HELP!! iS ANYONE HERE REALLY GOOD AT STATISTICS!!!! hURRY!!!?

i have my AP stats test tomorrow. I dont understand how to do this. Can you explain.





A large company is considering opening a franchise and wants to estimate the mean househol income for the area using a simple random sample of households. Based on information from a pilot study, the company assumes that the standard deviation of household income is $7,200. Of the following, which is the least number of households that should be surveyed to obtain as estimate that is within $200 of the true mean household income with 95 percent confidence??





a. 75


b.1300


c.5200


d.5500


e.7700





(i dont get how to get the margin of error) please explain too. Thank you SOOOO much!

HELP!! iS ANYONE HERE REALLY GOOD AT STATISTICS!!!! hURRY!!!?
So we know sigma = 7200


For a 95% CI, z = 1.96


CI = mu +/- z * sigma / sqrt(n)





z * sigma / sqrt(n) %26lt;= 200


sqrt(n) %26gt;= 1.96 * 7200 / (200) = 70.56


n %26gt;= 4978.7





n must be at least 4979.





The best answer seems to be 5200.
Reply:75 is the answer. Margin of error ?
Reply:Hi, this link should help you solve this problem. It explains every single step.





If I didnt have a deadline to beat today, i could've explained it myself.





Hope this helps.





http://www.isixsigma.com/library/content...