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

image945

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

image3534

By Property 4a of Expectation

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

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From these it follows that

image3538

We can now calculate the following expectations:

image3539

Also where i ≠ j

image3540image3541

By Property 2 of Expectation (case where i = j)

image3542

By Property 3 of Basic Concepts of Correlation when i ≠ j

image3543

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

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image3546

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18 Responses to Wilcoxon Rank Sum Test – Advanced

  1. Jack says:

    oh ,now i understand it!thanks your respond

  2. Jack says:

    It’s 2Σ i!=j to n1 [-(n+1)/12].I write wrong. There should’t ‘2’ in there i think .

    • Charles says:

      Jack,
      You need to sum all the terms cov(x_i,x_j) where i not equal to j. Note that each such covariance is repeated twice, once for cov(x_i,x_j) and once for cov(x_j,x_i). Thus, if you assume that the sum is where i < j, then you need to double the result. Another way to look at this is to determine how many pairs there are for the indices 1 to n1 where the indices are not equal. The answer is n1(n1-1), which is the value used in the proof. This is the same as 2 times n1(n-1)/2, the later being the number of pairs where the first index is less than the second index. To make this much clearer and more accurate, I have now replaced the lower limit of the summation symbol by i < j (instead of i not equal to j). Thanks for bringing this issue to my attention. Charles

  3. Marcel says:

    I have one question about this proof. You calculate the expectation E(rirj) for all j not equal i. I don’t understand why we could take this expectation as equivalent to E(rirj) for all j, i. In the covariance we have to use E(rirj) of all rangs, but you use the expectation for all j not equal i, why is it correct? Can you explain me this problem?

    Thanks for your answer!

    • Charles says:

      Marcel,
      I show E[ri rj] both where i = j (i.e. var(ri)) and where i is not equal to j.
      Charles

      • Marcel says:

        I am grateful to you for your answer. But I wanted to say that you take the expectation E(rirj) with i not equal to j by the covariance cov(rirj). I don’t understand, why we can do this. I thought we need the expectation of all i and j (also of the double sum i*j).
        I made a print-screen with both places in your proof: https://image.prntscr.com/image/7lBOfbg1RribBEjyIbN1Kw.png

        • Charles says:

          Marcel,
          Thanks for clarifying things. There are two case: (1) where i = j and (2) where i is not equal to j. In case (1) cov(rirj) = var(ri), which is the described in your print-screen. In case (2) the formula is the one shown in your print-screen. I have just updated the referenced webpage to try to make this a bit clearer. Does it help?
          Charles

  4. Marcel says:

    Your site is very good!
    Have you thougt about translate your work in other languages?

  5. willy kiptoo says:

    Thanks alot, you realy put it out on paper

  6. Orukpe Lewis says:

    I’m grateful for dis better understanding

  7. Nnamdi says:

    Thanks. Good one

  8. shark says:

    thanks 😀

  9. fra says:

    thanx! really helpful!

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