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The given problem is X follows $Rectangular(\theta,\theta^2)$ To test $H_0:\theta=2$ against $H_1:\theta=3$ They asked to calculate the power and size if given crictical region is $X\geq4.5$ Now they have calculated Size=$P_{H_0}(X\geq4.5)$ But my question is if we follow Neyman Pearson Lemma then $L(x|H_0)=0$ when $4\leq X\leq9$ Since under $H_0$ distribution is $Rectangular(2,4)$ and also $L(x|H_1)=\frac{1}{6}$ when $4\leq X\leq9$ Since under $H_1$ distribution is $Rectangular(3,9)$. Now Under $H_0$ there is probability 0 in $4\leq X\leq9$ hence $\frac{L(x|H_0)}{L(x|H_1)}=0$ in $4\leq X\leq9$. hence since neyman pearson lemma says $\frac{L(x|H_0)}{L(x|H_1)} \leq K$ is the critical region for any positive K, then in $4\leq X\leq9$ always $\frac{L(x|H_0)}{L(x|H_1)}=0\leq K$ since K is positive. Hence $4\leq X\leq9$ is always a critical region.So while calculation of size we must consider $(4,4.5)$ also as a crtical region apart from the given critical region $X\geq4.5$. Is my explanation correct???

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The Neyman-Pearson Lemma is about what the optimal rejection region is (which is apparently what you call "critical region", chances are both terms are in use), but you can define non-optimal tests with other rejection regions. If I understand your question right, you got $X\ge 4.5$ given as a rejection region, so this is a test different from the Neyman-Pearson optimal one, and if you are asked to analyse that one, you can leave the Neyman-Pearson Lemma alone.

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