# Wilcoxon Rank Sum Test – Advanced

Property 1: Suppose sample 1 has size n1 and rank sum R1 and sample 2 has size n2 and rank sum R2, then R1 R2 = n(n+1)/2 where n = n1 n2.

Proof: This is simply a consequence of the fact that the sum of the first n positive integers is $\frac{n(n+1)}{2}$. This can be proven by induction. For n = 1, we see that $\frac{n(n+1)}{2} = \frac{1(1+1)}{2}$ = 1 = n. Assume the result is true for n, then for n + 1 we have,  1 + 2 + … + n + (n+1) = $\frac{n(n+1)}{2}$ + (n + 1) = $\frac{n(n+1)+2(n+1)}{2}$$\frac{(n+1)(n+2)}{2}$

Property 2: When the two samples are sufficiently large (say of size > 10, although some say 20), then the W statistic is approximately normal N(μ, σ) where

Proof: We prove that the mean and variance of W = R1 are as described above. The normal approximation was proven in Mann & Whitney (1947) of Bibliography and we won’t repeat the proof here.

Let xi = the rank of the ith data element in the smaller sample. Thus, under the assumption of the null hypothesis, by Property 1

By Property 4a of Expectation

As we did in the proof of Property 1, we can show by induction on n that

From these it follows that

We can now calculate the following expectations:

Also where i ≠ j

By Property 2 of Expectation

By Property 3 of Basic Concepts of Correlation

By an extended version of Property 5 of Basic Concepts of Correlation

### 4 Responses to Wilcoxon Rank Sum Test – Advanced

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4. fra says: