Confidence interval for a sample from the norm dist

Question

Let $Y_1<Y_2$ be order statistics from a random sample of size $2$ from a normal distribution, $\mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. If $\bar{X}$ is the mean of this sample, how would I find the constant $c$ which solves this equation: $P(\bar{X} - c\sigma < \mu < \bar{X} + c\sigma)$. Thanks for the help.

Answer 1

The expression $P(\bar{X} - c\sigma < \mu < \bar{X} + c\sigma)$ is not actually an equation until you put "$=\text{something}$" after it. Presumably you want it to be equal to some number between $0$ and $1$. Let's call that number $p$.

Notice that $\bar{X} = (Y_1+Y_2)/2$, and $\bar{X} \sim \mathcal{N}(\mu, \sigma^2/2)$. So $$ \frac{\bar{X} - \mu}{\sigma/\sqrt{2}} \sim \mathcal{N}(0,1). $$ Now find $c$ such that $P(-c<Z<c)= p$ if $Z\sim\mathcal{N}(0,1)$. You get this from a table or from software unless you want to go through a moderately elaborate discussion of how to do it numerically.

Then notice that $$ -c < \frac{\bar{X}-\mu}{\sigma/\sqrt{2}} < c \text{ if and only if }\bar{X}-c\frac{\sigma}{\sqrt{2}} < \mu < \bar{X}+c\frac{\sigma}{\sqrt{2}}. $$