8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applican

Question

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. (b) Why is the normality assumption not a problem, despite the very small value of p?

Explanation / Answer

Confidence intervals are used to find a region in which we are 100 * ( 1 - a )% confident the true value of the parameter is in the interval. For large sample confidence intervals about the population proportion you have: pHat ± z * sqrt(phat * (1- phat) / n) where phat is the sample proportion z is the zscore for having a% of the data in the tails, i.e., P( |Z| > z) = a n is the sample size The sample proportion, phat = 0.01313929 The sample size n = 86991 The z score for a 0.95 confidence interval is the z score such that 0.025 is in each tail. z = 1.959964 The confidence interval is: ( phat ± z * sqrt(phat * (1 - phat) / n) ( 0.01238259 , 0.01389599 ) We are 0.95 confident the true value of the population proportion lies in this interval. normality is not an issue because the sample size is so large that the proportion will follow a normal distribution. we know this thanks to the central limit theorem

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