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York University (32,577)
MATH 2565 (19)
Lecture

# PPS9_Solution_.doc

4 Pages
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School
York University
Department
Mathematics and Statistics
Course
MATH 2565
Professor
Jagmohan Chawla
Semester
Winter

Description
owner Page 149 10/01/2013owner Page 149 10/01/2013123520813.doc PPS 9 (Solution) Problem19.23. Rickets. a) Independence Assumption: It is reasonable to think that the randomly selected children are mutually independent in regards to vitamin D deficiency. Randomization Condition: The 2,700 children were chosen at random. 10% Condition: 2,700 children are less than 10% of all English children. Success/Failure Condition: np= (2,700)(0.20) = 540 and nq= (2,700)(0.80) = 2160 are both greater than 10, so the sample is large enough. Since the conditions are met, we can use a one-proportion z- interval to estimate the proportion of the English children with vitamin D deficiency. ∗ pqˆ (0.2)0.80) p ± z = (0.2)± 2.326 = (18.2%, 2.8%) n 2700 b) We are 98% confident that between 18.2% and 21.8% of English children are deficient in vitamin D. c) About 98% of random samples of size 2,700 will produce confidence intervals that contain the true proportion of English children that are deficient in vitamin D. 24. Pregnancy. a) Independence Assumption: There is no reason to believe that one woman’s ability to conceive would affect others. Randomization Condition: These women are not chosen at random. Assume that they are representative of all women under 40 that had previously been unable to conceive. 10% Condition: 207 women is less than 10% of all such women. Success/Failure Condition: np= 49 and nqˆ= 158 are both greater than 10, so the sample is large enough. Since the conditions are met, we can use a one-proportion z- interval to estimate the proportion of the births to women at the clinic. 49 158 p ±z∗ pq=  49 ±1.645 207(207=(18 .8%,28.5%) n 207 207 b) We are 90% confident that between 18.8% and 28.5% of women under 40 who are treated at this clinic will give birth. c) About 90% of random samples of size 207 will produce confidence intervals that contain the true proportion of women under 40 who are treated at this clinic that will give birth. owner Page 150 10/01/2013owner Page 150 10/01/2013123520813.doc d) These data do not refute the clinics claim of a 25% success rate, since 25% is in the interval. Problem 20.17: Law school. a) H 0 The law school acceptance rate for LSATisfaction is 63% (p = 0.63) H A The law school acceptance rate for LSATisfaction is greater than 63% (p > 0.63) b) Randomization Condition: These 240 students may be considered representative of the population of law school applicants. 10% Condition: There are certainly more than 2,400 law school applicants. np nq Success/Failure Condition: = 151.2 and = 88.8 are both greater than 10, so the sample is large enough. The conditions have been satisfied, so a Normal model can be used to model the sampling distribution of the proportion, with μ = p = ˆ pq (0.63)(0.37) pˆ 0.63 and σ(p) = = ≈ 0.0312 n 240 We can perform a one-proportion z-test. The observed success 163 rate is p= = 0.6792 240
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