Question

An audit firm surveyed salespeople cheating on their expense reports and other unethical conduct. In the survey of 180 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 “bars” as “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 cheating on an expense report with a 94% confidence?

Answer 1

Solution :

Given that,

n = 180

Point estimate = sample proportion = = 0.58

1 - = 1 - 0.58 = 0.42

At 94% confidence level

= 1 - 94%

=1 - 0.94

= 0.06

Z = Z_0.06 = 1.55

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.55 (((0.58 * 0.42) / 180)

= 0.057

A 94% upper confidence interval for population proportion p is ,

+ E   

= 0.58 + 0.057 = 0.637

upper limit = 0.637