This is a self-study question. I am trying to complete this textbook problem.
Suppose that $X_{1}, X_{2}, \dots, X_{m}$ and $Y_{1}, Y_{2}, \dots, Y_{n}$ are independent random samples with $X_{i}$ and $Y_{i}$ being normally distributed with means $\mu_{1}$ and $\mu_{2}$ and variances $\sigma^2_{1}$ and $\sigma^2_{2}$, respectively. The difference between the sample means, $\bar{X} – \bar{Y}$, is then a linear combination of $m+n$ normally distributed random variables is also normally distributed. If $\sigma^2_{1} = 2, \sigma^2_{2} = 2.5$, 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 $0.95$.
I started by saying that $P(|\bar{X} – \bar{Y} – (\mu_{1} – \mu_{2})| \le 1) = 0.95$. I think I have to standardize the variables, but I am unsure on how to do that since I am working with two random variables at once.
The answer is $n = 17.29 \rightarrow 18$, but I am unsure how to get here.
Best Answer
Given independent random samples $X_{1}, X_{2}, \dots, X_{m}$ taken from a normal random variable with mean $\mu$ and variance $\sigma^2$, it is a standard result that the sample mean $\bar{X}$ is normally distributed with mean $\mu$ and variance $\sigma^2/m$. It follows that in your example $\bar{X}$ is normally distributed with mean $\mu_{1}$ and variance $\sigma^2_{1}/m$ and $\bar{Y}$ is normally distributed with mean $\mu_{2}$ and variance $\sigma^2_{2}/n$. It then follows that the difference $\bar{X} - \bar{Y}$ is normal with mean $(\mu_{1} -\mu_{2})$ and variance $(\sigma^2_{1}/m + \sigma^2_{2}/n)$. As $m=n$ and with the given values we find that $\bar{X} - \bar{Y}$ has variance $(4.5/n)$. After standardizing $P(-1<(\bar{X} - \bar{Y})-(\mu_{1} -\mu_{2})<1)=0.95$ we get $P((-1/\sqrt(4.5/n))- Z-(1/\sqrt(4.5/n)) = 0.95$ where $Z$ is the standard normal variable. The rest should be clear I hope.