Finding Uniformly Most Powerful(UMP) tests of size $\alpha$

Question

Hi I am working on the following problem:

Let $X_1,X_2,\ldots,X_n$ be a random sample from a distribution with PDF given by $$f(x\mid\theta)=\frac{c}{\theta^c}x^{c-1}e^{-(\frac{x}{\theta})^c}\,\,\,(\text{Weibull distribution})$$ where $c,\theta>0.$ and $c$ is a constant. Find the Uniformly Most Powerful(UMP) tests of size $\alpha$ for testing $$H_0: \theta\le\theta_0\,\,\,\text{VS.}\,\,\,\,H_1:\theta>\theta_0$$

If anyone wants to look at it I would be happy to provide it here but in short so far I got:

1. $T(\textbf{X})=\sum_{i=1}^n X_i^c$ is a complete sufficient statistic
2. $X_i^c\sim \exp(\theta^c)$ and therefore $T(\textbf{X})=\sum_{i=1}^n X_i^c\sim \mathrm{Gamma}(n,\theta^c)$
3. In this step I am stuck. From definition we have $$\alpha=\sup_{\theta\le\theta_0}\beta(\theta)$$ where \begin{align*} \beta(\theta)&=E_\theta[\delta(x)]\\ &=E_\theta[I(T\ge c)]\\ &=P_\theta[T\ge c]\,\,\,\,\,\,\,\text{where }T\sim \mathrm{Gamma}(n,\theta^c) \end{align*}

I could not find the value of $c$ from here.

Any help would be greatly appreciated. Thanks in advance.

Answer 1

To find the UMP test you should construct the likelihood ration test for the corresponding MP test, i.e., $$ \frac{L_(\theta_1;X)}{L_(\theta_0;X)} = \frac{\theta_0^c}{\theta_1 ^c}\exp\{\sum x_i^c(1/\theta_0 - 1/\theta_1\} > k $$ where $\theta_1 > \theta_0$. So the LR is monotonic strictly increasing function of the MMS $\sum x_i^c$, hence the UMP test is $$ \delta(X) = \mathrm{1}\{\sum x_i^c > k^*\} . $$ Then, $$ E\delta(X) = 1 - P(\sum X_i^c \le k^*) = 1- P(2\theta^c \sum X_i^c \le 2\theta^c k^*), $$ note that $2\theta^c \sum_{i=1}^n X_i^c \sim Gamma(n, 1/2) = \chi^2_{(2n)}$, thus $$ E_{\theta_0}\delta(X) = 1- F_{\chi^2(2n)}(2\theta_0^c k^*) =\alpha, $$ so we get that $$ k^* = \frac{\chi_{(2n)}^2(1-\alpha)}{2\theta_0^c}. $$