Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the hypothesis test:

H_0 : mu= mu_0 (mu_1 > mu_0)

H_1 : mu= mu_1

Applying N-P Lemma: {X: lambda(mu_0,mu_1;X)>= k} Where lambda(mu_0,mu_1;X) is the ratio of likelihood functions; with the likelihood of alternate on the numerator and the likelihood of the null in the denominator.

-Essentially my question is: why use the sign >= ? Is this related to (mu_1 > mu_0) assumption?

-Secondly, I would like to confirm that the essential method (N-P Lemma) involves changing {X: lambda(mu_0,mu_1;X)>= k} into => {X: function of data only >= s} while keeping an eye on the inequality sign.

-Third: In order to get only the 'function of data' on one side, the parameters must be moved to the other side of inequality. Since parameters are not constants but have specific distributions; I assume 's' is then a function. However 's' is irrelevant as its size is determined by choosing an appropriate sized critical region. Is this correct?

Thank you (...and apologies for the poor notations, are there any beginner guides on using appropriate notations?)

share|improve this question

1 Answer 1

up vote 1 down vote accepted
  1. k is chosen as a value sufficiently large that when the likelihood ratio is greater than or equal to k the p-value is less than or equal to the significance level and you will reject the null hypothesis. Hence the reason for >=.

  2. Using algebra the likelihood ratio simplifies to a function of the observed data and the parameters.

  3. Neyman and Pearson use the frequentist approach to statistical inference. Parameters are constants that are unknown. They are not assumed to have prior distributions. Those assumptions apply to the Bayesian approach not the frequentist approach.

share|improve this answer
    
Thanks, that points me in the right direction. –  student101 Sep 9 '12 at 13:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.