TITLE: Distribution of the sample mean of a exponential
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: Correction.
In discussing this question, I have discovered errors here.
Specifically if $n$ observations are sampled at random from
$\mathsf{Exp}(\text{rate} = \lambda),$ as shown in the Question above,
then $T \sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$
The proof is that the MGF of $X_i$ is $M_X(t) = \frac{\lambda}{1-t},$
so the MGF of $T$ is $M_T(t) = (\frac{\lambda}{1-t})^n,$ which is
the MGF of $\mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$
Consequently, $\bar X \sim \mathsf{Gamma}(n, n\lambda).$ 
(This relationship is illustrated in the link.)