**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 *x _{i} *= 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

*x*has a Bernoulli distribution and so

_{i}*μ*=

_{i}*E*[

*x*] = 1/2 and =

_{i}*var*(

*x*) = 1/2 ∙ 1/2 = 1/4 .

_{i}By Property 3a and 4a of Expectation it follows that

Since the *x*_{i} are independent it follows by Property 3b and 4b of Expectation that

Hi,

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.

Thanks,

Yan

Yan,

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.

Charles

Charles,

I sincerely appreciate your efforts and look forward to that release!

Yan

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