Hypothesis Testing for uniform distribution

Question

Let $X_i; i=1,2,3,\dots,n$ be iid samples from the uniform $U(0, θ)$ distribution, with $θ$ being the unknown parameter of interest. Consider the testing problem $H_0 : θ = θ_0$ versus $H_1 : θ > θ_0.$ One possible test is to reject $H_0$ when $X_{max} > C$ for some $C > 0.$

$1.$ Find $C$ in terms of the proposed $θ_0$ such that the probability of Type I error is 0.05.

$2.$ Use the duality between hypothesis testing and confidence interval to write down a level 0.95 lower confidence bound for $θ.$ That is, find $L(X_1,\dots , X_n)$ such that $P_θ(θ ≥ L(X_1, \dots , X_n)) = 0.95$ for all $θ > 0.$

$3.$ Choose $C$ as in (a). What is the power of the test in the case of $θ_0 = 1,\, θ = 2,\, n = 10?$

I have done $1$ using $P(X_{max}>C|θ = θ_0) = 1 - (C/θ_0)^n = 0.05;$ can anyone tell me if this is right? Also, how to do questions $2$ and $3$? Thanks so much!!!!

Answer 1

1. Yes, you are wright. Simply find $C$ from equality $1-(C/\theta_0)^n=0{.}05$.

2. Why not to use $X_{max}$ to construct $L$?

Find the probability $$P_\theta(X_{max}\leq t)=(t/\theta)^n=0.95.$$ Then $t/\theta=\sqrt[n]{0.95}$, $t=\sqrt[n]{0.95}\theta$. Finally, $$ P_\theta(X_{max}\leq \sqrt[n]{0.95}\theta)=P_\theta(\theta\geq X_{max}/\sqrt[n]{0.95})=0.95. $$ So $L(X_1,\ldots,X_n)=\dfrac{X_{max}}{\sqrt[n]{0.95}}$.

3. Use definition of a power of a test.