Problem: Investigating the lifetime of some product, we're interested in testing the hypotheses \begin{align} H_0: \mu=2\\ H_1: \mu<2 \end{align}
Where $\mu$ is the expectation value of the lifetime in some arbitrary unit. We assume that the lifetimes follow an exponential distribution with parameter $\lambda$, i.e. \begin{equation*} f_X(x)= \begin{cases} \lambda e^{-\lambda x} &\text{ for }x\geq 0\\ 0 &\text{ ellers } \end{cases} \end{equation*} and $E[X]=\mu=\frac{1}{\lambda}$.
Now assume we have sampled the lifetime of 10 sample products and found a sample mean $\bar{X}$. Let $c>0$ be some number such that, if $W\geq c$ we accept the null hypothesis, otherwise discard it.
Determine the number $c$ such that our test has signifiance level of 95%.
Solution (attempt)
Our book mostly covers how to find the sigifiance intervals for Normal distributions, or cases of known mean with n large, which is hardly applicable for $n=10$, and the assignment asks for an exact test of the nullhypothesis.
I've only noted the following, that if $X_1,X_2,....X_n$ are indenpendent Exponential($\lambda$) stochastic variables, then their sum follows a Gamma distribution. I.e. \begin{equation*} \bar{X}=\frac{X_1,X_2,...X_n}{n}=\frac{1}{n}\sum_{i=1}^{n}X_i\sim \frac{1}{n}Gamma(n,\lambda) \sim Gamma(n,n\lambda) \end{equation*}
That is, it follows some distribution \begin{equation} f_Y(y) = \frac{(n\lambda)^n y^{n-1}e^{-n\lambda y}}{\Gamma(n)}, y>0 \end{equation}
But how does this help me determine $c$?
P.S. The assignment problem also states that we might use the property: For $X\sim Gamma(\alpha_1,\lambda), Y\sim Gamma(\alpha_2,\lambda) \implies X+Y\sim Gamma(\alpha_1+\alpha_2,\lambda)$.
However, I'm not sure how to make use of this, as I don't see what is to be summed, any way.
Your hypothesis system can be rewritten as follows:
$$\begin{cases} \mathcal{H}_0: &\lambda=\frac{1}{2} \\ \mathcal{H}_1: &\lambda>\frac{1}{2} \end{cases}$$
Now let's fix $\lambda'<\lambda''$ and derive the likelihood ratio:
$$\frac{L(\lambda'|\mathbf{x})}{L(\lambda''|\mathbf{x})}=\frac{(\lambda')^n e^{-\lambda'\Sigma x}}{(\lambda'')^n e^{-\lambda''\Sigma x}}=\Bigg(\frac{\lambda'}{\lambda''}\Bigg)^nExp(\lambda''-\lambda')\Sigma x$$
It is evident that this ratio is a monotone increasing function in $T=\Sigma x$
Now it is enough to apply a well known theorem on UMP test for families having a monotone likelihood ratio to derive the critical region, that in this case is
$$\mathbb{P}\Bigg[T<k|\lambda=\frac{1}{2}\Bigg]=\alpha$$
In your case, you know that
$$T=\Sigma x\sim Gamma(10;\frac{1}{2})=\chi_{(20)}^2$$
Thus you can easy solve your exercise using the chi-square table
It results to me that we reject the Hypotheses that $\mu=2$ against the alternative that $\mu<2$ iff $\overline{X}_{10}<1.085$
The theorem I refer to is the following
taken from Mood Graybill Boes, Chapter IX, Tests of Hypotheses