Section 9.2 Inferences About Two Proportions Objective Compare the proportions of two populations using two samples from each population. Hypothesis Tests and Confidence Intervals of two proportions use the z-distribution 1 Notation First Population p1 First population proportion n1 First sample size x1 Number of successes in first sample p1 First sample proportion 2 Notation Second Population p2 Second population proportion n2 Second sample size x2 Number of successes in second sample p2 Second sample proportion 3 Definition The pooled sample proportion p x1 + x2 p= n +n 1 2 q =1–p 4 Requirements (1) Have two independent random samples (2) For each sample: The number of successes is at least 5 The number of failures is at least 5 Both requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval 5 Tests for Two Proportions The goal is to compare the two proportions H0 : p1 = p2 H1 : p1 p2 Two tailed H0 : p 1 = p 2 H0 : p1 = p2 H1 : p 1 < p 2 H1 : p1 > p2 Left tailed Right tailed Note: We only test the relation between p1 and p2 (not the actual numerical values) 6 Finding the Test Statistic z= ^ )–(p –p ) ( p^1 – p 2 1 2 Note: p1 – pq pq n 1 + n2 p2 =0 according to H0 This equation is an altered form of the test statistic for a single proportion (see Ch. 8-3) 7 Test Statistic Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics) 8 Steps for Performing a Hypothesis Test on Two Proportions • Write what we know • State H0 and H1 • Draw a diagram • Calculate the sample and pooled proportions • Find the Test Statistic • Find the Critical Value(s) • State the Initial Conclusion and Final Conclusion Note: Same process as in Chapter 8 9 Example 1 The table below lists results from a simple random sample of front-seat occupants involved in car crashes. Use a 0.05 significance level to test the claim that the fatality rate of occupants is lower for those in cars equipped with airbags. p1 : Proportion of fatalities with airbags p2 : Proportion of fatalities with no airbags What we know: x1 = 41 n1 = 11541 x2 = 52 n2 = 9853 Claim p1 < p2 α = 0.05 Claim: p1 < p2 Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements 10 Given: Example 1 x2 = 52 n2 = 9853 α = 0.05 Claim: p1 < p2 Diagram H0 : p1 = p2 H1 : p1 < p2 x1 = 41 n1 = 11541 Left-Tailed H1 = Claim Sample Proportions z = –1.9116 z-dist. –zα = –1.645 Pooled Proportion Test Statistic Critical Value (Using StatCrunch) Initial Conclusion: Since z is in the critical region, reject H0 Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts 11 Example 1 Given: α = 0.05 Claim: p1 < p2 x2 = 52 n2 = 9853 Diagram H0 : p1 = p2 H1 : p1 < p2 x1 = 41 n1 = 11541 Left-Tailed H1 = Claim z = –1.9116 z-dist. –zα = –1.645 Using StatCrunch Stat → Proportions → Two sample → With summary Sample 1: Number of successes: . 41 ● Hypothesis Test Number of observations: 11541 Null: prop. diff.= Sample 2: Number of successes: . 52 Alternative Number of observations: 9853 0 < P-value = 0.028 Initial Conclusion: Since P-value is less than α (with α = 0.05), reject H0 Final Conclusion: We Accept the claim the fatality rate of occupants is lower for those who wear seatbelts 12 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of p1–p2 CI = ( (p1–p2) – E, (p1–p2) + E ) Where 13 Example 2 Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2) x1 = 41 n1 = 11541 x2 = 52 n2 = 9853 p1 = 0.003553 p2 = 0.005278 CI = (-0.003232, -0.000218 ) Note: CI negative implies p1–p2 is negative. This implies p1<p2 14 Example 2 Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2) x1 = 41 n1 = 11541 x2 = 52 n2 = 9853 p1 = 0.003553 p2 = 0.005278 CI = (-0.003232, -0.000218 ) Note: CI negative implies p1–p2 is negative. This implies p1<p2 15 Example 2 Use the same sample data in Example 1 to construct a 90% Confidence Interval Estimate of the difference between the two population proportions (p1–p2) x1 = 41 n1 = 11541 x2 = 52 n2 = 9853 Using StatCrunch Stat → Proportions → Two sample → With summary Sample 1: Number of successes: . 41 ● Confidence Interval Number of observations: 11541 Level 0.9 Sample 2: Number of successes: . 52 Number of observations: 9853 CI = (-0.003232, -0.000218 ) Note: CI negative implies p1–p2 is negative. This implies p1<p2 16 Interpreting Confidence Intervals If a confidence interval limits does not contain 0, it implies there is a significant difference between the two proportions (i.e. p1 ≠ p2). Thus, we can interpret a relation between the two proportions from the confidence interval. In general: • If p1 = p2 then the CI should contain 0 • If p1 > p2 then the CI should be mostly positive • If p1 > p2 then the CI should be mostly negative 17 Example 3 Drug Clinical Trial Chantix is a drug used as an aid to stop smoking. The number of subjects experiencing insomnia for each of two treatment groups in a clinical trial of the drug Chantix are given below: Chantix Treatment Placebo Number in group 129 805 Number experiencing insomnia 19 13 (a) Use a 0.01 significance level to test the claim proportions of subjects experiencing insomnia is the same for both groups. (b) Find the 99% confidence level estimate of the difference of the two proportions. Does it support the result of the test? What we know: x1 = 41 n1 = 129 x2 = 52 n2 = 9853 α = 0.01 Claim: p1= p2 Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements 18 Example 3a Given: x1 = 19 n1 = 129 Diagram H0 : p1 = p2 H1 : p1 ≠ p2 x2 = 13 n2 = 805 Two-Tailed H0 = Claim Sample Proportions -zα/2 = -2.576 α = 0.01 Claim: p1= p2 z-dist. z = 7.602 zα/2 = 2.576 Pooled Proportion Test Statistic Critical Value (Using StatCrunch) Initial Conclusion: Since z is in the critical region, reject H0 Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups. 19 Example 3a Given: x1 = 19 n1 = 129 Diagram H0 : p1 = p2 H1 : p1 ≠ p2 α = 0.01 Claim: p1= p2 x2 = 13 n2 = 805 z-dist. Two-Tailed H0 = Claim Using StatCrunch Stat → Proportions → Two sample → With summary Sample 1: Number of successes: . 19 Number of observations: 129 Sample 2: Number of successes: . 13 Number of observations: 805 ● Hypothesis Test Null: prop. diff.= Alternative 0 ≠ P-value < 0.0001 i.e. the P-value is very small Initial Conclusion: Since the P-value is less than α (0.01), reject H0 Final Conclusion: We Reject the claim the proportions of the subjects experiencing insomnia is the same in both groups. 20 Example 3b Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference between the two population proportions (p1–p2) x1 = 19 n1 = 129 x2 = 13 n2 = 805 p1 = 0.14729 p2 = 0.01615 CI = (0.0500, 0.2123 ) Note: CI does not contain 0 implies p1 and p2 have significant difference. 21 Example 3b Use the same sample data in Example 3 to construct a 99% Confidence Interval Estimate of the difference between the two population proportions (p1–p2) x1 = 19 n1 = 129 x2 = 13 n2 = 805 Using StatCrunch Stat → Proportions → Two sample → With summary Sample 1: Number of successes: . 19 Number of observations: 129 Sample 2: Number of successes: . 13 Number of observations: 805 ● Confidence Interval Level 0.9 CI = (0.0500, 0.2123 ) Note: CI does not contain 0 implies p1 and p2 have significant difference. 22