Non-parametric tests that extend the sign tests. The single-sample version (observations x₁, x₂,…), suitable for a symmetric distribution, tests the null hypothesis that the population median has a specified value (m₀). The matched-pair (or paired-sample) version (observation pairs (x₁, y₁), (x₂, y₂),…) is concerned with the differences (x₁ − y₁), (x₂ − y₂),…. With the assumption that these differences are independent observations from a symmetric distribution, the null hypothesis is that this distribution has median zero.

To determine the value of the test statistic, z, the first step is to calculate the differences d₁, d₂, …, where d_j=x_j − m₀ (single sample) or d_j=x_j − y_j (matched pairs). After zero differences have been discarded, the remaining n are arranged in ascending order of |d_j|. The magnitudes are replaced by the corresponding ranks, with tied ranks where necessary. The signs of d₁, d₂,…, are now attributed to the ranks, resulting in signed ranks. Let P be the sum of the positive signed ranks and let T be the smaller of P and ½n(n+1)−P. The test statistic, given byis an observation from the upper half of an approximate standard normal distribution. The ½ is a continuity correction.

As an example, suppose that a symmetric distribution is believed to have median 100. A random sample of eight observations is reported as consisting of the values 92.3, 57.6, 88.8, 110.5, 100.0, 181.0, 96.0, 105.7. The supposed median is subtracted from each observation to give −7.7, −42.4, −11.2, 10.5, 0, 81.0, −4.0, 5.7. The value 0 is discarded and the remainder are arranged in order of ascending absolute magnitude: −4.0, 5.7, −7.7, 10.5, −11.2, −42.4, 81.0. Retaining the signs while replacing the values by ranks gives −1, 2, −3, 4, −5, −6, 7, so that P=13, ½n(n+1) − P=15 (since n=7) and T=13. The test statistic isComparing with the tables of upper-tail percentage points of the standard normal distribution (Appendix VI), we see that the null hypothesis that the median is 100 should not be rejected.