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.

Respuesta :

Answer:

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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

This means that [tex]n = 200, \pi = 0.58[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 - 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.5116[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 + 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.6484[/tex]

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).