Let X be a random sample of size one from $U(\theta,\theta+1)$ distribution, $\theta\in \mathbb{R}$. For testing $H_0:\theta=1$ against $H_1:\theta=2$, the critical region $ \left\{x : x>1 \right\}$ what is the power $1-\beta$ and size $\alpha$ of the test?
As we have single observation, to find $\alpha$ I did $$\int_{1}^{2}1dx=1$$ since, under null hypothesis $\theta=1$ so (1,2) is the critical region? (explain if not correct) And to find power of the test $$\int_{2}^3 1dx=1$$ since critical region is $x>1$ but the limits of the distribution under alternate hypothesis $\theta=2$ is (2,3) so, am I correct here finding the size and power of the test.
I doubt my solution because in calculating both size and power the whole region is taken as critical region what kind of test it is, P.S. this question was asked in an entrance examination for admission in an statistics course.
Since the size of the test is $P(reject H_0|H_0 true)$, we have, for the critical region $x>1$, that we reject the null hypothesis with probability 1 if $H_0$ is true (because the outcome $x=1$ has probability zero). Since the power is $P(reject H_0 | H_1 true)$, we have, for the same critical region, again a probability of 1.
Hence, your answer is correct. Maybe it is required to state an argument that by definition a uniform has no point masses to justify that the size is 1.
We conclude that we always reject $H_0$ with probability 1 no matter which hypothesis of the two is correct...not a very meaningful test.