Question

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? (Data are from Flying 120,

no. 11 [November 1993], p. 31.)

Answer 1

CI95 = p + or - (1.96)(√pq/n)
...where p = x/n, q = 1 - p.
Note: + or - 1.96 represents the 95% confidence interval using a z-table.

Now let's look at your data:
x = 1143; n = 86991
Therefore: p = 1143/86991
Convert to a decimal.
q = 1 - p

Plug the decimal values into the formula and calculate your interval.

For b), here's a hint: sample size.

I hope this will help get you started.