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?
To construct a confidence interval for the population proportion of positive drug tests, we can use the formula:
Confidence Interval = Sample Proportion ± Margin of Error
where the sample proportion is the number of positive drug tests divided by the total number of tests, and the margin of error takes into consideration the sample size and desired level of confidence.
(a) Calculation of the confidence interval:
Sample Proportion (p̂) = Number of positive drug tests / Total number of tests
= 1,143 / 86,991
≈ 0.01313
Margin of Error (ME) = Z * √(p̂(1-p̂) / n)
Here, we need to calculate the value of Z at a 95% confidence level. The standard value for a 95% confidence interval is 1.96 for large sample sizes.
ME = 1.96 * √(0.01313 * (1-0.01313) / 86,991)
Finally, the confidence interval can be calculated as:
Confidence Interval = 0.01313 ± ME
(b) Despite the very small value of p, the normality assumption is not a problem because the sample size is sufficiently large. According to the Central Limit Theorem, for a sample size larger than 30, the sampling distribution of the sample proportion will be approximately normal. In this case, the sample size is 86,991, which is well above 30, so we can assume that the distribution is close to normal.