 Solution

Put $Q={\sum }_i{X}_i^2$. For $\theta >0$, the log-likelihood is $\ell (\theta )=\text{constant}-(n/2)\log \theta -Q/(2\theta )$. Its derivative is $-n/(2\theta )+Q/(2{\theta }^2)$, changing from positive to negative at $Q/n$. ThusThe known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta )=n/(2{\theta }^2)$. The large-sample variance is consequently $2{\theta }^2/n$. With $z_{1-\alpha /2}$ the indicated standard-normal quantile, a Wald confidence interval isThe interval may be restricted to the parameter space $\theta >0$. The parameter is the variance itself, so the standard error is proportional to $\hat{\theta }$, not to its square root.