STAT 400 Lecture AL1 Answers for 7.3, 8.3 (Part 2) Spring 2015 Dalpiaz Consider two dichotomous populations, with “success” proportions p 1 and p 2 , respectively. Consider the sample proportions 111 ˆnxp = and 222 ˆnxp= where n 1 and n 2 are the sample sizes and x 1 and x 2 are the numbers of “successes” in the two samples from populations 1 and 2, respectively. If n 1 and n 2 are large, then ( ) 21ˆˆ pp − is approximately normal with mean p 1 – p 2 and standard deviation ( ) ( ) 222111 11nppnpp−⋅−⋅+. The confidence interval for the difference between two population proportions p 1 – p 2 is ( )( ) ( ) ˆˆˆˆ ˆˆ 222111221 11αnppnppzpp−⋅−⋅⋅+±− H 0 : p 1 = p 2 H 1 : p 1 < p 2 H 1 : p 1 > p 2 H 1 : p 1 ≠ p 2 Test Statistic: ( ) nnppppz=+−−⋅⋅ 21 21111 ˆˆˆˆ, where 2122112121 ˆˆˆnnpnpnnnxxp++=++=⋅⋅.1. In a comparative study of two new drugs, A and B, 120 patients were treated with drug A and 150 patients with drug B, and the following results were obtained. Drug A Drug B Cured 78 111 Not cured 42 39 Total 120 150 a) Construct a 95% confidence interval for the difference in the cure rates of the two drugs. 12078A ˆ=p = 0.65 150111B ˆ=p = 0.74 95% confidence level α = 0.05 α2 = 0.025. 2zα = 1.960. ( )( )( ) 12065.0165.015074.0174.0 96.1 65.074.0 −+−±−⋅⋅⋅ 0.09 ± 0.11 ( – 0.02 , 0.20 ) b) We wish to test whether drug B has a higher cure rate than drug A. Find the p-value of the appropriate test. Claim: p A < p B H 0 : p A = p B vs. H 1 : p A < p B 270189150120111782121 ˆ=++=++=nnxxp = 0.70.Test Statistic: Z = +−⋅⋅1501120130.070.0 74.065.0 = – 1.60. P-value: Left – tailed. P-value = P( Z < – 1.60 ) =
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