Property 1

Proof: By Property 1 of Wilcoxon Rank Sum Test, R1 + R= n(n+1)/2. Thus

Property 2: For n1 and n2 large enough the U statistic is approximately normal N(μ, σ) where

Proof: From Property 2 of Wilcoxon Rank Sum Test, the mean of R1 is $\frac {n_1(n_1+n_2+1)}{2}$. Similarly the mean of R2 is $\frac {n_2(n_1+n_2+1)}{2}$. Since

it follows from Property 3 of Expectation that the mean of U1 is

Similarly the mean of U2 is $\frac{n_1 n_2}{2}$. Since U = min(U1U2), it follows that the mean of U is also $\frac{n_1 n_2}{2}$.

By Property 3 of Expectation, the variance of U1 is the same as the variance of R1, which by Property 2 of Wilcoxon Rank Sum Test is $\frac {n_1 n_2(n_1+n_2+1)}{12}$.

Similarly the variance of U2 is the same as the variance of R2, which is again $\frac {n_1 n_2(n_1+n_2+1)}{12}$. Thus the variance of U is this same amount.

### 2 Responses to Mann-Whitney Test – Advanced

1. Pseudo says:

The materials on this site are excellent. Thank you for all of the work that has gone into generating them. I was wondering if you could provide a bit more detail on the normal approximation of U. The referenced proof for W invokes the central limit theorem, but I don’t see how that is applicable here. That would seem to reduce to showing that U (or W) is the mean of some distribution. The 1947 Mann Whitney paper presents a fairly complex derivation of the limit of U, without using the central limit theorem. Thanks!

• Charles says:

Yes, you are correct. I just changed the referenced webpage to reflect this. Thanks for catching this mistake.
Charles