Critical region for an uniform distribution Let $X_1, X_2, ... , X_n$ be a random sample from the uniform distribution over $[0, \theta]$. Suppose we wish to test $H_0 : \theta = 5$ versus $H_A : \theta < 5$ at significance level $\alpha = 0.05$. Use the LRT to find the critical region.
Here is my attempt:
\begin{align*}
  T = \frac{L(\theta_0 = 5)}{L(\theta_A < 5)} & = \frac{\Pi_{i = 1}^n \frac{1}{b-a}}{\Pi_{i = 1}^n \frac{1}{b-a}} \\
  & = \frac{\frac{1}{(5-a)^n}}{\frac{1}{(\theta_A-a)^n}} \\
  & = \frac{(\theta_A-a)^n}{(5-a)^n}
  \end{align*}
However, I am not sure how to proceed next. I know that we reject the null hypothesis if $T < c$ where $c$ is some constant. In my book they give an example where they now take $\ln$ on both sides and if I do so I get
$$ \ln(T) = - n \ln(\theta_A - a) + n \ln(5 - a))  < \ln(c) $$
but what am I supposed to solve for?
 A: The likelihood ratio is
$$
\lambda(x) = \frac{\sup_{\theta\in\Theta_0} L(\theta\mid x)}{\sup_{\theta\in\Theta} L(\theta\mid x)}.
$$
In this case $\Theta_0 = \{5\}$, $\Theta = [0,5]$, and your likelihood is
$$
L(\theta\mid x) = \theta^{-n}\mathbf 1_{x_\max \leq \theta }.
$$
On $[0,5]$ $\theta^{-n}$ is strictly decreasing so we want the smallest $\theta$ possible such that we don't violate the support, and this leads to $\sup_{\theta\in\Theta} L(\theta\mid x) = L(x_\max \mid x)$. This means our likelihood ratio is
$$
\lambda(x) = \frac{L(5 \mid x)}{L(x_\max \mid x)} = \left(\frac{5}{x_\max}\right)^{-n}.
$$
$x_\max$ is a sufficient statistic for $\theta$ here so it's nice to see that our LRT only depends on this sufficient statistic. You got this part already but I think your notation was a little unclear since $a$ and $b$ weren't really defined.
For the rejection region, we can get there by thinking about what we're trying to do. We'll reject the null hypothesis if $\lambda(x)$ is small, because this means there's a much higher likelihood outside of $\Theta_0$ than in it. To get our threshold then we know that we want the probability of a rejection under $H_0$ to be $\alpha$, so this means we need to solve
$$
P(\lambda(X) \leq c \mid \theta \in \Theta_0) = \alpha
$$
for $c$. The null hypothesis is simple here so things are a little easier:
$$
P\left(\lambda(X) \leq c \mid  \theta \in \Theta_0\right) = P\left(\left(\frac{5}{X_\max}\right)^{-n} \leq c\right) = \alpha.
$$
At this point can you finish the problem?
