Explanation In this context, Y 1 , Y 2 , ... , Y n denotes a random sample from the population having Poisson distribution with mean λ 1 and X 1 , X 2 , ... , X m denotes a random sample from the population having Poisson distribution with mean λ 2 . The likelihood function of the random variable X and Y is given below: L ( λ 1 , λ 2 ) = ∏ i = 1 n P ( Y i , λ 1 ) × ∏ j = 1 m P ( X i , λ 2 ) = ∏ i = 1 n e − λ 1 λ 1 Y i Y i ! × ∏ j = 1 m e − λ 2 λ 2 X j X j ! = e − n λ 1 λ 1 ∑ i = 1 n Y i ∏ i = 1 n Y i ! × e − m λ 2 λ 2 ∑ j = 1 m X j ∏ j = 1 m X j ! Neyman-Pearson Lemma: It is assumed to test H 0 : θ = θ 0 vs H a : θ = θ a , based on a random sample of Y 1 , Y 2 , ... , Y n from a distribution with parameter θ . Here, denote L ( θ ) the likelihood of the sample when the value of the parameter θ . For given α , the test maximizes the power at θ a has a rejection region is determined by, L ( θ 0 ) L ( θ a ) < k In this context, the value of k is chosen so that the test will have the desired value for α . Such a test is a most powerful α -level test for H 0 vs H a . The most powerful α -level test for testing H 0 : λ 1 = λ 2 = 2 versus H a : λ 1 = 1 2 , λ 2 = 3 is obtained as given below: Substitute the corresponding values in the above equation to get the most powerful test. L ( λ 01 , λ 02 ) L ( λ a 1 , λ a 2 ) < k L ( 2 , 2 ) L ( 1 2 , 3 ) < k e − 2 n 2 ∑ i = 1 n Y i ∏ i = 1 n Y i ! × e − 2 m 2 ∑ j = 1 m X j ∏ j = 1 m X j ! e − 0.5 n 0.5 ∑ i = 1 n Y i ∏ i = 1 n Y i ! × e − 3 m 3 ∑ j = 1 m X j ∏ j = 1 m X j ! < k e − 2 n − 2 m × 2 ∑ i = 1 n Y i + ∑ j = 1 m X j e − 0

Mathematical Statistics with Applications

7th Edition

ISBN: 9781133384380

Author: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

Publisher: Cengage Learning US

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Chapter 10.10, Problem 100E

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Derive the most powerful test for testing H0:λ1=λ2=2 versus Ha:λ1=12, λ2=3.

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Chapter 10.10, Problem 99E

Chapter 10.10, Problem 101E

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Chapter 10 Solutions

Mathematical Statistics with Applications

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Derive the most powerful test for testing H 0 : λ 1 = λ 2 = 2 versus H a : λ 1 = 1 2 , λ 2 = 3 .

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Derive the most powerful test for testing H0:λ1=λ2=2 versus Ha:λ1=12, λ2=3.

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