 Solution

Here $z_{\alpha }=1.96$, $\sigma (1/2)=1/(2\sqrt{n})$ and $\sigma (3/4)=\sqrt{3}/(4\sqrt{n})$. The desired approximate power is at least $0.975=\Phi (1.96)$ exactly when the argument of the upper tail in part (i) is at most $-1.96$:Rearranging and squaring the positive quantities yieldsThis is a normal approximation to a sample size calculation. It is not a guarantee of the exact finite-sample power for an arbitrary unrandomized binomial cutoff; the discrete test size and power can instead be checked using binomial probabilities.