# Kruskal-Wallis test

## Kruskal-Wallis test

The Kruskal-Wallis test is a generalized U-test for more than two groups. It tests H0 that data from k populations are not different.

## Requirements:

Data must be ordinal (rank-order) scaled. Distribution is free

## Idea:

The test works like the Mann-Whintey U-test. The data from all groups are brought together in one rank order. For each group the sum of ranks Ti and mean rank is then computed. Whereas the total sum of ranks is: with

k = number of groups

N = Total Number of measurements

The test value H is computed as follows: whereas

ni = sample size of group i

H is Chi-Square distributed with k-1 degrees of freedom

If there are tied ranks H is corrected as follows: whereas

ti = Number of subjects sharing rank i

p = number of tied ranks

and ## Post-hoc analysis

If the Kruskal-Wallis test is significant one probably wants to know which of the groups are different. BrightStat offers two different methods for post-hoc analysis:

The critical difference of the mean ranks after Conover (1971, 1980, 1999): whereas = critical difference of mean ranks of group i and j = critical t-value with N-k degrees of freedom

ni = sample size of group i

nj = sample size of group j

The critical difference of the mean ranks after Schaich and Hamerle (1984): whereas = critical difference of mean ranks of group i and j = critical Chi-Square-value with k-1 degrees of freedom

ni = sample size of group i

nj = sample size of group j

The method after Schaich and Hamerle is exact but lacks a bit of power, whereas the method of Conover is approximative and more liberal.

## Example of a Kruskal-Wallis test

A meteorologist has measured the amount of rain in four cities for six months. She wants to know if there are different amounts of rain in the four cities. The following table shows the raw data:

 Cities 1 RANK 2 RANK 68 8 119 22 93 16 116 21 123 24 101 17 83 14 103 18 108 19 113 20 122 23 84 15 SUM 104 113 MEAN 17.33 18.83

 Cities 3 RANK 4 RANK 70 10.5 61 5 68 8 54 1.5 54 1.5 59 3.5 73 12 67 6 81 13 59 3.5 68 8 70 10.5 SUM 53 30 MEAN 8.83 5

H is then computed as follows:   Because there are tied ranks H is corrected  the corrected H’ is then The critical 5% Chi-Square with 3 degrees of freedom is 7.81

The observed test-value is greater than the critical Chi-Square, so there must be some differences in the amount of rain between the four cities.

We might be interested in the critical difference of the mean ranks so we can check which cities are different from each other. Because each group has 6 measurements we get one critical difference for all comparisons:

After Conover we get  and after Schaich and Hamerle we get  we can now compare the differences of the group mean ranks with the two critical differences:

 CITY_1 CITY_2 CITY_3 CITY_2 -1.5 - - CITY_3 8.5 * 10 * - CITY_4 12.33 * ° 13.83 * ° 3.83

* significant difference after Conover

° significant difference after Schaich and Hamerle

BrightStat output of Kruskal-Wallis test example

This is a fictitious example.

## References

Bortz, J. (2005). Statistik für Human- und Sozialwissenschaftler (6th Edition). Heidelberg: Springer Medizin Verlag.

Conover, W.J. (1999). Practical nonparametric Statistics.(3rd edition). Wiley.

Kruskal, W.H. & Wallis, W.A. (1952). Use of ranks in one-criterion variance analysis. Journal of the American Statistical Association, 47 (260), 583 – 621.

Schaich, H.E. & Hamerle, A. (1984). Verteilungsfreie statistische Prüfverfahren, Berlin.

## Gallery

### PayPal Donation ### Donate IOTA 