Question

# A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for...

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1024 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.61 hours with a standard deviation of 0.55 hour. 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 mean time spent eating and drinking each day. A. Since the distribution of time spent eating and drinking each day is normally​ distributed, the sample must be large so that the distribution of the sample mean will be approximately normal. B. Since the distribution of time spent eating and drinking each day is not normally distributed​ (skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal. C. The distribution of the sample mean will always be approximately normal. D. The distribution of the sample mean will never be approximately normal. ​(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. A. The sample size is less than​ 5% of the population. B. The sample size is less than​ 10% of the population. C. The sample size is greater than​ 5% of the population. D. The sample size is greater than​ 10% of the population. ​(c) Determine and interpret a 90​% 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. There is a 90​% probability that the mean amount of time spent eating or drinking per day is between nothing and nothing hours. B. The nutritionist is 90​% confident that the mean amount of time spent eating or drinking per day is between nothing and nothing hours. C. The nutritionist is 90​% confident that the amount of time spent eating or drinking per day for any individual is between nothing and nothing hours. D. The requirements for constructing a confidence interval are not satisfied. ​(d) Could the interval be used to estimate the mean amount of time a​ 9-year-old spends eating and drinking each​ day? Explain. A. ​Yes; the interval is about the mean amount of time spent eating or drinking per day for people 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. B. ​Yes; the interval is about individual time spent eating or drinking per day and can be used to find the mean amount of time a​ 9-year-old spends eating and drinking each day. C. ​No; the interval is about people age 15 or older. The mean amount of time spent eating or drinking per day for​ 9-year-olds may differ. D. ​No; the interval is about individual time spent eating or drinking per day and cannot be used to find the mean time spent eating or drinking per day for specific age. E. A confidence interval could not be constructed in part ​(c). Click to select your answer.

(a)

B. Since the distribution of time spent eating and drinking each day is not normally distributed​ (skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal.

Option B. is correct.

(b)

B. The sample size is less than​ 10% of the population.

Option B. is correct.

(c)

B. The nutritionist is 90​% confident that the mean amount of time spent eating or drinking per day is between nothing and nothing hours.

Option B is correct.

(d)

C. ​No; the interval is about people age 15 or older. The mean amount of time spent eating or drinking per day for​ 9-year-olds may differ.

Option C. is correct.

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