4/7/11 Lecture 19 1 STOR 155 Introductory Statistics Lecture 19: Comparing two proportions Section 8.2 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL4/7/11 Lecture 19 2 Two populations: an extension of Lecture 18 • Population 1: with proportion • Population 2: with proportion • Interested in the difference • Sample 1: size count proportion • Sample 2: size count proportion • Consider the difference • Assume the two samples are independent, and both and are large. p1p2p1à p2D = pê1à pê2pê1= X1=n1pê2= X2=n2n1X1n2X2n1n24/7/11 Lecture 19 3 Useful probability facts The random variable D has approximately a normal distribution with mean and standard deviation SD(D) = An estimate of SD(D): p1à p2p1(1 à p1)=n1+ p2(1 à p2)=n2qSED= pê1(1 à pê1)=n1+ pê2(1 à pê2)=n2q4/7/11 Lecture 19 4 Confidence Interval for • Expression: where the margin of error – Confidence level C determines m = zãSEDzã[D à m; D + m]p1à p24/7/11 Lecture 19 5 Hypothesis testing for • We want to test versus some (1-sided or 2-sided) alternative. Recall the 4 steps … • Step 1: need to specify the alternative H0: p1= p2Hap1à p24/7/11 Lecture 19 6 Hypothesis testing for (continued) p1à p2z = D=SEDpS EDp= pê(1 à pê)(1=n1+ 1=n2)q• Step 2: Test statistic where and • Note: The pooled standard error is different from on page 3 pê =n1+n2X1+X2SEDpSED4/7/11 Lecture 19 7 Hypothesis testing (continued) • Step 3: The P-value will be equal to P(Z > z) for 1-sided (upper tail) P(Z < z) for 1-sided (lower tail) 2 P(Z > |z|) for 2-sided • Step 4: Compare the P-value with the significance level and draw your conclusion. Ha: p1> p2Ha: p1< p2Ha: p16=p2ë4/7/11 Lecture 19 8 Gender difference in frequent binge drinking ? Proportion of frequent binge drinkers • Population 1: (male college students) • Population 2: (female college students) • Sample 1: • Sample 2: • Total: p1p2pê1= 0:227pê2= 0:170n1= 7180;X1= 1630;n2= 9916;X2= 1684;n1+ n2= 17096;X1+ X2= 3314;pê = 0:1944/7/11 Lecture 19 9 Gender difference ? (continued) • Test vs • Test statistic: • P-value = P(Z > 9.34) = 0.00 • Reject • 95% CI for is (0.045, 0.069), where z = (0:227 à0:170) ä (0:194)(0:806)(1=7180 +1=9916)p= 9:34p1à p2H0: p1= p2Ha: p1> p2H0SED= (0:227)(0:773)=7180 +(0:170)(0:830)=9916p= 0:00622;m = zãSED= (1:96)(0:00622) = 0:0 1 24/7/11 Lecture 19 10 Take Home Message • CI for the difference • Hypothesis testing for comparing and … 4 steps • Note: different standard errors are used --- in CI in testing p1p1à
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