For $k=4$, the optimum is $q=2$. The supplied two-sided <sample size> formula gives
$$
n_k=\frac{1+2}{2}\left(\frac22\right)^2(0.84+1.96)^2
=\frac32(7.84)=11.76,
$$
so round upward to $12$ patients in each new-treatment arm and take $n_0=24$. The analyzed total is therefore
$$
n_{\mathrm{tot}}=4(12)+24=72.
$$
Allowing for ten percent attrition requires
$$
\left\lceil\frac{72}{0.9}\right\rceil=\boxed{80}
$$
patients to be recruited.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 3 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-207/3/a/ii/solution