Hypothesis testing for a Uniform distribution with both parameters unknown I need some help, I need to make a hypothesis testing for a Uniform distribution with both parameters $(a,b)$ unknown for
$\begin{align}  H_{0}:b-a>2    \end{align}$
So. I know that I need a minimal sufficient statistic which is going to be a vector $(X_{(1)},X_{(n)})$, where $X_{(1)}=\min\{X1,...,X_{n} \}$ and $X_{(n)}=\max\{X1,...,X_{n} \}$,
Is my reject region going to be $RR_{\alpha}=\{x_{1},...x_{n}:X_{(n)}-X_{(1)}<C_{\alpha} \}$ for a $0<C_{\alpha}<1$?
How can I make the hypothesis testing on which the probability error type 1 and 2 don't exceed $\alpha$ and $\beta$, where $\mathbb{P}(E.T.1)=\mathbb{P}$(Reject $H_{0}$|$H_{0}$ is true)=$\mathbb{P}(X_{(n)}-X_{(1)}\leq C_{\alpha}|b-a>2)$? Thanks
 A: Use your intuition.  What matters for the hypothesis $$H_0 : b - a > 2 \quad \text{vs.} \quad H_1 : b - a \le 2$$ is the range of the sample; if it is too large, you obviously cannot reject $H_0$.  The smaller the range of observed values, the more likely the alternative hypothesis.  For instance, if the sample were $(2, 1, 5, 3, 2, 4)$, you would immediately fail to reject $H_0$ since $5 - 1 = 4 > 2$.  But if the sample were $(2.1, 2.2, 2.3, 2.7, 1.9, 2.5)$, then the range is only $2.7 - 1.9 = 0.8$.  Is that small enough to provide enough evidence that $b - a \le 2$?  That's where you need to derive a rejection region that depends on $\alpha$.
So, your thought about using the first and last order statistics is on the right track but in fact, we only care about the statistic $Y = X_{(n)} - X_{(1)}$, not the specific values of these.  And your rejection region clearly should be in the direction of the event $Y \le C_\alpha$, per our discussion above.  So what is the sampling distribution of $Y$ under the (modified) null hypothesis $H_0 : b - a = 2$, and how do we find $C_\alpha$ such that $\Pr[Y \le C_\alpha \mid H_0] = \alpha$?  Note we use a modified $H_0$ because under all other cases where $b - a > 2$, the Type I error is strictly smaller than $\alpha$ for the same $C_\alpha$ criterion.
