How to calculate SD from pre SD and post SD? I have two values of SD calculated from paired samples of size $n=10$.
Results are:


*

*Pre test - mean=29.1 and SD=19.05

*Post test - mean=34.9 and SD=16.74


How do I calculate the SD?
 A: If I'm understanding you correctly, you're asking about the variance of the aggregate of two data sets. That is, if you have two data sets $X=(X_1,...,X_n)$ and $Y = (Y_1, ..., Y_m)$, you want to determine the standard deviation of the aggregated data set, $Z = (X_1, ..., X_n, Y_1, ..., Y_m)$, given knowledge of the means and standard deviations of $X$ and $Y$. 
It is a fact that the variance of an aggregated data set, $V$, is 
$$ V = \left( {\rm variance\ of \ the \ averages} \right) + \left( {\rm average\ of\ the\ variances} \right) $$ 
This is a consequence of the law of total variance. From the data you gave the variance of the averages is $8.41$ and the average of the variances is $321.565$. So, the standard deviation of the aggregate is approximately
$$ S = \sqrt{V} \approx \sqrt{8.41 + 321.565} \approx 18.165 $$
Note: $S$ is exactly the same as the sample standard deviation in the aggregated data set when you use the '$\frac{1}{n}$' sample variance rather than the unbiased sample variance (which scales by $\frac{1}{n-1}$). If you use the unbiased sample variance then $S$ slightly overestimates the standard deviation in the aggregated data set - I don't know which of these two standard deviations you've presented here. 
