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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

Richard
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1 Answers1

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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.

Michael Hardy
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