homework_12_ans_stud

homework_12_ans_stud - MTH 1210 Biostatistics Apr 1909...

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MTH 1210 Biostatistics Apr 19’09 Homework #12 ( Sampling Distribution of the Difference between Two Sample Proportions ) Use the handout # 13 and MS Excel program Three_distributions.xls, sheet 3 (see the textbook pg. 151) 5.6.1 . According to the 2000 US Census Bureau (A-8), in 2000, 9.5 percent of children in the state of Ohio were not covered by private or government health insurance. In the neighboring state of Pennsylvania, 4.9 percent of children were not covered by health insurance. Assume that these proportions are parameters for the child populations of the respective states. If a random sample of size 100 children is drawn from the Ohio population, and an independent random sample of size 120 children is drawn from the Pennsylvania population, what is the probability that the samples would yield a difference , 1 2 , of 0.09 or more ? p 1 = 0.09 5 p 2 = 0.049 σ 1 –  2 = 0.03532 8 [ σ 1 –  2 ] 2 = 0.001248 n 1 = 100 n 2 = 120 ( 1  2 ) 0 = 0.09 µ 1 –  2 = p 1 – p 2 = 0.046000 z 0 = [ ( 1 2 ) 0 – (p 1 – p 2 ) ]/σ 1 –  2 = 1.245467 P[ 1 2 ( 1 2 ) 0 ] = f(Z ≤ z 0 ) = 0.893520 P[ 1 2 > ( 1 2 ) 0 ] = f(Z > z 0 ) = 1 f(Z ≤ z 0 ) = 0.106480 ANS: 0.1065 5.6.2 . In the report cited in the problem 5.6.1 (A-8), the Census Bureau stated that for Americans in the age group 18 to 24 years, 64.8 percent had private health insurance. In the age group 25-34 years, the percentage was 72.1 . Assume that these percentages are the population parameters in
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This note was uploaded on 11/24/2011 for the course MATH 3000 taught by Professor Kzaer during the Spring '05 term at St. Johns College MD.

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homework_12_ans_stud - MTH 1210 Biostatistics Apr 1909...

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