Paired sample t-test and Repeated Measures ANOVA with a dichotomous predictor are two tests that answer the same question, and output the same p-value as far as I can tell.

This question has been asked before, but it was focused on the assumption of both tests. What strikes me more here is that the test statistics are different: one is based on the $t$ distribution while the other on the $F$ distribution. Is there any theoretical way to explain why these two distributions should be close to each other in the case of a dichotomous predictor?


1 Answer 1


If $X\sim t(\nu)$, i.e. $X$ has a $t$-distribution with $\nu$ degrees of freedom, then $X^{2}\sim F(1, \nu_{2} = \nu)$. It follows that in the case of two repeated measurements, the squared $t$ value of the paired $t$-test is equal to the $F$ value of the repeated measures ANOVA.

Here is an example using SPSS. There are two timepoints with repeated measurements. First the paired $t$-test:


The $t$ value is $-0.151$ with $173$ degrees of freedom. Now the repeated measures ANOVA:


The $F$ value is $0.023$ which is equal to $(-0.151)^{2}$, as expected. The degrees of freedom are $1$ and $173$. The $p$-values are identical.


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