Question

An audit firm surveyed salespeople cheating on their expense reports and other unethical conduct. In the survey of 200 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 listing “bars” as “restaurants” on an expense report with a 92% confidence?

Answer 1

Solution :

Given that,

n = 200

Point estimate = sample proportion = = 0.22

1 - = 1 - 0.22 = 0.78

At 92% confidence level

= 1 - 92%

=1 - 0.92

= 0.08

Z = Z_0.08 = 1.41

Margin of error = E = Z * (( * (1 - )) / n)

= 1.41 (((0.22 * 0.78) / 200)

= 0.041

A 92% upper confidence interval for population proportion p is ,

+ E  

= 0.22 + 0.041 = 0.261

upper bound = 0.261