(a) Find a 95\% confidence interval for $\mu$ and show that its width is about 3.16.
(d) Consider the event that the 95\% confidence interval for $\mu$ is narrower when $\s^2$ is known than it is when $\s^2$ is unknown, given that both intervals are computed from the same data, $X_1,\ldots,X_n$, where these are IID samples from $N(\mu,\s^2)$.
Show that if $n=5$ then this event has probability $1-F_4^{-1}(a)$, where $a=4(1.96/2.78)^2=1.988$, and $F_4$ is the cdf of the $\chi_4^2$ distribution. Verify from tables that this probability is a bit less than $0.75$ (actually, $0.748$). To what do you think this probability tends as $n\rightarrow\infty$?
Last modified: Tue Mar 11 10:26:12 1997