In: Statistics and Probability
An audit firm surveyed salespeople cheating on their expense reports and other unethical conduct. In the survey of 150 managers, 58% of the managers have caught salespeople cheating on an expense report, 50% have caught salespeople working a second job on company time, 22% have caught salespeople listing a “bars” as a “restaurants” on an expense report, and 19% have caught salespeople giving a kickback to a customer. What is the estimated upper limit for the proportion of managers who caught salespeople giving a kickback to a customer with a 99% confidence?
Solution :
Given that,
Point estimate = sample proportion = = 0.19
1 - = 1 - 0.19 = 0.81
Z/2 = Z0.01 = 2.33
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.33 (((0.19 * 0.81) / 150)
= 0.075
A 99% upper confidence interval for population proportion p is ,
p < + E
p < 0.19 + 0.075
p < 0.265