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Consider a population with three kinds of individuals labeled $1, 2$, and $3$ occuring in the Hardy–Weinberg proportions $f(1,\theta)=\theta^2,f(2,\theta)=2\theta(1−\theta),f(3,\theta)=(1−\theta)^2$. For a sample $X_1, . . . , X_n$ from this population, let $N_1$, $N_2$, and $N_3$ denote the number of $X_j$ equal to $1$, $2$, and $3$, respectively. Let $0<\theta_0<\theta_1 <1$. We know that the likelihood ratio is an increasing function of $2N_1+N_2$. (Likelihood ratio of Hardy–Weinberg proportions)

What I want to conclude now is that if $c>0$ and $\alpha\in(0,1)$ satisfy $P_{\theta_0}[2N_1+N_2\geq c]=\alpha$, then the test that rejects $H$ if, and only if, $2N_1 + N_2\geq c$ is most powerful (MP) for testing $H : \theta = \theta_0$ versus $K : \theta = \theta_1$.

Thanks for your help in advance!

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The Neyman–Pearson Lemma says the most powerful test for two point hypotheses is always the likelihood ratio test. You have a test comparing two point hypotheses, and you are looking at the likelihood ratio test, so it's the most powerful test.

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