Sampling distribution

Quote:

Originally Posted by mirrormirror

Suppose that X1, X2,....Xm and Y1,Y2,....,Yn are independent random samples, with the variables Xi normally distributed with mean $\mu _1$ and variance $\sigma{_1}^{2}$ and the variables Yi normally distributed with mean $\mu_2$ and variance $\sigma{_2}^{2}$ . The difference between the sample means, $\bar{X} - \bar{Y}$ is then a linear combination of m + n normally distributed random variables and is itself normally distributed.

Suppose that $\sigma{_1}^{2}$ = 2, $\sigma{_2}^{2}$ = 2.5, and m=n. Find the sample sizes so that ( $\bar{X} - \bar{Y}$ ) will be within 1 unit of ( $\mu _1 - \mu _2$ ) with probability .95.

I don't know how to solve this one. I've tried to set up P(( $\mu _1 - \mu _2) - 1$ $\preceq$ $(\bar{X} - \bar{Y} )$ $\preceq$ $(\mu _1 - \mu _2) + 1)$ = .95

but I don't if that's right. And, if so, how do I go on?

The variance of $\bar X-\bar Y$ is $\sigma_1^2/m+\sigma_2^2/n$ and its mean is $\mu_1-\mu_2$

RonL