Question 184292
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
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The sample proportion is 1143/86991 = 0.01314
 
(a) Construct a 95 percent confidence interval for the population proportion of positive drug tests.
E = z*sqrt(pq/n)
E = 1.96*sqrt(0.01314*0.98686/86991) = 0.000757
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95% CI: 0.01314-0.000757 < p < 0.01314+0.000757
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(b) Why is the normality assumption not a problem, despite the very small value of p? 
n is large
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Cheers,
Stan H.