Computing the bounds of a $\alpha$ level confidence interval I am solving this question (as practice) and I am stuck in the second question. I don't really understand how to compute the bounds of the confidence interval, and I am not sure how to approach it.
The question is given in the image (from Introduction to Mathematical Statistics by Hogg et al pg. 346 in my copy), but generally speaking given $Y$ which has a known distribution, how do we compute $c_1, c_2$ such that $$P(c_1 < Y < c_2) = 1 - \alpha,\quad 0 < \alpha < 1?$$
Since we know that $2 \theta W = \frac{2 \theta n}{\hat{\theta}}$ has a $\chi ^2 (2n)$ distribution, I was thinking that I could just normally compute the confidence interval and get $c_1$ and $c_2$ from that however I am unsure if that is the right technique for this question.

 A: Data: Suppose you have a sample of size $n = 10$ from $\mathsf{Beta}(3, 1),$ so that $\theta = 3.$
Sample simulated in R:
set.seed(305)
n = 10;  x = rbeta(n, 3, 1)
summary(x); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.6086  0.6916  0.8535  0.8052  0.9031  0.9581 
[1] 0.1241171  # sample SD

The MLE $\hat \theta = n/W =10/W = 4.388208,$
w = -sum(log(x))
[1] -2.278835
th.est = n/w;  th.est
[1] 4.388208

Given that $2\theta W \sim \mathsf{Chisq}(\nu=20),$ we have
$$P(L < 2\theta W < U) = 0.95,$$
where $L$ and $U$ cut probabilities $0.025$ from the lower and
upper tails of $\mathsf{Chisq}(20),$ respectively.
(th.est/20)*qchisq(c(.025,.975),20)
[1] 2.104316 7.497167

Given that
$$0.95 = P(L < 2\theta W < U) =P\left(L< \frac{20\theta}{\hat\theta} < U\right)\\ = 
P\left(\frac{\hat\theta L}{20}<\theta< \frac{\hat\theta U}{20}\right),$$
where $L$ and $U$ cut probability $0.025$ from the lower and upper
tails of $\mathsf{Chisq}(20).$
Then a 95% CI for $\theta$ is of the form
$\left(\frac{\hat\theta L}{20},\, \frac{\hat\theta U}{20}\right)
= (2,10, 7.50),$ which does happen cover $\theta = 3$ of our example.
(th.est/20)*qchisq(c(.025,.975),20)
[1] 2.104316 7.497167

