Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 8 / a

For known ${\sigma }^2>0$, maximizing the normal likelihood over $\mu$ is equivalent to minimizing ${\sum }_i(X_i-\mu {)}^2$. The identityshows that the unique minimum occurs at $\boxed{\hat{\mu }=\overline{X}}$. Equivalently, the log-likelihood has second derivative $-n/{\sigma }^2<0$.

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval