A student is trying to determine how much money a student in

A student is trying to determine how much money a student in her school spends when buying lunch in the cafeteria. She randomly selects 16 students who buy their lunch and determined that the mean amount for the sample is $4.26 with a standard deviation of $0.74. Construct a 95% confidence interval for the average amount spent on lunch.
             What is the point estimate for the mean amount spent on lunch at this school?  

What is the margin of error? The school cafeteria claims that the average amount is $3.70. What comment
can you make about this claim?

Solution

Point estimate of the mean amount = 4.26

The confidence interval :

Lower limit = 4.26 - (1.96 * 0.74 / sqrt(16) ) = 3.90

Upper Limit = 4.26 + (1.96 * 0.74 / sqrt(16) ) = 4.62

Margin of error = 1.96 * 0.74 / sqrt(16)

= 0.3626

The school cafeteria is understating the average amount of money spent by a student as the claim is more than 3 standard deviations away from the mean.

Hope this helps.