I got this question: The Weibull distribution with $\alpha>0$ has density function: $$f_{\alpha}(x)=\alpha x^{\alpha-1}e^{-x^{\alpha}}$$ with respect to standard Lebesgue measure. X follow this Weibull distribution.
I have to construct a 95 % two-sided confidence interval $C(X)$ for $\alpha$ based on the universal pivot. Can anyone help med with that? I think that in my notes I can see that the interval is given by $C=(\frac{X_n}{g^n_{0.975}},\frac{X_n}{g^n_{0.025}})$ where $g^n_{0.975}$ and $g^n_{0.025}$ denote the corresponding quantiles. But how can I find these and how to find the interval?