Combining samples based off mean and standard error I have two samples with a mean and SE for each. I want to combine them, so how do I calculate a combined standard error when combining two samples means? I can only find information about combining means and SD's at the moment. 
 A: If your first population has mean $\mu_1$ and variance $\sigma_1^2,$ then the sample mean ${\bar{x}_1}$ of your data has variance ${\sigma_1^2 \over n_1},$ where $n_1$ is the sample size. Similarly for your second sample the variance of the sample mean ${\bar{x}_2}$ is ${\sigma_2^2 \over n_2}.$
The variance of the combined sample mean ${\frac{1}{2}}\left(\bar{x}_1+\bar{x}_2\right)$ is then ${\frac{1}{4}}\left({\sigma_1^2 \over n_1}+{\sigma_2^2 \over n_2}\right).$ 
So its standard deviation is ${\frac{1}{2}}\sqrt{{\sigma_1^2 \over n_1}+{\sigma_2^2 \over n_2}}$
The standard error, which is an estimate of this standard deviation, is given by ${\frac{1}{2}}\sqrt{{s_1^2 \over n_1}+{s_2^2 \over n_2}},$ where $s_1$ and $s_2$ are the sample standard deviations. 
Note that this is for a simple average of the two sample means, not a weighted version. 
A: In your case it seems that you're trying to do a sort of meta-analysis where you want to act as if the two studies are one, with the commensurate larger N.
Keep in mind that the variance of a sample mean is tied to the N from which the same is taken. You can only average the variance of sample means across studies when the N's are equal. This averaged variance does not seem to be what you're looking for. What you want to do instead is get a weighted variance (not of the sample mean, but the variance estimate for each sample) across the studies and use that to then get the standard error of the sample mean considering the combined N as the sample. So for the weighted variance you can just combine the variances in the usual fashion. Given that you only have Ns and SEs then you first need to get the variance for each study.
var1 = SE1^2 * n1
var2 = SE2^2 * n2

Then pool those resulting variances
( var1 * (n1-1) + var2 * (n2 - 1) ) / (n2 + n1 - 2) 

and then calculate the new standard error.
SE = sqrt(varPooled / N)

Given the description in your comment of what you wanted to find out you should probably go this way.
