= Solution
{c}
The <pivotal quantity> $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb P(U\le t)=t^n$. Thus
$$
\mathbb P\left(\alpha^{1/n}\le\frac M\theta\le1\right)=1-\alpha.
$$
Rearranging these inequalities yields the exact <confidence interval>
$$
\boxed{\left[\widehat\theta,\frac{\widehat\theta}{\alpha^{1/n}}\right].}
$$
For every $\theta>0$, its coverage <probability> is exactly $1-\alpha$; the result uses the finite-sample <probability distribution>, with no <normal approximation>.
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