 Solution

Use a conventional two-sided test at significance level $0.05$. Let $n=4000$ per arm, $p_C=0.005$, $p_T=0.0025$ and $\Delta =p_C-p_T=0.0025$. Under the specified alternative, the normal approximation to the difference of independent binomial proportions has standard deviationFor planning the pooled null standard error, use $\bar{p}=(p_C+p_T)/2=0.00375$, giving ${\sigma }_0=\sqrt{2\bar{p}(1-\bar{p})/n}=0.00136674$. The rejection threshold is approximately $1.96{\sigma }_0=0.00267880$, exceeding the mean alternative difference. With $\Phi$ the standard normal distribution function, the approximate statistical power isNo: the usual two-sided 5% calculation gives about 45% power, below 50%. The expected death counts are only 20 and 10, so an exact discrete design calculation could refine this approximation.

The question does not specify the tail convention. If a beneficial direction is prespecified and a one-sided test at 5% is justified, replace $1.96$ by approximately $1.645$; the resulting statistical power is about $0.573$, and the answer is then yes. This change of testing convention must be declared.