Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1005 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.03 hours with a standard deviation of 0.65 hour. Complete parts ​(a) through ​(d) below.

​(c) Determine and interpret a

9595​%

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

C.There is a 95​% probability 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.

Answer 1

sample mean 'x̄=		1.030
sample size    n=		1005
std deviation s=		0.650
std error ='sx=s/√n=0.65/√1005=		0.0205

for 95% CI; and 1004 df, value of t=		1.9623
margin of error E=t*std error    =		0.0402
lower bound=sample mean-E =		0.9898
Upper bound=sample mean+E =		1.0702
from above 95% confidence interval for population mean =(0.990 <μ<1.070)

B.The nutritionist is 95​% confident that the mean amount of time spent eating or drinking per day is between 0.990 and 1.070 hours