= Solution
For each <bootstrap> replicate form the studentized pivot
$$
T^*=\frac{\widehat\theta^*-\widehat\theta}{\widehat s^*}.
$$
Let $t^*_{.025},t^*_{.975}$ be its empirical <quantiles>. The <bootstrap-t confidence interval> approximates
$$
\mathbb P\!\left(t^*_{.025}\leq\frac{\widehat\theta-\theta}{\widehat s}
\leq t^*_{.975}\right)\simeq0.95.
$$
Solving both inequalities for $\theta$ reverses the <quantile> order, giving
$$
\boxed{[\widehat\theta-t^*_{.975}\widehat s,\quad
\widehat\theta-t^*_{.025}\widehat s].}
$$
Using the same fixed <standard error> in every <bootstrap> numerator would not implement this studentized procedure; each replicate needs its own $\widehat s^*$.
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