How to do a Wilcoxon signed-rank test for matched pairs (in Python, using SciPy)

See all solutions.

Task

Assume we have two samples of data that come in matched pairs, $x_1,x_2,x_3,\dots x_k$ and $x_1^{\prime },x_2^{\prime },x_3^{\prime },\dots x_k^{\prime }$, which we might pair up as $(x_1,x_1^{\prime }),(x_2,x_2^{\prime }),\dots ,(x_k,x_k^{\prime })$. The two samples may be from different populations. Also assume that the sample sizes are small or the populations are not normally distributed.

Consider measuring the difference in each pair, $x_1-x_1^{\prime },x_2-x_2^{\prime },\dots ,x_k-x_k^{\prime }$. We want to perform tests that compare the median of those differences, $m_D$, to a hypothesized value (equal, greater, or less). One method is the Wilcoxon Signed-Rank Test for Matched Pairs.

Related tasks:

• How to do a Kruskal-Wallis test
• How to do a Wilcoxon rank-sum test
• How to do a Wilcoxon signed-rank test

Solution

The method we will use is equivalent to subtracting the two samples and then performing the signed-rank test. See how to do a Wilcoxon signed-rank test to compare the two methods.

We’re going to use fake data for illustrative purposes, but you can replace our fake data with your real data.

1
2
3
4
Copy
import numpy as np
# Replace sample1 and sample2 with your data
sample1 = np.array([156, 133, 90, 176, 119, 120, 40, 52, 167, 80])
sample2 = np.array([45, 36, 78, 54, 12, 25, 39, 48, 52, 70])

We choose a value, $0\le \alpha \le 1$, as the Type I Error Rate. We’ll let $\alpha$ be 0.05.

Two-tailed test

To test the null hypothesis $H_0:m_D=0$, we use a two-tailed test:

1
2
3
Copy
from scipy import stats
from scipy.stats import wilcoxon
wilcoxon(sample1 - sample2)

1
WilcoxonResult(statistic=0.0, pvalue=0.001953125)

Our p-value, 0.001953125, is less than $\alpha =0.05$, so we have sufficient evidence to reject the null hypothesis. The median difference is significantly different from zero.

Right-tailed test

To test the null hypothesis $H_0:m_D\le 0$, we use a right-tailed test:

1
Copy
wilcoxon(sample1 - sample2, alternative = 'greater')

1
WilcoxonResult(statistic=55.0, pvalue=0.0009765625)

Our p-value, 0.0009765625, is less than $\alpha =0.05$, so we have sufficient evidence to reject the null hypothesis. The median difference is significantly greater than zero.

Left-tailed test

To test the null hypothesis $H_0:m_D\ge 0$, we use a left-tailed test:

1
Copy
wilcoxon(sample1 - sample2, alternative = 'less')

1
WilcoxonResult(statistic=55.0, pvalue=1.0)

Our p-value, 1.0, is greater than $\alpha$, so we do not have sufficient evidence to reject the null hypothesis. We should continue to assume that the mean difference may be less than (or equal to) zero.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Contributed by Elizabeth Czarniak (CZARNIA_ELIZ@bentley.edu)