I would like to calculate what is $SE(\hat{x}-\hat{y})$ where $\hat{x}$ is the mean of the first sample and $\hat{y}$ is the mean of the second sample.

I know the answer should come out as

$\displaystyle\sqrt{\frac{(n_{x}-1)s_{x}^{2} + (n_{y}-1)s_{y}^{2}}{n_{x}+n_{y}-2}} \cdot \sqrt{\frac{1}{n_{x}}+\frac{1}{n_{y}}}$

but I don't see how to get there. I get that the first part of that equation is the square root of pooled standard deviation (pooled variance). However, I don't see where the second part comes from.

  • $\begingroup$ Please add the self-study tag, and read its tag-wiki (modifying your question to follow the guidelines on asking such questions if needed). $\endgroup$ – Glen_b Jul 13 '15 at 11:16

If two independent normally distributed random variables have variances $s_1^2$ and $s_2^2$ then their difference has variance $s_1^2+s_2^2$ and standard deviation $\sqrt{s_1^2+s_2^2}$.

Now suppose $s_1^2=\frac{s^2}{n_1}$ and $s_2^2=\frac{s^2}{n_2}$ since we are looking at the dispersion of the means with the assumption that the underlying distributions have the same variances. Then the standard deviation of the difference in the means is $\sqrt{\frac{s^2}{n_1}+\frac{s^2}{n_2}} = s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} $.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.