# Mann-Whitney null hypothesis under unequal variance

I'm just curious about the null hypothesis of a Mann-Whitney U test. I often see it stated that the null hypothesis is that two populations have equal distributions. But I'm thinking - if I had two normal populations with the same mean but extremely unequal variance, the Mann-Whitney test would probably not detect this difference.

I have also seen it stated that the null hypothesis of the Mann-Whitney test is $\Pr(X>Y)=0.5$ or the probability of an observation from one population ($X$) exceeding an observation from the second population ($Y$) (after exclusion of ties) is equal to 0.5. This seems to make a bit more sense but does not seem equivalent to the first null hypothesis I stated.

I'm hoping to get a bit of help untangling this. Thanks!

> out <- replicate( 100000, wilcox.test( rnorm(25, 0, 2), rnorm(25,0,10) )$p.value ) > hist(out) > mean(out < 0.05) [1] 0.07991 > prop.test( sum(out<0.05), length(out), p=0.05 ) 1-sample proportions test with continuity correction data: sum(out < 0.05) out of length(out), null probability 0.05 X-squared = 1882.756, df = 1, p-value < 2.2e-16 alternative hypothesis: true p is not equal to 0.05 95 percent confidence interval: 0.07824054 0.08161183 sample estimates: p 0.07991  So clearly this is rejecting more often than it should and the null hypothesis is false (this matches equality of distributions, but not prob=0.5). Thinking in terms of probability of X > Y also runs into some interesting problems if you ever compare populations that are based on Efron's Dice. • Hi Greg, thank you for the answer. It sounds like what you are saying is I found somewhat of a special case where the test does not work properly under the equal distributions null. And furthermore, the null hypotheses I stated are not equal. Is that correct? – Jimj Apr 20 '13 at 0:27 Mann-Whitney isn't sensitive to changes in variance with equal mean, but it can - as you see with the$P(X>Y)=0.5$form, detect differences that lead$P(X>Y)$to deviate from$0.5$(e.g. where both mean and variance increase together). Quite clearly if you had two normals with equal mean, their differences are symmetric about zero. Therefore$P(X>Y) = P(X-Y>0) = \frac{1}{2}$, which is the null situation. For example, if you have the distribution of$Y$being exponential with mean$1$while$X$has an exponential distribution with mean$k$(a scale change), the Mann-Whitney is sensitive to that (indeed, taking logs of both sides, its just a location-shift, and the Mann-Whitney is unaffected by monotonic transformation). -- If you're interested in tests which are conceptually very similar to the Mann-Whitney that are sensitive to differences in spread under equality of medians, there are several such tests. There's the Siegel-Tukey test and the Ansari-Bradley test, for example, both closely related to the Mann-Whitney-Wilcoxon two sample test. They are both based on the basic idea of ranking in from the ends. If you use R, the Ansari-Bradley test is built in ... ?ansari.test The Siegel-Tukey in effect just does a Mann-Whitney-Wilcoxon test on ranks computed from the sample differently; if you rank the data yourself, you don't really need a separate function for the p-values. Nevertheless, you can find some, as here: http://www.r-statistics.com/2010/02/siegel-tukey-a-non-parametric-test-for-equality-in-variability-r-code/ -- (in relation to ttnphns' comment under my original answer) You would be over-interpreting my response to read it as disagreeing with @GregSnow in any particularly substantive sense. There's certainly a difference in emphasis and to some extent in what we're talking about, but I'd be very surprised if there was much real disagreement behind it. Let's quote Mann and Whitney: "A statistic$U$depending on the relative ranks of the$x$'s and$y$'s is proposed for testing the hypothesis$f=g$." That's unequivocal; it utterly supports @GregSnow's position. Now, let's see how the statistic is constructed: "Let$U$count the number of times a$y$precedes an$x$." Now if their null is true, the probability of that event is$\frac{1}{2}$... but there are other ways to get a probability of 0.5, and in that sense one might construe that the test can work in other circumstances. To the extent that they're estimating a (re-scaled) probability that$Y$>$X$, it supports what I said. However, for the significance levels to be guaranteed to be exactly correct, you'd need the distribution of$U$to match the null distribution. That's derived on the assumption that all the permutations of the$X$and$Y$group-labels labels to the combined observations under the null were equally likely. This is certainly the case under$f=g$. Exactly as @GregSnow said. The question is the extent to which this is the case (i.e. that the distribution of the test statistic matches the one derived under the assumption that$f=g$, or approximately so), for the more generally expressed null. I believe that in many situations that it does; in particular for situations including but more general than the one you describe (two normal populations with the same mean but extremely unequal variance can be generalized quite a bit without altering the resulting distribution based on ranks), I believe that the distribution of the test statistic turns out to have the same distribution under which it was derived and so should be valid there. I did some simulations that seem to support this. However, it won't always be a very useful test (it may have poor power). I offer no proof that this is the case. I've applied some intuition/hand-wavy argument and also done some basic simulations that suggest it's true -- that the Mann-Whitney works (in that it has the 'right' distribution under the null) much more broadly than when$f=g$. Make of it what you will, but I don't construe this as substantive disagreement with @GregSnow Reference - Mann&Whitney's original paper • Did I get you right that you concur with this words from Wikipedia's Mann-Whitney talk page: the null hypothesis of Mann-Whitney U-test is not about the equality of distributions. Is is about the symmetry between two populations with respect to the probability of obtaining a larger observation. And so you don't agree with @Greg's answer, right? – ttnphns Apr 20 '13 at 5:52 • I have added some discussion in edit. – Glen_b -Reinstate Monica Apr 20 '13 at 7:14 • Very nice addition. I'll be studying it (I always felt as if there are nuances in MW test that continue to elude me). Meanwhile, will you agree if I'd say: "Because MW test statistic reflects just the (in)equality of mean ranks, there can be situations when f~=g [I understand f,g as original distributions, prior ranking] but the test is nevetherless fully relevant as it continues to deal with the same H0 as under f=g. An example of such situation is symmetric distributions fully identical except for spread parameter (variance)". – ttnphns Apr 20 '13 at 8:54 • In the notation (Mann and Whitney's by the way),$f$and$g$are the densities of$X$and$Y\$. I'd agree that to the extent that I have verified/understood the circumstances, your statement appears to be the case. I suspect there's still plenty about the Mann-Whitney that eludes me also. – Glen_b -Reinstate Monica Apr 20 '13 at 9:09