= 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 deviation>
$$
\sigma_1=\sqrt{\frac{p_C(1-p_C)+p_T(1-p_T)}n}=0.00136646.
$$
For 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> is
$$
1-\Phi\left(\frac{1.96\sigma_0-\Delta}{\sigma_1}\right)+\Phi\left(\frac{-1.96\sigma_0-\Delta}{\sigma_1}\right)\simeq0.448.
$$
\b[No: 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.
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