I've been trying to complete this question but have been struggling to see how to approach it.
Any help would be greatly appreciated.
Is there a standard way of approaching and answering questions like this? I've thought about Chi square and T- distributions but their definitions don't seem to be helping me. I know the first one follows a Chi (m+n-2) square distribution, but how do I solve the second one?
Thank you for your help!
For finding the distribution of the first one
$$(m+n-2)\frac{S^2}{\sigma^2}$$ let $S^2_1=\frac{1}{m-1}\sum_{i=1}^m(X_i-\overline X)^2$ and $S^2_2=\frac{1}{n-1}\sum_{j=1}^n(X_i-\overline X)^2$. Then
$$(m+n-2)\frac{S^2}{\sigma^2}=(m-1)\frac{S_1^2}{\sigma^2}\ +\ (n-1)\frac{S_2^2}{\sigma^2}$$
But as you correctly guessed, each summand of the last equation follows a $\chi_{m-1}^2$ and a $\chi_{n-1}^2$ respectively. And since $X_1\dots X_m$ and $Y_1\dots Y_n$ are independent random samples, we can add up those two distributions to get that
$$(m+n-2)\frac{S^2}{\sigma^2}(m+n-2)\sim \chi_{m+n-2}^2$$
For the last one, remember that if a sample of size $n$ distributes as a $N(\mu,\sigma^2)$, then $\overline X\sim N(\mu,\frac{\sigma^2}{\sqrt n})$. And that the difference of sample means $\overline X -\overline Y$, where $(X_1\dots X_m)$ and $(Y_1,\dots Y_n)$ are independent samples with distribution $N(\mu_1,\sigma^2)$ and $N(\mu_2,\sigma^2$) respectively, follows a $N(\mu_1-\mu_2,\sigma^2\sqrt{\frac{1}{m}+\frac{1}{n}})$. Using this and the previous distribution, you should be able to get the distribution of the last expression.