A transportation worker takes a simple random sample of 162 toll booth receipts for one day and finds that the

A transportation worker takes a simple random sample of 162 toll booth receipts for one day and finds that the mean is $7.38 with a standard deviation of $1.45. What is the 99% confidence interval for the mean amount of all the receipts that day?


($7.30, $7.46)
($7.27, $7.49)
($7.16, $7.60)
($7.09, $7.67)

Jack -

CI = mean +/- (z*)(standard error)

CI = 7.38 +/- (2.576)(1.45/sqrt 162) = ($7.09, $7.67)

Note, a t-distribution will give you the same result since n is so large.

Good Luck