Calculating Standard Deviation when given sample size, mean difference & p value I am trying to pool data in my meta-analysis and i need MEAN & SD. However the study has reported sample size (27), before (11.8) & after mean (11.9), and p value (0.540). 
I need the SD.
Thanks for your help.
 A: With before-after data, I presume this is a paired design, and that consequently the test actually being performed is a two-tailed paired t-test. You should clarify to be sure.
If you really only have the means to 3 figures, then 11.8 could represent anything between 11.75+ and 11.85-, while 11.9 could represent anything between 11.85+ and 11.95-.
As such the true difference in means is actually anything between about 0 and 0.2, but more likely to be near 0.1 than the end-values. 
Let's take the actual difference in sample means to be $d$.
Then a one-sample t-test statistic would be $\frac{d}{s_d/\sqrt{n}}$, and I presume you're after the standard deviation of the differences, $s_d$.
With 26 d.f., the (absolute value of) a two-tailed t-value that gives a p-value of 0.540 will be 0.621. So we have:
$s_d=d\sqrt{27}/0.621 = 8.37 d$
Now if $d$ was actually 0.1, that would imply $s_d$ is 0.837, but with the information given in the question it might be anything between 0 and 1.674.
If you're able to get $d$ more accurately than this, you can get $s_d$ to similar percentage accuracy.
Even a tiny bit of information - for example, knowing that the original observations must be integer - could help narrow it down (in that case, it would imply that the difference in means would be restricted to lie between 0.037 and 0.148).
A: Try ?sd. The sd() function in R will return a standard deviation if you have the original data. However, if you do not have the original data and are trying to determine the SD from only the reported sample size and mean values, then I'm pretty sure you are out of luck. The standard deviation and the mean are describing different things (distribution and central tendency respectively), so you can't get one from the other. 
HTH, 
Tim
A: If you can assume that before and after have the same s.d. then you might be able to get an estimate. For example:

s.d.assumed <- 1
print(
mean(replicate(1000,{
x.1 <- rnorm(n = 27,mean = 11.8,sd = s.d.assumed)
x.2 <- rnorm(n = 27,mean = 11.9,sd = s.d.assumed)
t.test(x.1,x.2)$p.value
}))
)

Gives a p.value of ~.48. That's assuming obviously the p-values were calculated using a t-test. You might be able to get an estimate of the sd by putting this idea into a form that you can run an optimization routine on. But a p-value of ~.5 corresponds to a s.d. of ~Inf so it might not be a very good estimate. It would of course be better to get the original data from the authors, if possible.
