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

Overview and Inferences about Two Proportions Section 8-1 & 8-2 Overview and Inferences about Two Proportions Created by Erin Hodgess, Houston, Texas

Overview There are many important and meaningful situations in which it becomes necessary to compare two sets of sample data. page 438 of text Examples in the discussion 450

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. page 458 of text 451

Notation for Two Proportions 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 population 2. The corresponding meanings are attached to p2, n2 , x2 , p2. and q2 , which come from ^ 451

Pooled Estimate of p1 and p2 The pooled estimate of p1 and p2   is denoted by p. = p n1 + n2 x1 + x2 q = 1 – p 452

Test Statistic for Two Proportions For H0: p1 = p2 , H0: p1  p2 , H0: p1 p2 H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2 + z = ( p1 – p2 ) – ( p1 – p2 ) ^ n1 pq n2 Example given at the bottom of page 460-462. 452

Test Statistic for Two Proportions For H0: p1 = p2 , H0: p1  p2 , H0: p1 p2 H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2 where p1 – p 2 = 0 (assumed in the null hypothesis) = p1 ^ x1 n1 p2 x2 n2 and Example given at the bottom of page 460-462. and q = 1 – p n1 + n2 p = x1 + x2 452

Example: For the sample data listed in Table 8-1, use a 0 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. 449

Example: For the sample data listed in Table 8-1, use a 0 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. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400 H0: p1 = p2, H1: p1 > p2 p = x1 + x2 = 24 + 147 = 0.106875 n1 + n2 200+1400 q = 1 – 0.106875 = 0.893125. 453

Example: For the sample data listed in Table 8-1, use a 0 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. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400 z = (0.120 – 0.105) – 0 (0.106875)(0.893125) + (0.106875)(0.893125) 200 1400 z = 0.64 454

Example: For the sample data listed in Table 8-1, use a 0 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. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 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. 454

Example: For the sample data listed in Table 8-1, use a 0 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. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 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. 454

Example: For the sample data listed in Table 8-1, use a 0 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. 200 n1 n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 ^ n2 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 1400 454

Confidence Interval Estimate of p1 - p2 ( p1 – p2 ) – E < ( p1 – p2 ) < ( p1 – p2 ) + E ^ n1 n2 p1 q1 p2 q2 + ^ where E = z Example at the bottom of page 463 Rationale for the procedures of this section on page 464-465. 456

Example: For the sample data listed in Table 8-1, find a 90% confidence interval estimate of the difference between the two population proportions. ^ ^ ^ ^ n1= 200 x1 = 24 p1 = x1 = 24 = 0.120 E = z p1 q1 p2 q2 + n1 n2 ^ (.12)(.88)+ (0.105)(0.895) E = 1.645 n1 200 200 1400 n2 = 1400 x2 = 147 p2 = x2 = 147 = 0.105 E = 0.400 ^ n2 1400 456

(0.120 – 0.105) – 0.040 < ( p1– p2) < (0.120 – 0.105) + 0.040 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 456