How to do a hypothesis test for a mean difference (matched pairs) (in Python, using SciPy)

Task

Say we have two sets of data that are not independent of each other and come from a matched-pairs experiment, $(x_{1}, x_{1}^{'}), (x_{2}, x_{2}^{'}), \dots, (x_{n}, x_{n}^{'})$ . We want to perform inference on the mean of the differences between these two samples, that is, the mean of $x_{1} - x_{1}^{'}, x_{2} - x_{2}^{'}, \dots, x_{n} - x_{n}^{'}$ , called $μ_{D}$ . We want to determine if it is significantly different from, greater than, or less than zero (or any other hypothesized value). We can do so with a two-tailed, right-tailed, or left-tailed hypothesis test for matched pairs.

Related tasks:

Solution

We choose a value, $0 \leq α \leq 1$ , as the Type I Error rate, and in this case we will set it to be 0.05.

We’re going to use fake fata here, but you can replace our fake data with your real data below. Because the data are matched pairs, the samples must be the same size.

# Replace the following example data with your real data
sample1 = [15, 10,  7, 22, 17, 14]
sample2 = [ 9,  1, 11, 13,  3,  6]

Two-tailed test

In a two-sided hypothesis test, the null hypothesis states that the mean difference is equal to 0 (or some other hypothesized value), $H_{0} : μ_{D} = 0$ .

from scipy import stats
stats.ttest_rel(sample1, sample2, alternative = "two-sided")

TtestResult(statistic=2.8577380332470415, pvalue=0.03550038112896236, df=5)

Our $p$ -value, 0.0355, is smaller than $α$ , so we have sufficient evidence to reject the null hypothesis and conclude that the mean difference between the two samples is significantly different from zero.

Note that the function above specifically tests whether the mean of $x_{i} - x_{i}^{'}$ is zero. If we want instead to test whether it is some other value $d \neq 0$ , then that’s equivalent to testing whether the mean of $(x_{i} - d) - x_{i}^{'}$ is zero. We could do so with the code below, which uses an example value of $d$ . The null hypothesis is now $H_{0} : μ_{D} = d$ .

d = 6  # as an example
stats.ttest_rel([ x - d for x in sample1 ], sample2, alternative = "two-sided")

TtestResult(statistic=0.4082482904638631, pvalue=0.6999865427788738, df=5)

The above $p$ -value is greater than $α = 0.05$ , so we could not conclude that the mean difference is significantly different from our chosen $d = 6$ .

Right-tailed test

If instead we want to test whether the mean difference is less than or equal to zero, $H_{0} : μ_{D} \leq 0$ , we can use a right-tailed test, as follows.

stats.ttest_rel(sample1, sample2, alternative = "greater")

TtestResult(statistic=2.8577380332470415, pvalue=0.01775019056448118, df=5)

Our $p$ -value, 0.01775, is smaller than $α$ , so we have sufficient evidence to reject the null hypothesis and conclude that the mean difference between the two samples is significantly greater than zero.

A similar change could be made to the code above to test $H_{0} : μ_{D} \leq d$ , as in the example code further above that uses $d = 6$ .

Left-tailed test

If instead we want to test whether the mean difference is greater than or equal to zero, $H_{0} : μ_{D} \geq 0$ , we can use a right-tailed test, as follows.

stats.ttest_rel(sample1, sample2, alternative = "less")

TtestResult(statistic=2.8577380332470415, pvalue=0.9822498094355188, df=5)

Our $p$ -value, 0.98225, is larger than $α$ , so we do not have sufficient evidence to reject the null hypothesis; we must continue to assume that the mean difference between the two samples is greater than or equal to zero.

A similar change could be made to the code above to test $H_{0} : μ_{D} \geq d$ , as in the example code further above that uses $d = 6$ .

Content last modified on 24 July 2023.

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Contributed by Elizabeth Czarniak (CZARNIA_ELIZ@bentley.edu)