1) a nutritionist wants to determine how much time nationally people spend eatin
ID: 3127828 • Letter: 1
Question
1) a nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1018 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.29 hours with a standard deviation of 0.62 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?
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 99% 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 spends eating and drinking each day? Explain....
2) The following data represent the asking price of a simple random sample of homes for sale. Construct a 99% confidence interval with and without the outlier included. Comment on the effect the outlier has on the confidence interval.
Here is the information: $232,000 $279,900 $219,900 $143,000 $205,800 $218,800 $459,900 $234,900, $187,500 $290,900 $147,800 $264,900
A) Construct a 99% confidence interval with the outlier included: ($_____ , $ _____)
B) Construct a 99% confidence interval with the outlier removed: ($_____ , $ _____)
C) Comment on the effect the outlier has on the confidence interval:
(pick one from below)
a) The outlier caused the width of the confidence interval to increase.
b) The outlier caused the width of the confidence interval to decrease.
c) The outlier had no effect on the width of the confidence interval.
3) an interactive poll found that 363 of 2,375 adults aged 18 or older have at least one tattoo.
a) obtain a point estimate for the proportion of adults who have at least one tattoo.
b) construct a 90% confidence interval for the proportion of adults with at least one tattoo. (lower bound _____) (upper bound ______)
c) construct a 99% confidence interval for the proportion of adults with at least one tattoo. (lower bound _____) (upper bound ______)
d) what is the effect of increasing the level of confidence on the width of the interval?
Explanation / Answer
1) a nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1018 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.29 hours with a standard deviation of 0.62 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?
The central limit theorem applies only if their is sufficiently large number of iterates, especially if the underlying distribution is not normal. Hence, we need a sufficiently large sample size here.
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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?
1018 is sufficiently large, but still remains a small part of the population, 200 million. Hence, that satisfies the condition of constructing a confidence interval.
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c) determine and interpret a 99% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day?
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
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 = 0.005
X = sample mean = 1.29
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 0.62
n = sample size = 1018
Thus,
Margin of Error E = 0.050053549
Lower bound = 1.239946451
Upper bound = 1.340053549
Thus, the confidence interval is
( 1.239946451 , 1.340053549 ) [ANSWER]
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d) could the interval be used to estimate the mean amount of time a 9 year old spends eating and drinking each day? Explain...
NO, because our sample is composed of people age 15 and older. [ANSWER, NO]
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