Question

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? Round final answer to 4 decimal places. Do not round intermediate calculations.

Answer 1

Solution :

Given that,

n = 150

Point estimate = sample proportion = = 0.19

1 - = 1 - 0.19 = 0.81

At 99% confidence level

= 1 - 99%

=1 - 0.99

= 0.01

Z = Z_0.01 = 2.326

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

= 2.326 (((0.19 * 0.81) / 150)

= 0.0745

A 99% upper confidence interval for population proportion p is ,

+ E  

= 0.19 + 0.0745 = 0.2645

upper limit = 0.2645