finding a constant in building a confidence interval I have a question about this. If I have $X_1,...,X_n$ be random sample from a uniform(0,$\theta$) where $\theta>0$. My question is how to find a constant c>1 such that (T,cT) is a 100$(1-\alpha)\%$ confidence interval for $\theta$.
I have calculated a complete sufficient statistics for $\theta$ is the $X_{(n)}$ i.e. the max ($X_1,X_2,..X_n)$. 
And so I think my T is $X_{(n)}$. But I am not sure how to find the c for the confidence interval. 
Could someone give some hints.  There is a question that is very similar here: https://math.stackexchange.com/questions/190436/confidence-interval-for-uniform  But I don't quite understand the solution. 
 A: By definition of confidence interval we want to find a $c>1$ such that 
$$P \left( T \leq \theta \leq cT \right) = 1 - \alpha,$$
where $T = X_{(n)} = \max \left( X_1, \ldots, X_n \right)$. 
Note that $X_i \sim \text{Unif}(0, \theta)$ for each $i$ implies that $P(X_i \leq \theta) = 1$ for each $i$ and furthermore that $P(T \leq \theta) = 1$. 
Next, note that $P \left( T \leq \theta \leq cT \right) = P(\theta \leq cT)$ because $P(T \leq \theta)=1$. To see this, 
\begin{align}
P \left( T \leq \theta \leq cT \right) &= P(\theta \leq cT) - P(\theta < T) \\
&= P \left(T \geq \frac{\theta}{c}\right) - P(T > \theta) \\
&= P \left(T \geq \frac{\theta}{c}\right) - 0, \text{ since } P(T \leq \theta) = 1
\end{align}
Now using the CDF of $T$, which is $P(T \leq t) = \left(\frac{t}{\theta} \right)^n$, we obtain
\begin{align}
P(\theta \leq cT) &= P\left(T \geq \frac{\theta}{c}\right) \\
&= 1 - P \left( T < \frac{\theta}{c} \right) \\
&= 1 - \frac{1}{c^n}.
\end{align}
Therefore, $$1 - \frac{1}{c^n} = 1 - \alpha$$ which implies that 
$$c = \frac{1}{\alpha^{\frac1n}} > 1$$
