Solution
= Solution
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.