# wilcoxon mann whitney or t-test?

I would like to compare the number of a particular type of disease that is prevalent in men vs women. The number of diseases is interval data (0,1,2,3,etc). Should I use the wilcoxon mann whitney or t-test?

This is a sample of what the data looks like.

patientID   gender  num_dz
1   F   0
2   F   2
3   F   4
4   F   2
5   F   1
6   F   3
7   F   2
8   M   1
9   M   2
10  M   0
11  M   0
12  M   0
13  M   1
14  M   0

• Can you clarify the structure of your data? I.e., does your dataset boil down to just a 2 x 2 table of numbers (a count for men and women, with and without the disease)? Or if not, can you give a few sample rows to explain what your data looks like? Treating counts as interval or ordinal data might not be ideal; depending on the situation, a procedure specifically designed for count data should be considered, if count data is what you are dealing with. – Brent Kerby Jul 30 '17 at 17:10
• @BrentKerby I have edited the post with a sample of what the data looks like. – ybao Jul 30 '17 at 17:18
• Thanks, so then what does num_dz mean? I understand it as some kind of count of number of diseases, but you indicated that you are interested in a particular type of disease, so what does it mean for a patient to have a number of a certain type of disease? Does this represent a count of how many times they had the disease over a certain period of time? Or is it something like how many instances/varieties of the disease they are simultaneously infected with? Or is this not a count but rather a code indicating which disease they have? – Brent Kerby Jul 30 '17 at 17:29
• @BrentKerby The dataset is cross sectional, so it is the number of particular diseases that the patient has been diagnosed with at a single time point. The dataset is a study of a few chronic diseases, so the patients have them for life. – ybao Jul 30 '17 at 17:33
• Thanks, one more question: how large are the samples? – Brent Kerby Jul 30 '17 at 20:53

This depends on whether your data is normally distributed or not, and if the relationship is dependent or independent. In data with normal distribution the t-test is preferred, but for non-normal distributions the Wilcoxon–Mann–Whitney is fine. Also, Wilcoxon–Mann–Whitney is for independent samples; the Wilcoxon signed-rank test is for dependent samples.

• The data is non-normally distributed, and independent samples. Thanks! – ybao Jul 30 '17 at 17:03

The Wilcoxon-Mann-Whitney makes no distributional assumptions and so could always be applied. It has the advantage of providing exceptionally good robustness; for instance, if there were to be a few extremely bad points in your data (e.g., individuals with 10000+ diseases) these wouldn't heavily influence the Wilcoxon-Mann-Whitney test, whereas even one of these would destroy the t-test. Having said that, the t-test is also quite robust, especially for large sample sizes, just not to the same extent as the Wilcoxon-Mann-Whitney test. For large sample sizes, such as in your scenario, the t-test does not require an assumption that the data be normally distributed; it only needs to come from a distribution to which the Central Limit Theorem applies (i.e., one with finite variance). The main advantage of the t-test is that it typically provides somewhat higher power.

The two tests also address somewhat different hypotheses. For the t-test, the null hypothesis is that the two populations have equal means ($H_0: \mu_1 = \mu_2$), whereas for the Wilcoxon-Mann-Whitney test, the null hypothesis is that given random representatives $X$ and $Y$ from the first and second populations respectively, there is an equal probability that $X$ is greater than $Y$ compared to the other way around ($H_0: P(X>Y) = P(Y>X)$).

In the end, there are justifications for going with either test. It might come down to what is expected for your field/audience. Personally, I would probably go with the t-test, on account of its slightly greater power and more straightforward interpretation, assuming that an inspection of the data (say, looking at the two histograms) does not reveal any red flags (i.e., heavy tails or implausibly large values).