# Most powerful test for Hardy–Weinberg proportions

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 $$K : \theta = \theta_1$$.