# Wilcoxon Test - non normality, non equal variances, sample size not the same

I know there are already a lot of posts out there, but I couldn't find this exact combination in any of them. Comparing two samples (Prices associated with men and with women), but I have neither the same sample size ($$n = 790$$ vs $$n=795$$) nor equal variance or normality.

My hypothesis is whether the prices for Women are greater than prices for Men. The Wilcoxon test is significant on a 4 % level.

Can I actually say anything helpful since so many assumptions are violated? Would another test be better?

Prices Women: Median 28.00, Mean 28.47,  Std. Dev. 17.17,  Skewness 0.91, Kurtosis 5.41
Prices Men:   Median 26.00, Mean, 29.08, Std. Dev. 22.43,  Skewness 2.39, Kurtosis 12.74

• The Wilcoxon test doesn’t assume normality or equal variances, so what’s the worry? Another way to think about the data would be to work on logarithmic scale. Log price may behave better than price. Dec 28 '18 at 10:12
• I found some other resources which state the opposite: data.library.virginia.edu/the-wilcoxon-rank-sum-test "This is where the Wilcoxon Rank Sum Test comes in. It only makes the first two assumptions of independence and equal variance." Or: kasuya.ecology1.org/stats/utest01e.html Dec 28 '18 at 11:09
• Independence, sure. Dec 28 '18 at 12:41
• The assumptions of the Wilcoxon rank-sum test are well described at en.wikipedia.org/wiki/…. They do not include any of the characteristics you mention.
– whuber
Dec 28 '18 at 16:02
• @whuber: the wiki page you linked seems to disagree with you! Specifically 3. Under the null hypothesis H0, the distributions of both populations are equal implies equal variance (if exists). Dec 29 '18 at 22:34

tl;dr if you want to interpret the rejection of the null hypothesis as evidence that prices for women are greater than those for men, then you do need the assumption of equal variance (in fact, equal distributions) between the two populations. If you are satisfied with showing that the distribution of prices for women differs in some way from that of men, then you don't need the extra assumption.

You don't need to worry about unequal sample size (this will affect the power of the test, but not its validity) or Normality.

For what it's worth, testing whether one group's values are larger on average than another group's when their variances also differ is a surprisingly deep question, even for Normally distributed data (where it's known as the Behrens-Fisher problem).

Under more strict assumptions than the general formulation above, e.g., if the responses are assumed to be continuous and the alternative is restricted to a shift in location, i.e., $$F_1(x) = F_2(x + δ)$$, we can interpret a significant Mann–Whitney U test as showing a difference in medians ...