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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 180 managers, 54% 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 cheating on an expense report with a 90% confidence?

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Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = 0.54

1 - = 1 - 0.54 = 0.46

Z = Z0.10 = 1.282  

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

= 1.282 (((0.54 * 0.46) / 180)

= 0.048

A 90% confidence interval for population proportion p is ,

+ E

= 0.54 + 0.048 = 0.588

upper limit = 0.588

orchestra answered 1 year ago
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