1. Chapter 8 Hypothesis Testing with Two Samples 1

2. Chapter Outline 8.1 Testing the Difference Between Means (Large Independent Samples) 8.2 Testing the Difference Between Means (Small Independent Samples) 8.3 Testing the Difference Between Means (Dependent Samples) 8.4 Testing the Difference Between Proportions 2

3. Section 8.1 Testing the Difference Between Means (Large Independent Samples) 3

4. Section 8.1 Objectives Determine whether two samples are independent or dependent Perform a two-sample z-test for the difference between two means μ 1 and μ 2 using large independent samples 4

5. Two Sample Hypothesis Test Compares two parameters from two populations. Sampling methods:  Independent Samples The sample selected from one population is not related to the sample selected from the second population.  Dependent Samples (paired or matched samples) Each member of one sample corresponds to a member of the other sample. 5

6. Independent and Dependent Samples Independent Samples Sample 1Sample 2 Dependent Samples Sample 1Sample 2 6

7. Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. Sample 1: Resting heart rates of 35 individuals before drinking coffee. Sample 2: Resting heart rates of the same individuals after drinking two cups of coffee. Solution: Dependent Samples (The samples can be paired with respect to each individual) 7

8. Example: Independent and Dependent Samples Classify the pair of samples as independent or dependent. Sample 1: Test scores for 35 statistics students. Sample 2: Test scores for 42 biology students who do not study statistics. Solution: Independent Samples (Not possible to form a pairing between the members of the samples; the sample sizes are different, and the data represent scores for different individuals.) 8

9. Two Sample Hypothesis Test with Independent Samples 1.Null hypothesis H 0  A statistical hypothesis that usually states there is no difference between the parameters of two populations.  Always contains the symbol , =, or . 2.Alternative hypothesis H a  A statistical hypothesis that is true when H 0 is false.  Always contains the symbol >, , or <. 9

10. Two Sample Hypothesis Test with Independent Samples H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 Regardless of which hypotheses you use, you always assume there is no difference between the population means, or μ 1 = μ 2. 10

11. Two Sample z - Test for the Difference Between Means Three conditions are necessary to perform a z-test for the difference between two population means μ 1 and μ 2. 1.The samples must be randomly selected. 2.The samples must be independent. 3.Each sample size must be at least 30, or, if not, each population must have a normal distribution with a known standard deviation. 11

12. Two Sample z - Test for the Difference Between Means If these requirements are met, the sampling distribution for (the difference of the sample means) is a normal distribution with Sampling distribution for : Mean: Standard error: 12

13. Two Sample z - Test for the Difference Between Means Test statistic is The standardized test statistic is When the samples are large, you can use s 1 and s 2 in place of  1 and  2. If the samples are not large, you can still use a two-sample z-test, provided the populations are normally distributed and the population standard deviations are known. 13

14. Using a Two - Sample z - Test for the Difference Between Means (Large Independent Samples) 1.State the claim mathematically. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Sketch the sampling distribution. 4.Determine the critical value(s). 5.Determine the rejection region(s). State H 0 and H a. Identify . Use Table 4 in Appendix B. In WordsIn Symbols 14

15. Using a Two - Sample z - Test for the Difference Between Means (Large Independent Samples) 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols 15

16. Example: Two - Sample z - Test for the Difference Between Means A consumer education organization claims that there is a difference in the mean credit card debt of males and females in the United States. The results of a random survey of 200 individuals from each group are shown below. The two samples are independent. Do the results support the organization’s claim? Use α = 0.05. Females (1)Males (2) s 1 = $750s 2 = $800 n 1 = 200n 2 = 200 16

17. Z 0-1.96 0.025 1.96 0.025 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :   n 1 =, n 2 = Rejection Region: Test Statistic: 0.05 200 μ 1 = μ 2 μ 1 ≠ μ 2 -1.961.96 -1.03 Decision: At the 5% level of significance, there is not enough evidence to support the organization’s claim that there is a difference in the mean credit card debt of males and females. Fail to Reject H 0 17

18. Example: Using Technology to Perform a Two-Sample z-Test The American Automobile Association claims that the average daily cost for meals and lodging for vacationing in Texas is less than the same average costs for vacationing in Virginia. The table shows the results of a random survey of vacationers in each state. The two samples are independent. At α = 0.01, is there enough evidence to support the claim? Texas (1)Virginia (2) s 1 = $15s 2 = $22 n 1 = 50n 2 = 35 18

19. Solution: Using Technology to Perform a Two-Sample z-Test H 0 : H a : μ 1 ≥ μ 2 μ 1 < μ 2 TI-83/84set up: Calculate: Draw: 19

20. z 0 0.01 Solution: Using Technology to Perform a Two-Sample z-Test Decision: At the 1% level of significance, there is not enough evidence to support the American Automobile Association’s claim. Fail to Reject H 0 Rejection Region: -0.93 -2.33 20

21. Section 8.1 Summary Determined whether two samples are independent or dependent Performed a two-sample z-test for the difference between two means μ 1 and μ 2 using large independent samples 21

22. Slide 8- 22 © 2012 Pearson Education, Inc. Comparing Two Means http://www.learner.org/courses/againstallodds/unitpages/unit27.html

23. Section 8.2 Testing the Difference Between Means (Small Independent Samples) 23

24. Section 8.2 Objectives Perform a t-test for the difference between two means μ 1 and μ 2 using small independent samples 24

25. Two Sample t - Test for the Difference Between Means If samples of size less than 30 are taken from normally- distributed populations, a t-test may be used to test the difference between the population means μ 1 and μ 2. Three conditions are necessary to use a t-test for small independent samples. 1.The samples must be randomly selected. 2.The samples must be independent. 3.Each population must have a normal distribution. 25

26. Two Sample t - Test for the Difference Between Means The standardized test statistic is The standard error and the degrees of freedom of the sampling distribution depend on whether the population variances and are equal. 26

27.  The standard error for the sampling distribution of is Two Sample t - Test for the Difference Between Means Variances are equal  Information from the two samples is combined to calculate a pooled estimate of the standard deviation.  d.f.= n 1 + n 2 – 2 27

28. Variances are not equal  If the population variances are not equal, then the standard error is  d.f = smaller of n 1 – 1 or n 2 – 1 Two Sample t - Test for the Difference Between Means 28

29. Normal or t-Distribution? Are both sample sizes at least 30? Are both populations normally distributed? You cannot use the z-test or the t-test. No Yes Are both population standard deviations known? Use the z-test. Yes No Are the population variances assumed to be equal? Use the z-test. Use the t-test with d.f = smaller of n 1 – 1 or n 2 – 1. Use the t-test with Yes No Yes d.f = n 1 + n 2 – 2. 29

30. Two-Sample t-Test for the Difference Between Means (Small Independent Samples) 1.State the claim mathematically. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom and sketch the sampling distribution. 4.Determine the critical value(s). State H 0 and H a. Identify . Use Table 5 in Appendix B. d.f. = n 1 + n 2 – 2 or d.f. = smaller of n 1 – 1 or n 2 – 1. In WordsIn Symbols 30

31. Two-Sample t-Test for the Difference Between Means (Small Independent Samples) 5.Determine the rejection region(s). 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols 31

32. Example: Two - Sample t - Test for the Difference Between Means The braking distances of 8 Volkswagen GTIs and 10 Ford Focuses were tested when traveling at 60 miles per hour on dry pavement. The results are shown below. Can you conclude that there is a difference in the mean braking distances of the two types of cars? Use α = 0.01. Assume the populations are normally distributed and the population variances are not equal. (Adapted from Consumer Reports) GTI (1)Focus (2) s 1 = 6.9 fts 2 = 2.6 ft n 1 = 8n 2 = 10 32

33. t 0-3.499 0.005 3.499 0.005 Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a :   d.f. = Rejection Region: Test Statistic: 0.01 8 – 1 = 7 μ 1 = μ 2 μ 1 ≠ μ 2 -3.4993.499 -3.496 Decision: At the 1% level of significance, there is not enough evidence to conclude that the mean braking distances of the cars are different. Fail to Reject H 0 33

34. Example: Two - Sample t - Test for the Difference Between Means A manufacturer claims that the calling range (in feet) of its 2.4-GHz cordless telephone is greater than that of its leading competitor. You perform a study using 14 randomly selected phones from the manufacturer and 16 randomly selected similar phones from its competitor. The results are shown below. At α = 0.05, can you support the manufacturer’s claim? Assume the populations are normally distributed and the population variances are equal. Manufacturer (1)Competition (2) s 1 = 45 fts 2 = 30 ft n 1 = 14 n 2 = 16 34

35. Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a :   d.f. = Rejection Region: Test Statistic: 0.05 14 + 16 – 2 = 28 μ 1 ≤ μ 2 μ 1 > μ 2 Decision: t 01.701 0.05 35

36. Solution: Two - Sample t - Test for the Difference Between Means 36

37. Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a :   d.f. = Rejection Region: Test Statistic: 0.05 14 + 16 – 2 = 28 μ 1 ≤ μ 2 μ 1 > μ 2 1.811 Decision: At the 5% level of significance, there is enough evidence to support the manufacturer’s claim that its phone has a greater calling range than its competitors. Reject H 0 t 01.701 0.05 37

38. Slide 8- 38 © 2012 Pearson Education, Inc.

39. Section 8.2 Summary Performed a t-test for the difference between two means μ 1 and μ 2 using small independent samples 39

40. Section 8.3 Testing the Difference Between Means (Dependent Samples) 40

41. Section 8.3 Objectives Perform a t-test to test the mean of the difference for a population of paired data 41

42. The test statistic is the mean of these differences.  t - Test for the Difference Between Means To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found:  d = x 1 – x 2 Difference between entries for a data pair Mean of the differences between paired data entries in the dependent samples 42

43. t - Test for the Difference Between Means Three conditions are required to conduct the test. 1.The samples must be randomly selected. 2.The samples must be dependent (paired). 3.Both populations must be normally distributed. If these conditions are met, then the sampling distribution for is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs. -t0-t0 t0t0 μdμd 43

44. Symbols used for the t - Test for μ d The number of pairs of data The difference between entries for a data pair, d = x 1 – x 2 The hypothesized mean of the differences of paired data in the population n d Symbol Description 44

45. Symbols used for the t - Test for μ d Symbol Description The mean of the differences between the paired data entries in the dependent samples The standard deviation of the differences between the paired data entries in the dependent samples sdsd 45

46. t - Test for the Difference Between Means The test statistic is The standardized test statistic is The degrees of freedom are d.f. = n – 1 46

47. t - Test for the Difference Between Means (Dependent Samples) 1.State the claim mathematically. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom and sketch the sampling distribution. 4.Determine the critical value(s). State H 0 and H a. Identify . Use Table 5 in Appendix B if n > 29 use the last row (∞). d.f. = n – 1 In WordsIn Symbols 47

48. t - Test for the Difference Between Means (Dependent Samples) 5.Determine the rejection region(s). 6.Calculate and Use a table. 7.Find the standardized test statistic. In WordsIn Symbols 48

49. t - Test for the Difference Between Means (Dependent Samples) 8.Make a decision to reject or fail to reject the null hypothesis. 9.Interpret the decision in the context of the original claim. If t is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols 49

50. Example: t - Test for the Difference Between Means Golfer12345678 Score (old)8984968274928591 Score (new)83 928476918091 A golf club manufacturer claims that golfers can lower their scores by using the manufacturer’s newly designed golf clubs. Eight golfers are randomly selected, and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are again asked to give their most recent score. The scores for each golfer are shown in the table. Assuming the golf scores are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? 50

51. t 01.415 0.10 Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a :   d.f. = Rejection Region: Test Statistic: 0.10 8 – 1 = 7 μ d ≤ 0 μ d > 0 Decision: d = (old score) – (new score) 51

52. Solution: Two - Sample t - Test for the Difference Between Means OldNewdd2d2 8983636 848311 9692416 8284–24 7476–24 929111 8580525 91 00 Σ = 13Σ = 87 d = (old score) – (new score) 52

53. Slide 8- 53 © 2012 Pearson Education, Inc.

54. t 01.415 0.10 Solution: Two - Sample t - Test for the Difference Between Means H 0 : H a :   d.f. = Rejection Region: Test Statistic: 0.10 8 – 1 = 7 μ d ≤ 0 μ d > 0 1.415 Decision: d = (old score) – (new score) 1.498 At the 10% level of significance, the results of this test indicate that after the golfers used the new clubs, their scores were significantly lower. Reject H 0 54

55. Section 8.3 Summary Performed a t-test to test the mean of the difference for a population of paired data 55

56. Section 8.4 Testing the Difference Between Proportions 56

57. Section 8.4 Objectives Perform a z-test for the difference between two population proportions p 1 and p 2 57

58. Two-Sample z-Test for Proportions Used to test the difference between two population proportions, p 1 and p 2. Three conditions are required to conduct the test. 1.The samples must be randomly selected. 2.The samples must be independent. 3.The samples must be large enough to use a normal sampling distribution. That is, n 1 p 1  5, n 1 q 1  5, n 2 p 2  5, and n 2 q 2  5. 58

59. Two-Sample z-Test for the Difference Between Proportions If these conditions are met, then the sampling distribution for is a normal distribution Mean: A weighted estimate of p 1 and p 2 can be found by using Standard error: 59

60. Two-Sample z-Test for the Difference Between Proportions The test statistic is The standardized test statistic is where 60

61. Two-Sample z-Test for the Difference Between Proportions 1.State the claim. Identify the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the critical value(s). 4.Determine the rejection region(s). 5.Find the weighted estimate of p 1 and p 2. State H 0 and H a. Identify . Use Table 4 in Appendix B. In WordsIn Symbols 61

62. Two-Sample z-Test for the Difference Between Proportions 6.Find the standardized test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. If z is in the rejection region, reject H 0. Otherwise, fail to reject H 0. In WordsIn Symbols 62

63. Example: Two - Sample z - Test for the Difference Between Proportions In a study of 200 randomly selected adult female and 250 randomly selected adult male Internet users, 30% of the females and 38% of the males said that they plan to shop online at least once during the next month. At α = 0.10 test the claim that there is a difference between the proportion of female and the proportion of male Internet users who plan to shop online. Solution: 1 = Females2 = Males 63

64. Z 0-1.645 0.05 1.645 0.05 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :   n 1 =, n 2 = Rejection Region: Test Statistic: 0.10 200 250 p 1 = p 2 p 1 ≠ p 2 Decision: 64

65. Solution: Two - Sample z - Test for the Difference Between Means 65

66. Solution: Two - Sample z - Test for the Difference Between Means 66

67. z 0-1.645 0.05 1.645 0.05 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :   n 1 =, n 2 = Rejection Region: Test Statistic: 0.10 200 250 p 1 = p 2 p 1 ≠ p 2 -1.6451.645 -1.77 Decision: At the 10% level of significance, there is enough evidence to conclude that there is a difference between the proportion of female and the proportion of male Internet users who plan to shop online. Reject H 0 67

68. Example: Two - Sample z - Test for the Difference Between Proportions A medical research team conducted a study to test the effect of a cholesterol reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At α = 0.01 can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? (Adapted from New England Journal of Medicine) Solution: 1 = Medication2 = Placebo 68

69. z 0-2.33 0.01 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :   n 1 =, n 2 = Rejection Region: Test Statistic: 0.01 4700 4300 p 1 ≥ p 2 p 1 < p 2 Decision: 69

70. Solution: Two - Sample z - Test for the Difference Between Means 70

71. Solution: Two - Sample z - Test for the Difference Between Means 71

72. z 0-2.33 0.01 Solution: Two - Sample z - Test for the Difference Between Means H 0 : H a :   n 1 =, n 2 = Rejection Region: Test Statistic: 0.01 4700 4300 p 1 ≥ p 2 p 1 < p 2 -2.33 -3.46 Decision: At the 1% level of significance, there is enough evidence to conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo. Reject H 0 72

73. Section 8.4 Summary Performed a z-test for the difference between two population proportions p 1 and p 2 73

74. Slide 4- 74 © 2012 Pearson Education, Inc. Elementary Statistics: Picturing the World Fifth Edition by Larson and Farber Chapter 8: Hypothesis Testing with Two Samples

75. Slide 8- 75 © 2012 Pearson Education, Inc. Classify the pair of samples as dependent or independent. Sample 1: Exam scores for 25 students in Dr. Smith’s morning statistics class. Sample 2: Exam scores for 28 students in Dr. Smith’s afternoon statistics class. A. Dependent B. Independent

76. Slide 8- 76 © 2012 Pearson Education, Inc. Classify the pair of samples as dependent or independent. Sample 1: Exam scores for 25 students in Dr. Smith’s morning statistics class. Sample 2: Exam scores for 28 students in Dr. Smith’s afternoon statistics class. A. Dependent B. Independent

77. Slide 8- 77 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The mean cell phone bill for females (1) is higher than the mean cell phone for males (2). A. H 0 : μ 1 > μ 2 H a : μ 1 ≤ μ 2 B. H 0 : μ 1 < μ 2 H a : μ 1 ≥ μ 2 C. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 D. H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2

78. Slide 8- 78 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The mean cell phone bill for females (1) is higher than the mean cell phone for males (2). A. H 0 : μ 1 > μ 2 H a : μ 1 ≤ μ 2 B. H 0 : μ 1 < μ 2 H a : μ 1 ≥ μ 2 C. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 D. H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2

79. Slide 8- 79 © 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: μ 1 = μ 2 ; s 1 = 3.7 n 1 = 40 s 2 = 2.1 n 2 = 35 A. z = 3.25 B. z = 3.36 C. z = 2.45 D. z = 5.89

80. Slide 8- 80 © 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: μ 1 = μ 2 ; s 1 = 3.7 n 1 = 40 s 2 = 2.1 n 2 = 35 A. z = 3.25 B. z = 3.36 C. z = 2.45 D. z = 5.89

81. Slide 8- 81 © 2012 Pearson Education, Inc. Find the pooled estimate of the standard deviation for the following situation: s 1 = 1.3 n 1 = 12 s 2 = 0.8 n 2 = 15 A. B. C. D.

82. Slide 8- 82 © 2012 Pearson Education, Inc. Find the pooled estimate of the standard deviation for the following situation: s 1 = 1.3 n 1 = 12 s 2 = 0.8 n 2 = 15 A. B. C. D.

83. Slide 8- 83 © 2012 Pearson Education, Inc. Find the standardized test statistic t for the following situation (assume the populations are normally distributed and the population variances are equal): Claim: μ 1 = μ 2 ; s 1 = 1.3 n 1 = 12 s 2 = 0.8 n 2 = 15 A. t = –1.72 B. t = –1.63 C. t = –0.63 D. t = –0.69

84. Slide 8- 84 © 2012 Pearson Education, Inc. Find the standardized test statistic t for the following situation (assume the populations are normally distributed and the population variances are equal): Claim: μ 1 = μ 2 ; s 1 = 1.3 n 1 = 12 s 2 = 0.8 n 2 = 15 A. t = –1.72 B. t = –1.63 C. t = –0.69 D. t = –0.63

85. Slide 8- 85 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The mean score before and after a treatment are the same. A. H 0 : μ 1 = μ 2 H a : μ 1 ≠ 0 B. H 0 : μ 1 ≠ μ 2 H a : μ 1 = μ 2 C. H 0 : μ d = 0 H a : μ d ≠ 0 D. H 0 : μ d ≠ μ 2 H a : μ d = 0 Before758364 After544273

86. Slide 8- 86 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The mean score before and after a treatment are the same. A. H 0 : μ 1 = μ 2 H a : μ 1 ≠ 0 B. H 0 : μ 1 ≠ μ 2 H a : μ 1 = μ 2 C. H 0 : μ d = 0 H a : μ d ≠ 0 D. H 0 : μ d ≠ μ 2 H a : μ d = 0 Before758364 After544273

87. Slide 8- 87 © 2012 Pearson Education, Inc. Find the mean of the difference between the paired data entries in the dependent samples, A. 4.17 B. 4.83 C. 1.33 D. 5.5 Before758364 After544273

88. Slide 8- 88 © 2012 Pearson Education, Inc. Find the mean of the difference between the paired data entries in the dependent samples, A. 4.17 B. 4.83 C. 1.33 D. 5.5 Before758364 After544273

89. Slide 8- 89 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The proportion of female accountants (1) is less than the proportion of male accountants (2). A. H 0 : p 1 < p 2 H a : p 1 ≥ p 2 B. H 0 : p 1 ≤ p 2 H a : p 1 > p 2 C. H 0 : p 1 = p 2 H a : p 1 ≠ p 2 D. H 0 : p 1 ≥ p 2 H a : p 1 < p 2

90. Slide 8- 90 © 2012 Pearson Education, Inc. State the null and alternative hypotheses. The proportion of female accountants (1) is less than the proportion of male accountants (2). A. H 0 : p 1 < p 2 H a : p 1 ≥ p 2 B. H 0 : p 1 ≤ p 2 H a : p 1 > p 2 C. H 0 : p 1 = p 2 H a : p 1 ≠ p 2 D. H 0 : p 1 ≥ p 2 H a : p 1 < p 2

91. Slide 8- 91 © 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: p 1 > p 2 ; x 1 = 100 n 1 = 250 x 2 = 90 n 2 = 300 A. z = 2.46 B. z = 1.45 C. z = 0.35 D. z = 3.37

92. Slide 8- 92 © 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: p 1 > p 2 ; x 1 = 100 n 1 = 250 x 2 = 90 n 2 = 300 A. z = 2.46 B. z = 1.45 C. z = 0.35 D. z = 3.37