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

Since $n_0=q{n}_k$ and the $k$ experimental arms each contain $n_k$ patients,ConsequentlyDifferentiation gives $V^{\prime }(q)=(1-k/q^2)/n_{\mathrm{tot}}$, so the stationary point is $q=\sqrt{k}$. Sincethis is the unique minimum. Thus a shared standard arm should be $\sqrt{k}$ times the size of each new-treatment arm.

For $k=4$, the optimum is $q=2$. The supplied two-sided sample size formula givesso round upward to $12$ patients in each new-treatment arm and take $n_0=24$. The analyzed total is thereforeAllowing for ten percent attrition requirespatients to be recruited.

A group sequential design can stop early for efficacy or futility, reducing expected sample size, cost, and time and limiting exposure to an inferior treatment. Its disadvantages include greater design and operational complexity, a potentially larger maximum sample size, and the need for adjusted stopping boundaries and inference to preserve the type I error after repeated looks.

The canonical joint distribution for group sequential test statistics is

Let ${\phi }_2(z;\mu ,\Sigma )$ be the bivariate normal density with the mean and covariance from part ii. Continuing and then not rejecting means $f_1<Z_1<e_1$ and $Z_2<e_2$, so its probability is

The trial uses $N_1$ observations per arm if it stops at the first analysis and $N_2$ if it continues. Since $Z_1\sim N(\delta \sqrt{I_1},1)$,Therefore the expected per-arm sample size is