An exact one-sided normal sample size and its effect-variance sensitivity

Since $1\mathrm{/}2<\mathrm{\Phi }(1)<1$ by [F2], $\alpha ={\beta }_{\mathrm{I}\mathrm{I}}=1-\mathrm{\Phi }(1)$ belong to $(0,1)$. Thus $z_{1-\alpha }=z_{1-{\beta }_{\mathrm{I}\mathrm{I}}}=z_{\mathrm{\Phi }(1)}=1$, and the quantile sum in [F1] is $2$.

The theorem's formula gives $n_{\ast }=\max \{1,\lceil (\sigma \mathrm{/}\sigma {)}^2(2{)}^2\rceil \}=4$ by [F3]. At $n=3$, power is $\mathrm{\Phi }(\sqrt{3}-1)<\mathrm{\Phi }(1)$ because $\sqrt{3}<2$ and $\mathrm{\Phi }$ is strictly increasing. Therefore the target fails at $n=3$, while [F1] and the formula certify that $n=4$ is the least positive feasible size.

At $n=4$, halving $\delta$ changes $\sqrt{n}\delta \mathrm{/}\sigma$ from $2$ to $1$, so the power becomes $\mathrm{\Phi }(1-1)=\mathrm{\Phi }(0)=1\mathrm{/}2<\mathrm{\Phi }(1)=1-{\beta }_{\mathrm{I}\mathrm{I}}$.

If instead $\delta =\sigma$ and $n=4$ are fixed while the standard deviation becomes $2\sigma$, then the standardized effect also becomes $1$; the power is again $1\mathrm{/}2$, below target. This standard-deviation change quadruples the variance, so the example records variance sensitivity without treating the variance itself as the standard deviation.