Past exam of the options II course of the University of Oxford / 2012 / O / O5

Independence and normality give $\overline{X}\sim N(\mu ,1/n)$. Consequently $Z=\sqrt{n}(\overline{X}-\mu )$ is standard normal, andRearranging the event gives $\mu >\overline{X}-z_{\alpha }/\sqrt{n}$. Thus the random interval $(\overline{X}-z_{\alpha }/\sqrt{n},\infty )$ covers the fixed parameter with probability $1-\alpha$. After observing the sample, its realized confidence interval isThe coverage refers to repeated samples, not to a posterior probability for the fixed parameter.

The Neyman-Pearson lemma concerns testing two simple hypotheses with densities $f_0$ and $f_1$ relative to a common measure. A test that rejects when $f_1(x)>k{f}_0(x)$, accepts when $f_1(x)<k{f}_0(x)$, and randomizes on equality if necessary to achieve size $\alpha$, is most powerful among all tests with size at most $\alpha$:Here $\phi$ is the described likelihood-ratio test and test functions take values in $[0,1]$. In the continuous normal setting below the boundary has probability zero, so randomization is unnecessary.

The likelihood ratio between the two normal means isSince ${\mu }_1>{\mu }_0$, it is strictly increasing in $\overline{x}$. The Neyman–Pearson test therefore rejects for large sample means. Under the null, $\sqrt{n}(\overline{X}-{\mu }_0)\sim N(0,1)$, so choosing a right-tail probability $\alpha$ givesThis has size exactly $\alpha$ because the sampling distribution is continuous.

For every particular alternative mean ${\mu }_1>{\mu }_0$, the likelihood ratio is increasing in the same statistic $\overline{x}$. Its size-$\alpha$ threshold is fixed solely by the null distribution, so it is always the same $c_{\alpha }$ found in part (c), independent of ${\mu }_1$.

The single rejection region is therefore most powerful against every member of the composite alternative. Hence the test is uniformly most powerful for $H_1:\mu >{\mu }_0$.

For the observed mean, the one-sided $p$-value is the null probability of a sample mean at least as large:Standardization under the null gives

The standard normal distribution function is strictly increasing. Thus $p>\alpha$ impliesRearranging givesThis is the confidence-interval/test duality for the matching one-sided procedures. The strict inequalities agree with the printed open interval.