I please ask someone to check if my calculations are right.
I have $X_1, ..., X_n$ from a $\mathcal{E}(\lambda): f(x, \lambda) = \lambda e^{-\lambda x}$.
I have to find the $k$ such that $P(\bar{X} \le k) = \alpha$, where $\bar{X}$ is the sample mean; i did: $Y=\sum_{i = 1}^{n} X_i$ $Y \sim \Gamma (n, \lambda)$ $\bar{X} = \frac{1}{n} Y \sim \Gamma(n, \frac{\lambda}{n})$ $T = 2\frac{n}{\lambda} \bar{X} \sim \Gamma(\frac{2n}{2}, 2) \stackrel{d}{=}\chi^2 (2n) $ $P(\bar{X} \le k) = P(T \le k' = 2\frac{n}{\lambda} k) = \alpha $
Then i can find the value of $k'$ from the table, and finally find $k$. I'm i missing something? (I can't reach the result stated by the book).
Thank you very much.