= Solution
A numerical power calculation needs a significance level, desired power, sidedness and allocation; these are not specified in this part. For a concrete planning illustration, assume independent batches, equal samples per supplier, a two-sided 5% test with 80% power, and true rates $p_B=0.02$ and $p_A=0.04$. The observed 2% from B and C is being used as a planning value for B, not as proof that B's population rate is exactly known. There is also a numerical inconsistency in the stated frame: at six batches per working day, B and C together produce only 120 batches in two weeks, and their proposed schemes would test 24. The asserted 3,000 sampled batches cannot literally come from that frame. The calculation treats $60/3000$ as a stipulated planning estimate; its collection would require a larger frame or longer period.
For two independent <sample proportions>, the approximate null <variance> of their difference is $2\bar p(1-\bar p)/n$ and its <variance> at the alternative is $[p_A(1-p_A)+p_B(1-p_B)]/n$, with $\bar p=(p_A+p_B)/2$. Separating the null critical value from the alternative mean by the required power quantile gives the <sample size for comparing two proportions>:
$$
n\simeq\frac{\left[z_{1-\alpha/2}\sqrt{2\bar p(1-\bar p)}
+z_{1-\beta}\sqrt{p_A(1-p_A)+p_B(1-p_B)}\right]^2}
{(p_A-p_B)^2}.
$$
With $z_{0.975}=1.960$, $z_{0.8}=0.842$ and $\bar p=0.03$, this gives $n\simeq1140.83$, so the normal-approximation calculation rounds to \b[1,141 batches per supplier], or approximately 1,150 for a practical planning target. This is per supplier, not the combined total. A different power or a one-sided test changes the answer. Positive clustering of sampled batches requires a cluster-aware calculation or inflation, so the independent-batch calculation should not simply be applied to C's one-day clusters.
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