Consider a random sample of observations of a $U(0,θ)$ distribution. Find a lower bound of 95% confidence for θ
My idea was:
Let $X_ {1}, X_ {2}, ..., X_ {N}$ be a random sample of a uniform population in the interval $(0, θ)$ and $X_ {n}$ the maximum. So $(\frac{1}{a}) ^ n - (\frac{1}{b}) ^ n$ is the confidence coefficient of the parameter.
But I do not know how to continue to solve this problem, can anybody help me?
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_n\le m) = P(X\le m)^n = (m/\theta)^n.$$ Intuitively, we know the $\theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-\alpha$ coverage (where $\alpha=0.05)$: $$ 1-\alpha=P(\theta < (1+c^*)M_n) = 1- P\left(M_n\le \frac{\theta}{1+c^*}\right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = \alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-\alpha$ confidence (i.e. a minimal statistic $x$ such that $P(\theta>x) \le 1-\alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-\alpha = P((1+c^*)M_n< \theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-\alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.