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

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.