Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 2

Let ${\hat{p}}_k=X_k/n_k$, where $X_k\sim \mathrm{Binomial}(n_k,p_k)$ independently, and putThe Wald statistic is $Z=({\hat{p}}_1-{\hat{p}}_0)/\sqrt{\hat{V}}$. The central limit theorem and Slutsky theorem give $Z\dot{\sim }N(0,1)$ under $H_0$. When $p_1-p_0={\delta }^{\ast }>0$,

For equal arm size $n$, write $v^{\ast }=p_1(1-p_1)+p_0(1-p_0)$ at the clinically relevant alternative. A one-sided level-$\alpha$ Wald test rejects when $Z>z_{1-\alpha }$, and its approximate statistical power isEquating this to $1-\beta$ gives the per-arm sample sizerounded up. If the design uses a null-based critical standard error $v_0$ but an alternative standard error $v^{\ast }$, the corresponding more general formula is

A group sequential design can reduce the expected sample size by stopping at an interim analysis once efficacy or futility is sufficiently clear. It does not generally reduce the prespecified maximum sample size: repeated opportunities to reject inflate the Type I error, so valid sequential stopping boundaries usually require a modest increase in maximum information relative to a fixed-sample design with the same power. The benefit is a smaller expected sample size under alternatives that often cross an early boundary, and sometimes under the null through early futility stopping.

Letwhere $m_j$ is the cumulative sample size per arm. The canonical joint distribution for group sequential test statistics isThus $\mathrm{Corr}(Z_1,Z_2)=\sqrt{m_1/m_2}$. This correlation arises because the second statistic reuses all first-stage observations.

Under $H_0$, $(Z_1,Z_2)$ is a bivariate standard normal distribution with correlation $\rho =\sqrt{m_1/m_2}$. Rejection occurs either at stage 1 through $Z_1\ge u_1$, or at stage 2 through $Z_2\ge u_2$ after continuation $l_1<Z_1<u_1$. Hence the Type I error isThe final lack-of-benefit boundary $l_2$ affects acceptance, but not the probability of crossing an efficacy boundary.

Response-adaptive randomization can assign a larger proportion of later participants to the treatment currently estimated to be better, improving outcomes for participants within the trial. Its allocation probabilities depend on earlier outcomes, which complicates statistical inference; delayed responses and calendar-time trends can also make adaptation ineffective or biased, and an allocation aimed at patient benefit need not maximize power.

Under Neyman allocation, sample sizes are proportional to the arm standard deviations. HereFor total size $n_{\max }$, the minimized asymptotic variance isEqual allocation givesThe Neyman allocation therefore reduces the large-sample variance by $0.04/n_{\max }$, about $5.9\%$ of the equal-allocation variance.