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The task is from book Mathematical Statistics and Data Analysis by John. A. Rice and I solved it but the solutions are different, yet I believe the solutions are wrong.

We have one observation of variable $X$ that is distributed uniformly on $[0, \theta]$. Null hypothesis is $H_0 \colon \theta = 1$ and alternative hypothesis is $H_1 \colon \theta = 2$. For $0 < t< 1$, what is the significance level and power of the test that rejects $H_0$ when $X\leq t$? For $0 < t< 1$, what is the significance level and power of the test that rejects $H_0$ when $X \in [1-t, 1]$?

For the test $X \leq t$ we have significance level as $$\alpha = P(H_0 ~\text{rejected}\mid H_0~\text{is true}) = P(X \leq t \mid \theta=1) = \frac{t-0}{1-0} = t$$ (which is the same as in the solutions) and power of the test is $$1-\beta = P(H_0~\text{rejected} \mid H_1~\text{is true}) = P(X \leq t \mid \theta = 2) = \frac{t-0}{2-0} = \frac{t}{2}$$ but in solutions it is $1-\frac{t}{2}$.

For the test $X \geq 1-t$, the significance level is $$\alpha = P(H_0 ~\text{rejected}\mid H_0~\text{is true}) = P(X \in [1-t,1] \mid \theta=1) = \frac{1-(1-t)}{1-0} = t$$ (again the same as in solutions) and the power is $$1-\beta = P(H_0~\text{rejected} \mid H_1~\text{is true}) = P(X \in [1-t,1] \mid \theta = 2) = \frac{1-(1-t)}{2-0} = \frac{t}{2}$$ but in solutions it is again $1-\frac{t}{2}$. Am I getting something wrong as per definition of power of the test or are really the solutions just wrong?

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  • $\begingroup$ Perhaps a problem with notation. Often $\beta$ is used for probability of Type II error and $\pi$ is used for power. Then probability of Type II error is $\beta = P( H_0\; acc | H_1\; true) = 1 - \frac{t}{2}.$ $\endgroup$ – BruceET Aug 12 '18 at 7:24
  • $\begingroup$ @BruceET Sorry, there should be $1-\beta$...I edited the question. I agree that $\beta = 1 - \frac{t}{2}$, but that as you said, is probability of Type II error, but power is $1-\beta$. Also, book uses $1-\beta$ in the definition of power. $\endgroup$ – campovski Aug 12 '18 at 7:26

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