Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 949 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.65 hours with a standard deviation of 0.54 hour. Complete parts ​(a) through ​(d) below.

c) Determine and interpret a 95 % confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. Select the correct choice below and fill in the answer​ boxes, if​ applicable, in your choice. ​(Type integers or decimals rounded to three decimal places as needed. Use ascending​ order.)

A. The nutritionist is 95​% confident that the amount of time spent eating or drinking per day for any individual is between ------ and ------ hours.

B. There is a 95 ​% probability that the mean amount of time spent eating or drinking per day is between ------- and ----- hours.

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

D. The requirements for constructing a confidence interval are not satisfied.

Im trying to complete this section but i am having trouble calculating the interval for this section i know the answer is C but im not sure how to calculate the z value for this problem

Answer 1

Answer is

B.    There is a 95 ​% probability that the mean amount of time spent eating or drinking per day is between 1.616 and 1.684 hours.

Working

Given
X̅ = 1.65           ....... Sample Mean
n = 949           ....... Sample Size
σ = 0.54           ....... Population Standard Deviation

For 95% Confidence interval

α = 0.05,      α/2 = 0.025
From z tables of Excel function NORM.S.INV(α/2) we find the z value
z = NORM.S.INV(0.025) = 1.96
We take the positive value of z

Confidence interval is given by

= (1.616, 1.684)

95% Confidence interval is (1.616, 1.684) hours