Question

5. (10 points) Let X1,... , Xio be a random sample of size 10 from a Poisson distribution with mean θ. The rejection region for testing Ho :-0.1 vs. 1.1: θ-0.5 is given by Σ"i z > 4. Determine the significance level α and the power of the test at θ : 05.

Answer 1

10 Xi ~ Poisson(10 * θ) Xi ~ Poisson(0)

For testing the hypothesis

$Ho : \theta = 0.1 Vs Ha: \theta = 0.5$

rejection region (critical region ) is given by

10 Ti > 4

i) alpha = level of significance = size of type I error

Since rejection of true null hypothesis is known type I error.

10 .Tİ

Under Ho : theta = 0.1,

10 Xi~Poisson (1)

10 10

$\alpha = 1- \left \{ P( \sum_{i=1}^{10}Xi=0) + P( \sum_{i=1}^{10}Xi=1)+P( \sum_{i=1}^{10}Xi=2)+P( \sum_{i=1}^{10}Xi=3)+P( \sum_{i=1}^{10}Xi=4)\right \}$

0! 1!

a = 1-(0.3679 0.0153/ 0.3679 + 0.1839 0.0613

o0,0037

ii) prob . of Rejection of false null hypothesis is known as Power

Power = P ( Reject Ho / Ha is true).

$Power = P(\sum_{i=1}^{10}x_{i} > 4 / \theta = 0.5)$

Under Ha : theta = 0.5 , $\sum_{i=1}^{10}Xi \sim Poisson(5)$

$Power = P(\sum_{i=1}^{10}x_{i} > 4 / \theta = 0.5) = 1 - P(\sum_{i=1}^{10}x_{i} \leq 4 / \theta =0.5)$

Power = 0! 2! 4!

$Power = 1- \left \{ 0.0067 + 0.0337 + 0.0842 + 0.1404 +0.1755 \right \}$

Power = 0.5595

Power of the test at theta = 0.5 is 0.5595.