.0

10.2: 44.6 48.5 41.0 47.3 46.1

Use the F test at level a ¼ .05 to test for any

differences in true average yield due to the differ-

ent salinity levels.

23. Apply the modified Tukey’s method to the data in

Exercise 22 to identify significant differences

among the m

’s.

24. The following partial ANOVA table is taken from

the article “Perception of Spatial Incongruity” (J.

Nerv. Ment. Dis., 1961: 222) in which the abilities

of three different groups to identify a perceptual

incongruity were assessed and compared. All indi-

viduals in the experiment had been hospitalized to

undergo psychiatric treatment. There were 21

individuals in the depressive group, 32 individuals

in the functional “other” group, and 21 individuals

in the brain-damaged group. Complete the

ANOVA table and carry out the F test at level

a ¼ .01.

Source df Sum of Squares Mean Square f

Groups 76.09

Error

Total 1123.14

25. Lipids provide much of the dietary energy in the

bodies of infants and young children. There is a

growing interest in the quality of the dietary lipid

supply during infancy as a major determinant of

growth, visual and neural development, and long-

term health. The article “Essential Fat Require-

ments of Preterm Infants” (Amer. J. Clin. Nutrit.,

2000: 245S–250S) reported the following data on

total polyunsaturated fats (%) for infants who were

randomized to four different feeding regimens:

breast milk, corn-oil-based formula, soy-oil-based

formula, or soy-and-marine-oil-based formula:

Regimen

Sample

Size

Sample

Mean

Sample

Breast milk 8 43.0 1.5

CO 13 42.4 1.3

SO 17 43.1 1.2

SMO 14 43.5 1.2

a. What assumptions must be made about the four

total polyunsaturated fat distributions before

carrying out a single-factor ANOVA to decide

whether there are any differences in true aver-

age fat content?

b. Carry out the test suggested in part (a). What

can be said about the P-value?

26. Samples of six different brands of diet/imitation

margarine were analyzed to determine the level of

physiologically active polyunsaturated fatty acids

(PAPFUA, in percentages), resulting in the fol-

lowing data:

Imperial 14.1 13.6 14.4 14.3

Parkay 12.8 12.5 13.4 13.0 12.3

Blue Bonnet 13.5 13.4 14.1 14.3

Chiffon 13.2 12.7 12.6 13.9

Mazola 16.8 17.2 16.4 17.3 18.0

Fleischmann’s 18.1 17.2 18.7 18.4

(The preceding numbers are fictitious, but the

sample means agree with data reported in the Jan-

uary 1975 issue of Consumer Reports.)

a. Use ANOVA to test for differences among the

true average PAPFUA percentages for the dif-

ferent brands.

b. Compute CIs for all (m

– m

)’s.

c. Mazola and Fleischmann’s are corn-based,

whereas the others are soybean-based. Compute

a CI for

þ m



þ m

[Hint: Modify the expression for Vð

yÞ that led to

(11.5) in the previous section.]

27. Although tea is the world’s most widely consumed

beverage after water, little is known about its

nutritional value. Folacin is the only B vitamin

11.3 More on Single-Factor ANOVA 581

present in any significant amount in tea, and

recent advances in assay methods have made

accurate determination of folacin content feasible.

Consider the accompanying data on folacin con-

tent for randomly selected specimens of the four

leading brands of green tea.

Brand Observations

1 7.9 6.2 6.6 8.6 8.9 10.1 9.6

2 5.7 7.5 9.8 6.1 8.4

3 6.8 7.5 5.0 7.4 5.3 6.1

4 6.4 7.1 7.9 4.5 5.0 4.0

(Data is based on “Folacin Content of Tea,” J.

Amer. Dietetic Assoc. , 1983: 627–632.) Does this

data suggest that true average folacin content is the

same for all brands?

a. Carry out a test using a ¼ .05 via the P-value

method.

b. Assess the plausibility of any assumptions

required for your analysis in part (a).

c. Perform a multiple comparisons analysis to

identify significant differences among brands.

28. In single-factor ANOVA with sample sizes J

(i ¼

1, ..., I), show that SSTr ¼

ðX

i

 X



i

 nX



,wheren ¼

29. When sample sizes are equal (J

¼ J), the para-

meters a

, a

,..., a

of the alternative parameteri-

zation are restricted by Sa

¼ 0. For unequal

sample sizes, the most natural restriction is

¼ 0. Use this to show that

E(MSTr) ¼ s

I  1

What is E(MSTr) when H

is true? [This expec-

tation is correct if SJ

¼ 0 is replaced by the

restriction Sa

¼ 0 (or any other single linear

restriction on the a

’s used to reduce the model

to I independent parameters), but SJ

¼ 0

simplifies the algebra and yields natural estimates

for the model parameters (in particular,



i







).]

30. Reconsider Example 11.7 involving an investiga-

tion of the effects of different heat treatments on

the yield point of steel ingots.

a. If J ¼ 8 and s ¼ 1, what is b for a level

.05 F test when m

¼ m

, m

¼ m

– 1, and m

¼ m

+ 1?

b. For the alternative of part (a), what value of J is

necessary to obtain b ¼ .05?

c. If there are I ¼ 5 heat treatments, J ¼ 10, and

s ¼ 1, what is b for the level .05 F test when

four of the m

’s are equal and the fifth differs by

1 from the other four?

31. For unequal sample sizes,, the noncentrality para-

meter is

and f

¼ð1=IÞ

Referring to Exercise 22, what is the power of the

test when m

¼ m

, m

¼ m

– s,andm

¼ m

+ s?

32. In an experiment to compare the quality of four

different brands of reel-to-reel recording tape, five

2400-ft reels of each brand (A–D) were selected

and the number of flaws in each reel was deter-

mined.

A: 10 5 12 14 8

B: 14 12 17 9 8

C: 13 18 10 15 18

D: 17 16 12 22 14

It is believed that the number of flaws has approx-

imately a Poisson distribution for each brand. Ana-

lyze the data at level .01 to see whether the

expected number of flaws per reel is the same for

each brand.

33. Suppose that X

is a binomial variable with para-

meters n and p

(so it is approximately normal

when np

 10 and nq

 10). Then since m

, VðX

Þ¼s

¼ np

ð1  p

Þ¼m

ð1  m

=nÞ.

How should the X

’s be transformed so as to

stabilize the variance? [Hint: g(m

) ¼ m

(1 – m

/n).]

34. Simplify E(MSTr) for the random effects model

when J

¼ J

¼  ¼ J

¼ J.

11.4

Two-Factor ANOVA with K

¼ 1

In many experimental situations there are two factors of simultaneous interest. For

example, suppose an investigator wishes to study permeability of woven material

used to construct automobile air bags (related to the ability to absorb energy).

582 CHAPTER 11 The Analysis of Variance

An experiment might be carried out using I ¼ 4 temperature levels (10



C, 15



C, 25



C) and J ¼ 3 levels of fabric denier (420-D, 630-D, 840-D).

When factor A consists of I levels and factor B consists of J levels, there are IJ

different combinations (pairs) of levels of the two factors, each called a treatment.

With K

¼ the number of observations on the treatment consisting of factor A at

level i and factor B at level j, we focus in this section on the case K

¼ 1, so that the

data consists of IJ observations. We will first discuss the fixed effects model, in

which the only levels of interest for the tw o factors are those actually represented in

the experiment. The case in which one or both factors are random is discussed

briefly at the end of the section.

Example 11.11 Is it really as easy to remove marks on fabrics from erasable pens as the word

erasable might imply? Consider the following data from an experiment to compare

three different brands of pens and four different wash treatments with respect to

their ability to remove marks on a particular type of fabric (based on “An Ass ess-

ment of the Effects of Treatment, Time, and Heat on the Removal of Erasable Pen

Marks from Cotton and Cotton/Polyester Blend Fabrics,” J. Test. Eval., 1991:

394–397). The response variable is a quantitative indicator of overall specimen

color change; the lower this value, the more marks were removed.

Washing treatment

1 2 3 4 Total

1 .97 .48 .48 .46 2.39

Brand of Pen 2 .77 .14 .22 .25 1.38

3 .67 .39 .57 .19 1.82

Total 2.41 1.01 1.27 .90 5.59

Is there any difference in the true average amount of color change due either to the

different brand s of pen or to the different washing treatments?

■

As in single-factor ANOVA, double subscripts are used to identify random

variables and observed values. Let

¼ the random variable (rv) denoting the measurement when factor A is

held at level i and factor B is held at level j

¼ the observed value of X

The x

’s are usually presented in a two-way table in which the ith row contains the

observed values when factor A is held at level i and the jth column cont ains the

observed values when factor B is held at level j. In the erasable-pen experiment of

Example 11.11, the number of levels of factor A is I ¼ 3, the number of levels of

factor B is J ¼ 4, x

¼ .48, x

¼ .14, and so on.

Whereas in single-factor ANOVA we were interested only in row means and

the grand mean, here we are interested also in colu mn means. Let

i

the average of data obtained

when factor A is held at level i

j¼1

11.4 Two-Factor ANOVA with K

¼ 1 583

j

the average of data obtained

when factor B is held at level j

i¼1



¼ the grand mean ¼

i¼1

j¼1

with observed values



i



j

,and





. Totals rather than averages are denoted by

omitting the horizontal bar (so x

j

, etc.). Intuitively, to see whether there is

any effect due to the levels of factor A, we should compare the observed



i

’s with each

other, and information about the different levels of factor B should come from the



j

’s.

The Model

Proceeding by analogy to single-factor ANOVA, one’s first inclination in specify-

ing a model is to let m

¼ the true average response when factor A is at level i and

factor B at level j, giving IJ mean parameters. Then let

¼ m

þ e

where e

is the random amount by which the obser ved value differs from its

expectation and the e

’s are assumed normal and independent with common vari-

ance s

. Unfortunately, there is no valid test procedure for this choice of parameters.

The reason is that under the alternative hypothesis of interest, the m

’s are free to

take on any values whatsoever, whereas s

can be any value greater than zero, so that

there are IJ + 1 freely varying parameters. But there are only IJ observations, so

after using each x

as an estimate of m

, there is no way to estimate s

To rectify this problem of a model having more parameters than observed

values, we must specify a model that is realistic yet involves relatively few

parameters.

Assume the existence of I parameters a

, a

,..., a

and J parameters

, b

,..., b

such that

¼ a

þ b

þ e

ði ¼ 1; ...; I; j ¼ 1; ...; JÞð11:9Þ

so that

¼ a

þ b

ð11:10Þ

Including s

, there are now I + J + 1 model parameters, so if I  3 and J  3,

there will be fewer parameters than observations [in fact, we will shortly modify

(11.10) so that even I ¼ 2 and/or J ¼ 2 will be accommodated].

The model specified in (11.9) and (11.10) is called an additive model

because each mean response m

is the sum of an effect due to factor A at level i

) and an effect due to factor B at level j (b

). The difference between mean

584 CHAPTER 11 The Analysis of Variance

responses for factor A at level i and level i

when B is held at level j is m

– m

When the model is additive,

 m

¼ða

þ b

Þða

þ b

Þ¼a

 a

which is independent of the level j of the second factor. A similar result holds for

– m

. Thus additivity means that the difference in mean responses for two levels

of one of the factors is the same for all levels of the other factor. Figure 11.7(a) shows a

set of mean responses that satisfy the condition of additivity (which implies parallel

lines), and Figure 11.7(b) shows a nonadditive configuration of mean responses.

Example 11.12

(Example 11.11

continued)

When we plot the observed x

’s in a manner analogous to that of Figure 11.7,we

get the result shown in Figur e 11.8. Although there is some “crossing over” in the

observed x

’s, the configuration is reasonably representative of what would be

expected under additivity with just one observation per treatment.

Levels of A

Levels of B

Mean response

Levels of B

Mean response

Figure 11.7 Mean responses for two types of model: (a) additive; (b) nonadditive

Color change

1 2

Washin

treatment

Brand 1

3 4

1.0

Brand 2

Brand 3

Figure 11.8 Plot of data from Example 11.11 ■

11.4 Two-Factor ANOVA with K

¼ 1 585

Expression (11.10) is not quite the final model description because the a

’s

and b

’s are not uniquely determined. Following are two different configur ations of

the a

’s and b

’s that yield the same additive m

’s.

¼ 1 b

¼ 4 b

¼ 2 b

¼ 5

¼ 1 m

¼ 2 m

¼ 5 a

¼ 0 m

¼ 2 m

¼ 5

¼ 2 m

¼ 3 m

¼ 6 a

¼ 1 m

¼ 3 m

¼ 6

By subtracting any constant c from all a

’s and adding c to all b

’s, other config-

urations corresponding to the same additive model are obtained. This nonunique-

ness is eliminated by use of the following model.

¼ m þ a

þ b

þ e

ð11:11Þ

where

i¼1

¼ 0,

j¼1

¼ 0 and the e

’s are assumed independent,

normally distributed, with mean 0 and common variance s

This is analogous to the alternative choice of parame ters for single-factor ANOVA

discussed in Section 11.3. It is not difficult to verify that (11.11) is an additive

model in which the parameters are uniquely determined (e.g., for the m

’s men-

tioned previously, m ¼ 4, a

¼ –.5, a

¼ .5, b

¼1.5, and b

¼ 1.5). Notice that

there are only I – 1 independently determined a

’s and J – 1 independently

determined b

’s, so (including m)(11.11) specifies I + J – 1 mean parameters.

The interpretation of the parameters of (11.11) is straightforward: m is the true

grand mean (mean response averaged over all levels of both factors), a

is the effect

of factor A at level i (measured as a deviation from m), and b

is the effect of factor B

at level j. Unbiased (and maximum likelihood) estimators for these parameters are

m ¼



¼ X

i

 X



¼ X

j

 X



There are two different hypotheses of interest in a two-factor experiment with

¼ 1. The first, denoted by H

, states that the different levels of factor A have

no effect on true average response. The second, denoted by H

, asserts that there is

no factor B effect.

: a

¼ a

¼¼a

¼ 0

versus H

: at least one a

6¼ 0

: b

¼ b

¼¼b

¼ 0

versus H

: at least one b

6¼ 0

ð11:12Þ

(No factor A effect implies that all a

’s are equal, so they must all be 0 since they

sum to 0, and similarly for the b

’s.)

586 CHAPTER 11 The Analysis of Variance

Test Procedures

The description and analysi s now follow closely that for single-factor ANOVA.

The relevant sums of squares and their computing forms are given by

SST ¼

i¼1

j¼1

ðX

 X



i¼1

j¼1





df ¼ IJ  1

SSA ¼

i¼1

j¼1

ðX

i

 X



i¼1

i





df ¼ I  1

SSB ¼

i¼1

j¼1

ðX

j

 X



j¼1

j





df ¼ J  1

SSE ¼

i¼1

j¼1

ðX

 X

i

 X

j

þ X



df ¼ðI  1ÞðJ  1Þ

ð11:13Þ

and the fundamental identity

SST ¼ SSA þ SSB þ SSE ð11:14Þ

allows SSE to be determined by subtraction.

The expression for SSE results from replacing m, a

, and b

S½X

ðm þ a

þ b

Þ

by their respective estimators. Error df is IJ – number of

mean parameters estimated ¼ IJ –[1+(I –1)+(J – 1)] ¼ (I – 1)(J – 1). As in

single-factor ANOVA, total variation is split into a part (SSE) that is not explained

by either the truth or the falsity of H

or H

and two parts that can be explained by

possible falsi ty of the two null hypotheses.

Forming F ratios as in single-factor ANOVA, we can show as in Section 11.1

that if H

is true, the corresponding F ratio has an F distribution with numerator

df ¼ I – 1 and denominator df ¼ (I –1)(J – 1); an analogous result applies when

testing H

Hypotheses Test Statistic Value Rejection Region

versus H

MSA

MSE

 F

;I1;ðI1ÞðJ1Þ

versus H

MSB

MSE

 F

;J1;ðI1ÞðJ1Þ

Example 11.13

(Example 11.12

continued)

The x

i

’s (row totals) and x

j

’s (column totals) for the color change data are displayed

along the right and bottom margins of the data table in Example 11.11. In addition,

¼ 3:2987 and the correction factor is x



= IJðÞ¼5:59ðÞ

=12 ¼ 2:6040.

The sums of squares are then

11.4 Two-Factor ANOVA with K

¼ 1 587