 Solution

For $S={\sum }_i{x}_i^2$, the likelihood ratio in favor of the larger variance isBecause ${\sigma }_1>{\sigma }_0$, this is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper-tail rejection region. Under the null, $S/{\sigma }_0^2\sim {\chi }_n^2$, so size $\alpha$ is obtained withContinuity of the chi-squared distribution makes the boundary have probability zero. This is a fixed-mean normal variance test.