I'd like help with the following problem:
Consider a random sample $X_1, ..., X_n$ from the Exponential distribution $\epsilon(\theta)$ with density $\theta e^{-\theta x}$, $x>0$. Suppose we wish to test the hypothesis $H_0: \theta=1$ vs $H_1: \theta \neq 1$.
(1) Calculate the likelihood ratio statistic for this testing problem and (2) show that the LRT is based on the statistic $T=\sum_{i=1}^nX_i$ and rejects if $T\in(0,c_1]\cup[c_2,\infty)$ for some suitably chosen $c_1, c_2$.
My work in a nutshell: Knowing that the likelihood function is given by $L(\theta;x_1,...,x_n)=\theta^ne^{-\theta n\bar{x}}$ and that the likelihood ratio statistic is given by $\lambda(x_1,...,x_n)=\frac{\sup_{\Theta=1}{L(\theta;x_1,...,x_n)}}{\sup_{\theta\in\Theta}L(\theta;x_1,...,x_n)}$, I find the MLE estimator to be $\hat{\theta}=\frac{1}{\bar{x}}$, plug it into the denominator, ending up with $\lambda(x_1,...,x_n)=\bar{x}^n e^{-n\bar{x}+1}=(\frac{\sum_{i=0}^n X_i}{n})^ne^{-\sum_{i=0}^n X_i+n}$, which I believe is enough to show that the LRT is based on $T=\sum_{i=0}^nX_i$. Now, knowing that the rejection region is given by $R=\{T:\lambda\leq c\}$ for some set critical value of $c$ (chosen as a function of $\alpha$). I then end up with $(\frac{\sum_{i=0}^n X_i}{n})^ne^{-\sum_{i=0}^n X_i+n}\leq c$, which I'm not sure is correct (and would like confirmation before I proceed with straight computation to find the two rejection regions $(0,c_1]\cup[c_2,\infty)$.
Thanks a bunch for your time. For context, I'm a first-year math major.
more or less right, the sum goes from 1 to n, not from zero. The $n$ in the exponent is obviously out of the summation... in a more suitable notation, you get that
$$\lambda(\mathbf{x})\propto T^n e^{n-T}$$
Where $T=\Sigma_iX_i\sim \text{Gamma}(n;\theta)$
Doing a drawing of your likelihood ratio (I fixed some $n$ to have the graphics) you see this
thus the probability for $\lambda(\mathbf{x})<c$ is equivalent to $T\leq c_1$ or $T\geq c_2$ for appropriately chosen constants $c_1,c_2$
Of course the two quantiles $c_1,c_2$ can be derived by a $\chi_{(2n)}^2$ distribution and consequently found using the standard table