Reject $H_0$ in favor of $H_1$ if $\hat p > p_0 + z_\alpha \sqrt{\frac{p(1-p)}{n}}$ A hypothesis test for a proportion is the same as a [[test for a mean]] when you consider that $\hat p \sim N \Big (p, \frac{p(1-p)}{n} \Big )$ provided the sample size is sufficiently large to meet the criteria $\Big (\hat p - 3 \sqrt{\frac{\hat p(1- \hat p)}{n}}, \ \hat p + 3 \sqrt{\frac{\hat p(1- \hat p)}{n}}$ is completely contained with $[0,1]$. For composite hypotheses, we need calculate the maximum of the probability of a [[Type I Error]]. We can find that the quantity $\frac{c-p}{\sqrt{\frac{p(1-p)}{n}}}$ is increasing as a function of $p$, which allows us to use the endpoint of the range of values in the parameter space where the null hypothesis is true to maximize the probability of a Type I error.