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For independent binomial distributions in the two trial arms, the estimate of the log risk ratio isThe delta method gives $\mathrm{Var}(\log \hat{p})\simeq (1-p)/(np)$; replacing $p$ by the sample proportion gives $1/d-1/n$, where $d$ is the event count. Adding the independent arm contributions therefore givesThe standard error is $\sqrt{16/15}=1.0328$. Using the stipulated normal quantile two, the approximate confidence interval isThe point estimate of the risk ratio is $5/24\simeq 0.208$, corresponding to approximately 79% lower mortality risk in the transfusion arm. Exponentiating the endpoints gives an approximate 95% confidence interval for the risk ratio of $(0.0264,1.6437)$. It includes one, so the trial is compatible with no difference as well as substantial benefit or some harm. The point estimate favours transfusion, but the data are too imprecise to establish a difference at the 5% level. These are large-sample approximations, especially rough with only one event in an arm. A frequentist confidence interval describes repeated-sampling coverage, not a 95% posterior probability for this fixed parameter.

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:With $z_{0.975}=1.960$, $z_{0.8}=0.842$ and $\bar{p}=0.03$, this gives $n\simeq 1140.83$, so the normal-approximation calculation rounds to 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.