# Mann-Whitney vs. t-test to compare two samples with different distributions

How should I test the statistical difference of a mean (or median?) concentration of an enzyme between two groups of patients where the distribution in one group is normal, whereas in other it is non-normal? The distribution of values for the entire study population is non-normal. Do I use a Mann-Whitney test, or Student's T-test?

• What are your sample sizes? – Underminer Dec 20 '13 at 19:48
• On what basis can you assert the distribution of one of the populations? – Glen_b -Reinstate Monica Dec 21 '13 at 0:37

The first thing you should do is to decide on what question is really important to answer. You say that the 2 groups have different distributions, is the difference in mean or median really the most important thing when you already know of a distributional difference?

Consider that you are looking at 2 different treatments to lower cholesterol. The first will actually raise the cholesterol in most patients, but a few will have very drastic reductions (along with medical complications either due to or resulting in the drop in cholesterol), the second treatment will result in a moderate reduction in cholesterol in all patients with some small variation between the patients. Given the above information, do you really care which has higher mean/median reduction?

If you decide that the mean or median difference really is a question of interest (and what will you do if the mean favors one group and the median favors the other group?) then there are some additional questions to ask. How non-normal is the 2nd group? (and how non-normal is the 1st group, are you really sure it is normal? or just close enough)? If the non-normality is not severe (how severe can be a function of the sample size and shape) then the t-test could be reasonable. Other approaches that do not require the normality and let you compare means or medians are permutation and bootstrap tests (but make sure that you are comfortable with the assumptions required for those). The Mann-Whitney/Wilcoxon test is actually a special case of the permutation test (but the statistic being compared is not the median (or mean) without further assumptions).

One assumption of the two sample t-test is that both samples come from a normal population. If you are convinced your samples both do not come from normal populations, you could use the Wilcoxon rank-sum test to test for difference in medians.

• This is not true. This is a necessary requirement for the t-test to be exact. However, we often use asymptotic tests, such as the maximum likelihood test for inference in regression models with specified working probability models. – AdamO Apr 23 '13 at 20:00
• @AdamO, $P$-values from the $t$-test can be horribly inaccurate if its assumptions are not satisfied. Non-normality messes up the standard deviation more than it messes up the mean. – Frank Harrell Jul 23 '13 at 11:44
• @Frank I used to think that, too, until I did some simulations and was surprised at some of the "bad" situations in which the t-test worked well. The issue is how much non-normality messes up the standard error (not the SD) relative to how the sampling mean varies. Now I adopt a more nuanced attitude: the t-test can be remarkably resistant to many kinds of non-normality. It appears to get into most trouble with strongly skewed distributions (and I'm not persuaded that some of the efforts to incorporate skewness adjustments are any good except in very special circumstances). – whuber Aug 22 '13 at 13:04
• Rand Wilcox has shown that a tiny bit of contamination from a normal with higher variance, not even noticeable when plotting the density, can make the $t$ statistic have a distribution that is very far from the $t$ distribution. – Frank Harrell Aug 22 '13 at 13:39
• For the sake of others who may read this thread in the future: The question refers to the Mann-Whitney test, and @user27008 refers to the Wilcoxon rank-sum test. These are two different names for the same test. – Harvey Motulsky Aug 22 '13 at 14:14