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15 Distribution-Free Procedures

Copyright © Cengage Learning. All rights reserved.

15.4 Distribution-Free ANOVA

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Distribution-Free ANOVAThe single-factor ANOVA model for comparing I population or treatment means assumed that for i = 1, 2, . . . , I, a random sample of size Ji was drawn from a normal population with mean i and variance 2.

This can be written as

Xij = I + ij j = 1, . . . , Ji; i = 1, . . . , I

where the ij’s are independent and normally distributed with mean zero and variance 2. The next procedure for testing equality of the i’s requires only that the ij’s have the same continuous distribution.

(15.14)

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The Kruskal-Wallis Test

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The Kruskal-Wallis TestLet N = Ji, the total number of observations in the data set, and suppose we rank all N observations from 1 (the smallest Xij) to N (the largest Xij).

When H0: 1 = 2 = · · · = 1

is true, the N observations all come from the same distribution, in which case all possible assignments of the ranks 1, 2, . . . , N to the I samples are equally likely and we expect ranks to be intermingled in these samples.

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The Kruskal-Wallis TestIf, however, H0 is false, then some samples will consist mostly of observations having small ranks in the combined sample, whereas others will consist mostly of observations having large ranks.

More specifically, if Rij denotes the rank of Xij among the N observations, and Ri and Ri denote, respectively, the total and average of the ranks in the ith sample, then when H0 is true,

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The Kruskal-Wallis TestThe Kruskal-Wallis test statistic is a measure of the extent to which the Ri’s deviate from their common expected value (N + 1)/2, and H0 is rejected if the computed value of the statistic indicates too great a discrepancy between observed and expected rank averages.

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The Kruskal-Wallis TestTest Statistic

K

The second expression for K is the computational formula; it involves the rank totals (Ri’s) rather than the averages and requires only one subtraction.

If H0 is rejected when k c, then c should be chosen so that the test has level .

(15.15)

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The Kruskal-Wallis TestThat is, c should be the upper-tail critical value of the distribution of K when H0 is true.

Under H0, each possible assignment of the ranks to the I samples is equally likely, so in theory all such assignments can be enumerated, the value of K determined for each one, and the null distribution obtained by counting the number of times each value of K occurs.

Clearly, this computation is tedious, so even though there are tables of the exact null distribution and critical values for small values of the Ji’s, we will use the following “large-sample” approximation.

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The Kruskal-Wallis TestPropositionWhen H0 is true and either

I = 3 Ji 6 (i = 1, 2, 3)

or

I > 3 Ji 5 (i = 1, . . . , I )

then K has approximately a chi-squared distribution withI – 1 df.

This implies that a test with approximate significance level rejects H0 if k

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Example 9The accompanying observations (Table 15.6) on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths—4 in., 6 in., 8 in., 10 in., and 12 in.—were used (“Modeling Joints Made with Light-Gauge Metal Connector Plates,” Forest Products J., 1979: 39–44).

Table 15.6Data and Ranks for Example 9

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Example 9

The computed value of K is

K

At level .01, = 13.277, and since 20.12 13.277, H0 is

rejected and we conclude that expected axial stiffness does depend on plate length.

cont’d

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Friedman’s Test for a Randomized Block Experiment

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Friedman’s Test for a Randomized Block Experiment

Suppose Xij = + i + j + ij,

where i is the ith treatment effect, j is the jth block effect, and the ij’s are drawn independently from the same continuous (but not necessarily normal) distribution.

Then to test H0: 1 = 2 = · · · = 1 = 0,

the null hypothesis of no treatment effects, the observations are first ranked separately from 1 to I within each block, and then the rank average ri is computed for each of the I treatments.

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Friedman’s Test for a Randomized Block Experiment

When H0 is true, the ri’s should be close to one another, since within each block all I! assignments of ranks to treatments are equally likely.

Friedman’s test statistic measures the discrepancy between the expected value (I + 1)/2 of each rankaverage and the ri’s.

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Friedman’s Test for a Randomized Block Experiment

Test Statistic

As with the Kruskal-Wallis test, Friedman’s test rejects H0 when the computed value of the test statistic is too large.

For the cases I = 3, J = 2, . . . , 15 and I = 4, J = 2, . . . , 8,Lehmann’s book gives the upper-tail critical values for the test.

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Friedman’s Test for a Randomized Block Experiment

Alternatively, for even moderate values of J, the test statistic Fr has approximately a chi-squared distribution with I – 1 df when H0 is true, so H0 can be rejected if fr

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Example 10The article “Physiological Effects During Hypnotically Requested Emotions” (Psychosomatic Med., 1963: 334–343) reports the following data (Table 15.7) on skin potential (mV) when the emotions of fear, happiness, depression, and calmness were requested from each of eight subjects.

Table 15.7Data and Ranks for Example 10

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Example 10Thus

fr = (1686) – 3(8)(5)

= 6.45

At level .05, = 7.815, and because 6.45 < 7.815, H0 is not rejected.

There is no evidence that average skin potential depends on which emotion is requested.

cont’d