# Upper Level of Confidence interval. [duplicate]

Let $$X_1,X_2,...,X_n$$ be a sample from the uniform distribution on $$\theta$$ points. Find an upper $$(1−α)$$ level confidence bound for $$\theta$$, based on $$\max(X_1,X_2,...,X_n)$$.

My Approach:

$$X \sim U[0, \theta]$$ The probability distribution of $$X_{(n)}$$ is: $$f_{X_{(n)}}(x) = \frac{n x^{n-1}}{\theta^n}$$ I want to apply the method of Chebychev’s inequality hence I calculate few required quantities: $$\mathbb{E}[X_{(n)}] = \frac{n\theta}{n+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbb{E}[(X_{(n)} - \theta)^2] = \frac{2\theta^2}{(n+1)(n+2)}$$ By Chebychev's inequality, we can write: $$P\bigg(|X_{(n)} - \theta| \geq \epsilon\sqrt{\mathbb{E}[(X_{(n)} - \theta)^2]}\bigg) \leq \frac{1}{\epsilon^2}$$ $$P\bigg(\frac{|X_{(n)} - \theta|}{\theta} < \epsilon\sqrt{\frac{2}{(n+1)(n+2)}}\bigg) > 1 - \frac{1}{\epsilon^2}$$ Since $$X_{(n)} \xrightarrow{P} \theta$$, we replace $$\theta$$ by $$X_{(n)}$$ in the denominator, and, for moderately large $$n$$, $$P\bigg(\frac{|X_{(n)} - \theta|}{X_{(n)}} < \epsilon\sqrt{\frac{2}{(n+1)(n+2)}}\bigg) > 1 - \frac{1}{\epsilon^2}$$ $$P\bigg(X_{(n)} - \epsilon X_{(n)}\frac{\sqrt{2}}{\sqrt{(n+1)(n+2)}} < \theta < X_{(n)} + \epsilon X_{(n)}\frac{\sqrt{2}}{\sqrt{(n+1)(n+2)}}\bigg) > 1 - \frac{1}{\epsilon^2}$$

Replace $$1/\epsilon^2$$ with $$\alpha$$

$$P\bigg(X_{(n)} - \frac{X_{(n)}\sqrt{2}}{\sqrt{\alpha (n+1)(n+2)}} < \theta < X_{(n)} + \frac{X_{(n)}\sqrt{2}}{ \sqrt{\alpha(n+1)(n+2)}}\bigg) > 1 - \alpha$$

Hence this is the confidence interval at $$1-\alpha$$ confidence level.

How can I get the only the upper-level confidence bound at same confidence level from it or from some other method?