A sales and marketing management magazine conducted a survey on salespeople cheating on their expense reports and other unethical conduct. In the survey on 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 "strip bar" as a restaurant on an

2 expense report, and 19% have caught salespeople giving a kickback to a customer.

construct a 95% confidence interval estimate of the population

3 proportion of managers who have caught salespeople cheating on an expense report.

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

In the survey on 200 managers, 58% of the managers have caught salespeople cheating on an expense report.

This means that

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: