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I feel a bit stupid, but I seem unable to find a way to calculate a paired sample t-test having the group means and standard deviations (not the individual data) of one group before and after an intervention:

mean_pre: 11.7, SD_pre: 2.4 mean_post: 16.7, SD_post:3.1

Claudio
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1 Answers1

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This can't be done. You need either the standard deviation of the difference scores or the correlation between the pre and post scores. Consider the following examples.

Xpre  <- c(13.196, 12.092, 11.410, 14.006,  7.796)
Xpost <- c(14.870, 16.819, 15.237, 14.552, 22.022)

mean(Xpre) is 11.7, sd(Xpre) is 2.4, mean(Xpost) is 16.7, and sd(Xpost) is 3.1. All good so far. We can compute the t-statistic for the paired t-test using the formula $\frac{\bar X_D}{s_D/\sqrt{n_D}}$, where $\bar X_D$ is the mean of the difference scores, $s_D$ is the standard deviation of the difference scores, and $n_D$ is the number of difference scores. This give us a t-statistic of $2.063$.

Now consider the following new dataset:

Xpre  <- c(11.187, 15.639,  9.191, 10.720, 11.764)
Xpost <- c(14.442, 21.387, 13.651, 16.038, 17.982)

These have the same means and standard deviations as the first dataset, but the t-statistic is $9.551$.

So, you can see that knowing the means and standard deviations of the original scores is not enough. We can wildly varying t-statistics from two datasets that both have the means and standard deviations you quoted.

Noah
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  • If you know the context, sometimes you can give reasonable advice about the covariance (and so say whether the un-paired test is likely to be too small or too large). So in the pre-post case, many datasets it is reasonable to expect a positive covariance, so the unpaired t-test will probably be smaller than if you calculated with the original data taking into account the covariance. – Andy W Nov 03 '22 at 17:30