# Kruskall-Wallis test

#### EDIEZ40

##### New Member
Hello,

I am trying to apply the Kruskall-Wallis test to compare the means of 5 different groups. Could anybody confirm if I need to have at least 5 observations per group in order to obtain accurate results? Is the test sensitive to the number of observations/ group then?

#### ohms_law

##### New Member
Hi EDIEZ40,

I'm no expert on this by any means (I actually just started reading about the Kruskal-Wallis one-way analysis of variance since you posted this), so hopefully someone will come along who knows more about this than I do.

Based on what I'm reading here: Kruskal-Wallis one-way analysis of variance, it looks like the answer to your question is yes:
4. Finally, the p-value is approximated by \Pr(\chi^2_{g-1} \ge K). If some ni's are small (i.e., less than 5) the probability distribution of K can be quite different from this chi-square distribution. If a table of the chi-square probability distribution is available, the critical value of chi-square, \chi^2_{\alpha: g-1}, can be found by entering the table at g − 1 degrees of freedom and looking under the desired significance or alpha level. The null hypothesis of equal population medians would then be rejected if K \ge \chi^2_{\alpha: g-1}. Appropriate multiple comparisons would then be performed on the group medians.
That sounds to me like, if there are less than 5 observations it's pretty much guaranteed that the data won't fit into a chi-square distribution. That being the case, you then couldn't trust the value.

Anyway, thanks for posting this. Non-parametric statistics gives me an interesting avenue to pursue with the question I posted earlier.