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I'm having trouble to find a critical region for the MP test. I know the r.v $X_1,...,X_6$ are iid. following the uniform distribution $U(0,\theta)$, $\theta>0$. The uniformy most powerful test was built for the hypothesis $$H_0: \theta = 1 \quad H_1: \theta \neq 1, \quad \alpha = 0.125.$$ We want to find the critical area for this test.

My approach: I found the testing statistic: $T(x) = max(X_1,...,X_n) = X_{(6)}$, and then tried to find the ciritical value by using the significance level $\alpha = 0.125$ and the property that $$\alpha = P_{H_0}(T(x) >k) = P_{H_0}(X_{(6)} >k) = 1 - P_{H_0}(X_{(6)} <k) = 1 - P(X <k)^n = 1- k^6,$$ because the $F_x(t) = \frac{t}{\theta}$, so under null hypothesis $F_x(t) = t$. Then my $k = (\frac{7}{8})^{(\frac{1}{6})}$ and the critical area is $(k,\infty)$, but I knwo that is correct answer is $X_{(6)} \in (0, \frac{\sqrt{2}}{2}) \cup (1, \infty)$.

The thing is I think my critical area is wrong, because it doesn't take under consideration the fact that $x \in (0,\theta)$. I read somwhere that I should use $P(X_{(6)} \in (0,k)) = \alpha$ but I have no idea where it comes from and if it is correct.

user13696679
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  • The UMP test here is unique; see https://math.stackexchange.com/q/1736322/321264. One way to see this is to verify that it is UMP for both the alternatives $H_1':\theta>1$ and $H_2':\theta<1$. – StubbornAtom Feb 10 '22 at 12:56
  • Your solution has the same $\alpha$ as the correct answer, i.e. when $\theta=1$ the probability of being in the critical region is the same. But it has a lower power, most obviously when $\theta <1$ as the probability of being in your critical region goes down while the correct answer's goes up – Henry Feb 10 '22 at 13:14
  • Note that $\frac{\sqrt{2}}{2}=\left(\frac18\right)^{1/6}$ – Henry Feb 10 '22 at 13:15
  • @Henry yes indeed but from the equation we get $(1-\frac{1}{8})^{\frac{1}{6}}$ and not 1/8^(1/6) – user13696679 Feb 10 '22 at 13:42
  • @user13696679 - In the correct answer, the critical region for the maximum $X_{(6)}$ is about $[0,0.7071) \cup (1,\infty)$, compared to your $(0.97799, \infty)$. So under $H_0$ these give $\alpha$ of $0.7071^6 -0^6$ or $1^6-0.97799^6$ both being $0.125$ – Henry Feb 10 '22 at 13:57

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