= Solution
Here the <exponential distribution> is parametrized by its <mean>, so $f(x;\theta)=\theta^{-1}e^{-x/\theta}$ for $x\geq0$ and $\theta>0$. Its <log-likelihood> and <maximum-likelihood estimate> are
$$
\ell(\theta)=-n\log\theta-\frac{\sum_iX_i}{\theta},\qquad \widehat\theta=\overline X.
$$
For one observation, $\partial_\theta\log f=-1/\theta+X/\theta^2$. Since $\operatorname{Var}(X)=\theta^2$, its <Fisher information> is $I_1(\theta)=1/\theta^2$. Consequently the estimated <standard error> is $\overline X/\sqrt n$, giving
$$
\boxed{\left[\overline X\left(1-\frac{z_{1-\alpha/2}}{\sqrt n}\right),\quad
\overline X\left(1+\frac{z_{1-\alpha/2}}{\sqrt n}\right)\right].}
$$
This is a large-sample <confidence interval>, not an exact one. If its lower endpoint is negative, intersecting with the <parameter space> $(0,\infty)$ removes impossible values without excluding any possible true <statistical parameter>.
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