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How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hour. (a) A histogram of time spent ea ting 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. (b) There are over 200 million Americans 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 Americans 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 American spends eating and drinking each day? Explain.

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Answer:

Explanation:

a)

To construct a confidence interval ,the histogram should be bell shaped.It means that the it should be normally distributed with no skewness.To eliminate the skewness a large sample size is required so that the sample is normally distribute about the mean Also ,in order to construct a t-interval the sample data must come from the population that is normally distributed or the sample size is larger than 30.Since the question stated that the population distribution is skewed to the right \bar{x} is guaranteed to be normally distributed if
n\geq 30

b)The sample satisfies the normal distribution because the sample sixe is greater than 30 which is 1001.

c) [
\bar{x}=1.22 \,s=.65\, n=1001

Area in the right tail =2.5% or .025

degree of freedom =1001-1 =1000


t_(\alpha /2)=1.96

95% confidence interval is :
\bar{x}\pm t_(\alpha /2)s/√(n)

lower bound is :
1.22-1.96*.65/√(1001) =1.18

upper bound is :
-1.22+1.96*.65/√(1001)=1.26

d)No because the sample data was obtined from Americans age 15 or older which cannot be applied to a different age group

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