Question

H0: theta <= theta0 vs. H1: theta > theta0. Let’s say we are testing this based on one observation of X from a Beta distribution with PARAMETERS (1,theta). Reject H0 if X<= c for some c.

1. Write an expression for the power function of the test and sketch a plot of it.

2. Find c so that the test has size alpha0.

3. alpha0=0.05 and theta0=. Find c, the probability of Type 1 error if theta = 0.9, and the power of the test if theta = 2.

Answer 1

$Beta(1,\theta)\text{ has density }f(x|\theta)=\frac{\Gamma(\theta+1)}{\Gamma(\theta)\Gamma(1)}(1-x)^{\theta -1}\\ \text{i.e }f(x|\theta)=\theta(1-x)^{\theta-1}\\ \text{So, we have }X\sim f(x|\theta)\\ \text{The required power function is: }\beta(\theta)=P_{\theta}(X\leq c)\\ \implies \beta(\theta)=\int_{0}^c\theta(1-x)^{1-\theta}dx=1-(1-c)^\theta\text{ (Ans.)}\\ \text{ The graph is given below: }\\$

ni cook

$\text{Note, }\beta\text{ is increasing in }\theta\\ \text{We want size of the test to be }\alpha\\ \text{i.e }\sup_{\theta\leq \theta_{0}}\beta(\theta)=\alpha_{0}\\ \implies \beta(\theta_{0})=\alpha_{0}\\ \implies 1-(1-c)^{\theta_{0}}=\alpha_{0}\\ \implies c=1-(1-\alpha_{0})^{1/\theta_{0}}\text{ (Ans.)}$$\text{ Since the value of }\theta_{0}\text{ is not given, I'll leave the answer in terms of }\theta_{0}\\ \text{We have }\alpha_{0}=0.05\\ \implies c =1-(1-0.05)^{1/\theta_{0}}=1-0.95^{1/\theta_{0}}\\ \text{For }\theta=0.9\\ P(\text{Type I error})=\beta(0.9)=1-(1-c)^{0.9}=1-0.95^{0.9/\theta_{0}}\text{ (Ans.)}\\ \text{For }\theta = 2\\ P(\text{ Type II error})=1-\beta(2)=0.95^{2/\theta_{0}}\text{ (Ans.)}$