Wilcoxon Rank Sum (Mann Whitney) test-SAS Results Interpretation

I need a clarification on interpreting Wilcoxon Rank Sum (Mann Whitney) test results using SAS. Here is a sample results look like. I am trying to form a hypothesis and interpret the results. In this case, is this the test for average or medians, since the results display the mean score. Also should I use the normal approximation or t approximation for the p-value.

• Actually the Mann-Whitney-Wilcoxon is neither a test for means or medians without additional assumptions. In full generality it's a test for whether one variable is stochastically larger than the other (in a one-tailed test on continuous variates, it's a test for the alternative $P(X>Y)>\frac{1}{2}$; in a two tailed test, it tests both that case and the one with one inequality - but not both - flipped). Under some assumptions (e.g. alternative is a location shift) it can be a test for equality of both median and mean. We have many questions on site which discuss this issue. Oct 6, 2016 at 0:18
• See some of the discussion here and here You may also get some value from this, this ... ctd Oct 6, 2016 at 0:40
• and maybe this Oct 6, 2016 at 0:40
• While it's not a test for means, note that the "sum of scores" and "mean score" that your output is talking about would be the sum of ranks and mean of ranks for each sample. The test does compare mean ranks but this doesn't relate to a comparison of means for the populations whose observations were ranked. Oct 6, 2016 at 0:59
• Here's a demonstration that it doesn't test a difference in medians -- a pair of samples with the same median but where the proportion of $x_i>y_i$ is considerably smaller than $\frac12$ (enough to lead to rejection at the 5% level). x: 9.1, 9.15, 9.2, 9.25, 9.3, 9.35, 9.4, 9.9, 10.1, 11.1, 11.15, 11.2, 11.25, 11.3, 11.35, 11.4 and y: 9.5, 9.55, 9.6, 9.65, 9.7, 9.75, 9.8, 9.85, 10, 11.5, 11.55, 11.6, 11.65, 11.7, 11.75, 11.8, 11.85. (A roughly corresponding way to set up two populations should be clear enough - mixtures of uniforms should suffice) Oct 6, 2016 at 5:42

Interpretation: As a test of location, one tests whether the data location is the same for two independent populations. The null hypothesis implies that there is a significant difference in location of the populations, and the alternative hypothesis implies that a significant difference was not detected under the conditions of the test. When the probability of this test is low [most often $p<0.05$, with a Type I error (AKA alpha, false positive rate) set at 0.05] we would accept the null (H0) hypothesis as the likelihood of the alternative hypothesis (H1) is low.
The Mann-Whitney U test uses the U-statistic as a measure of location. "U" stands for Unbiased. Such statistics arise in the context of minimum variance unbiased estimators, or UVME for short. The U-statistic has the property that the average over sample values $ƒ_n(x\phi)$ is exactly equal to the population value $ƒ_N(x)$ for a simple random sample $\phi$ of size $n$ taken from a population of size $N$.