1. Chapter 10 Comparing Two Treatments Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved.

2. Experimental Design Matched Paired Samples –Subjects are chosen in “pairs” that are alike. In each pair, one subject gets one treatment, the other gets the second treatment Independent Samples –Groups are randomly divided, given two different treatments Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved.

3. Figure 10.1a (p. 377) Independent samples, each of size 4.

4. Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved. Figure 10.1b (p. 377) Matched pairs design with four pairs of subjects. Separate random assignment of Drug 1 each pair.

5. Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved. Box 2 on Page 379 Statistical Model: Independent Random Samples

6. We know look to compare the means of these two treatments or population. When the sample sizes are large, we do not need to make additional assumptions. When both sample sizes are large, the sample means are both approximately normal, and their difference is approximately normal. Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved.

7. 10.5 Introduction When comparing two populations, we’ll be considering the parameter  1 –  2. When we have two independent randomly-chosen large samples, we’ll use the two means and two standard deviations (or variances) in our point estimates of expected value and standard error.

8. Confidence Interval As we did before, we can construct a confidence interval: Estimate of parameter +/- (z-value)*estimated standard of error Statistics, 5/E by Johnson and Bhattacharyya Copyright © 2006 by John Wiley & Sons, Inc. All rights reserved.

9. Box on Page 381 Large Samples Confidence Interval for  1 -  2

10. 10.1-10.3 Example A – confidence interval In doing a survey of textbook costs, Rodney found that at Whatsammatta U., in a sample of 42, the mean cost was $63.42 with s = $8.13. At Matriarch College (affectionately known as U. Mama) the mean textbook cost was $67.19 in a sample of 35 with s = $7.29. Construct a 95% confidence interval.

11. 10.1-10.3 Example A – confidence interval In doing a survey of textbook costs, Rodney found that at Whatsammatta U., in a sample of 42, the mean cost was $63.42 with s = $8.13. At Matriarch College (affectionately known as U. Mama) the mean textbook cost was $67.19 in a sample of 35 with s = $7.29. Construct a 95% confidence interval. IMPORTANT! Note that we are given the standard deviation s. We’ll need the variance s 2 for the various formulae.

12. 10.1-10.3 Example A – confidence interval | ← 1 –  →| |←  1 –  2 →| z  / 2 – z  / 2 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ; 95% confidence interval.

13. 10.1-10.3 Example A – confidence interval | ← 1 –  →| |←  1 –  2 →| z  / 2 – z  / 2 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ; 95% confidence interval.

14. 10.1-10.3 Example A – confidence interval | ← 1 –  →| |←  1 –  2 →| z  / 2 – z  / 2 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ; 95% confidence interval.

15. 10.1-10.3 Example A – confidence interval | ← 1 –  →| |←  1 –  2 →| z  / 2 – z  / 2 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ; 95% confidence interval.

16. 10.1-10.3 Example A – hypothesis test In doing a survey of textbook costs, Rodney found that at Whatsammatta U., in a sample of 42, the mean cost was $63.42 with s = $8.13. At Matriarch College (affectionately known as U. Mama) the mean textbook cost was $67.19 in a sample of 35 with s = $7.29. Test the hypothesis (  = 0.05 ) that textbooks cost less at Whatsammatta U. than they do at U. Mama.

17. 10.1-10.3 Example A – hypothesis test In doing a survey of textbook costs, Rodney found that at Whatsammatta U., in a sample of 42, the mean cost was $63.42 with s = $8.13. At Matriarch College (affectionately known as U. Mama) the mean textbook cost was $67.19 in a sample of 35 with s = $7.29. Test the hypothesis (  = 0.05 ) that textbooks cost less at Whatsammatta U. than they do at U. Mama. IMPORTANT! Note that we are given the standard deviation s. We’ll need the variance s 2 for the various formulae.

18. 10.1-10.3 Example A – hypothesis test (1) Set null hypothesis (“status quo”) and alternate hypothesis (“challenge”). group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

19. 10.1-10.3 Example A – hypothesis test (1) Set null hypothesis (“status quo”) and alternate hypothesis (“challenge”). (2) Since n 1 = 42 and n 2 = 35 are both greater than 30, we can treat this as a large-sample case and use the normal distribution table to approximate probabilities. group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

20. 10.1-10.3 Example A – hypothesis test (3) For  = 0.05, the boundary value for the rejection region is – z 0.05 = – 1.645. group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

21. 10.1-10.3 Example A – hypothesis test (4) Compute the test statistic. – z 0.05 = – 1.645 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

22. 10.1-10.3 Example A – hypothesis test (4) Compute the test statistic. – z 0.05 = – 1.645 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

23. 10.1-10.3 Example A – hypothesis test (4) Compute the test statistic. – z 0.05 = – 1.645 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

24. 10.1-10.3 Example A – hypothesis test (5) Draw a conclusion. Since z falls in the rejection region, reject H 0. Based on this sample, there is evidence to say that text books cost less at Whatsammatta U. – z 0.05 = – 1.645 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

25. 10.1-10.3 Example A – hypothesis test (5) Draw a conclusion using the P- value. P- value = P(z < – 2.14) = 0.0162 group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0

26. 10.1-10.3 Example A – hypothesis test group 1: Whatsammatta U., n = 42, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 35, mean = $67.19, s = $7.29 ;  = 0.05, test  1 <  2, that is  1 –  2 < 0 (5) Draw a conclusion using the P- value. P- value = P(z < – 2.14) = 0.0162 For  = 0.05 we reject H 0 ; This is a fairly strong conclusion. We’d have needed  = 0.01 to conclude “fail to reject H 0 ”.

27. 10.1-10.3 Example B – hypothesis test In doing a survey of hours spent watching educational television per month, Regina found that in Pleasantville, in a sample of 40 households, the mean was 60 hours with a sample variance of 64. In Nicetown the mean was 63 hours in a sample of 50 households with a sample variance of 81. Is there a statistically significant difference between the two towns’ educational television viewing habits? (  = 0.05 )

28. 10.1-10.3 Example B – hypothesis test In doing a survey of hours spent watching educational television per month, Regina found that in Pleasantville, in a sample of 40 households, the mean was 60 hours with a sample variance of 64. In Nicetown the mean was 63 hours in a sample of 50 households with a sample variance of 81. Is there a statistically significant difference between the two towns’ educational television viewing habits? (  = 0.05 ) IMPORTANT! Note that we are given the variance s 2, which is what we’ll need the for the various formulae. A second “squaring” would be wrong!

29. 10.1-10.3 Example B – hypothesis test (1) Set null hypothesis (“status quo”) and alternate hypothesis (“challenge”). group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

30. 10.1-10.3 Example B – hypothesis test (1) Set null hypothesis (“status quo”) and alternate hypothesis (“challenge”). (2) Since n 1 = 40 and n 2 = 50 are both greater than 30, we can treat this as a large-sample case and use the normal distribution table to approximate probabilities. group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

31. 10.1-10.3 Example B – hypothesis test (3) For  = 0.05,  /2= 0.025, the boundary value for the rejection region is z 0.025 = 1.960. group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

32. 10.1-10.3 Example B – hypothesis test (4) Compute the test statistic. z 0.025 = 1.960 group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

33. 10.1-10.3 Example B – hypothesis test (4) Compute the test statistic. z 0.025 = 1.960 group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

34. 10.1-10.3 Example B – hypothesis test (4) Compute the test statistic. z 0.025 = 1.960 group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

35. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion. Since z does not fall in the rejection region, fail to reject H 0. Based on this sample, there is not enough evidence to say that the two towns view different amounts of educational television. z 0.025 = 1.960 group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

36. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion using the P- value. group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

37. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion using the P- value. P- value = P(z 1.67) group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

38. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion using the P- value. P- value = P(z 1.67) = 2 (0.0475) group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

39. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion using the P- value. P- value = P(z 1.67) = 2 (0.0475) = 0.0950 group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

40. 10.1-10.3 Example B – hypothesis test (5) Draw a conclusion using the P- value. P- value = P(z 1.67) = 2 (0.0475) = 0.0950 For  = 0.05 we fail to reject H 0 ; Note that for  = 0.10 we would have concluded “reject H 0 ”. group 1: Pleasantville, n = 40, mean = 60, s 2 = 64 ; group 2: Nicetown, n = 50, mean = 63, s 2 = 81 ;  = 0.05, test  1 ≠  2, that is  1 –  2 ≠ 0

41. 10.5 Example A In doing a survey of textbook costs, Rodney found that at Whatsammatta U., in a sample of 19, the mean cost was $63.42 with s = $8.13. At Matriarch College (affectionately known as U. Mama) the mean textbook cost was $67.19 in a sample of 23 with s = $7.29. Using the conservative approach determine the 95% confidence interval for the difference of means. Calculate the confidence interval in the same way as before, but now use t instead of z. The degrees of freedom will be the smaller of n 1 -1 and n 2 -1.

42. 10.5 Example A – confidence interval group 1: Whatsammatta U., n = 19, mean = $63.42, s = $8.13 ; group 2: U. Mama, n = 23, mean = $67.19, s = $7.29 ; 95% confidence interval. d.f=19-1=18