# Finding exact hypothesis test for exponential distribution

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 $$p=P(\bar{X}>53)$$.