Margin of error of a sample?

I am so lost! please hellp!!!





A National Retail Foundation survey found households intended to spend an average of $649 during the December holiday season (The Wall Street Journal, December 2, 2002). Assume that the survey included 600 households and that the sample standard deviation was $175.





a. With 95% confidence, what is the margin of error?








b. What is the 95% confidence interval estimate of the population mean?











c. The prior year, the population mean expenditure per household was $632. Discuss the change in holiday season expenditures over the one-year period.

Margin of error of a sample?
n=600


X=649


s175





a. ME is 14.1744


b.637.8256, 663.1744


c. i dont understand what this question is asking





but to find the ME you take a confidence leve of 95% with a degree of freedom of 599 which is aprox. 1.984 then you mulitiply that (1.984) by std dev/sq root of the sample size


so..


1.984 X 175/24.495


for b. you take the mean and ± the ME so 649±14.174





i hope that helps