Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from an exponential distribution with mean \(\theta\). Then \(100(1 - \alpha)\%\) asymptotic confidence interval for \(\theta\) based on sample mean \(\overline{X}_{n}\) :
Answer: B
The interval is \(\Bigg(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}},\) \(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}}\Bigg)\), with \(z_{p}\) the lower p-quantile of N(0, 1).
For an exponential with mean \(\theta\), the variance is \(\theta^{2}\), so by the central limit theorem \(\frac{\sqrt{n}(\overline{X}_{n} - \theta)}{\theta} \approx N(0, 1)\). Hence, approximately,
\[ \begin{aligned} z_{\alpha/2} &\lt \sqrt{n}\left(\frac{\overline{X}_{n}}{\theta} - 1\right) \lt z_{1-\alpha/2} \\ \frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}} &\lt \theta \lt \frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}} \end{aligned} \]
with probability about \(1 - \alpha\). Since \(z_{\alpha/2} \lt 0 \lt z_{1-\alpha/2}\), the larger quantile gives the lower limit.
Option A has the limits in reverse order (its lower end exceeds its upper end), and option C drops the factor \(\sqrt{n}\) from the numerators. The interval can be determined, so D is wrong.
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