Given $\hat{\theta}=\frac{-n}{\sum\limits_{i=1}^nlog(X_i)}$ is the mle of $\theta$ for a $\beta(\theta,1)$ distribution and $W=-\sum\limits_{i=1}^nlog(X_i)$ has a $\Gamma\left(n,\frac{1}{\theta}\right)$ distribution. Answer the following:
i) Show that $2\theta W$ has a $\chi^2_{(2n)}$ distribution.
ii) Using i), find $c_1$ and $c_2$ such that,
$$ Pr\left(c_1<\frac{2\theta n}{\hat{\theta}}<c_2\right)=1-\alpha $$
for $0<\alpha<1$. Then, obtain a $(1-\alpha)100\%$ confidence interval for $\theta$.
iii) For $\alpha=0.05$ and $n=10$ compare the length of this interval with the length of the interval given by; $\hat{\theta} \pm z_{\alpha/2}\frac{\hat{\theta}}{\sqrt{n}}$.
I'm not having trouble with i) anymore, but I'm hoping to get some guidance for ii) and iii). For ii), since $\frac{2\theta n}{\hat{\theta}}=2\theta W$ are $c_1$ and $c_2$ equal to $\chi^2_{(2n),\frac{\alpha}{2}}$ and $\chi^2_{(2n),1-\frac{\alpha}{2}}$ respectively? And then is the $(1-\alpha)100\%$ confidence interval for $\theta$ given by;
$$ Pr\left(\frac{\chi^2_{(2n),\frac{\alpha}{2}}}{2W}<\theta<\frac{\chi^2_{(2n),1-\frac{\alpha}{2}}}{2W}\right), $$
even though we're dividing by the random variable $W$? And iii) is still confusing to me. Any help is appreciated.