Mann-Whitney Test – Advanced

Property 1
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Proof: By Property 1 of Wilcoxon Rank Sum Test, R1 + R2 = n(n+1)/2. Thus

image3548 image3549

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

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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

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

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Similarly the mean of U2 is \frac{n_1 n_2}{2}. Since U = min(U1, U2), 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.

7 thoughts on “Mann-Whitney Test – Advanced”

  1. sir,
    i have sample size 86 and 82 a have calculated U statistic 348.5
    now how can I find critical value for my large sample size .How can I analyse null hypothesis
    my data is as under
    SUM OF RANK COUNT U STATISTIC
    EXP GROUP 10444.5 86 348.5
    CONTROL GROUP 3751.5 82 6703.5

    Reply

  2. I’d like to be able to construct an Excel spreadsheet that could compute its own Mann Whitney critical values. I can find lots of existing tables online, but I can’t seem to find the formula by which those critical values have been generated. Any help would be greatly appreciated.

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  4. 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!

    Reply

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

      Reply

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