Approximate confidence interval; Rayleigh Distribution

Question

This is a HW problem in my undergrad Statistics course. I am unsure of how to proceed. We were given a text file of data. I assumed that we would need $\overline{X}$; which I found to be $1.312670963$. In a past assignment we showed that as
\begin{equation} n\rightarrow \infty, \tilde{\theta} = \sqrt{\dfrac{1}{2n}\sum_{i=1}^nX_i^2}\xrightarrow{d}N\bigg(\theta,\dfrac{\theta^2}{4n}\bigg) \end{equation} We are asked to construct an approximate $95\%$ confidence interval for $\theta$. If I have the Fisher function as $J(\theta)=\dfrac{4n}{\theta^2}$. Can I state
\begin{equation} \bigg(\hat{\theta}-\dfrac{1.96\hat{\theta}}{2\sqrt{n}},\hat{\theta}-\dfrac{1.96\hat{\theta}}{2\sqrt{n}}\bigg) \end{equation} Using the score function and setting it to $0$. I found: \begin{equation} \hat{\theta}=\sqrt{\dfrac{1}{2n}\sum_{i=1}^nX_i^2} \end{equation}

Would I be correct in using this in the above equation? I am not sure how to start this problem. Any help would be appreciated. Thanks

Answer 1

If your CI is "approximate", then assuming your $n$ is sufficiently large, you can use the asymptotic distribution of $\tilde\theta$. ie, for 95% confidence,

$$CI = \hat{\tilde{\theta}} \pm 1.96\times\sqrt\frac{\tilde{\hat{\theta}}^2}{4n}$$