Use $2\sum_{i=1}^n Y_i/\beta$ which is a pivotal quantity to derive a 95% confidence interval for $\beta$

Question

Suppose $Y_1$,...$Y_n$ is a random sample from a gamma distribution with $\alpha = 2$ and unknown $\beta$.

GOAL: Use $2\sum_{i=1}^n Y_i/\beta$ which is a pivotal quantity to derive a 95% confidence interval for $\beta$

First question, could there be more than one 95% confidence interval satisfies this question?

The one I am given is $$P(\chi_.975^2 \le 2\sum_{i=1}^n Y_i/\beta \le \chi_.025^2) = .95$$

After several intermediate steps which will get us to $$\left[\frac{2\sum_{i=1}^n Y_i}{\chi_.975^2}, \frac{2\sum_{i=1}^n Y_i}{\chi_.025^2}\right]$$ is the 95% confidence interval for $\beta$

Is the the right interval as I am not sure $$P(\chi_.975^2 \le 2\sum_{i=1}^n Y_i \le \chi_.025^2) = .95$$ or $$P(\chi_.025^2 \le 2\sum_{i=1}^n Y_i \le \chi_.975^2) = .95$$ or both are the same thing?

Answer 1

I am suprised that you say $\chi_{.975}^2 \lt \chi_{.025}^2$, but if it is true then $\displaystyle P\left(\chi_.025^2 \le 2\sum_{i=1}^n Y_i \le \chi_.975^2\right) = 0$ which is not what you want.

There will be other 95% confidence interval. Examples would include the one-sided cases$$ \displaystyle \left[\frac{2\sum_{i=1}^n Y_i}{\chi_.95^2}, \infty\right) \text{ or } \displaystyle \left(-\infty,\frac{2\sum_{i=1}^n Y_i}{\chi_.05^2}\right]$$