Waiting times in hours at a popular restaurant are believed

Waiting times (in hours) at a popular restaurant are believed to be approximately normally distributed during busy periods. A sample of 20 customers revealed a mean waiting time of 1.58 hours with a sample standard deviation of 1.5 hours. Construct a 95% confidence interval for the population mean waiting time. [Hint: use the t confidence interval formula because the sample size is not large and the sample standard deviation is given.]

Solution

n=20

Mean = 1.58

SD = 1.5

t value for 95% CI = 2.086

Margin of error ME = (t*SD)/sqrt(n)

= (2.086 * 1.5)/sqrt(20)

= 0.6997

Confidence Interval

= ( Mean - ME , Mean + ME)

= ( 1.58 - 0.6997 , 1.58 + 0.6997 )

= ( 0.8803 , 2.2797 ) Answer