 Solution

The Neyman-Pearson lemma says that for testing a simple null density $f_0$ against a simple alternative $f_1$, a size-$\alpha$ test that rejects for the largest values of $f_1/f_0$ is most powerful. Randomization on a likelihood-ratio boundary can be used to achieve the exact size.

The source writes the same ${\lambda }_0$ in both hypotheses but immediately introduces ${\lambda }_1>{\lambda }_0$. Use the intended alternative $\lambda ={\lambda }_1$. If read literally as two identical hypotheses, there is no such distinct alternative to test.

For Weibull shape two, the sample likelihood is proportional toThuswhich is increasing in $S={\sum }_i{X}_i^2$. The most powerful critical region is thereforeIndeed, $2S/{\lambda }_0^2\sim {\chi }_{2n}^2$, so $C=({\lambda }_0^2/2)q_{2n,1-\alpha }$. The critical value depends only on the null and size, not on the chosen larger alternative. Hence the same test is uniformly most powerful for $\lambda >{\lambda }_0$, and in particular for the printed restricted range $\lambda >{\lambda }_1$.