standard error of the difference in means first off this my first post on here so if I missed out on a guideline tell me!!
Okay so this is the problem I'm facing:
I have 2 means of populations we want to compare. I also have the standard error for both. I'm then asked to calculate the standard error of the difference in means.
It seems to me that to do this I would need the amount of samples taken but I don't have this number. Can someone point me the way? 
Thanks!
 A: Let $X_1, \dots, X_{n_1}$ be a random sample from $\mathsf{Norm}(\mu_1, \sigma_1);$ with $\mu_1$ estimated by $\bar X = \frac 1 n \sum_i X_i$ and $\sigma_1^2$ estimated by $S_1^2 = \frac 1 {n-1}\sum(X_i -\bar X)^2.$
And let $Y_1, \dots, Y_{n_2}$ be a random sample from $\mathsf{Norm}(\mu_2, \sigma_2)$ with $\mu_2$ and $\sigma_2^2$ estimated similarly by the $Y_i$s.
Then $Var(\bar X) = \sigma_1^2/n_1,$ estimated by $S_1^2/n_1.$ Similarly,
$Var(\bar Y) = \sigma_1^2/n_2,$ estimated by $S_2^2/n_2.$ Because $\bar X$ and $\bar Y$ come from
independent samples 
$$Var(\bar X - \bar Y) = Var(\bar X) + (-1)^2Var(\bar Y)
= \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}.$$
Thus, the standard error of $\bar X - \bar Y$ is
$$SD(\bar X - \bar Y) = \sqrt{ \frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2}},$$
which can be estimated by 
$$\sqrt{ \frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}},$$
which is called the "estimated standard error" of the difference in sample means. (The word estimated is often dropped when the estimation is obvious.)
In the special case where one assumes that $\sigma_1^2 = \sigma_2^2 = \sigma^2,$ one estimates $\sigma^2$ by
$$S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}.$$
(This is a 'weighted average' of $S_1^2$ and $S_2^2,$ in which the weights are the respective
degrees of freedom.)
Then $SD(\bar X - \bar Y)$ is estimated by
$$\sqrt{ \frac{S_p^2}{n_1} +\frac{S_p^2}{n_2}}
= S_p\sqrt{ \frac{1}{n_1} +\frac{1}{n_2}},$$
the pooled (estimated) standard error of the difference in means.
