# 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!

• The variance of the difference between two independent random variables is the sum of their variances – Henry Apr 15 at 19:28
• You do need to know the sizes of the two samples. You should try to match formulas shown in my Answer with corresponding formulas shown in your text. The notation might be slightly different, but you should be able to recognize them. – BruceET Apr 16 at 1:35

## 1 Answer

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.