Finding 95% upper confidence bound for $\theta$ if your sample follows the uniform distribution

Question

If you have a random sample of size $n = 20$ from a uniform distribution on the interval $(0,\theta)$, find a $95$% upper confidence bound for $\theta$

So I tried using Chi square but it didn't work. Please help!

Answer 1

Note that $$ F_{X_{(n})}(t) = (F_{X}(t))^n = \frac{t ^ n}{\theta ^ n} $$ hence, as $F(F^{-1}(x))=x$, $$ {n}\sqrt{\frac{X_{(n)}}{\theta}} \sim U[0,1], $$ namely, $$ \mathbb{P}\left( {n}\sqrt{\frac{X_{(n)}}{\theta}}\le 0.95\right) = 0.95, $$ $$ \mathbb{P}\left( \theta \ge X_{(n)}/0.95^{1/n}\right) = 0.95. $$

Answer 2

Hint

You know for sure that $\theta \ge m:=\max_{i=1}^{20} x_i,$ so consider an interval of the form $[m,m+\delta].$ You need to choose $\delta$ just large enough so that $P(\theta\in [m,m+\delta]) =0.95.$

$\theta\in[m,m+\delta]$ if and only if $\max x_i > \theta -\delta,$ so you need to compute $P(\max x_i > \theta-\delta).$