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 working a second job on company time with a 96% confidence?

Answer 1

Upper limit = 0.5726

Explanation:

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 200

P = x/n = 0.50

Confidence level = 96%

Critical Z value = 2.0537

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.50 ± 2.0537* sqrt(0.50*(1 – 0.50)/200)

Confidence Interval = 0.50 ± 2.0537* 0.0354

Confidence Interval = 0.50 ± 0.0726

Lower limit = 0.50 - 0.0726 = 0.4274

Upper limit = 0.50 + 0.0726 = 0.5726

Confidence interval = (0.4274, 0.5726)