= Solution
{c}
Let $z=z_{1-\alpha/2}$ be the $(1-\alpha/2)$ <standard normal quantile>. Since $\widehat\mu$ is a <consistent estimator>, <Slutsky's theorem> gives
$$
\frac{\sqrt{3n}(\widehat\mu-\mu)}{\widehat\mu}\xrightarrow dN(0,1).
$$
Its estimated <standard error> is $\widehat\mu/\sqrt{3n}$, giving the approximate $(1-\alpha)$ <Wald confidence interval>
$$
\boxed{\left[\widehat\mu\left(1-\frac z{\sqrt{3n}}\right),\quad
\widehat\mu\left(1+\frac z{\sqrt{3n}}\right)\right]\cap(0,\infty).}
$$
For each fixed $\mu>0$, this procedure's <coverage probability> tends to $1-\alpha$. Intersecting with the positive parameter space preserves coverage of the true positive value.
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