Statistics Confidence Interval Estimators I'm working on this problem and don't know how to approach it:
problem http://puu.sh/5B36g.png
Could anyone help steer me in the right direction?
Thanks
 A: That $\sigma_1$ and $\sigma_2$ are known is unrealistic; hence included here only to illustrate theory.  The hint reminds you that
$$
\bar X_n \sim N\left(\mu_1,\frac{\sigma_1^2}{n}\right)\text{ and }\bar Y_m\sim N\left(\mu_2,\frac{\sigma_2^2}{m}\right),
$$
(and there's an obvious typo in the question, where it says $\bar X_m$ instead of $\bar Y_m$).  Then we have
$$
\bar X_n - \bar Y_m \sim N\left(\mu_1-\mu_2,\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m}\right).
$$
So
$$
\Pr\left(-A<\frac{(\bar X_n-\bar Y_m)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m}}}<A\right) = 0.95,
$$
where $A$ is a number you find in a table.  This can be rearranged into
$$
\Pr\left(\bar X_n-\bar Y_m - A\sqrt{\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m}} < \mu_1-\mu_2< \bar X_n-\bar Y_m + A\sqrt{\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m}}\right)=0.95.
$$
If $\sigma_1$ and $\sigma_2$ were unknown but assumed equal, you'd get something involving a $t$-distribution with $n+m-2$ degrees of freedom.  Drop the assumption that they're equal and then you'd have a hard problem called the Behrens--Fisher problem.
