Question

Based on a survey conducted by Greenfield Online, 25–34-year-olds spend the most each week on fast food. The average weekly amount of $44 (based on 115 participants) was reported in a May 2009 USA Today Snapshot. Assuming that weekly fast-food expenditures are normally distributed with a known standard deviation of $14.50, construct a 90% confidence interval for the mean weekly amount that 25–34-year-olds spend each week on fast food.
(Please show work)

Answer 1

Okay, so the formula for a confidence interval is 
test statistic +/- (critical value)(standard deviation of statistic
44 +/- (invNorm(area: 0.05, u=0, std:1))(14.5)
44 +/- 23.85
(20.15, 67.85)
hope this helps :)