= Solution
{c}
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 to
$$
\lambda^{-2n}\exp\left(-\frac1{\lambda^2}\sum_iX_i^2\right).
$$
Thus
$$
\frac{L(\lambda_1)}{L(\lambda_0)}
=\left(\frac{\lambda_0}{\lambda_1}\right)^{2n}
\exp\left[\left(\lambda_0^{-2}-\lambda_1^{-2}\right)\sum_iX_i^2\right],
$$
which is increasing in $S=\sum_iX_i^2$. The most powerful critical region is therefore
$$
\boxed{S\geq C,\qquad
\mathbb P_{\lambda_0}(S\geq C)=\alpha.}
$$
Indeed, $2S/\lambda_0^2\sim\chi^2_{2n}$, 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 \b[uniformly most powerful for $\lambda>\lambda_0$], and in particular for the printed restricted range $\lambda>\lambda_1$.
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