 Solution

Treat the two retrieved sets of positive tests as independent binomial samples, each of size 500, and use a two-sided test at significance level $0.05$. The normal-approximation sample size for comparing two proportions, with equal numbers $n$ in each group, isFor $p_0=0.4$, $p_1=0.3$, this gives $n\simeq 355.94$, or 356 per group. Thus 500 positives in each period are sufficient for 80% power under these assumptions.

More explicitly, a pooled two-proportion test rejects whenwhere $\hat{p}$ pools the maximum-readout successes. At the proposed alternative, the difference has mean $0.10$ and standard deviation $\sqrt{0.45/500}=0.03$. The planning null critical difference is $1.96\sqrt{0.455/500}\simeq 0.0591$, giving approximate powerThis compares the readout percentage among positive tests. It assumes comparable measurement rules and independent observations; a change in that percentage alone does not identify its causal explanation or the prevalence of drug use in the full tested population.