1. Slide 1 Copyright © 2004 Pearson Education, Inc.

2. Slide 2 Copyright © 2004 Pearson Education, Inc. Chapter 8 Inferences from Two Samples 8-1 Overview 8-2 Inferences about Two Proportions 8-3 Inferences about Two Means: Independent Samples 8-4 Inferences about Matched Pairs 8-5 Comparing Variation in Two Samples

3. Slide 3 Scope The previous chapter presented hypothesis tests and confidence intervals for a single population parameter (mean, variance, or proportion). This chapter extends those results to the case of two independent populations. Most of the practical applications of the procedures in this chapter arise in the context of simple comparative experiments in which the objective is to study the difference in the parameters of the two populations.

4. Slide 4 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 8-1 & 8-2 Overview and Inferences about Two Proportions

5. Slide 5 Copyright © 2004 Pearson Education, Inc. Overview There are many important and meaningful situations in which it becomes necessary to compare two sets of sample data.

6. Slide 6 Copyright © 2004 Pearson Education, Inc. Inferences about Two Proportions Assumptions 1. We have proportions from two independent simple random samples. 2. For both samples, the conditions np  5 and nq  5 are satisfied.

7. Slide 7 Copyright © 2004 Pearson Education, Inc. population 2. The corresponding meanings are attached to p 2, n 2, x 2, p 2. and q 2, which come from ^^ Notation for Two Proportions For population 1, we let: p 1 = population proportion n 1 = size of the sample x 1 = number of successes in the sample p 1 = x 1 (the sample proportion) q 1 = 1 – p 1 ^^ n1n1 ^

8. Slide 8 Copyright © 2004 Pearson Education, Inc.  The pooled estimate of p 1 and p 2 is denoted by p. Pooled Estimate of p 1 and p 2  q = 1 – p = p n 1 + n 2 x 1 + x 2

9. Slide 9 Copyright © 2004 Pearson Education, Inc. Test Statistic for Two Proportions For H 0 : p 1 = p 2, H 0 : p 1  p 2, H 0 : p 1  p 2 H 1 : p 1  p 2, H 1 : p 1 p 2 + z =z = ( p 1 – p 2 ) – ( p 1 – p 2 ) ^ ^ n1n1 pq n2n2

10. Slide 10 Copyright © 2004 Pearson Education, Inc. Test Statistic for Two Proportions where p 1 – p 2 = 0 (assumed in the null hypothesis) = p1p1 ^ x1x1 n1n1 p2p2 ^ x2x2 n2n2 = and q = 1 – p n 1 + n 2 p = x 1 + x 2 For H 0 : p 1 = p 2, H 0 : p 1  p 2, H 0 : p 1  p 2 H 1 : p 1  p 2, H 1 : p 1 p 2

11. Slide 11 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.

12. Slide 12 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200n1n1 n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 ^ n2n2 n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 1400 ^ H 0 : p 1 = p 2, H 1 : p 1 > p 2 p = x 1 + x 2 = 24 + 147 = 0.106875 n 1 + n 2 200+1400 q = 1 – 0.106875 = 0.893125.

13. Slide 13 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200n1n1 n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 ^ n2n2 n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 1400 ^ z = (0.120 – 0.105) – 0 (0.106875)(0.893125) + (0.106875)(0.893125) 200 1400 z = 0.64

14. Slide 14 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200n1n1 n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 ^ n2n2 n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 1400 ^ z = 0.64 This is a right-tailed test, so the P- value is the area to the right of the test statistic z = 0.64. The P-value is 0.2611. Because the P-value of 0.2611 is greater than the significance level of  = 0.05, we fail to reject the null hypothesis.

15. Slide 15 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200n1n1 n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 ^ n2n2 n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 1400 ^ z = 0.64 Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support the claim that the proportion of black drivers stopped by police is greater than that for white drivers. This does not mean that racial profiling has been disproved. The evidence might be strong enough with more data.

16. Slide 16 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. 200n1n1 n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 ^ n2n2 n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 1400 ^

17. Slide 17 Copyright © 2004 Pearson Education, Inc. n1n1 n2n2 p 1 q 1 p 2 q 2 + ^ ^ ^ ^ where E = z   Confidence Interval Estimate of p 1 - p 2 ( p 1 – p 2 ) – E < ( p 1 – p 2 ) < ( p 1 – p 2 ) + E ^^^^

18. Slide 18 Copyright © 2004 Pearson Education, Inc. Example: For the sample data listed in Table 8-1, find a 90% confidence interval estimate of the difference between the two population proportions. n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 n1n1 200 ^ n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 n2n2 1400 ^ n1n1 n2n2 + p 1 q 1 p 2 q 2 ^ ^ ^ ^ E = z   E = 1.645 200 1400 (.12)(.88)+ (0.105)(0.895) E = 0.400

19. Slide 19 Copyright © 2004 Pearson Education, Inc. n 1 = 200 x 1 = 24 p 1 = x 1 = 24 = 0.120 n1n1 200 ^ n 2 = 1400 x 2 = 147 p 2 = x 2 = 147 = 0.105 n2n2 1400 ^ (0.120 – 0.105) – 0.040 < ( p 1 – p 2 ) < (0.120 – 0.105) + 0.040 –0.025 < ( p 1 – p 2 ) < 0.055 Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.

20. Slide 20 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 8-3 Inferences about Two Means: Independent Samples

21. Slide 21 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Inferences About Two Means: a)Independent Samples b)Dependent Samples (paired)

22. Slide 22 INFERENCE FOR A DIFFERENCE IN MEANS OF TWO NORMAL DISTRIBUTIONS, VARIANCES KNOWN

23. Slide 23 Assumptions

24. Slide 24

25. Slide 25 We are interested in testing that the difference in means μ 1 – μ 2 of two normal populations is equal to a specified value Δ o. Thus the null hypothesis is: H o : μ 1 – μ 2 = Δ o In many cases, we will specify Δ o = 0, so that we are testing the equality of the two means; i.e. H o : μ 1 = μ 2

26. Slide 26

27. Slide 27 Example 10-1 (Douglas)

28. Slide 28 …Example 10-1 (Douglas)

29. Slide 29 …Example 10-1 (Douglas)

30. Slide 30 Copyright © 2004 Pearson Education, Inc. Notes on the previous procedure In reality, the population standard deviations σ 1 and σ 2 are almost never known, The best approach is to assume that σ 1 and σ 2 are unknown, Do not assume that σ 1 = σ 2, although some books do this. Use the t-test statistic as shown in Figure 8-3 i.e. we should always use the t-distribution unless otherwise instructed.

31. Slide 31

32. Slide 32 Copyright © 2004 Pearson Education, Inc. Definitions Two Samples: Independent The sample values selected from one population are not related or somehow paired with the sample values selected from the other population (the previous paint example) If the values in one sample are related to the values in the other sample, the samples are dependent. Such samples are often referred to as matched pairs or paired samples.

33. Slide 33 Copyright © 2004 Pearson Education, Inc.

34. Slide 34 Copyright © 2004 Pearson Education, Inc. Assumptions (For independent samples) 1. The two samples are independent. 2. Both samples are simple random samples. 3. Either or both of these conditions are satisfied: The two sample sizes are both large (with n 1 > 30 and n 2 > 30) or both samples come from populations having normal distributions.

35. Slide 35 Copyright © 2004 Pearson Education, Inc. (x 1 – x 2 ) – (µ 1 – µ 2 ) t = n1n1 n2n2 + s1.s1. s2s2 22 Hypothesis Tests Test Statistic for Two Means:

36. Slide 36 Copyright © 2004 Pearson Education, Inc. Degrees of freedom: In this book we use this estimate: df = smaller of n 1 – 1 and n 2 – 1. P-value:Refer to Table A-3. Use the procedure summarized in Figure 7-6. Critical values:Refer to Table A-3. Hypothesis Tests Test Statistic for Two Means:

37. Slide 37 Copyright © 2004 Pearson Education, Inc. Data Set 30 in Appendix B includes the distances of the home runs hit in record-setting seasons by Mark McGwire and Barry Bonds. Sample statistics are shown. Use a 0.05 significance level to test the claim that the distances come from populations with different means. McGwire Versus Bonds

38. Slide 38 Copyright © 2004 Pearson Education, Inc. McGwire Versus Bonds Data Set 30 in Appendix B includes the distances of the home runs hit in record-setting seasons by Mark McGwire and Barry Bonds. Sample statistics are shown. Use a 0.05 significance level to test the claim that the distances come from populations with different means. McGwire Bonds n 7073 x 418.5 403.7 s 45.5 30.6

39. Slide 39 Copyright © 2004 Pearson Education, Inc. McGwire Versus Bonds

40. Slide 40 Copyright © 2004 Pearson Education, Inc. McGwire Versus Bonds

41. Slide 41 Copyright © 2004 Pearson Education, Inc. Claim:  1   2 H o :  1 =  2 H 1 :  1   2  = 0.05 n 1 – 1 = 69 n 2 – 1 = 72 df = 69 t.025 = 1.994 McGwire Versus Bonds

42. Slide 42 Copyright © 2004 Pearson Education, Inc. Test Statistic for Two Means: (x 1 – x 2 ) – (µ 1 – µ 2 ) t = n1n1 n2n2 + s1.s1. s2s2 2 2 McGwire Versus Bonds

43. Slide 43 Copyright © 2004 Pearson Education, Inc. Test Statistic for Two Means: (418.5 – 403.7 ) – 0 t = 70 + 45.5 2 30.6 2 73 = 2.273 McGwire Versus Bonds

44. Slide 44 Copyright © 2004 Pearson Education, Inc. Figure 8-2 McGwire Versus Bonds Claim:  1   2 H o :  1 =  2 H 1 :  1   2  = 0.05

45. Slide 45 Copyright © 2004 Pearson Education, Inc. Figure 8-2 McGwire Versus Bonds Claim:  1   2 H o :  1 =  2 H 1 :  1   2  = 0.05 There is significant evidence to support the claim that there is a difference between the mean home run distances of Mark McGwire and Barry Bonds. Reject Null

46. Slide 46 Copyright © 2004 Pearson Education, Inc. Confidence Intervals (x 1 – x 2 ) – E < (µ 1 – µ 2 ) < (x 1 – x 2 ) + E + n1n1 n2n2 s1s1 s2s2 where E = z   2 2

47. Slide 47 Copyright © 2004 Pearson Education, Inc. Using the sample data given in the preceding example, construct a 95%confidence interval estimate of the difference between the mean home run distances of Mark McGwire and Barry Bonds. n1n1 n2n2 + s1s1 s2s2 E = t   2 2 E = 1.994 70 73 + 45.530.6 2 2 E = 13.0 McGwire Versus Bonds

48. Slide 48 Copyright © 2004 Pearson Education, Inc. Using the sample data given in the preceding example, construct a 95%confidence interval estimate of the difference between the mean home run distances of Mark McGwire and Barry Bonds. (418.5 – 403.7) – 13.0 < (  1 –  2 ) < (418.5 – 403.7) + 13.0 1.8 < (  1 –  2 ) < 27.8 We are 95% confident that the limits of 1.8 ft and 27.8 ft actually do contain the difference between the two population means. McGwire Versus Bonds

49. Slide 49 Copyright © 2004 Pearson Education, Inc. McGwire Versus Bonds

50. Slide 50 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 8-4 Inferences from Matched Pairs

51. Slide 51 Copyright © 2004 Pearson Education, Inc. Assumptions 1. The sample data consist of matched pairs. 2. The samples are simple random samples. 3. Either or both of these conditions is satisfied: The number of matched pairs of sample data is ( n > 30) or the pairs of values have differences that are from a population having a distribution that is approximately normal.

52. Slide 52 Copyright © 2004 Pearson Education, Inc. s d = standard deviation of the differences d for the paired sample data n = number of pairs of data. µ d = mean value of the differences d for the population of paired data d = mean value of the differences d for the paired sample data (equal to the mean of the x – y values) Notation for Matched Pairs

53. Slide 53 Copyright © 2004 Pearson Education, Inc. t =t = d – µ d sdsd n where degrees of freedom = n – 1 Test Statistic for Matched Pairs of Sample Data

54. Slide 54 Copyright © 2004 Pearson Education, Inc. P-values and Critical Values Use Table A-3 (t-distribution).

55. Slide 55 Copyright © 2004 Pearson Education, Inc. Confidence Intervals degrees of freedom = n –1 where E = t  / 2 sdsd n d – E < µ d < d + E

56. Slide 56 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate? Using Table A-2 consists of five actual low temperatures and the corresponding low temperatures that were predicted five days earlier. The data consist of matched pairs, because each pair of values represents the same day. Use a 0.05 significant level to test the claim that there is a difference between the actual low temperatures and the low temperatures that were forecast five days earlier.

57. Slide 57 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate?

58. Slide 58 Copyright © 2004 Pearson Education, Inc. d = –13.2 s = 10.7 n = 5 t  /2 = 2.776 (found from Table A-3 with 4 degrees of freedom and 0.05 in two tails) Are Forecast Temperatures Accurate?

59. Slide 59 Copyright © 2004 Pearson Education, Inc. H 0 :  d = 0 H 1 :  d  0 t =t = d – µ d n sdsd = –13.2 – 0 = –2.759 10.7 5 Are Forecast Temperatures Accurate?

60. Slide 60 Copyright © 2004 Pearson Education, Inc. H 0 :  d = 0 H 1 :  d  0 t =t = d – µ d n sdsd = –13.2 – 0 = –2.759 10.7 5 Because the test statistic does not fall in the critical region, we fail to reject the null hypothesis. Are Forecast Temperatures Accurate?

61. Slide 61 Copyright © 2004 Pearson Education, Inc. H 0 :  d = 0 H 1 :  d  0 t =t = d – µ d n sdsd = –13.2 – 0 = –2.759 10.7 5 The sample data in Table 8-2 do not provide sufficient evidence to support the claim that actual and five-day forecast low temperatures are different. Are Forecast Temperatures Accurate?

62. Slide 62 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate?

63. Slide 63 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate?

64. Slide 64 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate? Using the P-Value Method

65. Slide 65 Copyright © 2004 Pearson Education, Inc. Using the same sample matched pairs in Table 8-2, construct a 95% confidence interval estimate of  d, which is the mean of the differences between actual low temperatures and five-day forecasts. Are Forecast Temperatures Accurate?

66. Slide 66 Copyright © 2004 Pearson Education, Inc. E = t/2E = t/2 sdsd n E = (2.776) ( ) 10.7 5 = 13.3 Are Forecast Temperatures Accurate?

67. Slide 67 Copyright © 2004 Pearson Education, Inc. d – E <  d < d + E –13.2 – 13.3 <  d < –13.2 + 13.3 –26.5 <  d < 0.1 Are Forecast Temperatures Accurate?

68. Slide 68 Copyright © 2004 Pearson Education, Inc. Are Forecast Temperatures Accurate?

69. Slide 69 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 8-5 Comparing Variation in Two Samples

70. Slide 70 Copyright © 2004 Pearson Education, Inc. Measures of Variation s = standard deviation of sample  = standard deviation of population s 2 = variance of sample  2 = variance of population

71. Slide 71 Copyright © 2004 Pearson Education, Inc. Assumptions 1. The two populations are independent of each other. 2. The two populations are each normally distributed.

72. Slide 72 Copyright © 2004 Pearson Education, Inc. s 1 =larger of the two sample variances n 1 =size of the sample with the larger variance 2 Notation for Hypothesis Tests with Two Variances

73. Slide 73 Copyright © 2004 Pearson Education, Inc. Notation for Hypothesis Tests with Two Variances s 1 =larger of the two sample variances n 1 =size of the sample with the larger variance  1 =variance of the population from which the sample with the larger variance was drawn The symbols s 2, n 2, and  2 are used for the other sample and population. 2 2 22

74. Slide 74 Copyright © 2004 Pearson Education, Inc. Critical Values: Using Table A-5, we obtain critical F values that are determined by the following three values: s1 s1 F = s2 s2 2 2 Test Statistic for Hypothesis Tests with Two Variances 1. The significance level . 2. Numerator degrees of freedom (df 1 ) = n 1 – 1 3. Denominator degrees of freedom (df 2 ) = n 2 – 1

75. Slide 75 Copyright © 2004 Pearson Education, Inc. F Distribution

76. Slide 76 F Distribution

77. Slide 77 Copyright © 2004 Pearson Education, Inc.  All one-tailed tests will be right-tailed.  All two-tailed tests will need only the critical value to the right.  When degrees of freedom is not listed exactly, use the critical values on either side as an interval. Use interpolation only if the test statistic falls within the interval.

78. Slide 78 Copyright © 2004 Pearson Education, Inc. If the two populations do have equal variances, then F = will be close to 1 because and are close in value. s1 s1 s 2 2 2 s1s1 s2s2 2 2

79. Slide 79 Copyright © 2004 Pearson Education, Inc. If the two populations have radically different variances, then F will be a large number. Remember, the larger sample variance will be s 1. 2

80. Slide 80 Copyright © 2004 Pearson Education, Inc. Consequently, a value of F near 1 will be evidence in favor of the conclusion that  1 =  2. 2 2 But a large value of F will be evidence against the conclusion of equality of the population variances.

81. Slide 81 Copyright © 2004 Pearson Education, Inc. Data Set 17 in Appendix B includes the weights (in pounds) of samples of regular Coke and regular Pepsi. Sample statistics are shown. Use the 0.05 significance level to test the claim that the weights of regular Coke and the weights of regular Pepsi have the same standard deviation. Regular Coke Regular Pepsi n3636 x0.816820.82410 s0.0075070.005701 Coke Versus Pepsi

82. Slide 82 Copyright © 2004 Pearson Education, Inc. Claim:  1 =  2 H o :  1 =  2 H 1 :  1   2  = 0.05 Coke Versus Pepsi 22 22 22 Value of F = s1 s1 s2 s2 2 2 0.005701 2 0.007507 2 = = 1.7339

83. Slide 83 Copyright © 2004 Pearson Education, Inc. Coke Versus Pepsi

84. Slide 84 Copyright © 2004 Pearson Education, Inc. Claim:  1 =  2 H o :  1 =  2 H 1 :  1   2  = 0.05 Coke Versus Pepsi 22 22 22 There is not sufficient evidence to warrant rejection of the claim that the two variances are equal.