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Comparing Proportions of Drivers Stopped by the Police

Explore hypothesis testing for two proportions using sample data of black and white drivers being stopped by the police with significance level 0.05. Assess if the proportion of black drivers stopped is greater than white drivers.

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Comparing Proportions of Drivers Stopped by the Police

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  1. 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 Chapter 8 Inferences from Two Samples Chapter 8, Triola, Elementary Statistics, MATH 1342

  2. Section 8-1 & 8-2 Overview and Inferences about Two Proportions Created by Erin Hodgess, Houston, Texas Chapter 8, Triola, Elementary Statistics, MATH 1342

  3. There are many important and meaningful situations in which it becomes necessary to compare two sets of sample data. Overview (p.438) Chapter 8, Triola, Elementary Statistics, MATH 1342

  4. Assumptions (p.439) 1. We have proportions from two     independent simple random samples. 2. For both samples, the conditions np 5 and nq 5 are satisfied. Inferences about Two Proportions Chapter 8, Triola, Elementary Statistics, MATH 1342

  5. For population 1, we let: p1 = population proportion n1 = size of the sample x1 = number of successes in the sample ^ p1 = x1 (the sample proportion) q1 = 1 – p1 n1 ^ ^ ^ ^ Corresponding meanings are attached to p2, n2 , x2 , p2. and q2 , which come from population 2. Notation for Two Proportions Chapter 8, Triola, Elementary Statistics, MATH 1342

  6. The pooled estimate of p1 andp2 is denoted by p. x1 + x2 p = n1 + n2 • q =1 – p Pooled Estimate of p1 andp2 Chapter 8, Triola, Elementary Statistics, MATH 1342

  7. x2 x1 ^ ^ p1 p2 and = = n1 n2 x1 + x2 q =1–p p = and n1 + n2 Test Statistic for Two Proportions (p.441) For H0: p1 = p2 , H0: p1 = p2 , H0: p1=p2 H1: p1 p2 , H1: p1 < p2 , H1: p 1> p2 wherep1 –p 2 = 0 (assumed in the null hypothesis) Chapter 8, Triola, Elementary Statistics, MATH 1342

  8. ^ ^ ( p1 –p2 ) – ( p1 –p2 ) z= pq pq + n2 n1 Test Statistic for Two Proportions (p.441) For H0: p1 = p2 , H0: p1 = p2 , H0: p1=p2 H1: p1 p2 , H1: p1 < p2 , H1: p 1> p2 Chapter 8, Triola, Elementary Statistics, MATH 1342

  9. 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. (p.441) Chapter 8, Triola, Elementary Statistics, MATH 1342

  10. H0: p1 = p2, H1: p1 > p2 p = x1 + x2 = 24 + 147 = 0.106875 n1 + n2 200+1400 q = 1 – 0.106875 = 0.893125. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  11. z = (0.120 – 0.105) – 0 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 (0.106875)(0.893125) + (0.106875)(0.893125) 200 1400 ^ z = 0.64 n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  12. 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. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 (0.120 – 0.105) – 0.040 < ( p1– p2) < (0.120 – 0.105) + 0.040 –0.025 < ( p1– p2) < 0.055 ^ n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 Chapter 8, Triola, Elementary Statistics, MATH 1342

  13. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  14. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 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. 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  15. n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n1 200 n2= 1400 x2= 147 p2=x2= 147 = 0.105 ^ n2 1400 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. 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  16. ^ ^ ^ ^ ( p1 –p2 ) – E < ( p1 –p2 ) < ( p1 –p2 ) + E ^ ^ ^ ^ p1q1 p2 q2 whereE = z + n2 n1 Confidence Interval Estimate of p1 - p2 Chapter 8, Triola, Elementary Statistics, MATH 1342

  17. Example: For the sample data listed in Table 8-1, find a 90% confidence interval estimate of the difference between the two population proportions. (p.444) ^ ^ ^ ^ p1q1 p2 q2 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 E = z + n2 n1 ^ (0.105)(0.895) (.12)(.88)+ E = 1.645 n1 200 200 1400 n2= 1400 x2= 147 p2=x2= 147 = 0.105 E = 0.400 ^ n2 1400 Chapter 8, Triola, Elementary Statistics, MATH 1342

  18. Section 8-3 Inferences about Two Means: Independent Samples Created by Erin Hodgess, Houston, Texas Chapter 8, Triola, Elementary Statistics, MATH 1342

  19. 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. 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. Definitions Chapter 8, Triola, Elementary Statistics, MATH 1342

  20. 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 n1 > 30 and n2 > 30) or both samples come from populations having normal distributions. Assumptions (p.453) Chapter 8, Triola, Elementary Statistics, MATH 1342

  21. Test Statistic for Two Means: Hypothesis Tests (x1 –x2) – (µ1 – µ2) t= s2 s1. 2 2 + n2 n1 Chapter 8, Triola, Elementary Statistics, MATH 1342

  22. Test Statistic for Two Means: Hypothesis Tests Degrees of freedom: In this book we use this estimate: df = smaller of n1 – 1 and n2 – 1. P-value: Refer to Table A-3. Use the procedure summarized in Figure 7-6. Critical values: Refer to Table A-3. Chapter 8, Triola, Elementary Statistics, MATH 1342

  23. 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 70 73 x 418.5 403.7 s 45.5 30.6 McGwire Versus Bonds (p.455) Chapter 8, Triola, Elementary Statistics, MATH 1342

  24. McGwire Versus Bonds Chapter 8, Triola, Elementary Statistics, MATH 1342

  25. Claim: 12 Ho : 1 = 2 H1 : 12  = 0.05 McGwire Versus Bonds n1 – 1 = 69 n2 – 1 = 72 df = 69 t.025 = 1.994 Chapter 8, Triola, Elementary Statistics, MATH 1342

  26. Test Statistic for Two Means: McGwire Versus Bonds (x1 –x2) – (µ1 – µ2) t= s1. s2 2 2 + n2 n1 Chapter 8, Triola, Elementary Statistics, MATH 1342

  27. Test Statistic for Two Means: McGwire Versus Bonds (418.5 – 403.7) – 0 t= 30.62 45.52 + 70 73 = 2.273 Chapter 8, Triola, Elementary Statistics, MATH 1342

  28. McGwire Versus Bonds Claim: 12 Ho : 1 = 2 H1 : 12  = 0.05 Figure 8-2 Chapter 8, Triola, Elementary Statistics, MATH 1342

  29. McGwire Versus Bonds Claim: 12 Ho : 1 = 2 H1 : 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 Figure 8-2 Chapter 8, Triola, Elementary Statistics, MATH 1342

  30. (x1 –x2) – E < (µ1 – µ2) < (x1 –x2) + E s2 s1 2 2 whereE = t + n2 n1 Confidence Intervals Chapter 8, Triola, Elementary Statistics, MATH 1342

  31. 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. McGwire Versus Bonds (p.457) s2 s1 2 2 E = t + n2 n1 2 2 45.5 30.6 E= 1.994 + 73 70 E= 13.0 Chapter 8, Triola, Elementary Statistics, MATH 1342

  32. 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. McGwire Versus 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  33. Section 8-4 Inferences from Matched Pairs Created by Erin Hodgess, Houston, Texas Chapter 8, Triola, Elementary Statistics, MATH 1342

  34. 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. Assumptions (p.467) Chapter 8, Triola, Elementary Statistics, MATH 1342

  35. sd= standard deviation of the differences d for the paired sample data n = number of pairs of data. µd= 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 Chapter 8, Triola, Elementary Statistics, MATH 1342

  36. d –µd t= sd n where degrees of freedom = n – 1 Test Statistic for Matched Pairs of Sample Data (p.467) Chapter 8, Triola, Elementary Statistics, MATH 1342

  37. P-values and Critical Values Use Table A-3 (t-distribution) on p.736. Chapter 8, Triola, Elementary Statistics, MATH 1342

  38. degrees of freedom =n–1 d – E < µd < d + E sd whereE=t/2 n Confidence Intervals Chapter 8, Triola, Elementary Statistics, MATH 1342

  39. 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. (p.468) Are Forecast Temperatures Accurate? Chapter 8, Triola, Elementary Statistics, MATH 1342

  40. Are Forecast Temperatures Accurate? Chapter 8, Triola, Elementary Statistics, MATH 1342

  41. 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? Chapter 8, Triola, Elementary Statistics, MATH 1342

  42. H0: d = 0 H1: d  0 Are Forecast Temperatures Accurate? d –µd = –13.2 – 0 = –2.759 t= sd 10.7 5 n Chapter 8, Triola, Elementary Statistics, MATH 1342

  43. H0: d = 0 H1: d  0 Are Forecast Temperatures Accurate? d –µd = –13.2 – 0 = –2.759 t= sd 10.7 5 n Because the test statistic does not fall in the critical region, we fail to reject the null hypothesis. Chapter 8, Triola, Elementary Statistics, MATH 1342

  44. H0: d = 0 H1: d  0 Are Forecast Temperatures Accurate? d –µd = –13.2 – 0 = –2.759 t= sd 10.7 5 n 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  45. Are Forecast Temperatures Accurate? (p.469) Chapter 8, Triola, Elementary Statistics, MATH 1342

  46. Are Forecast Temperatures Accurate? (p.470) 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. Chapter 8, Triola, Elementary Statistics, MATH 1342

  47. sd n Are Forecast Temperatures Accurate? E=t/2 E= (2.776)( ) 10.7 5 = 13.3 Chapter 8, Triola, Elementary Statistics, MATH 1342

  48. Are Forecast Temperatures Accurate? d – E < d < d + E –13.2 – 13.3 < d < –13.2 + 13.3 –26.5 < d < 0.1 Chapter 8, Triola, Elementary Statistics, MATH 1342

  49. Are Forecast Temperatures Accurate? (p.470) In the long run, 95% of such samples will lead to confidence intervals that actually do contain the true population mean of the differences. Chapter 8, Triola, Elementary Statistics, MATH 1342

  50. Section 8-5 Comparing Variation in Two Samples Created by Erin Hodgess, Houston, Texas Chapter 8, Triola, Elementary Statistics, MATH 1342

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