= Solution
For independent <binomial distributions> in the two trial arms, the estimate of the <log risk ratio> is
$$
\widehat\ell=\log\frac{1/12}{4/10}=\log\frac5{24}=-1.5686.
$$
The <delta method> gives $\operatorname{Var}(\log\widehat p)\simeq(1-p)/(np)$; replacing $p$ by the <sample proportion> gives $1/d-1/n$, where $d$ is the event count. Adding the independent arm contributions therefore gives
$$
\widehat{\operatorname{Var}}(\widehat\ell)
=\left(\frac11-\frac1{12}\right)+\left(\frac14-\frac1{10}\right)
=\frac{16}{15}=1.0667.
$$
The <standard error> is $\sqrt{16/15}=1.0328$. Using the stipulated normal quantile two, the approximate <confidence interval> is
$$
\boxed{\widehat\ell\pm2\operatorname{SE}(\widehat\ell)
=(-3.6342,\ 0.4970).}
$$
The point estimate of the <risk ratio> is $5/24\simeq0.208$, 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. \b[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.
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