A nutritionist wants to determine how much time nationally p

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 917 people age 15 r older, the mean amount of time spent eating or drinking oer day is 1.03 hours with astandard deviation of 0.56 hour.

a) Determin and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.

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

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

There is a 95% probability that the mean about time spent eating or drinking is between ... and .... hours

The requirements for constructing a confidence interval are not satisfied.

b) Could the interval be used to estimate the mean amount of time a 9 year old spends eating and drinking each day?

No,the interval is about people of age 15 or older. The amount of time spend eating or drinking for a 9 year old may differ.

Yes, the interval is about individuals time spend eating or drinking per day and can be used to find the mean amount of time a 9 year old spends eating or drinking each day.

A confidence interval could not be constructed in part a.

No, the interval is about time spend eating or drinking per day and cannot be used to find the mean time spent eating or drinking per day for a specific age

Yes the interval is about the mean amount of time spent eating or drinking per day for people age 15 or older and can be used to find the mean amount of time spent eating or drinking per day for 9 year olds.

Solution

a)

The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day is between .... and .... hours. [OPTION B]

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    1.03          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.56          
n = sample size =    917          
              
Thus,              
              
Lower bound =    0.993754721          
Upper bound =    1.066245279          
              
Thus, the confidence interval is              
              
(   0.993754721   ,   1.066245279   ) [ANSWER]

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b) Could the interval be used to estimate the mean amount of time a 9 year old spends eating and drinking each day?

No,the interval is about people of age 15 or older. The amount of time spend eating or drinking for a 9 year old may differ. [ANSWER]