How to back-calculate change from baseline from a p-value for a paired t test I have a result from a study in a research paper which gives a mean/SD for 2 time points (pre & post treatment), and a p-value based on a paired-samples t-test.  What I want to calculate is the SD of the change from pre to post.  Is there any way I can back-calculate this using the p-value and mean/SD for each time point?
Thanks in advance-
Julie
 A: This is possible, and relatively straightforward given that a paired t-test simplifies to a one-sample t-test of the differences where:
$t = \frac{\delta}{se} $
where $se = \frac{s}{\sqrt{n}}$  
where $\delta$ is the difference in means (equivalently, the mean difference), and $se$ is the standard error of the difference in means, and $s$ is the standard deviation of the difference (which is what you want).
I think the standard deviations of the two timepoints are not needed for this particular solution.
I'll just add here that this will only be a useful exercise if the p-value is reported in some detail (e.g. if just reported p < 0.01, then you'll only be able to get an upper bound on the value of the sample standard deviation: and also the rounding inherent in the initial summary statistics might have a big impact. But for getting a rough idea about their sample standard deviation, I think the following will suffice.)
SOLUTION:
For a given $t$ with $x$ degrees of freedom (df, here = sample size - 1), the original authors would have derived the p-value by examining the two-tailed t-distribution (most likely two-tailed, anyway).
You need to reverse this last process: take the inverse of the two-tailed t-distribution for this studies df at the reported p-value.
Let's assume that reported from the study is that the sample size is 17 (so df = 16), difference in means is 2.4 units, and p-value is 0.039.
In R to solve for $t$:
p.val <- .039
qt((1-p.val/2), 16)
# returns 2.248337

In Excel, one could use =T.INV.2T(0.039, 16) which returns 2.248336901
Now solve the first equation to get the standard error:
$t = \frac{\delta}{se} $ which rearranges to:     $se = \frac{\delta}{t} $
which here is
$se = \frac{2.4}{2.248337} = 1.067456$
and then to find standard deviation we take the standard error and solve:
$se = \frac{s}{\sqrt{n}}$ rearranged as: $s = se \times \sqrt{n}$
which here is then $s = 1.067456 \times \sqrt{17} = 4.401233$
