= Solution
{c}
The <standard error> suggested by part (iii) is $\theta/\sqrt n$. Substituting the <consistent estimator> $T_2$ gives $T_2/\sqrt n$; <Slutsky's theorem> gives
$$
\frac{\sqrt n(T_2-\theta)}{T_2}\xrightarrow dN(0,1).
$$
Using the central <standard normal quantiles> $\pm1.96$, an approximate 95% <Wald confidence interval> is
$$
\boxed{\left[T_2\left(1-\frac{1.96}{\sqrt n}\right),\quad
T_2\left(1+\frac{1.96}{\sqrt n}\right)\right]\cap(0,\infty)}.
$$
Its <coverage probability> approaches 0.95 for each fixed $\theta>0$ as $n\to\infty$. The intersection enforces positivity and does not change whether the true positive <statistical parameter> is covered. The value 1.64 would give the wrong central two-sided <coverage probability>.
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