I am getting stuck with the following practical statistics problem:
The lifetime duration of one type of lightbulb has exponential distribution. A factory guarantees that the expected lifetime of the lightbulbs they produce is greater than 50 days. In order to achieve high quality production, a worker picks randomly a sample of 40 lightbulbs and notes than in average, they have a lifetime of 53 days.
The factory want to have a 95 percent of probability of not selling if the requirements are not fulfilled
Propose an exact test statistic with simple null hypothesis, what decision should be taken ?
Here is where I got so far:
We have X ~ $\mathcal{E}(\lambda)$ and our null hypothesis is that $H_0$ the null hypothesis is : $$\frac{1}{\lambda} = 50 \space \text{vs} \space \frac{1}{\lambda} > 50$$
Now, I am having some difficulty constructing the exact test, if it was approximate, what I would propose is:
$$T = \dfrac{\sqrt(n)(X_n - E[X])}{\sigma}$$
and since T converges in distribution to $\mathcal{N}(0,1)$, I think that the probability they are talking about is the same as calculating
$P(T > z_{\alpha}) = \alpha$ with $\alpha = 0.5$, so to know what decision should be taken, if $T > z_{\alpha}$ then we reject the null hypothesis and we sell.
But since they are asking for an exact test, I don't know what to do, I think it can be useful the fact that $2\lambda\sum_{i=1}^n X_i$ has distribution $\mathcal{X_{2n}}^2$ but I have no idea how to construct $T$ providing that information. Any help would be greatly appreciated.
Let $n=40$. Under the null $X_1,\dotsc, X_n\stackrel{\text{i.i,d}}{\sim} \text{Exp}(\lambda)$. Then note that $$ \bar{X}=n^{-1}\sum_{i=1}^nX_i\sim\text{Gamma}(n, \lambda n)\tag{0} $$ where we say that $Y\sim \text{Gamma}(p,\lambda)$ (where $p,\lambda>0)$ if $Y$ has the density $$ f_{Y}(y)=\frac{\lambda^{p}}{\Gamma(p)} y^{p-1 }e^{-\lambda y};\quad (y>0). $$ Hence $\bar{X}$ is a test-statistic and we can compute the p-value $p=P(\bar{X}>53)$.