= Solution
Here $z_\alpha=1.96$, $\sigma(1/2)=1/(2\sqrt n)$ and $\sigma(3/4)=\sqrt3/(4\sqrt n)$. 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 $-1.96$:
$$
-\frac{\sqrt n}{\sqrt3}+\frac{2(1.96)}{\sqrt3}\leq-1.96.
$$
Rearranging and squaring the positive quantities yields
$$
\boxed{n\geq[1.96(2+\sqrt3)]^2\approx53.5066,\qquad n=54\text{ as the approximate design}.}
$$
This 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.
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