= 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, is
$$
n\simeq\frac{\left[z_{0.975}\sqrt{2\bar p(1-\bar p)}+
z_{0.8}\sqrt{p_0(1-p_0)+p_1(1-p_1)}\right]^2}{(p_0-p_1)^2},
\qquad\bar p=\frac{p_0+p_1}{2}.
$$
For $p_0=0.4$, $p_1=0.3$, this gives $n\simeq355.94$, or 356 per group. Thus \b[500 positives in each period are sufficient for 80% power under these assumptions].
More explicitly, a pooled two-proportion test rejects when
$$
\left|\frac{\widehat p_0-\widehat p_1}
{\sqrt{\widehat p(1-\widehat p)(1/500+1/500)}}\right|>1.96,
$$
where $\widehat p$ 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 $1.96\sqrt{0.455/500}\simeq0.0591$, giving approximate power
$$
\Phi\left(\frac{0.10-0.0591}{0.03}\right)
+\Phi\left(\frac{-0.10-0.0591}{0.03}\right)
\simeq0.913.
$$
This 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.
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