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Section 9.2 Inferences About Two Proportions

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. p 1 First population proportion n 1 First sample size

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Section 9.2 Inferences About Two Proportions

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  1. Section 9.2Inferences 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

  2. p1 First population proportion n1 First sample size x1 Number of successes in first sample p1 First sample proportion Notation First Population

  3. p2 Second population proportion n2 Second sample size x2 Number of successes in second sample p2 Second sample proportion Notation Second Population

  4. The pooled sample proportion p x1 + x2 p = n1 + n2 Definition q =1 – p

  5. (1) Have two independentrandom samples (2) For each sample: The number of successes is at least 5 The number of failures is at least 5 Requirements Both requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

  6. Tests for Two Proportions The goal is to compare the two proportions H0:p1 =p2 H1:p1 p2 H0:p1 =p2 H1:p1 <p2 H0:p1 =p2 H1:p1 >p2 Two tailed Left tailed Right tailed Note: We only test the relation between p1 and p2 (not the actual numerical values)

  7. ^ ^ ( p1 –p2 ) – ( p1 –p2 ) z= pq pq + n2 n1 Finding the Test Statistic Note: p1 –p2 =0 according to H0 This equation is an altered form of the test statistic for a single proportion (see Ch. 8-3)

  8. Test Statistic Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)

  9. 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

  10. 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 airbagsp2: Proportion of fatalities with no airbags Claimp1 <p2 What we know: x1 =41 x2 =52 α=0.05 n1 =11541 n2 =9853 Claim:p1 <p2 Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements

  11. Given: x1 =41 x2 =52 α=0.05 n1 =11541 n2 =9853 Claim:p1 <p2 Example 1 Diagram z-dist. H0:p1 =p2 H1:p1 <p2 –zα = –1.645 z = –1.9116 Left-Tailed H1 =Claim Sample Proportions 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

  12. Stat → Proportions → Two sample → With summary Given: x1 =41 x2 =52 α=0.05 n1 =11541 n2 =9853 Claim:p1 <p2 Example 1 Sample 1: Number of successes: . Number of observations: Sample 2: Number of successes: . Number of observations: ●Hypothesis Test Diagram z-dist. H0:p1 =p2 H1:p1 <p2 Null: prop. diff.= Alternative –zα = –1.645 z = –1.9116 Left-Tailed H1 =Claim Using StatCrunch 41 11541 0 52 <  9853 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

  13. 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

  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 x2 =52 p1 = 0.003553n1 =11541 n2 =9853 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 x2 =52 p1 = 0.003553n1 =11541 n2 =9853 p2 = 0.005278 CI = (-0.003232, -0.000218 ) Note: CI negative implies p1–p2 is negative. This implies p1<p2

  16. Stat → Proportions → Two sample → With summary 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) Sample 1: Number of successes: . Number of observations: Sample 2: Number of successes: . Number of observations: ●Confidence Interval Level x1 =41 x2 =52n1 =11541 n2 =9853 Using StatCrunch 41 11541 0.9 52 9853 CI = (-0.003232, -0.000218 ) Note: CI negative implies p1–p2 is negative. This implies p1<p2

  17. 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

  18. 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: • Use a 0.01 significance level to test the claim proportions of subjects experiencing insomnia is the same for both groups. • Find the 99% confidence level estimate of the difference of the two proportions. Does it support the result of the test? Placebo 805 13 Chantix Treatment 129 19 Number in group Number experiencing insomnia • What we know: x1 =41 x2 =52 α=0.01 • n1 =129 n2 =9853 Claim:p1=p2 Note: Each sample has more than 5 successes and failures, thus fulfilling the requirements

  19. Given: x1 =19 x2 =13 α=0.01 • n1 =129 n2 =805 Claim:p1=p2 Example 3a Diagram z-dist. z = 7.602 H0:p1 =p2 H1:p1 ≠p2 -zα/2 = -2.576 zα/2 = 2.576 Two-Tailed H0 =Claim Sample Proportions 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.

  20. Stat → Proportions → Two sample → With summary • Given: x1 =19 x2 =13 α=0.01 • n1 =129 n2 =805 Claim:p1=p2 Example 3a Sample 1: Number of successes: . Number of observations: Sample 2: Number of successes: . Number of observations: ●Hypothesis Test Diagram z-dist. H0:p1 =p2 H1:p1 ≠p2 Null: prop. diff.= Alternative Two-Tailed H0 =Claim Using StatCrunch 19 129 0 13 ≠ 805 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.

  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 x2 =13 p1 = 0.14729n1 =129 n2 =805 p2 = 0.01615 CI = (0.0500, 0.2123 ) Note: CI does not contain 0 implies p1 and p2 have significant difference.

  22. Stat → Proportions → Two sample → With summary 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) Sample 1: Number of successes: . Number of observations: Sample 2: Number of successes: . Number of observations: ●Confidence Interval Level x1 =19 x2 =13n1 =129 n2 =805 Using StatCrunch 19 129 0.9 13 805 CI = (0.0500, 0.2123 ) Note: CI does not contain 0 implies p1 and p2 have significant difference.

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