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Testing the Difference between Proportions Section 11.3 Assumptions Randomly Selected Samples Approximately normal since 1 1 ≥ 10 1 1 − 1 ≥ 10 and 2 2 ≥ 10 2 1 − 2 ≥ 10 Independent (at least 10n in population) Normally the Standard Deviation of Statistic is p1q1 p2 q2 n1 n2 But, Since we claim p1 p2 in the Ho We can combine the values to form one proportion: x1 x2 pc n1 n2 And the Standard Deviation of Statistic becomes 1 1 pc qc n1 n2 Can you find this on the formula Sheet? They use a combined p. A sample of 50 randomly selected men with high triglyceride levels consumed 2 tablespoons of oat bran daily for six weeks. After six weeks, 60% of the men had lowered their triglyceride level. A sample of 80 men consumed 2 tablespoons of wheat bran for six weeks. After six weeks, 25% had lower triglyceride levels. Is there a significant difference in the two proportions at the 0.01 significance level? To calculate pc we need to find x1 and x2. x1 0.60(50) 30 x2 0.25(80) 20 So….. 30 20 50 pc 0.385 50 80 130 Parameter: p1 proportion of men with high triglyceride levels who ate oat bran. p2 proportion of men with high triglyceride levels who ate wheat bran. p p difference in prop. of men w/high triglycerides between those who ate oat & bran 2 1 Hypothesis: H o : p1 p2 H A : p1 p2 Assumptions: * Randomly Selected Samples * Approximately Normal since n1 p1 50(0.6) 30 5 n1q1 50(.4) 20 5 and n2 p2 80(0.25) 20 5 n2 q2 80(0.75) 60 5 * Independent – (at least 500 men eat oat and 800 eat wheat bran Name of Test: 2-Proportion Z-Test ) ^ ^ p1 p 2 p1 p2 z 1 1 pc qc n1 n2 z .6 .25 0 1 1 .385 * .615 * 50 80 z 3.99 P Value P( z 3.99)* 2 P Value ncdf (3.99, )* 2 P Value 0 Reject the Ho since the P-Value(0) < (0.05) There is sufficient evidence to support the claim that there is a difference in the proportion of men who lowered their triglycerides by eating oat bran and the proportion who lowered their triglycerides by eating wheat bran. In a sample of 100 store customers, 43 used a Mastercard. In another sample of 100, 58 used a Visa card. Is the proportion of customers who use Mastercard less than those using Visa? p1 prop using mastercard p2 prop using visa p1 p2 diff in prop using mastercard & visa H o : p1 p2 H A : p1 p2 43 58 100 100 101 0.505 200 pc Assumptions: 1. Randomly Selected Samples 2. Approx. Normal n1 p1 100(.43) 43 5 n1q1 100(.57) 57 5 n2 p2 100(.58) 58 5 n q 100(.42) 42 5 2 2 3. Independent (at least 1000 of each) ^ ^ p1 p 2 p1 p2 .43 .58 0 z 2.12 1 1 1 1 . 505 * . 495 * pc qc 100 100 n1 n2 P Val P( z 2.12) ncdf ( , 2.12) 0.017 Reject the Ho since the p-val(.017) < (0.05) There is sufficient evidence to support the claim that the proportion using mastercard is less than the proportion using visa. So how would we find a confidence interval? PANIC! In a sample of 80 Americans, 55% wished that they were rich. In a sample of 90 Europeans, 45% wished that they were rich. Is there a difference in the proportions. Find and interpret the 95% confidence interval for the difference of the two proportions. p1 prop Americans who wish to be rich p2 prop Europeans who wish to be rich p1 p2 diff in prop Am & Eurp who wish to be rich Assumptions: 1. Randomly Selected Samples 2. Approx Norm n1 p1 80(.55) 44 5 n1q1 80(.45) 36 5 n2 p2 90(.45) 41 5 n q 90(.55) 50 5 2 2 3. Independent (at least 800 Am and 900 Europeans. ^ ^ p1 p 2 z 2 p1q1 p2 q2 n1 n2 .55(.45) .45(.55) .55 .45 1.96 80 90 .10 .150 .05, .25 We’re 95% confident that the difference in proportion of Americans who wish to be rich and the proportion of Europeans who wish to be rich is between -.05 and .25. In fact, since this interval contains 0, there is no significant difference. Homework Worksheet