Wilcoxon Signed-Ranks Test – Advanced

Property 1: For n is sufficiently large, the T statistic (or even T+ or T-) has an approximately normal distribution N(μ, σ) where


Proof: We prove that the mean and variance of T+ are as described above. The proof for T- is the same and since T = min(T+, T-) it is clear that all T have the mean and variance described above. The approximation comes from the Central Limit Theorem. We now show that the mean and variance are as indicated.

Let xi = 1 if the sign of the data element in the sample with rank i is positive and = 0 if it is negative. Thus, under the assumption of the null hypothesis, each xi has a Bernoulli distribution and so μi = E[xi] = 1/2 and \sigma_i^2 = var(xi) = 1/2  ∙ 1/2 = 1/4 .

By Property 3a and 4a of Expectation it follows that


Since the xi are independent it follows by Property 3b and 4b of Expectation that

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4 Responses to Wilcoxon Signed-Ranks Test – Advanced

  1. Yan Zhang says:

    The proof here seems not to take ties — data elements that share the same rank — into consideration. In my textbook, there are correlation term for variance but none for the expectation. I do want to learn more about how to generalize this proof to situation like this.

    • Charles says:

      You are correct. The proof doesn’t take ties into account. In the next release of the Real Statistics Resource Pack, I do plan to change some of the non-parametic tests to take ties into account.

  2. Pingback: ウィルコクソンの符号順位統計量の平均・分散 | 有意に無意味な話

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