There is something that I am doing wrong in the exercise below and I would appreciate some help figuring it out.
Let $X_1, X_2, \ldots, X_5$ be a random sample from a $\Gamma \left(3,3 \right)$ distribution . Define $W=\sum_{i=1}^5 X_i $.
Obtain the distribution of $W$ and of $2W/3$
Find $c_1$ and $c_2$ such that $P \left( c_1 < W <c_2 \right)=0.95 $
Since $W$ is the sum of iid gamma variables, $W \sim \Gamma \left(15,3 \right)$ and now a straightforward transformation argument shows that $Y=\frac{2}{3} W \sim \chi^2 \left(30 \right) $. This completes question 1.
Now for question 2, my thoughts were to transform these probabilities using the result from question 1, namely that $\frac{2}{3} W \sim \chi^2 \left(30 \right) $. Hence, unless I have made a horrible mistake here:
$$P \left( \frac{2}{3} c_1 < Y < \frac{2}{3} c_2 \right)=0.95 $$
We can find the constants from a chisquare table for $30$ degrees of freedom. In my case, they are $16.791$ for the lower bound and $46.979$ for the upper one.
Then if we equate $\frac{2}{3}c_1=16.791$ and solve for $c_1$ we should be able to find the constant for the above gamma confidence interval, right?
The problem is that according to $R$, $P \left( W<c_1 \right) =1 $, instead of the desired $0.025$ which means I have gone wrong somewhere. I do not think it is in question though $1$ as I have verified the result many times.
Thank you.

