Question

In: Statistics and Probability

An audit firm surveyed salespeople cheating on their expense reports and other unethical conduct. In the...

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?

Solutions

Expert Solution

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= Z0.08 = 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 limit = 0.261


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