 Solution

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.