Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1071 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.73 hours with a standard deviation of .51 hour.

1) Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.

The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day is between _____ and _____ hours.

Answer 1

Solution :

Given that,

Sample size = n = 1071

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z _0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (0.51 / 1071)

Margin of error = E =  0.03

At 95% confidence interval estimate of the population mean is,

- E < < + E

1.73 - 0.03 < < 1.73 + 0.03

1.70 < < 1.76

The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day is between 1.70 and 1.76 hours.