The Wilcoxon Signed-Rank Test
The Wilcoxon Signed-Rank Test is a distribution free test or non-parametric test which is used to test the population mean when the samples do not come from any specific distribution. The idea behind The Wilcoxon Signed-Rank Test is to median-center the samples and give signed-ranks to the individual values. If the sum of the negative ranks (S-) is larger, we can conclude that the population mean is lesser than the mentioned value. If the sum of the positive ranks(S+) is larger, the population mean is greater than the mentioned value. For detailed explanation, the post The Wilcoxon Signed-Rank Test can be followed.
If the sample size is larger, the test statistic S+ is approximately normally distributed with mean \(n(n+1)/4\) and variance \(n(n+1)(2n+1)/4\). In this case, the Wilcoxon signed-rank test can be performed by computing the z-score of S+ and then the normal table can be used to find the p-value. The z-score is:
\[z = \frac{(S+) - n(n+1)/4}{\sqrt{n(n+1)(2n+1)/24}}\]Let us look at the derivation for the distribution parameters.
#### Derivation of distribution mean :The statistic used in the Wilcoxon Signed-Rank Test(let us say it \(W\) which is S+) can be denoted as:
\[W = \sum _{i=1}^{n} I _{i}R _{i}\]where \(I_i\) is the indicator variable defined as 0 when \(x_i - y_i\) is negative and 1 otherwise and \(R_i\) is their ranks. When the sample size is \(n\), the distribution of the sum of positive ranks is also equivalent to distribute the ranks from \(1\) to \(n\) randomly into two subsets, one denoting the positive ranks and other the negative ranks. Hence, this can be denoted as:
\[U = \sum _{i=1}^{n} U _{i}\]where \(U _{i}=i\), when the rank belongs to the positive ranks and \(U _{i}=0\) otherwise (when the rank belongs to the negative ranks). As these two distributions are similar:
\[E(W) = E(U)\]Now, as we are randomly distributing the ranks amongst the positive and negative ranks, the probability of each of the rank being negative or positive will be same and will be equal to 1/2, i.e.
\[P(U_i = 0) = P(U_i = i) = 0.5\]The expected value of \(U\) can be calculated as:
\[E(U) = \sum _{i=1}^{n} E(U_i) = \sum _{i=1}^{n} \big[\frac{1}{2}0 + \frac{1}{2}i \big] = \frac{1}{2} \sum _{i=1}^{n} i = \frac{n(n+1)}{4}\]Hence,
\[E(W) = E(U) = \frac{n(n+1)}{4}\]#### Derivation of distribution variance :For the calculation of variance, we need to calculate the individual variances of each of the \(U_i\)s and sum them all. As, \(Var(X) = E[X^2] - E^{2}[X]\), the variance of \(U_i\) can be given as:
\[Var(U_i) = E[U _{i}^2] - E^{2}[U_i]\]Expected value of \(U_i\) and \(U_i^2\) can be calculated as:
\[E[U_i] = 0.\frac{1}{2} + i.\frac{1}{2} = \frac{i}{2}\]\[E[U_i^2] = 0^2.\frac{1}{2} + i^2.\frac{1}{2} = \frac{i^2}{2}\]Hence,
\[Var(U_i) = \frac{i^2}{2} - \big(\frac{i}{2} \big)^2 = \frac{i^2}{4}\]The variance of \(W\) can be given as:
\[Var(W) = Var(U) = \sum _{i=1}^{n} Var(U_i) = \sum _{i=1}^{n} \frac{i^2}{4} = \frac{n(n+1)(2n+1)}{24}\]#### Reference :