= Solution
For a <bootstrap-t confidence interval>, estimate the original <standard error> $\widehat s$ of $\widehat r$, for example by the standard deviation of the outer <paired bootstrap> correlations. For each outer resample $b$, estimate its own <standard error> $s_b^*$ by drawing many inner paired resamples from that outer sample and taking the standard deviation of their correlations. The resulting studentized statistics are
$$
T_b^*=\frac{r_b^*-\widehat r}{s_b^*}.
$$
Let $t_p^*$ be their empirical quantiles. Approximating the distribution of $(\widehat r-r)/\widehat s$ by that of $T^*$ and solving the two inequalities for $r$ gives
$$
\boxed{[\widehat r-t_{0.975}^*\widehat s,\ \widehat r-t_{0.025}^*\widehat s].}
$$
The reversal of quantiles is essential. The pivot requires positive <standard errors> and nondegenerate correlations; its claimed 95% coverage is an approximation under <bootstrap> regularity, not an exact finite-sample guarantee. One may intersect the final interval with the parameter space $[-1,1]$.
The same percentile and studentized procedures can be applied to the <sample mean> of $X$, and then only the $X$ observations need be resampled. They are useful when a normal approximation is doubtful. However the mean already has the simple standard-error estimate $S_X/10$ here. For normal observations, the classical exact interval is
$$
\boxed{\overline X\pm t_{99,0.975}\frac{S_X}{10},}
$$
and the same form is an asymptotic approximation for independent finite-variance observations under appropriate large-sample conditions. Thus <bootstrap> intervals are valid alternatives, but are often unnecessary for a mean with a reliable classical pivot; the correlation has no corresponding distribution-free elementary pivot. An infinite-variance distribution would require other theory rather than this ordinary standard-error argument.
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