Let $X_1,X_2,\cdots ,X_n$ be a random sample from a continuous distribution with median $\mu$. If $[X_{min}, X_{max}]$ is used as a confidence interval for $\mu$, what is its confidence level? What is the confidence level if $n=10$?
The formula for the confidence interval is $\overline{x}-z(\alpha /2)(\sigma)\leq \mu \leq \overline{x}+z(\alpha /2)(\sigma)$. I know that I need to solve for $\alpha$, but I'm not sure what the fact that the question involves a random sample means for the solution.
For any random variable $Y$ with continuous distribution, we have $\Pr(Y\le a)=\Pr(Y\lt a)$, so we can be a little casual about inequalities.
For any $i$, since $\mu$ is the median of $X_i$, we have $\Pr(X_i\lt \mu)=\frac{1}{2}$. The probability that $X_{\text{max}}$ is $\lt \mu$ is the probability that all the $X_i$ are $\lt \mu$. By independence, this is $\frac{1}{2^n}$.
Similarly, the probability that $X_{\text{min}}$ is $\gt \mu$ is equal to the probability that all the $X_i$ are $\gt \frac{1}{2}$. This is $\frac{1}{2^n}$.
If we use the interval $\left(X_{\text{min}}, X_{\text{max}}\right)$ as a confidence interval for the median $\mu$, then with probability $\frac{1}{2^n}$ the upper estimate will be too small, and with probability $\frac{1}{2^n}$ the lower estimate will be too large. Thus the confidence level is $\frac{1}{2^n}+\frac{1}{2^n}$, or more simply $\frac{1}{2^{n-1}}$.