Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. The record shows that the mean was 1.5 hours. Suppose for a random sample of 921 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.62 hours with a standard deviation of 0.71 hour.

What is the upper boundary of the 95% C.I.?

Answer 1

sample std dev ,    s =    0.710
Sample Size ,   n =    921
Sample Mean,    x̅ =   1.62

α=1-0.95=0.05

degree of freedom=   DF=n-1=   920  
't value='   tα/2=   1.6465   [Excel formula =t.inv(α,df) ]

Standard Error , SE =   s/√n =   0.0234
margin of error ,   E=t*SE = 1.6465*0.0234 = 0.039

confidence interval is       
Interval Upper Limit=   x̅ + E = 1.62 + 0.039 = 1.659
upper boundary of the 95% C.I. = 1.659