 Solution

For equal allocation, let $n$ be the required sample size per arm, $p_C=0.0045$, $p_T=0.003$ and $\Delta =0.0015$. In the standard sample size for comparing two proportions calculation, a two-sided test at 5% with 80% statistical power requiresUsing $z_{0.975}=1.95996$ and $z_{0.8}=0.84162$ gives $n\simeq 26063.64$, which is rounded upward for actual recruitment. ThusSince the risks are small, dropping the $(1-p)$ corrections and using $1.96+0.84$ gives $N_{\mathrm{total}}\simeq 2(2.8{)}^2(0.0075)/(0.0015{)}^2=52266.7$, conventionally about 52.3 thousand. These are slightly different approximations to the same planning requirement. Add an allowance if outcome ascertainment is incomplete or the allocation is clustered. A prespecified one-sided test would need a different, smaller sample size; the calculation here uses the conventional two-sided interpretation consistently with part (b).