A sales and marketing management magazine conducted a survey

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

What is the sampling error of a 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report? Explain/show how you obtain your answer.

If you have the answer to this problem please break it down as simply as possible so I wil be able to understand it. I am really clueless in formulas etc but I want to be able to do problems like this in the future.

Solution

Standard Error = Sqrt(p*(1-p)/n))
     x = Mean
     n = Sample Size,
58% of the managers have caught salespeople cheating on an expense report = 58% of 200 =116
Sample Size(n)=200
Sample proportion =0.58
Standard Error = Sqrt ( (0.58*0.42) /200) )
= 0.0349