1. Chap 9-1 Two-Sample Tests

2. Chap 9-2 Two Sample Tests Population Means, Independent Samples Means, Related Samples Population Variances Group 1 vs. independent Group 2 Same group before vs. after treatment Variance 1 vs. Variance 2 Examples: Population Proportions Proportion 1 vs. Proportion 2

3. Chap 9-3 Difference Between Two Means Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 unknown Goal: Test hypotheses or form a confidence interval for the difference between two population means, μ 1 – μ 2 The point estimate for the difference is X 1 – X 2 *

4. Chap 9-4 Independent Samples Population means, independent samples  Different data sources  Unrelated  Independent  Sample selected from one population has no effect on the sample selected from the other population  Use the difference between 2 sample means  Use Z test or pooled variance t test * σ 1 and σ 2 known σ 1 and σ 2 unknown

5. Chap 9-5 Difference Between Two Means Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 unknown * Use a Z test statistic Use S to estimate unknown σ, use a t test statistic and pooled standard deviation

6. Chap 9-6 Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 Known Assumptions:  Samples are randomly and independently drawn  population distributions are normal or both sample sizes are  30  Population standard deviations are known * σ 1 and σ 2 unknown

7. Chap 9-7 Population means, independent samples σ 1 and σ 2 known …and the standard error of X 1 – X 2 is When σ 1 and σ 2 are known and both populations are normal or both sample sizes are at least 30, the test statistic is a Z-value… (continued) σ 1 and σ 2 Known * σ 1 and σ 2 unknown

8. Chap 9-8 Population means, independent samples σ 1 and σ 2 known The test statistic for μ 1 – μ 2 is: σ 1 and σ 2 Known * σ 1 and σ 2 unknown (continued)

9. Chap 9-9 Hypothesis Tests for Two Population Means Lower-tail test: H 0 : μ 1  μ 2 H 1 : μ 1 < μ 2 i.e., H 0 : μ 1 – μ 2  0 H 1 : μ 1 – μ 2 < 0 Upper-tail test: H 0 : μ 1 ≤ μ 2 H 1 : μ 1 > μ 2 i.e., H 0 : μ 1 – μ 2 ≤ 0 H 1 : μ 1 – μ 2 > 0 Two-tail test: H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 i.e., H 0 : μ 1 – μ 2 = 0 H 1 : μ 1 – μ 2 ≠ 0 Two Population Means, Independent Samples

10. Chap 9-10 Two Population Means, Independent Samples Lower-tail test: H 0 : μ 1 – μ 2  0 H 1 : μ 1 – μ 2 < 0 Upper-tail test: H 0 : μ 1 – μ 2 ≤ 0 H 1 : μ 1 – μ 2 > 0 Two-tail test: H 0 : μ 1 – μ 2 = 0 H 1 : μ 1 – μ 2 ≠ 0  /2  -z  -z  /2 zz z  /2 Reject H 0 if Z < -Z  Reject H 0 if Z > Z  Reject H 0 if Z < -Z  /2  or Z > Z  /2 Hypothesis tests for μ 1 – μ 2

11. Chap 9-11 Population means, independent samples σ 1 and σ 2 known The confidence interval for μ 1 – μ 2 is: Confidence Interval, σ 1 and σ 2 Known * σ 1 and σ 2 unknown

12. Chap 9-12 Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 Unknown Assumptions:  Samples are randomly and independently drawn  Populations are normally distributed or both sample sizes are at least 30  Population variances are unknown but assumed equal * σ 1 and σ 2 unknown

13. Chap 9-13 Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 Unknown (continued) * σ 1 and σ 2 unknown Forming interval estimates:  The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ  the test statistic is a t value with (n 1 + n 2 – 2) degrees of freedom

14. Chap 9-14 Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 Unknown The pooled standard deviation is (continued) * σ 1 and σ 2 unknown

15. Chap 9-15 Population means, independent samples σ 1 and σ 2 known σ 1 and σ 2 Unknown Where t has (n 1 + n 2 – 2) d.f., and The test statistic for μ 1 – μ 2 is: * σ 1 and σ 2 unknown (continued)

16. Chap 9-16 Population means, independent samples σ 1 and σ 2 known The confidence interval for μ 1 – μ 2 is: Where * σ 1 and σ 2 unknown Confidence Interval, σ 1 and σ 2 Unknown

17. Chap 9-17 Pooled S p t Test: Example You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQ Number 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16 Assuming both populations are approximately normal with equal variances, is there a difference in average yield (  = 0.05)?

18. Chap 9-18 Calculating the Test Statistic The test statistic is:

19. Chap 9-19 Solution H 0 : μ 1 - μ 2 = 0 i.e. (μ 1 = μ 2 ) H 1 : μ 1 - μ 2 ≠ 0 i.e. (μ 1 ≠ μ 2 )  = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 2.0154 Test Statistic: Decision: Conclusion: Reject H 0 at  = 0.05 There is evidence of a difference in means. t 0 2.0154-2.0154.025 Reject H 0.025 2.040

20. Chap 9-20 Related Samples Tests Means of 2 Related Populations  Paired or matched samples  Repeated measures (before/after)  Use difference between paired values:  Eliminates Variation Among Subjects  Assumptions:  Both Populations Are Normally Distributed  Or, if Not Normal, use large samples Related samples D = X 1 - X 2

21. Chap 9-21 Mean Difference, σ D Known The i th paired difference is D i, where Related samples D i = X 1i - X 2i The point estimate for the population mean paired difference is D : Suppose the population standard deviation of the difference scores, σ D, is known n is the number of pairs in the paired sample

22. Chap 9-22 The test statistic for the mean difference is a Z value: Paired samples Mean Difference, σ D Known (continued) Where μ D = hypothesized mean difference σ D = population standard dev. of differences n = the sample size (number of pairs)

23. Chap 9-23 Confidence Interval, σ D Known The confidence interval for D is Paired samples Where n = the sample size (number of pairs in the paired sample)

24. Chap 9-24 If σ D is unknown, we can estimate the unknown population standard deviation with a sample standard deviation: Related samples The sample standard deviation is Mean Difference, σ D Unknown

25. Chap 9-25 The test statistic for D is now a t statistic, with n-1 d.f.: Paired samples Where t has n - 1 d.f. and S D is: Mean Difference, σ D Unknown (continued)

26. Chap 9-26 The confidence interval for D is Paired samples where Confidence Interval, σ D Unknown

27. Chap 9-27 Lower-tail test: H 0 : μ D  0 H 1 : μ D < 0 Upper-tail test: H 0 : μ D ≤ 0 H 1 : μ D > 0 Two-tail test: H 0 : μ D = 0 H 1 : μ D ≠ 0 Paired Samples Hypothesis Testing for Mean Difference, σ D Unknown  /2  -t  -t  /2 tt t  /2 Reject H 0 if t < -t  Reject H 0 if t > t  Reject H 0 if t < -t   or t > t  Where t has n - 1 d.f.

28. Chap 9-28  Assume you send your salespeople to a “customer service” training workshop. Is the training effective? You collect the following data: Paired Samples Example Number of Complaints: (2) - (1) Salesperson Before (1) After (2) Difference, D i C.B. 6 4 - 2 T.F. 20 6 -14 M.H. 3 2 - 1 R.K. 0 0 0 M.O. 4 0 - 4 -21 D =  DiDi n = -4.2

29. Chap 9-29  Has the training made a difference in the number of complaints (at the 0.01 level)? - 4.2D = H 0 : μ D = 0 H 1 :  μ D  0 Test Statistic: Critical Value = ± 4.604 d.f. = n - 1 = 4 Reject  /2 - 4.604 4.604 Decision: Do not reject H 0 (t stat is not in the reject region) Conclusion: There is not a significant change in the number of complaints. Paired Samples: Solution Reject  /2 - 1.66  =.01

30. Chap 9-30 Two Population Proportions Goal: test a hypothesis or form a confidence interval for the difference between two population proportions, p 1 – p 2 The point estimate for the difference is Population proportions Assumptions: n 1 p 1  5, n 1 (1-p 1 )  5 n 2 p 2  5, n 2 (1-p 2 )  5

31. Chap 9-31 Two Population Proportions Population proportions The pooled estimate for the overall proportion is: where X 1 and X 2 are the numbers from samples 1 and 2 with the characteristic of interest Since we begin by assuming the null hypothesis is true, we assume p 1 = p 2 and pool the two p s estimates

32. Chap 9-32 Two Population Proportions Population proportions The test statistic for p 1 – p 2 is a Z statistic: (continued) where

33. Chap 9-33 Confidence Interval for Two Population Proportions Population proportions The confidence interval for p 1 – p 2 is:

34. Chap 9-34 Hypothesis Tests for Two Population Proportions Population proportions Lower-tail test: H 0 : p 1  p 2 H 1 : p 1 < p 2 i.e., H 0 : p 1 – p 2  0 H 1 : p 1 – p 2 < 0 Upper-tail test: H 0 : p 1 ≤ p 2 H 1 : p 1 > p 2 i.e., H 0 : p 1 – p 2 ≤ 0 H 1 : p 1 – p 2 > 0 Two-tail test: H 0 : p 1 = p 2 H 1 : p 1 ≠ p 2 i.e., H 0 : p 1 – p 2 = 0 H 1 : p 1 – p 2 ≠ 0

35. Chap 9-35 Hypothesis Tests for Two Population Proportions Population proportions Lower-tail test: H 0 : p 1 – p 2  0 H 1 : p 1 – p 2 < 0 Upper-tail test: H 0 : p 1 – p 2 ≤ 0 H 1 : p 1 – p 2 > 0 Two-tail test: H 0 : p 1 – p 2 = 0 H 1 : p 1 – p 2 ≠ 0  /2  -z  -z  /2 zz z  /2 Reject H 0 if Z < -Z  Reject H 0 if Z > Z  Reject H 0 if Z < -Z   or Z > Z  (continued)

36. Chap 9-36 Example: Two population Proportions Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?  In a random sample, 36 of 72 men and 31 of 50 women indicated they would vote Yes  Test at the.05 level of significance

37. Chap 9-37  The hypothesis test is: H 0 : p 1 – p 2 = 0 (the two proportions are equal) H 1 : p 1 – p 2 ≠ 0 (there is a significant difference between proportions)  The sample proportions are:  Men: p s1 = 36/72 =.50  Women: p s2 = 31/50 =.62  The pooled estimate for the overall proportion is: Example: Two population Proportions (continued)

38. Chap 9-38 The test statistic for p 1 – p 2 is: Example: Two population Proportions (continued).025 -1.961.96.025 -1.31 Decision: Do not reject H 0 Conclusion: There is not significant evidence of a difference in proportions who will vote yes between men and women. Reject H 0 Critical Values = ±1.96 For  =.05

39. Chap 9-39 Hypothesis Tests for Variances Tests for Two Population Variances F test statistic H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 Two-tail test Lower-tail test Upper-tail test H 0 : σ 1 2  σ 2 2 H 1 : σ 1 2 < σ 2 2 H 0 : σ 1 2 ≤ σ 2 2 H 1 : σ 1 2 > σ 2 2 *

40. Chap 9-40 Hypothesis Tests for Variances Tests for Two Population Variances F test statistic The F test statistic is: = Variance of Sample 1 n 1 - 1 = numerator degrees of freedom n 2 - 1 = denominator degrees of freedom = Variance of Sample 2 * (continued)

41. Chap 9-41  The F critical value is found from the F table  The are two appropriate degrees of freedom: numerator and denominator  In the F table,  numerator degrees of freedom determine the column  denominator degrees of freedom determine the row The F Distribution where df 1 = n 1 – 1 ; df 2 = n 2 – 1

42. Chap 9-42 F0 Finding the Rejection Region rejection region for a two-tail test is:  FLFL Reject H 0 Do not reject H 0 F 0  FUFU Reject H 0 Do not reject H 0 F0  /2 Reject H 0 Do not reject H 0 FUFU H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 H 0 : σ 1 2  σ 2 2 H 1 : σ 1 2 < σ 2 2 H 0 : σ 1 2 ≤ σ 2 2 H 1 : σ 1 2 > σ 2 2 FLFL  /2 Reject H 0 Reject H 0 if F < F L Reject H 0 if F > F U

43. Chap 9-43 Finding the Rejection Region F0  /2 Reject H 0 Do not reject H 0 FUFU H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 FLFL  /2 Reject H 0 (continued) 2. Find F L using the formula: Where F U* is from the F table with n 2 – 1 numerator and n 1 – 1 denominator degrees of freedom (i.e., switch the d.f. from F U ) 1. Find F U from the F table for n 1 – 1 numerator and n 2 – 1 denominator degrees of freedom To find the critical F values:

44. Chap 9-44 F Test: An Example You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data : NYSE NASDAQ Number 2125 Mean3.272.53 Std dev1.301.16 Is there a difference in the variances between the NYSE & NASDAQ at the  = 0.05 level?

45. Chap 9-45 F Test: Example Solution  Form the hypothesis test: H 0 : σ 2 1 – σ 2 2 = 0 ( there is no difference between variances) H 1 : σ 2 1 – σ 2 2 ≠ 0 ( there is a difference between variances)  Numerator:  n 1 – 1 = 21 – 1 = 20 d.f.  Denominator:  n 2 – 1 = 25 – 1 = 24 d.f. F U = F.025, 20, 24 = 2.33  Find the F critical values for  =.05:  Numerator:  n 2 – 1 = 25 – 1 = 24 d.f.  Denominator:  n 1 – 1 = 21 – 1 = 20 d.f. F L = 1/F.025, 24, 20 = 1/2.41 =.41 FU:FU:FL:FL:

46. Chap 9-46  The test statistic is: 0  /2 =.025 F U =2.33 Reject H 0 Do not reject H 0 H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 F Test: Example Solution  F = 1.256 is not in the rejection region, so we do not reject H 0 (continued)  Conclusion: There is not sufficient evidence of a difference in variances at  =.05 F L =0.41  /2 =.025 Reject H 0 F