Let $X$ be a random variable with pdf $f(x; \theta) = \theta x^{\theta - 1}$ where $ 0 < x < 1 $ and $\theta > 0$. Consider $H_0: \theta = 1/2$ versus $H_A: \theta = 1/4$
Then I am asked to derive the most powerful test using $\alpha = 0.05$.
What I have done is the following:
$$ \begin{align*} T = \frac{ L(1/2) }{ L(1/4) } & = \frac{\frac{1}{2}x^{-\frac{1}{2}}}{\:\frac{1}{4}x^{-\frac{3}{4}}} \\ & = 2x^{\frac{1}{4}} \end{align*} $$
Thus we reject $H_0$ if
$$ 2x^{\frac{1}{4}} < c \Leftrightarrow x < \left( \frac{c}{2} \right)^4 = c_2 $$
However, I am not certain whether or not the critical region is
$$ R = \{ X \in (0, 1) \ | \ X > c_2 \} $$
or
$$ R = \{ X \in (0, 1) \ | \ X < c_2 \} $$
How do I determine whether or not I must have $<$ or $>$?. Because when I then have to solve for c by calculating
$$ 0.05 = P(X \in (0,1) \in R \ | \ H_0) = P(X > c_2) $$
the result differs whether that $X > c_2$ or $X < c_2$.
TIA for any help.
The two hypothesis are both simple thus the UMP test is based on Neyman-Pearson's Lemma. You calculated it correctly finding
$$x\leq c^*$$
Thus the critical region is
$$\mathbb{P}[X\leq c^*|\theta=0.5]=0.05$$
that is
$$\int_0^{c^*}0.5x^{-0.5}dx=0.05$$
$$c^*=0.0025$$
Concluding: you can reject $H_0$ if your single observation $X\leq 0.0025$