Constructing a symmetrical $100(1-a)\%$ confidence interval for $\theta$. Exercise :

Let $X_1, \dots, X_n$ be a random sample from the distribution function $F(x) =1 - \frac{\theta^3}{x^3}, \; x \geq \theta$ where $\theta >0$ unknown parameter.
(i) Find a maximum likelihood estimator for $\theta$
(ii) Find a sufficient and complete statistics function $T$, for $\theta$.
(iii) Check if the maximum likelihood estimator you found is unbiased for $\theta$.
(iv) Find the distribution of $Y = T/\theta$.
(v) Construct a symmetrical $100(1-a)\%$ confidence interval for $\theta$.

Question :
I have successfully solved each one of the parts (i) to (iv) but I find myself stuck on (v). I am not very familiar with confidence intervals, so it could be something simple. I would really appreciate a thorough explanation or solution so as I can grasp the idea of such question parts, since this is a usual exams question.
 A: This is a Pareto population with pdf $$f_{\theta}(x)=\frac{3\theta^3}{x^4}\mathbf1_{x\ge\theta}\quad,\,\theta>0$$
A sample $(X_1,X_2,\ldots,X_n)$ is drawn from this population. 
A sufficient statistic for $\theta$ is then $X_{(1)}=\min(X_1,\ldots,X_n)$, which itself has a Pareto distribution with support $[\theta,\infty)$. This suggests that an appropriate pivot for $\theta$ is given by $Z=\frac{\theta}{X_{(1)}}$.
Indeed, for any $z\in (0,1)$, we have $$P_{\theta}(Z\le z)=z^{3n}\quad,\,\forall\,\theta$$
So the pdf of $Z$ is $$f_Z(z)=3n z^{3n-1}\mathbf1_{0< z< 1}$$
In other words, $Z$ has a $\mathsf{Beta}(3n,1)$ distribution. 
Now there are several options available to find a confidence interval for $\theta$. You could, for example, find $z_1,z_2$ such that $P_{\theta}\left[z_1\le Z\le z_2\right]=1-\alpha\,\,\ldots(*)$ holds for every $\theta$. Then a confidence interval for $\theta$ with confidence coefficient $1-\alpha$ will be $[z_1 X_{(1)},z_2 X_{(1)}]$ subject to the restriction $(*)$.
You could apply the same procedure working with the pivot $X_{(1)}/\theta$ instead.
But knowing the distribution of $Z$, here I would use the fact that $-2\cdot 3n\ln Z=-6n\ln Z\sim \chi^2_2$.
So if $\chi^2_{\alpha/2,2}$ is the $(1-\frac{\alpha}{2})$th fractile of $\chi^2_2$ distribution, we have $$P_{\theta}\left[\chi^2_{1-\alpha/2,2}\le -6n\ln Z\le \chi^2_{\alpha/2,2}\right]=1-\alpha\quad,\,\forall\,\theta$$
Or, $$P_{\theta}\left[\exp\left(\frac{\chi^2_{\alpha/2,2}}{-6n}\right)\le Z\le \exp\left(\frac{\chi^2_{1-\alpha/2,2}}{-6n}\right)\right]=1-\alpha\quad,\,\forall\,\theta$$
So a $100(1-\alpha)\%$ confidence interval for $\theta$ is 
$$\left[\exp\left(\frac{\chi^2_{\alpha/2,2}}{-6n}\right)X_{(1)}\,,\,\exp\left(\frac{\chi^2_{1-\alpha/2,2}}{-6n}\right)X_{(1)}\right]$$
I could not find an exact confidence interval that would be symmetric. If my calculations are correct, then the shortest (expected) length confidence interval for $\theta$ with confidence coefficient $1-\alpha$ based on the pivot $Z$ is $\left[\alpha^{1/3n}X_{(1)}\,,\, X_{(1)}\right]$, which is not symmetric either.
