= Solution
The <standard normal distribution> function is strictly increasing. Thus $p>\alpha$ implies
$$
1-\Phi(\sqrt n(\overline x-\mu_0))>\alpha
\quad\Longleftrightarrow\quad
\sqrt n(\overline x-\mu_0)<z_\alpha.
$$
Rearranging gives
$$
\boxed{\mu_0>\overline x-\frac{z_\alpha}{\sqrt n},\quad\text{so }\mu_0\in\left(\overline x-\frac{z_\alpha}{\sqrt n},\infty\right).}
$$
This is the confidence-interval/test duality for the matching one-sided procedures. The strict inequalities agree with the printed <open> <interval>.
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