# Kruskal-Wallis one-way analysis of variance

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**Kruskal-Wallis test**, which can be applied when you can not make the assumption that the groups follow a gaussian distribution.

This test is similar to the Wilcoxon test for 2 samples.

Suppose you want to see if the means of the following 4 sets of values are statistically similar:

Group A: 1, 5, 8, 17, 16

Group B: 2, 16, 5, 7, 4

Group C: 1, 1, 3, 7, 9

Group D: 2, 15, 2, 9, 7

Group B: 2, 16, 5, 7, 4

Group C: 1, 1, 3, 7, 9

Group D: 2, 15, 2, 9, 7

To use the test of Kruskal-Wallis simply enter the data, and then organize them into a list:

a = c(1, 5, 8, 17, 16) b = c(2, 16, 5, 7, 4) c = c(1, 1, 3, 7, 9) d = c(2, 15, 2, 9, 7) dati = list(g1=a, g2=b, g3=c, g4=d)

Now we can apply the

`kruskal.test()`

function:kruskal.test(dati) Kruskal-Wallis rank sum test data: dati Kruskal-Wallis chi-squared = 1.9217, df = 3, p-value = 0.5888

The value of the test statistic is 1.9217. This value already contains the fix when there are ties (repetitions). The p-value is greater than 0.05; also the value of the test statistic is lower than the chi-square-tabulation:

qchisq(0.950, 3) [1] 7.814728

The conclusion is therefore that I accept the null hypothesis H0: the means of the 4 groups are statistically equal.

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