91-501

Compare Several Populations with

Unknown Distributions

(the Kruskal-Wallis test)

The Kruskal-Wallis (KW) Test for Comparing

Populations with Unknown Distributions

A nonparametric test for comparing population

medians by Kruskal and Wallis (KW Test)

The KW procedure tests the null hypothesis that k samples from possibly different populations actually originate from similar populations, at least as far as their central tendencies, or medians, are concerned. The test assumes that the variables under consideration have underlying continuous distributions.

In what follows assume we have k samples, and the sample size of the i-th sample is ni where i = 1, 2, . . ., k.

Test based on Ranks of Combined Data

In the computation of the KW statistic, each observation is replaced by its rank in an ordered combination of all the k samples.

By this we mean that the data from the k samples combined are ranked in a single series. The minimum observation is replaced by a rank of 1, the next-to-the-smallest by a rank of 2, and the largest or maximum observation is replaced by the rank of N, where N is the total number of observations in all the samples (N is the sum of the ni).

Compute the sum of the ranks for

each sample

The next step is to compute the Sum of

the Ranks for each of the original samples.

The KW test determines whether these

sums of ranks are so different by sample

that they are not likely to have all come

from the same population.

Test statistic follows a

2 distribution It can be shown that if the k samples come from the same

population, that is, if the null hypothesis is true, then the

test statistic, H, used in the KW procedure is distributed

approximately as a chi-square statistic with df = k - 1,

provided that the sample sizes of the k samples are not

too small (say, ni>4, for all i). H is defined as follows:

where

k = number of samples (groups)

ni = number of observations for the i-th sample or group

N = total number of observations (sum of all the ni)

Ri = sum of ranks for group i

=12

( + 1)

2

=1 3( + 1)

Example

The following data are from a comparison of four (4)

investment firms. The observations represent

percentage of growth during a three month period

for recommended funds.

A B C D

4.2 3.3 1.9 3.5

4.6 2.4 2.4 3.1

3.9 2.6 2.1 3.7

4.0 3.8 2.7 4.1

2.8 1.8 4.4

Step 1: Express the data in terms

of their ranks

A B C D

17 10 2 11

19 4.5 4.5 9

14 6 3 12

15 13 7 16

8 1 18

SUM 65 41.5 17.5 66

Compute the test statistic

The corresponding H test statistic is

=12

19(20)

652

4+41.52

5+17.52

5+662

5 3 20 = 13.678

=12

( + 1)

2

=1 3( + 1)

From the Chi-Square table, the critical value for =0.05 with d.o.f. = k-1=3 is 7.81. Since 13.678 > 7.81, we reject

the null hypothesis (H0). Note that the rejection region for

the KW-test procedure is One-Sided, since we only reject

the null hypothesis when the H-statistic is too large.

Example: Four heat-treatment schedules are tested against a certain

metal material, the internal stress is measured. There are

totally 24 samples tested, with 6 samples treated under a

specific schedule. The results are shown in the following

table. Using Kruskal-Wallis method, determine if the four

treatment schedule produce the same result in terms of the

internal stress.

Schedule A Schedule B Schedule C Schedule D

value rank value rank value rank value rank

4.5 21.5 3.8 12 3.5 8.0 3.0 4

5.0 24 4.0 16.5 4.5 21.5 2.8 3

3.5 8 3.9 13.5 3.2 5 2.2 2

3.7 11 4.2 19 2.1 1 3.4 6

4.8 23 3.6 10 3.5 8.0 4.0 16.5

4.0 16.5 4.4 20 4.0 16.5 3.9 13.5

104 91 60 45 SUM

=12

24(25)

1042

6+912

6+602

6+452

6 3 25 = 7.41

From the Chi-Square table, the critical value for =0.05 with d.o.f. = k-1=3 is 7.81. Since 7.41 < 7.81, we accept

the null hypothesis (H0), i.e., we do not have strong

evidence to reject the null hypothesis that the mean

internal stress are all equal under the four different heat

treatment schedule.