Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1124 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 2.09 hours with a standard deviation of 0.76 hours. Complete parts ​(a) through ​(d) below.

​(a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the meantime spent eating and drinking each day.

(b) In​ 2010, there were over 200 million people nationally age 15 or older. Explain why​ this, along with the fact that the data were obtained using a random​ sample, satisfies the requirements for constructing a confidence interval.

(c) Determine and interpret a 95​% confidence interval for the mean amount of time people age 15 or older spend eating and drinking each day.

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

Answer 1

A)

*Since the distribution is skewed right (not normally distributed) the sample must be large so that the distribution of the same mean will be approximately normal.

b)

*The sample size is less than 5% of the population (n < 0.05N)

c)

The 95% confidence interval is 2.046<μ<2.134.

Which means the nutritionist can be 95% confident that the mean amount of time Americans spend eating/drinking per day is between this confidence interval

d)

*No, the interval is about people age 15 and older. The amount for 9-year-olds could easily be very different.