Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 19C (Statistics)

Put $S={\sum }_i{X}_i$, and let $z_{.95}$ be the exact standard-normal 95th percentile. Under parameter $\theta$, $S=n\theta +\sqrt{n}Z$ with $Z$ standard normal. This distribution can be justified directly: the joint density of the centered observations is proportional to $\exp [-{\sum }_i(X_i-\theta {)}^2/2]$. An orthogonal change of coordinates with first row $(1,\dots ,1)/\sqrt{n}$ has unit absolute Jacobian and preserves the sum of squares. Its density factors into $n$ one-dimensional standard-normal densities, making the first coordinate $Z=(S-n\theta )/\sqrt{n}$ standard normal.

Consider the test ${\varphi }_{\ast }=1_{\{S>\sqrt{n}{z}_{.95}\}}$. Its rejection probability isincreasing in $\theta$. Therefore its maximum rejection probability over the null $\theta \le 0$ is achieved at zero and equals $\alpha =1/20$. This proves it has the required test size.

To prove the uniformly most powerful test property, rather than quote the Neyman-Pearson lemma, fix any ${\theta }_1>0$. The likelihood ratio of the full sample against $\theta =0$ isstrictly increasing in $S$. Set $K=\exp ({\theta }_1\sqrt{n}{z}_{.95}-n{\theta }_1^2/2)$. For any competing possibly randomized test $0\le \varphi \le 1$ with composite-null size at most $\alpha$, we have ${\mathbb{E}}_0\varphi \le \alpha ={\mathbb{E}}_0{\varphi }_{\ast }$. Pointwise,because both factors are nonnegative above the threshold and both are nonpositive below it. Integrating givesThis comparison proves the likelihood-ratio optimality result needed here in full. Since the same critical region works for every ${\theta }_1>0$, it proves the uniformly most powerful test property over the entire alternative.

For $n=9$, the exact critical region and its quoted-table approximation areUsing the supplied rounded percentile $1.65$ gives $\sum X_i>4.95$. The exact cutoff uses $z_{.95}\approx 1.644854$ and hence $\overline{X}>0.548285$; the rounded cutoff is the intended numerical approximation, not a literally exact size-$0.05$ threshold. The Gaussian distribution is continuous, so equality at the cutoff has probability zero and no randomization there is needed.