The trial uses $N_1$ observations per arm if it stops at the first analysis and $N_2$ if it continues. Since $Z_1\sim N(\delta\sqrt{I_1},1)$,
$$
\mathbb P_\delta(\text{continue})
=\Phi(e_1-\delta\sqrt{I_1})-\Phi(f_1-\delta\sqrt{I_1}).
$$
Therefore the expected per-arm sample size is
$$
\boxed{
\mathbb E_\delta N
=N_1+(N_2-N_1)
\left\{\Phi(e_1-\delta\sqrt{I_1})-\Phi(f_1-\delta\sqrt{I_1})\right\}}.
$$


Solution

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

 Solution

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