Let $X_1,X_2,X_3 $ denote a random sample of size $3$ from population $X$ with probability density function.
$f(x;\theta)=\dfrac{1}{\theta}e^{\frac{-x}{\theta}}$ , $x>0$
The null hypothesis $H_0:\theta=3$ is to be rejected in favor of the alternative $H_\alpha:\theta\ne3$ if and only if $\overline{X} >6.296.$ What is the significance level of the test?
I tried to do it as follows:
$1.\ \ \ X\sim exp(\alpha) , Y=\sum_{i}X_i \sim G(\alpha,n) $
Using gamma PDF $f_y(y)=\dfrac{a^{\lambda}e^{-ax}x^{n-1}}{\Gamma{n}}$
$\overline{X}\sim(na,n)$
$P(\overline{X}>6.296)=\int_{6.296}^{\infty}\dfrac{e^{-x}x^{2}}{2}dx$
$2. \text{If} \ \ \overline{X}\sim(na,n) ,2\overline{X}a \sim \chi_{n}^2 $
$P(Reject H_0|\theta=3)=P( \overline{X} >6.296)=P( 2\overline{X}\cdot\frac{1}{3} >6.296)$
Using both methods I get $\alpha=0.01$ but the answer given is $0.05$ I don't know what am I doing wrong please correct me and is there any other quick method to solve this problem?
Please try to use the same notations as me because I have an exam coming I don't want to get confused.
For convenience, let $c = 6.296$. Then $$\begin{align}\Pr[\bar X > c] &= \int_{x=c}^\infty \frac{1}{2} x^2 e^{-x} \, dx \\ &= \left[ -\frac{1}{2} x^2 e^{-x} \right]_{x=c}^\infty + \int_{x=c}^\infty x e^{-x} \, dx \\ &= \frac{1}{2} c^2 e^{-c} + \left[ -x e^{-x} \right]_{x=c}^\infty + \int_{x=c}^\infty e^{-x} \, dx \\ &= \frac{1}{2} c^2 e^{-c} + c e^{-c} + e^{-c} \\ &\approx 0.0499925. \end{align}$$
I do not know how you got $0.01$. It is worth noting (this is not a coincidence) that if $X \sim \operatorname{Gamma}(n, \theta)$, then $$\Pr[X > \lambda \theta] = \Pr[Y < n],$$ where $Y \sim \operatorname{Poisson}(\lambda)$. Note that the gamma parametrization is by shape and scale.