 Solution

Put $u=\sqrt{m}\bar{X}$ and $v=\sqrt{n}\bar{Y}$. Test (a) accepts inside the discwhereas $Z$ is the projection of $(u,v)$ onto the unit vectorIf $q_1=[{\Phi }^{-1}(1-\alpha /2){]}^2$, then $q_1$ is the corresponding ${\chi }_1^2$ quantile. Since a ${\chi }_2^2$ variable is stochastically larger than a ${\chi }_1^2$ variable, $q_2>q_1$. Choose a point on that unit-vector line with squared radius strictly between $q_1$ and $q_2$. A whole open neighborhood then makes (b) reject while (a) accepts. The joint normal density of $(\bar{X},\bar{Y})$ is strictly positive everywhere for every true $(\lambda ,\mu )$, so this neighborhood always has positive probability.