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I have a very straight-forward question:
If the difference of scores is normally distributed, the sample distributions don't matter in a paired t-test, right?
Therefore, If the samples are not normally-distributed, but their differences are (Scores 2 - Scores 1), then it's ok to run the parametric test?

I've seen some related questions, but not this specifically one (here here , for example).

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    Just to add a small comment. Even if the differences are not normal (although it is an assumption), the paired t-test often performs well anyway. – bzki Sep 05 '22 at 17:01
  • Correct, though you can't know that the differences are normal (not that this is important). – Glen_b Sep 06 '22 at 07:01
  • @Glen_b exactly, which is why the robustness of the paired t-test to non-normality is beneficial (and makes it useful). – bzki Sep 10 '22 at 17:20

1 Answers1

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You are correct. A paired t-test is conducted on the differences of the paired scores. It doesn't look at the individual scores in any way.

A paired t-test is precisely the same as a one-sample t-test on the differences of the pairs.

Consider the following observations, A and B. While the distributions of each are skewed, the differences are relatively symmetric and bell-shaped in distribution.

The results of the paired t-test and one-sample t-test are the same.

A = (1, 2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7)

B = (0.98, 2.10, 3.84, 4.11, 3.02, 3.01, 5.67, 5.07, 6.20, 5.67, 6.77, 7.61, 6.15, 9.32, 8.41, 7.71, 8.64, 8.49, 8.15, 9.00, 8.80, 7.85, 8.90)

In R:

A = c(1, 2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7)

B = c(0.98, 2.10, 3.84, 4.11, 3.02, 3.01, 5.67, 5.07, 6.20, 5.67, 6.77, 7.61, 6.15, 9.32, 8.41, 7.71, 8.64, 8.49, 8.15, 9.00, 8.80, 7.85, 8.90)

hist(A)

hist(B)

Difference = A - B

hist(Difference)

t.test(A, B, paired=TRUE)

t.test(Difference)

Paired t-test

t = -3.3339, df = 22, p-value = 0.00301

One Sample t-test

t = -3.3339, df = 22, p-value = 0.00301

Sal Mangiafico
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