# Wilcoxon signed-rank test or One Sample t-test?

I conducted a training session and for each of my users, I measured pre and post score (before and after training session). Each of my users (n=28) is measured twice, so my data is paired:

Data:

user pre-score post-score
1    10        30
....


Hypothesis: $$H_0: \mu_{Post-Pre} \leq 0\\ H_1: \mu_{Post-Pre} > 0$$

In order to check the hypothesis, whether the post-score is significantly higher than the pre-score (whether the training had a positive effect on users' score), I would conduct a paired t-test (t.test(Post, Pre, mu=0, conf.level=0.95, alt="greater") in R).

Problem: The post data is not normally distributed. That's why I would go with a non-parametric test, i.e. Wilcoxon signed-rank, instead of the paired t-test.

However, if a calculate the difference between the pairs (Post-Pre), I end up with one sample which is normally distributed. So I would have to test the following hypothesis in a one sample t-test (t.test(Post-Pre, mu=0, conf.level=0.95, alt="greater") in R): $$H_0: \mu_{Diff.} \leq 0\\ H_1: \mu_{Diff.} > 0\\$$

Question: Which method is appropriate? Should I use the non-parametric paired Wilcoxon text because my post data is not normally distributed? Or can I work with the difference of the pairs (one sample) which is normally distributed and use a one sample t-test?

• The one sample t-test assumes that the differerences are normally distributed, not the scores. – Jeremy Miles Jan 14 '19 at 17:54
• So, both samples (pre and post) each must follow a normal distribution? If one does not (like my post data), I'd use a non-parametric test? – Ioannis K. Jan 14 '19 at 18:07
• Only the differences would need to be drawn from a normal population (or something sufficiently close to a normal population). Normality of the differences is used in deriving the t-distribution (you need it for the distribution of the numerator and the distribution of the denominator and for the two being independent). Outside that, you may be able to justify an asymptotic normal distribution for the statistic (via CLT and Slutsky's theorem). – Glen_b -Reinstate Monica Jan 15 '19 at 0:04