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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.

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$T$ is actually $\chi^2(2n)$ distributed, since $\Gamma(k/2,2)=\chi^2(k)$. Does that give you the right answer? :) –  MånsT Jun 8 '12 at 9:19
    
Ok, I wrote it wrong. I actually considered $\chi^2(2n)$, but the answer is wrong :( ... –  Aslan986 Jun 8 '12 at 9:24
    
Having read your solution more carefully, I think that you got the distribution of $Y$ wrong. It should be $\Gamma(n,1/\lambda)$; see en.wikipedia.org/wiki/Gamma_distribution#Special_cases –  MånsT Jun 8 '12 at 9:43
    
You are right. I followed Italian Wikipedia which is wrong ( it.wikipedia.org/wiki/Distribuzione_esponenziale#Distribuzioni ). I'f you want to write an answer i will accept it. Thank you very much. –  Aslan986 Jun 8 '12 at 10:08
    
This is a common mistake: the different parametrisations used for the exponential distribution often cause confusion. I hope that you corrected the error in Italian Wikipedia :) –  MånsT Jun 8 '12 at 10:22
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My comment from above, so that the question can be marked as answered:

Having read your solution more carefully, I think that you got the distribution of Y wrong. It should be Γ(n,1/λ); see Wikipedia.

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