I have the following question: Find the best size $\alpha$ test of $H_0:\theta=\theta_0$ vs $H_1:\theta =\theta_1$, with $\theta_1>\theta_0$, write down the expression for the power function when $X_1,\dots,X_n$ are IID exponential random variables, with parameter $\theta$.
Use the $\chi^2$ tables to find the least sample size which will allow us to test $H_0:\theta=1$ against $H_1:\theta=3$ with $\alpha=5\%$ and $\beta\leq 10\%$. Describe the appropiate critical region numerically.
I have managed to find the best test size in the following way:
First of all, finding the likelihood ratio:
$\begin{gather}\frac{f_1(x|\theta_1)}{f_0(x|\theta_0)}= \left(\frac{\theta_1}{\theta_0}\right)^n \cdot e^{-(\sum x_i)(\theta_1-\theta_0)} \end{gather}$.
Then, for the sufficient statistic $T = \sum X_i$, we condition that the likelihood ratio be bigger than $k(\alpha)$. Computing the inequality, we get that we need to have $T<c_s$, where $c_s=\frac{n(ln\theta_1 - ln\theta_0)-lnk(\alpha)}{\theta_1-\theta_0}$.
Now, we know further that $T\sim Gamma(n,\theta)$. Hence, by Neyman-Pearson Lemma we have that :
$\begin{gather}\alpha = \int_{c_s}^{\infty}\frac{1}{\Gamma(n)}t^{n-1}e^{-t}dt \end{gather}$
We can compute $c_s$. However, I do not know what to do for the next part.