1. The Neymann-Pearson Lemma Suppose that the data x 1, …, x n has joint density function f(x 1, …, x n ;  ) where  is either  1 or  2. Let g(x 1, …, x n ) = f(x 1, …, x n ;  1 ) and h(x 1, …, x n ) = f(x 1, …, x n ;  2 ) We want to test H 0 :  =  1 (g is the correct distribution) against H A :  =  2 (h is the correct distribution)

2. The Neymann-Pearson Lemma states that the Uniformly Most Powerful (UMP) test of size  is to reject H 0 if: and accept H 0 if: where k  is chosen so that the test is of size .

3. Neyman Pearson Lemma g(x)g(x) h(x)h(x) x

4. g(x)g(x) h(x)h(x) x

5. g(x)g(x) h(x)h(x) x

6. Definition The hypothesis H:    is called simple if  consists only of one point. If the set  consists of more than one point the hypothesis H is called composite. The Neymann Pearson lemma finds the UMP (uniformly most powerful) test of size  when testing a simple Null Hypothesis (H 0 ) vs a simple Alternative Hypothesis (H A )

7. A technique for finding the UMP (uniformly most powerful) test of size  for testing a simple Null Hypothesis (H 0 ) vs a composite Alternative Hypothesis (H A ). 1.Pick an arbitrary value of the parameter,  1, when the Alternative Hypothesis (H A ) is true. (convert H A into a simple hypothesis 2.Use the Neymann Pearson lemma to find UMP (uniformly most powerful) test of size  for testing a simple Null Hypothesis (H 0 ) vs a simple Alternative Hypothesis 3.If this test does not depend on the choice of  1 then the test found is the uniformly most powerful test of size  for testing a simple Null Hypothesis (H 0 ) vs a composite Alternative Hypothesis (H A ).

8. Likelihood Ratio Tests A general technique for developing tests using the Likelihood function This technique works for composite Null hypotheses (H 0 ) and composite Alternative hypotheses (H A )

9. Likelihood Ratio Tests Suppose that the data x 1, …, x n has joint density function f(x 1, …, x n ;  1, …,  p ) where  (  1, …,  p ) are unknown parameters assumed to lie in  (a subset of p-dimensional space). Let  denote a subset of . We want to test H 0 : (  1, …,  p )   against H A : (  1, …,  p )  

10. The Likelihood Ratio Test of size  rejects H 0 : (  1, …,  p )   in favour of H A : (  1, …,  p )   if: where k  is chosen so that the test is of size  and L (  1, …,  p ) = f(x 1, …, x n ;  1, …,  p ) Is the Likelihood function the maximum of L (  1, …,  p ) subject to the restriction (  1, …,  p )   Also

11. Example Suppose that x 1, …, x n is a sample from the Normal distribution with mean  (unknown) and variance  2 (unknown). Then x 1, …, x n have joint density function f(x 1, …, x n ;  ) Suppose that we want to test H 0 :  =  0 against H A :  ≠  0 L(  ) = f(x 1, …, x n ;  )

12. Note: We want to test H 0 :  =  0 against H A :  ≠  0 L(  2 ) = f(x 1, …, x n ;  2 ) Note: and if H 0 :  =  0 is true then

13. We have already shown that is at a maximum when L(  2 ) = f(x 1, …, x n ;  2 ) Thus

14. Now consider maximizing when L(  2 ) = f(x 1, …, x n ;  2 ) This is equivalent to choosing v to maximize or

15. Hence if or

16. Thus Is maximized subject to L(  2 ) = f(x 1, …, x n ;  2 ) when thus

17. The Likelihood Ratio Test of size  rejects H 0 :  =  0 in favour of H A :  ≠  0 if: i.e. if

18. or Now and

19. thus and

20. The Likelihood Ratio Test of size  rejects H 0 :  =  0 in favour of H A :  ≠  0 if: or where k  (or K  ) are chosen so that the test is of size . The value that achieves this is

21. Conclusion: The Likelihood Ratio Test of size  for testing H 0 :  =  0 against H A :  ≠  0 is the Students t-test

22. Example Suppose that x 1, …, x n is a sample from the Uniform distribution from 0 to  (unknown) Then x 1, …, x n have joint density function f(x 1, …, x n ;  ) Suppose that we want to test H 0 :  =  0 against H A :  ≠  0 Note: L(  ) = f(x 1, …, x n ;  )

23. We have already shown that is at a maximum when L(  ) = f(x 1, …, x n ;  ) Thus

24. Also it can be shown that is maximized subject to    = {  0 } when L(  ) = f(x 1, …, x n ;  ) Thus

25. Hence We will reject H 0 if < k 

26. Hence will reject H 0 if: or if i. e. or

27. Summarizing: We reject H 0 if: or if Where K  (equivalently k  ) is chosen so that Again to find K  we need to determine the sampling distribution of : and when H 0 is true

28. then The sampling distribution of: We want when H 0 is true Let when H 0 is true thus or

29. Final Summary: We reject H 0 if: or ifand

30. Example: Suppose we have a sample of n = 30 from the Uniform distribution or if and We want to test H 0 :  = 10 (  0 ) against H 0 :  ≠ 10 We are going to reject H 0 :  = 10 if: hence we accept H 0 :  = 10.

31. Comparing Populations Proportions and means

32. Sums, Differences, Combinations of R.V.’s A linear combination of random variables, X, Y,... is a combination of the form: L = aX + bY + … where a, b, etc. are numbers – positive or negative. Most common: Sum = X + YDifference = X – Y

33. Means of Linear Combinations The mean of L is:  L = a  X + b  Y + … Most common:  X+Y =  X +  Y  X – Y =  X -  Y IfL = aX + bY + …

34. Variances of Linear Combinations If X, Y,... are independent random variables and L = aX + bY + … then Most common:

35. If X, Y,... are independent normal random variables, then L = aX + bY + … is normally distributed. In particular: X + Y is normal with X – Y is normal with Combining Independent Normal Random Variables

36. Comparing proportions Situation We have two populations (1 and 2) Let p 1 denote the probability (proportion) of “success” in population 1. Let p 2 denote the probability (proportion) of “success” in population 2. Objective is to compare the two population proportions

37. We want to test either: or

38. is an estimate of p i (i = 1,2) Recall: has approximately a normal distribution with

39. Where: A sample of n 1 is selected from population 1 resulting in x 1 successes A sample of n 2 is selected from population 2 resulting in x 2 successes

40. is an estimate of p 1 – p 2 We want to estimate and test p 1 – p 2 has approximately a normal distribution with

41. The statistic:

42. If then is true Hence

43. The test statistic:

44. Where: A sample of n 1 is selected from population 1 resulting in x 1 successes A sample of n 2 is selected from population 2 resulting in x 2 successes

45. The Alternative Hypothesis H A The Critical Region

46. Example In a national study to determine if there was an increase in mortality due to pipe smoking, a random sample of n 1 = 1067 male nonsmoking pensioners were observed for a five-year period. In addition a sample of n 2 = 402 male pensioners who had smoked a pipe for more than six years were observed for the same five-year period. At the end of the five-year period, x 1 = 117 of the nonsmoking pensioners had died while x 2 = 54 of the pipe-smoking pensioners had died. Is there a the mortality rate for pipe smokers higher than that for non-smokers

47. We want to test:

48. The test statistic:

49. Note:

50. The test statistic:

51. We reject H 0 if: Not true hence we accept H 0. Conclusion: There is not a significant (  = 0.05) increase in the mortality rate due to pipe-smoking

52. Estimating a difference proportions using confidence intervals Situation We have two populations (1 and 2) Let p 1 denote the probability (proportion) of “success” in population 1. Let p 2 denote the probability (proportion) of “success” in population 2. Objective is to estimate the difference in the two population proportions  = p 1 – p 2.

53. Confidence Interval for  = p 1 – p 2 100P% = 100(1 –  ) % :

54. Example Estimating the increase in the mortality rate for pipe smokers higher over that for non- smokers  = p 2 – p 1