# Mann Whitney Wilcoxon test

My stats background is not very strong so I was hoping I'd be able to get some help. In my MSc thesis I am using R to test for significant differences in means between populations. My data is not normally distributed (based on QQplots, Shapiro-Wilk's and histograms), therefore I cannot use parametric tests. So I am using a Mann Whitney Wilcoxon test instead (wilcox.test in R).

Now from my understanding Mann Whitney Wilcoxon test for significant differences in medians, not means. I have a table with my data reported as means, but I report significance testing results in my thesis based on medians as calculated by R. Is this an appropriate approach? Or do I need to convert the means to medians in my table?

Thanks.

• It doesn't test for difference in medians. It corresponds to a test for the median pairwise difference (in the population). You can get a rejection when the sample medians are equal. Many posts on site discuss this issue. – Glen_b Mar 18 '18 at 9:37

The Wilcoxon test is acutally not a test of medians but of ranks. There is a median test, but that apparently has low power(1) ist thus rarely used (R obviously has a package - you know! (2)). You might also want to look in the wilcox.test- help pages to read about a "pseudomedian": help(wilcox.test)
Chances are, that you can actually use parametric tests in the absence of normality, if you have enough data. It is hard to tell, how much data is enough for how big a deviance from normality. Some say $n>30$ or $n>50$ might bie good rules of thumb.