SOLUTION: A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 901 people age 15 or older, the mean amount of time s
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Question 1159809: A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 901 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.02 hours with a standard deviation of 0.66 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 by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 99%CI is mean+/- t*s/sqrt(n)
t(0.995, df=900)=2.581
SE is 0.66/sqrt(901)
half-interval or margin of error is 0.0567
99% CI is (0.963, 1.077) units hours
We are highly confident that the true mean (parameter) is in the given interval. The true mean is unknown and unknowable, but we strongly believe it is in this interval. (other interpretations as well).
