Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 7C / Solution

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$, $Q={\sum }_i(X_i-\bar{X}{)}^2$, and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal likelihood is proportional to $({\sigma }^2{)}^{-n/2}\exp [-{\sum }_i(X_i-\mu {)}^2/(2{\sigma }^2)]$. Its unrestricted maximizers are $\hat{\mu }=\bar{X}$, ${\hat{\sigma }}^2=Q/n$. Under the null the variance maximizer is $Q_0/n$. Substitution gives the generalized likelihood-ratio test statisticThe ratio decreases as $|T|$ increases, so the likelihood-ratio rejection region is two-sided in $T$. Under $H_0$, $\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal and $Q/{\sigma }^2$ is independent chi-squared with $n-1$ degrees of freedom. Independence follows by resolving the isotropic Gaussian sample vector into its projection on the constant vector and its orthogonal complement. Hence $T$ has the exact Student t-distribution with $n-1$ degrees of freedom.

For a chosen significance level $\alpha$, compute the sample mean and the sample standard deviation with denominator $n-1$, and reject precisely whenEquivalently, report the two-sided $p$-value $2[1-F_{t_{n-1}}(|T|)]$. This requires $n\ge 2$ and positive population variance; $S>0$ holds almost surely under that model. The nuisance variance is removed by the exact null distribution, so one should not substitute a normal critical value or a large-sample chi-squared approximation. This is the normal-mean likelihood ratio with unknown variance.