Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1064 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.28 hours with a standard deviation of 0.75 hour.

Determine and interpret a 99​% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.

Answer 1

Solution :

Given that,

= 1.28

= 0.75

n = 1064

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (0.75 / 1064)

= 0.0592

At 99% confidence interval estimate of the population mean is,

- E < < + E

1.28 - 0.0592 < < 1.28+ 0.0592

1.2208 < < 1.3392

(1.2208 , 1.3392 )