The Daytona Beach Tourism Commission is interested in the av

The Daytona Beach Tourism Commission is interested in the average amount of money a typical college student spends per day during spring break. It is presumed that daily spending habits are normally distributed. They survey 35 students and find that the mean daily spending is $63.57 with a sample standard deviation of $17.32.

What is the margin of error for a 95% confidence interval estimate for the population mean daily spending?

Solution

Note that              
              
E (margin of error) = z(alpha/2) * s / sqrt(n)                      
where              
alpha/2 = (1 - confidence level)/2 =    0.025          

z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    17.32          
n = sample size =    35          
              
Thus,              
              
E = 5.738018664 [ANSWER]