= Solution
For $S=\sum_i x_i^2$, the <likelihood ratio> in favor of the larger <variance> is
$$
\frac{f_{\sigma_1^2}(x)}{f_{\sigma_0^2}(x)}
=\left(\frac{\sigma_0}{\sigma_1}\right)^n
\exp\left[\frac S2\left(\frac1{\sigma_0^2}-\frac1{\sigma_1^2}\right)\right].
$$
Because $\sigma_1>\sigma_0$, 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 with
$$
\boxed{C=\{S\geq k_\alpha\},\qquad k_\alpha=\sigma_0^2F_n^{-1}(1-\alpha).}
$$
Continuity of the chi-squared distribution makes the boundary have <probability> zero. This is a <fixed-mean normal variance test>.
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