RES 341 chapter exercise 8.62

# RES 341 chapter - p z*sqrt(p(1-p/N = 0.013139 1.96*sqrt(0.013139*0.986861/86991 = 0.01390(0.01238 0.01390(b Why is the normality assumption not a

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8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a 95 percent confidence interval for the population proportion of positive drug tests. p - z*sqrt(p(1-p)/N) to p + z*sqrt(p(1-p)/N) N = 86991 (sample size) z = 1.96 p = 1143/86991 = 0.013139 1-p = 0.986861 p - z*sqrt(p(1-p)/N) = 0.013139 - 1.96*sqrt(0.013139*0.986861/86991)= 0.01238
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Unformatted text preview: p + z*sqrt(p(1-p)/N) = 0.013139 + 1.96*sqrt(0.013139*0.986861/86991) = 0.01390 (0.01238, 0.01390) (b) Why is the normality assumption not a problem, despite the very small value of p ? (Data are from Flying 120, no. 11 [November 1993], p. 31.) N is such a large number pn and qn are both greater than 5, it is a normality assumption. np = 1,143 > 5 n(1-p) = 85848 > 5...
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## This note was uploaded on 06/25/2011 for the course RES 341 taught by Professor Hermis during the Spring '10 term at University of Phoenix.

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