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

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SST ¼ 3:2987  2:6040 ¼ :6947

SSA ¼

½2:39

þ 1:38

þ 1:82

2:6040 ¼ :1282

SSB ¼

½2:41

þ 1:01

þ 1:27

þ :90

2:6040 ¼ :4797

SSE ¼ :6947 ð:1282 þ :4797Þ¼:0868

The accompanying ANOVA table (Table 11.5) summarizes further calculations.

The critical value for testing H

at level of significance .05 is F

.05,2,6

¼ 5.14.

Since 4.43 < 5.14, H

cannot be rejected at significance level .05. Based on this

(small) data set, we cannot conclude that true average color change depends on

brand of pen. Because F

.05,3,6

¼ 4.76 and 11.05  4.76, H

is rejected at signifi-

cance level .05 in favor of the assertion that color change varies with washing

treatment. A statistical computer package gives P-values of .066 and .007 for these

two tests.

How can plausibility of the normality and constant variance assumptions be

investigated graphically? Define the predicted valu es (also called fitted values)

m þ





þð



i







Þþð



j







Þ¼



i



j







, and the residuals

(the differences between the observations and predicted values)



¼ x





i





j





. We can check the normality assumption with a nor-

mal plot of the residuals, Figure 11.9(a), and we can check the constant variance

assumption with a plot of the residuals against the fitt ed values, Figure 11.9(b).

Table 11.5 ANOVA table for Example 11.13

Source of Variation df Sum of Squares Mean Square f

Factor A (pen brand) I1 ¼ 2 SSA ¼ .1282 MSA ¼ .0641 f

¼ 4.43

Factor B (wash

treatment)

J1 ¼ 3 SSB ¼ .4797 MSB ¼ .1599 f

¼ 11.05

Error (I1)(J1) ¼ 6 SSE ¼ .0868 MSE ¼ .01447

Total IJ1 ¼ 11 SST ¼ .6947

Residual

Normal Probability Plot of the Residuals

Fitted Value

0.15

0.10

0.05

0.0

−0.5

−0.10

Residuals Versus the Fitted Values

Percent

Residual

0.20.10.0−0.1−0.2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure 11.9 Plots from MINITAB for Example 11.13

588

CHAPTER 11 The Analysis of Variance

The normal plot is reasonably straight, so there is no reason to question

normality for this data set. On the plot of the residua ls against the fitted values, we

are looking for differences in vertical spread as we move horizontally across the

graph. For example, if there were a narrow range for small fitted values and a

wide range for high fitted values, this would suggest that the variance is higher for

larger responses (this happens often, and it can sometimes be cured by replacing

each observation by its logarithm). No such problem occurs here, so there is no

evidence against the constant variance assumption.

■

Expected Mean Squares

The plausibility of using the F tests just described is demonstrated by determining

the expected mean squares. After some tedious algebra,

E(MSE) ¼ s

(when the model is additive)

E(MSA) ¼ s

I  1

i¼1

E(MSB) ¼ s

J  1

j¼1

When H

is true, MSA is an unbiased estimator of s

,soF is a ratio of two unbiased

estimators of s

. When H

is false, MSA tends to overestimate s

,soH

should be

rejected when the ratio F

is too large. Similar comments apply to MSB and H

Multiple Comparisons

When either H

or H

has been rejected, Tukey’s procedure can be used to

identify significant differences between the levels of the factor under investigation.

The steps in the analysis are identical to those for a single-fa ctor ANOVA:

1. For comparing levels of factor A, obtain Q

a,I,(I1)(J1)

For comparing levels of factor B, obtain Q

a, J,(I1)(J1)

2. Compute

w ¼ Q·(estimated standard deviation of the sample means being compared)

a;I;ðI1ÞðJ1Þ



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

MSE J

for factor A comparisons

a; J;ðI1ÞðJ1Þ



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

MSE I

for factor B comparisons

(

(because, e.g., the sta ndard deviation of

i

is s=

ﬃﬃﬃ

3. Ar range the sample means in increasing order, underscor e those pairs differing

by less than w, and identify pairs not underscored by the same line as

corresponding to significantly different levels of the given factor.

Example 11.14

(Example 11.13

continued)

Identification of significa nt differences among the four washing treatments requires

.05,4,6

¼ 4.90 and w ¼ 4.90

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

:01447=3

¼ :340. The four factor B sample means

(column averages) are now listed in increasing order, and any pair differing by less

than .340 is underscored by a line segment:

11.4 Two-Factor ANOVA with K

¼ 1 589



4



2



3



1

.300 .337 .423 .803

Washing treatment 1 is significantly worse than the other three treatments, but no

other significant differences are identified. In particular, it is not apparent which

among treatments 2, 3, and 4 is best at removing marks.

■

Randomized Block Experiments

In using single-factor ANOVA to test for the presence of effects due to the

I different treatments under study, once the IJ subjects or experimental units have

been chosen, treatments should be allocated in a completely random fashion. That

is, J subjects should be chosen at random for the first treatment, then another

sample of J chosen at random from the remaining IJ – J subjects for the second

treatment, and so on.

It frequently happens, though, that subjects or exper imental units exhibit

differences with respect to other characteristics that may affect the observed

responses. For example, some patients might be healthier than others. When this

is the case, the presence or absence of a significant F value may be due to these

differences rather than to the presence or absence of factor effects. This was the

reason for introducing a paired experim ent in Chapter 10. The generalization of the

paired experiment to I > 2 is called a randomized block experiment. An extrane-

ous factor, “blocks,” is constructed by dividi ng the IJ units into J groups with I units

in each group. This grouping or blocking is done in such a way that within each

block, the I units are homogeneous with respect to other factor s thought to affect the

responses. Then within each homogeneous block, the I treatments are rando mly

assigned to the I units or subjects in the block.

Example 11.15 A consumer product-testing organization wished to compare the annual power

consumption for five different brand s of dehumidifier. Because power consumption

depends on the prevailing humi dity level, it was decided to monitor each brand at

four different levels ranging from moderate to heavy humidity (thus blocking on

humidity level). Within each level, brand s were randomly assigned to the five

selected locations. The resulting amount of power consumption (annual kWh)

appears in Table 11.6.

Table 11.6 Power consumption data for Example 11.15

Blocks (humidity level)

Treatments (brands) 1 2 3 4 x

i



i

1 685 792 838 875 3190 797.50

2 722 806 893 953 3374 843.50

3 733 802 880 941 3356 839.00

4 811 888 952 1005 3656 914.00

5 828 920 978 1023 3749 937.25

·j

3779 4208 4541 4797 17,325

590

CHAPTER 11 The Analysis of Variance

Since

¼ 15;178;901:00 and x



=ðIJÞ¼15;007;781:25

SST ¼ 15;178;901:00  15;007;781:25 ¼ 171;119:75

SSA ¼

½60;244;04915;007;781:25 ¼ 53;231:00

SSB ¼

½75;619;99515;007;781:25 ¼ 116;217 :75

and

SSE ¼ 171;119:75  53;231:00  116;217:75 ¼ 1671:00

The ANOVA calculations are summarized in Table 11.7

Since F

.05,4,12

¼ 3.26 and f

¼ 95.57  3.26, H

is rejected in favor of H

, and

we conclude that power consumption does depend on the brand of humidifier.

To identify significantly different brands, we use Tukey’s procedure. Q

.05,5,12

4.51 and w ¼ 4.51

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

139:25=4

¼ 26:6.



1



3



2



4



5

797.50 839.00 843.50 914.00 937.25

The underscoring indicates that the brands can be divided into three groups with

respect to power consumption.

Because the block factor is of secondary interest, F

.05,3,12

is not needed,

though the computed value of F

is cle arly highly significant. Figure 11.10

shows SAS output for this data. Notice that in the first part of the ANOVA table,

the sums of squares (SS’s) for treatments (brands) and blocks (humidity levels) are

combined into a single “model” SS.

In many experimental situations in which treatments are to be applied to

subjects, a single subject can receive all I of the treatments. Blocking is then often

done on the subjects themselves to control for variability between subjects; each

subject is then said to act as its own control. Social scientists sometimes refer to

such experiments as repeated-measures designs. The “units” within a block are then

the different “instances” of treatment application. Similarly, blocks are often taken

as different time periods, locations, or observers.

Table 11.7 ANOVA table for Example 11.15

Source of Variation df Sum of Squares Mean Square f

Treatments (brands) 4 53,231.00 13,307.75 f

¼ 95.57

Blocks 3 116,217.75 38,739.25 f

¼ 278.20

Error 12 1671.00 139.25

Total 19 171,119.75

11.4 Two-Factor ANOVA with K

¼ 1 591

In most randomized block experiments in which subjects serve as blocks, the

subjects actually participating in the experiment are selected from a large popula-

tion. The subjects then contribute random rather than fixed effects. This does not

affect the procedure for comparing treatments when K

¼ 1 (one observation per

“cell,” as in this section), but the procedure is altered if K

¼ K > 1. We will

shortly consid er two-factor models in which effects are random.

More on Blocking When I ¼ 2, either the F test or the paired differences t test can

be used to analyze the data. The resulting conclusion will not depend on which

procedure is used, since T

¼ F and t

a=2;n

¼ F

a;1;n

Just as with pairing, blocking entails both a potential gain and a potential loss

in precision. If there is a great deal of heterogeneity in experimental units, the value

of the variance parameter s

in the one-way model will be large. The effect of

blocking is to filter out the variation represented by s

in the two-way model

appropriate for a randomized block experiment. Other things being equal, a smaller

Analysis of Variance Procedure

Dependent Variable: POWERUSE

Sum of Mean

Source DF Squares Square F Value Pr > F

Model 7 169448.750 24206.964 173.84 0.0001

Error 12 1671.000 139.250

Corrected Total 19 171119.750

R-Square C.V. Root MSE POWERUSE Mean

0.990235 1.362242 11.8004 866.25000

Source DF Anova SS Mean Square F Value Pr > F

BRAND 4 53231.000 13307.750 95.57 0.0001

HUMIDITY 3 116217.750 38739.250 278.20 0.0001

Alpha = 0.05 df = 12 MSE = 139.25

Critical Value of Studentized Range = 4.508

Minimum Significant Difference = 26.597

Means with the same letter are not significantly different.

Tukey Grouping

Mean

937.250

914.000

843.500

839.000

797.500

BRAND

Figure 11.10 SAS output for consumption data ■

592 CHAPTER 11 The Analysis of Variance

value of s

results in a test that is more likely to detect departures from H

(i.e., a

test with greater power).

However, other things are not equal here, since the single-factor F test is

based on I(J – 1) degrees of freedom (df) for error, whereas the two-factor F test is

based on (I – 1)(J – 1) df for error. Fewer degrees of freedom for error results in a

decrease in power, essentially because the denominator estimator of s

is not as

precise. This loss in degrees of freedom can be especially serious if the experi-

menter can afford only a small number of observations. Nevertheless, if it appears

that blocking will significantly reduce variability, it is probably worth the loss in

degrees of freedom.

Models for Random Effects

In many experiments, the actual levels of a factor used in the experiment, rather

than being the only ones of interest to the experimenter, have been selected from a

much larger population of possible levels of the factor. In a two-factor situation,

when this is the case for both factors, a random effects model is appropriate. The

case in which the levels of one factor are the only ones of interest and the levels of

the other factor are selected from a population of levels leads to a mixed effects

model. The two-factor random effects model when K

¼ 1is

¼ m þ A

þ B

þ e

i ¼ 1; :::; I; j ¼ 1; :::; JðÞ

where the A

’s, B

’s, and e

’s are all independent, normally distributed rv’s

with mean 0 and variances s

, and s

, respectively.

The hypotheses of interest are then H

: s

¼ 0 (level of factor A does not contrib-

ute to variation in the response) versus H

: s

> 0 and H

: s

¼ 0 versus

: s

> 0. Whereas E(MSE) ¼ s

as before, the expected mean squares for

factors A and B are now

E MSAðÞ¼s

þ Js

E MSBðÞ¼s

þ Is

Thus when H

) is true, F

) is still a ratio of two unbiased estimators of s

It can be shown that a test with significance level a for H

versus H

still rejects

if f

 F

a,I1,(I1)(J1)

, and, similarly, the same proce dure as before is used to

decide between H

and H

For the case in which factor A is fixed and factor B is random, the mixed

model is

¼ m þ a

þ B

þ e

i ¼ 1; :::; I; j ¼ 1; :::; JðÞ

where

¼ 0, and the B

’s, and e

’s are all independent, normally

distributed rv’s with mean 0 and variances s

and s

, respective ly.

11.4 Two-Factor ANOVA with K

¼ 1 593

Now the two null hypotheses are

: a

¼¼a

¼ 0 and H

: s

¼ 0

with expected mean squar es

E MSEðÞ¼s

E MSAðÞ¼s

I  1

E MSBðÞ¼s

þ Is

The test procedures for H

versus H

and H

versus H

are exactly as before.

For example, in the analysis of the color change data in Example 11.11, if the

four wash treatments were randomly selected, then because f

¼ 11.05 and

.05,3,6

¼ 4.76, H

: s

¼ 0 is rejected in favor of H

: s

> 0. An estimate of

the “variance component” s

is then given by (MSB – MSE)/I ¼ .0485.

Summarizing, when K

¼ 1, although the hypotheses and expected

mean squares differ from the case of both effects fixed, the test procedures are

identical.

Exercises Section 11.4 (35–48)

35. The number of miles of useful tread wear (in

1000’s) was determined for tires of each of five

different makes of subcompact car (factor A,

with I ¼ 5) in combination with each of four

different brands of radial tires (factor B, with

J ¼ 4), resulting in IJ ¼ 20 observations. The

values SSA ¼ 30.6, SSB ¼ 44.1, and SSE ¼

59.2 were then computed. Assume that an addi-

tive model is appropriate.

a. Test H

: a

¼ a

¼ 0 (no

differences in true average tire lifetime due

to makes of cars) versus H

: at least one

6¼ 0 using a level .05 test.

b. H

: b

¼ b

¼ 0 (no differences

in true average tire lifetime due to brands of

tires) versus H

: at least one b

6¼ 0 using a

level .05 test.

36. Four different coatings are being considered for

corrosion protection of metal pipe. The pipe will

be buried in three different types of soil. To

investigate whether the amount of corrosion

depends either on the coating or on the type of

soil, 12 pieces of pipe are selected. Each piece is

coated with one of the four coatings and buried in

one of the three types of soil for a fixed time,

after which the amount of corrosion (depth of

maximum pits, in .0001 in.) is determined. The

depths are shown in this table:

Soil Type (B)

123

Coating (A)

1 64 49 50

2 53 51 48

3 47 45 50

4 51 43 52

a. Assuming the validity of the a dditive model,

carry out the ANOVA analysis using an

ANOVA table to see whether the amount of

corrosion depends on either the type of coat-

ing used or the type of soil. Use a ¼ .05.

b. Compute

;

; and

37. The data set shown below is from the article

“Compounding of Discriminative Stimuli from

the Same and Different Sensory Modalities”

(J. Exp. Anal. Behav., 1971: 337–342). Rat

response was maintained by fixed interval sche-

dules of reinforcement in the presence of a tone or

two separate lights. The lights were of either mod-

erate (L1) or low (L2) intensity. Observations are

given as the mean number of responses emitted by

594

CHAPTER 11 The Analysis of Variance

each subject during single and compound stimuli

presentations over a 4-day period. Carry out an

appropriate analysis.

38. In an experiment to see whether the amount of

coverage of light-blue interior latex paint

depends either on the brand of paint or on the

brand of roller used, 1 gallon of each of four

brands of paint was applied using each of three

brands of roller, resulting in the following data

(number of square feet covered).

Roller Brand

123

1 454 446 451

Paint 2 446 444 447

Brand 3 439 442 444

4 444 437 443

a. Construct the ANOVA table. [Hint: The com-

putations can be expedited by subtracting 400

(or any other convenient number) from each

observation. This does not affect the final

results.]

b. State and test hypotheses appropriate for

deciding whether paint brand has any effect

on coverage. Use a ¼ .05.

c. Repeat part (b) for brand of roller.

d. Use Tukey’s method to identify significant

differences among brands. Is there one brand

that seems clearly preferable to the others?

e. Check the normality and constant variance

assumptions graphically.

39. In an experiment to assess the effect of the angle

of pull on the force required to cause separation

in electrical connectors, four different angles

(factor A) were used and each of a sample of

five connectors (factor B) was pulled once at

each angle (“A Mixed Model Factorial Experi-

ment in Testing Electrical Connectors,” Indust.

Qual. Control, 1960: 12–16). The data appears in

the accompanying table.

12345



45.3 42.2 39.6 36.8 45.8



44.1 44.1 38.4 38.0 47.2



42.7 42.7 42.6 42.2 48.9



43.5 45.8 47.9 37.9 56.4

Does the data suggest that true average separa-

tion force is affected by the angle of pull? State

and test the appropriate hypotheses at level .01

by first constructing an ANOVA table (SST

¼ 396.13, SSA ¼ 58.16, and SSB ¼ 246.97).

40. A particular county employs three assessors who

are responsible for determining the value of resi-

dential property in the county. To see whether

these assessors differ systematically in their

assessments, 5 houses are selected, and each

assessor is asked to determine the market value

of each house. With factor A denoting assessors

(I ¼ 3) and factor B denoting houses (J ¼ 5),

suppose SSA ¼ 11.7, SSB ¼ 113.5, and SSE

¼ 25.6.

a. Test H

: a

¼ a

¼ 0 at level .05. (H

states that there are no systematic differences

among assessors.)

b. Explain why a randomized block experiment

with only 5 houses was used rather than a

one-way ANOVA experiment involving a

total of 15 different houses with each assessor

asked to assess 5 different houses (a different

group of 5 for each assessor).

Subject

Stimulus 1 2 3 4 x

i



i

L1 8.0 17.3 52.0 22.0 99.3 24.8

L2 6.9 19.3 63.7 21.6 111.5 27.9

Tone (T) 9.3 18.8 60.0 28.3 116.4 29.1

L1 + L2 9.2 24.9 82.4 44.9 161.4 40.3

L1 + T 12.0 31.7 83.8 37.4 164.9 41.2

L2 + T 9.4 33.6 96.6 40.6 180.2 45.1

j

54.8 145.6 438.5 194.8 833.7

11.4 Two-Factor ANOVA with K

¼ 1 595

41. The article “Rate of Stuttering Adaptation

Under Two Electro-Shock Conditions” (Behav.

Res. Therapy, 1967: 49–54) gives adaptation

scores for three different treatments: (1) no

shock, (2) shock following each stuttered

word, and (3) shock during each moment of

stuttering. These treatments were used on each

of 18 stutterers.

a. Summary statistics include x

1

¼ 905; x

2

913; x

3

¼ 936; x



¼ 2754;

j

¼ 430; 295

and

¼ 143; 930. Construct the

ANOVA table and test at level .05 to see

whether true average adaptation score

depends on the treatment given.

b. Judging from the F ratio for subjects (factor

B), do you think that blocking on subjects was

effective in this experiment? Explain.

42. The article “The Effects of a Pneumatic Stool

and a One-Legged Stool on Lower Limb Joint

Load and Muscular Activity During Sitting and

Rising” (Ergonomics, 1993: 519–535) gives the

accompanying data on the effort required of a

subject to arise from four different types of stools

(Borg scale). Perform an analysis of variance

using a ¼ .05, and follow this with a multiple

comparisons analysis if appropriate.

Subject

1 23456789



i

Type

Stool

1 121077898798.56

2 15 14 14 11 11 11 12 11 13 12.44

3 12 13 13 10 8 11 12 8 10 10.78

4 10 12 9 9 7 10 11 7 8 9.22

43. The strength of concrete used in commercial

construction tends to vary from one batch to

another. Consequently, small test cylinders of

concrete sampled from a batch are “cured” for

periods up to about 28 days in temperature- and

moisture-controlled environments before

strength measurements are made. Concrete is

then “bought and sold on the basis of strength

test cylinders” (ASTM C 31 Standard Test

Method for Making and Curing Concrete Test

Specimens in the Field). The accompanying

data resulted from an experiment carried out to

compare three different curing methods with

respect to compressive strength (MPa). Analyze

this data.

Batch Method A Method B Method C

1 30.7 33.7 30.5

2 29.1 30.6 32.6

3 30.0 32.2 30.5

4 31.9 34.6 33.5

5 30.5 33.0 32.4

6 26.9 29.3 27.8

7 28.2 28.4 30.7

8 32.4 32.4 33.6

9 26.6 29.5 29.2

10 28.6 29.4 33.2

44. Check the normality and constant variance assump-

tions graphically for the data of Example 11.15.

45. Suppose that in the experiment described in Exer-

cise 40 the five houses had actually been selected

at random from among those of a certain age and

size, so that factor B is random rather than fixed.

Test H

: s

¼ 0 versus H

: s

> 0 using a level

.01 test.

46. a. Show that a constant d can be added to (or

subtracted from) each x

without affecting

any of the ANOVA sums of squares.

b. Suppose that each x

is multiplied by a nonzero

constant c. How does this affect the ANOVA

sums of squares? How does this affect the

values of the F statistics F

and F

? What

effect does “coding” the data by y

¼ cx

have on the conclusions resulting from the

ANOVA procedures?

47. Use the fact that EX



¼ m þ a

þ b

with Sa

¼ 0 to show that E(X

i

 X



) ¼ a

, so that

¼ X

i

 X



is an unbiased estimator for a

48. The power curves of Figures 11.5 and 11.6 can be

used to obtain b ¼ P(type II error) for the F test in

two-factor ANOVA. For fixed values of a

, a

, ...,

, the quantity f

¼ðJ=IÞ

is computed.

Then the figure corresponding to v

¼ I–1is

entered on the horizontal axis at the value f, the

power is read on the vertical axis from the curve

labeled v

¼ (I–1)(J–1), and b ¼ 1 – power.

a. For the corrosion experiment described in

Exercise 36, find b when a

¼ 4, a

¼ 0, a

¼2, and s ¼ 4. Repeat for a

¼ 6,

¼ 0, a

¼ a

¼3, and s ¼ 4.

b. By symmetry, what is b for the test of H

versus

in Example 11.11 when b

¼ .3, b

¼ b

¼ –.1, and s ¼ .3?

596 CHAPTER 11 The Analysis of Variance

11.5

Two-Factor ANOVA with K

> 1

In Section 11.4, we analyzed data from a two-factor exper iment in which there

was one observation for each of the IJ combinations of levels of the two factors. To

obtain valid test procedures, the m

’s were assumed to have an

additive struct ure with m

¼ m þ a

þ b

, Sa

¼ Sb

¼ 0. Additivity means that

the difference in true average responses for any two levels of the factors is the

same for each level of the other factor. For example, m

 m

ðm þ a

þ b

Þðm þ a

þ b

Þ¼a

 a

independent of the level j of the second

factor. This is shown in Figure 11.7(a), in which the lines connecting true average

responses are parallel.

Figure 11.7(b) depicts a set of true average responses that does not have

additive structure. The lines connecting these m

’s are not parallel, which

means that the difference in true average responses for different levels of one

factor does depend on the level of the other factor. When additivity does not

hold, we say that there is interaction between the different levels of the factors.

The assumption of additivity allowed us in Section 11.4 to obtain an estimator of

the random error variance s

(MSE) that was unbiased whether or not either null

hypothesis of interest was true. When K

> 1 for at least one (i, j) pair, a valid

estimator of s

can be obtained without assuming additivity. In specifying the

appropriate model and deriving test procedures, we will focus on the case K

¼ K

> 1, so the number of observations per “cell” (for each combination of levels)

is constant.

Parameters for the Fixed Effects Model with Interaction

Rather than use the m

’s themselves as model parameters, it is usual to use an

equivalent set that reveals more clearly the role of interaction. Let

m ¼



i



j

ð11:15Þ

Thus m is the expected response averaged over all levels of both factors (the true

grand mean),



i

is the expected resp onse averaged over leve ls of the second factor

when the first factor A is held at level i, and similarly for



j

. Now d efine



i

 m ¼ the effect of factor A at level i



j

 m ¼ the effect of factor B at level j

¼ m

ðm þ a

þ b

ð11:16Þ

from which

¼ m þ a

þ b

þ g

ð11:17Þ

11.5 Two-Factor ANOVA with K

> 1 597