HELP! STATISTICS

Question

A sample of 250 orders for? take-out food at a local pizzeria found that the average cost of an order was ?$29 with s=?$15.
 Find the margin of? error, if we use a? 95% confidence? interval, for the average cost of an order.

Answer 1

Hello,

 

Assuming "margin of error" refers tot he confidence interval, we first need to find the standard error (SE) and the degrees of freedom (df).

 

SE = s/sqrt(n)

in your example, this would be 15/sqrt(250) = 0.949

 

df = n - 1

in your example, this would be 250 - 1 = 249.

 

using this information, we can also figure out the critical t value. for this, you'll have to use a t table (your teacher may have given you one or you can find them online). the critical t value for df = 249 and 95% confidence is 1.969.

 

now we have everything we need. the margin of error will be the mean plus or minus, SE*t critical value. in this case, it would be 0.949*1.969 = 1.869

 

and that's it. the average cost was $29 plus or minus 1.869 (might as well round to 1.87 since it's money)