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 1,043 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.)
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
To calculate the sample proportion, divide the number of positive drug tests (1,043) by the total number of tests conducted (86,991):
Sample Proportion (p̂) = 1,043 / 86,991
Next, we need to calculate the margin of error, which is based on the standard error (SE). The formula for the standard error is:
SE = sqrt((p̂ * (1 - p̂)) / n)
Where n is the sample size, which in this case is 86,991.
Substituting the values into the formula, we get:
SE = sqrt((p̂ * (1 - p̂)) / n) = sqrt(((1043/86991) * (1 - 1043/86991)) / 86991)
Now, to construct a 95% confidence interval, we need to find the critical z-value for a two-tailed test. Since the sample size is large (86,991) and the normality assumption is not a problem, we can use the standard normal distribution.
For a 95% confidence level, the z-value is approximately 1.96.
Finally, we can calculate the margin of error using the formula:
Margin of Error = z * SE
Margin of Error = 1.96 * SE
Now, we can substitute the values we have calculated into the confidence interval formula:
Confidence Interval = Sample Proportion ± Margin of Error
Confidence Interval = p̂ ± (1.96 * sqrt((p̂ * (1 - p̂)) / n))
(a) Let's calculate the confidence interval:
p̂ = 1,043 / 86,991 ≈ 0.011975
SE = sqrt(((1043/86991) * (1 - 1043/86991)) / 86991) ≈ 0.00003462
Margin of Error = 1.96 * 0.00003462 ≈ 0.00006778
Confidence Interval = 0.011975 ± 0.00006778
Therefore, the 95% confidence interval for the population proportion of positive drug tests is approximately (0.011908, 0.012042).
(b) The normality assumption is not a problem despite the very small value of p because we have a large sample size (86,991). According to the Central Limit Theorem, when the sample size is large, the sampling distribution of the sample proportion becomes approximately normally distributed, regardless of the underlying population distribution. This allows us to use the standard normal distribution to calculate the critical z-value and construct the confidence interval for the population proportion.