The expected Treatment A deaths are $1/2$ at month 1, $1/2$ at month 2, and $3(2/6)=1$ at the three tied deaths at month 5. Thus
$$
E_A=2,
\qquad O_A=3.
$$
This is not half of the five deaths because <right censoring> changes the treatment proportions in successive <risk sets>. For Treatment B, $O_B=2$ and $E_B=3$. One observed-to-expected relative-risk estimate is therefore
$$
\frac{O_A/E_A}{O_B/E_B}
=\frac{3/2}{2/3}=\frac94.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-207/4/a/solution