Transcript ACTIVE-WITH-ANSast3e_08

Active Learning Lecture Slides
For use with Classroom Response Systems
Statistical Inference: Confidence Intervals
8.1 A good point estimate has which of the
following characteristics?
a) Bias: None
b) Bias: High
c) Bias: None
d) Bias: High
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Standard Error: High
Standard Error: High
Standard Error: Low
Standard Error: Low
8.1 A good point estimate has which of the
following characteristics?
a) Bias: None
b) Bias: High
c) Bias: None
d) Bias: High
Copyright © 2013 Pearson Education, Inc.
Standard Error: High
Standard Error: High
Standard Error: Low
Standard Error: Low
8.2 When the sampling distribution is
approximately normal, what is the margin of error
equal to for a 95% confidence interval?
a) 1.96
b) 1.96*standard error
c) Standard error
d) Point estimate  1.96*standard error
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8.2 When the sampling distribution is
approximately normal, what is the margin of error
equal to for a 95% confidence interval?
a) 1.96
b) 1.96*standard error
c) Standard error
d) Point estimate  1.96*standard error
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8.3 In 2006 the GSS had a special topic that
investigated disabilities. They asked respondents if
they had difficulty fully participating in school,
housework or other daily activities and 265 out of
2,749 said “yes”. What is the point estimate of the
population proportion of Americans that have
difficulty completing these tasks?
a) 0.10
b) 0.097
c) 0.265
d) p
e) Unknown
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8.3 In 2006 the GSS had a special topic that
investigated disabilities. They asked respondents if
they had difficulty fully participating in school,
housework or other daily activities and 265 out of
2,749 said “yes”. What is the point estimate of the
population proportion of Americans that have
difficulty completing these tasks?
a) 0.10
b) 0.097
c) 0.265
d) p
e) Unknown
Copyright © 2013 Pearson Education, Inc.
8.4 True or False: A point estimate is better than
an interval estimate because it gives you the exact
value for which you are looking.
a) True
b) False
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8.4 True or False: A point estimate is better than
an interval estimate because it gives you the exact
value for which you are looking.
a) True
b) False
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8.5 True or False: An interval estimate gives you a
region that the parameter has to fall within.
a) True
b) False
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8.5 True or False: An interval estimate gives you a
region that the parameter has to fall within.
a) True
b) False
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8.6 The formula below gives a region of plausible
values of :
pˆ  z
pˆ 1  pˆ 
n
a) the population proportion
b) the population mean
c) the sample mean
d) the sample proportion
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8.6 The formula below gives a region of plausible
values of :
pˆ  z
pˆ 1  pˆ 
n
a) the population proportion
b) the population mean
c) the sample mean
d) the sample proportion
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8.7 In 2006 the GSS asked 2,986 people if they were
very happy, pretty happy, or not too happy and 920
people said that they were very happy. Is the sample
“large” enough to calculate the 95% confidence
interval to estimate the proportion of all Americans that
are very happy?
a) Yes, there are more than 30 observations.
b) Yes, there are more than 15 successes and 15
failures.
c) No, there are not more than 15 successes and 15
failures.
d) Cannot be determined.
Copyright © 2013 Pearson Education, Inc.
8.7 In 2006 the GSS asked 2,986 people if they were
very happy, pretty happy, or not too happy and 920
people said that they were very happy. Is the sample
“large” enough to calculate the 95% confidence
interval to estimate the proportion of all Americans that
are very happy?
a) Yes, there are more than 30 observations.
b) Yes, there are more than 15 successes and 15
failures.
c) No, there are not more than 15 successes and 15
failures.
d) Cannot be determined.
Copyright © 2013 Pearson Education, Inc.
8.8 In 2006 the GSS asked 2,986 people if they
were very happy, pretty happy, or not too happy
and 920 people said that they were very happy.
Find the 95% confidence interval to estimate the
proportion of all Americans that are very happy.
a) (0, 0.02)
b) (0.25, 0.37)
c) (0.27, 0.35)
d) (0.29, 0.32)
Copyright © 2013 Pearson Education, Inc.
8.8 In 2006 the GSS asked 2,986 people if they
were very happy, pretty happy, or not too happy
and 920 people said that they were very happy.
Find the 95% confidence interval to estimate the
proportion of all Americans that are very happy.
a) (0, 0.02)
b) (0.25, 0.37)
c) (0.27, 0.35)
d) (0.29, 0.32)
Copyright © 2013 Pearson Education, Inc.
8.9 What is the correct z-score for a 97%
confidence interval for the population proportion?
a) 1.88
b) 1.96
c) 2.05
d) 2.17
e) None of the above
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8.9 What is the correct z-score for a 97%
confidence interval for the population proportion?
a) 1.88
b) 1.96
c) 2.05
d) 2.17
e) None of the above
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8.10 Based off of the same sample, which of the
confidence intervals for the population mean
would be the widest?
a) A 90% confidence interval
b) A 95% confidence interval
c) A 99% confidence interval
d) Cannot be determined
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8.10 Based off of the same sample, which of the
confidence intervals for the population mean
would be the widest?
a) A 90% confidence interval
b) A 95% confidence interval
c) A 99% confidence interval
d) Cannot be determined
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8.11 The margin of error of a confidence interval
of the population mean decreases as…
a) the sample size decreases.
b) the sample size increases.
c) the sample mean increases.
d) the sample mean decreases.
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8.11 The margin of error of a confidence interval
of the population mean decreases as…
a) the sample size decreases.
b) the sample size increases.
c) the sample mean increases.
d) the sample mean decreases.
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8.12 The General Social Survey included a
question about how many hours the respondent
spent doing religious activities outside of their own
home. For the 1,414 respondents the sample mean
was 6.15 hours and the sample standard deviation
was 16.53 hours. Find the 95% confidence interval
for the population mean amount of time spent doing
religious activities outside of their own home.
a) (5.43, 6.87)
b) (-26.24, 38.55)
c) (5.29, 7.01)
d) (5.03, 7.27)
Copyright © 2013 Pearson Education, Inc.
8.12 The General Social Survey included a
question about how many hours the respondent
spent doing religious activities outside of their own
home. For the 1,414 respondents the sample mean
was 6.15 hours and the sample standard deviation
was 16.53 hours. Find the 95% confidence interval
for the population mean amount of time spent doing
religious activities outside of their own home.
a) (5.43, 6.87)
b) (-26.24, 38.55)
c) (5.29, 7.01)
d) (5.03, 7.27)
Copyright © 2013 Pearson Education, Inc.
8.13 The General Social Survey included a
question about how many hours the respondent
spent doing religious activities outside of their own
home. For the 1,414 respondents the sample mean
was 6.15 hours and the sample standard deviation
was 16.53 hours. What can we say about the
distribution of hours spent doing religious
activities?
a) It is bell shaped.
b) It is right skewed.
c) It is left skewed.
d) Nothing can be determined.
Copyright © 2013 Pearson Education, Inc.
8.13 The General Social Survey included a
question about how many hours the respondent
spent doing religious activities outside of their own
home. For the 1,414 respondents the sample mean
was 6.15 hours and the sample standard deviation
was 16.53 hours. What can we say about the
distribution of hours spent doing religious
activities?
a) It is bell shaped.
b) It is right skewed.
c) It is left skewed.
d) Nothing can be determined.
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8.14 Which of the following is NOT a property of
the t distribution?
a) It is symmetric.
b) It is indexed by a degree of freedom equal to
n - 1.
c) It has more spread in the tails than the normal
distribution.
d) The shape becomes closer and closer to the
normal distribution as n decreases.
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8.14 Which of the following is NOT a property of
the t distribution?
a) It is symmetric.
b) It is indexed by a degree of freedom equal to
n - 1.
c) It has more spread in the tails than the normal
distribution.
d) The shape becomes closer and closer to the
normal distribution as n decreases.
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8.15 A marketing researcher is interested in estimating the
mean amount of money spent on lunch by college
students. The average amount spent on lunch by a random
sample of 10 students is $6.30 with a standard deviation of
$2.21. Find the 95% confidence interval for the population
mean amount spent on lunch every day.
a)
6 . 30  1 . 96
2 . 21
10
b)
6 . 30  2 . 228
2 . 21
10
c)
6 . 30  2 . 262
2 . 21
10
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8.15 A marketing researcher is interested in estimating the
mean amount of money spent on lunch by college
students. The average amount spent on lunch by a random
sample of 10 students is $6.30 with a standard deviation of
$2.21. Find the 95% confidence interval for the population
mean amount spent on lunch every day.
a) 6 .3 0  1 .9 6
2 .2 1
10
b) 6 .3 0  2 .2 2 8
2 .2 1
10
c)
6 . 30  2 . 262
2 . 21
10
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8.16 In 2002 a local survey (using a SRS) found
that 13.4% of people strongly agreed with the
statement that “More parking meters should be
installed downtown.” Using the 2002 data as a
guideline, determine the sample size needed to
estimate the proportion of people that would agree
for the current year within 0.01 at 95% confidence.
a) 4458
b) 3140
c) 45
d) 23
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8.16 In 2002 a local survey (using a SRS) found
that 13.4% of people strongly agreed with the
statement that “More parking meters should be
installed downtown.” Using the 2002 data as a
guideline, determine the sample size needed to
estimate the proportion of people that would agree
for the current year within 0.01 at 95% confidence.
a) 4458
b) 3140
c) 45
d) 23
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8.17 Suppose that you were interested in determining the
proportion of Americans that agreed with the statement
that “Taxes should not be raised for any reason.”
Assuming that you have no idea what proportion will
agree with this statement, determine the sample size
needed to estimate the proportion of people that agree for
the current year within 0.01 at 95% confidence.
a) 16,513
b) 9,604
c) 4,458
d) 4,900
e) 49
Copyright © 2013 Pearson Education, Inc.
8.17 Suppose that you were interested in determining the
proportion of Americans that agreed with the statement
that “Taxes should not be raised for any reason.”
Assuming that you have no idea what proportion will
agree with this statement, determine the sample size
needed to estimate the proportion of people that agree for
the current year within 0.01 at 95% confidence.
a) 16,513
b) 9,604
c) 4,458
d) 4,900
e) 49
Copyright © 2013 Pearson Education, Inc.
8.18 Suppose that you are interested in estimating the
population mean entry salary of engineers. You think that
entry salaries probably range from $30,000 to $100,000
and the distribution of salaries is bell shaped. You want
to be accurate to within $3,000 of the population mean
entry salary and be 95% confident. What size sample do
you need?
a) 100
b) 87
c) 59
d) 8
e) Cannot be determined
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8.18 Suppose that you are interested in estimating the
population mean entry salary of engineers. You think that
entry salaries probably range from $30,000 to $100,000
and the distribution of salaries is bell shaped. You want
to be accurate to within $3,000 of the population mean
entry salary and be 95% confident. What size sample do
you need?
a) 100
b) 87
c) 59
d) 8
e) Cannot be determined
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8.19 Suppose that you want to estimate the entry
level salaries of high school teachers for the state
of Nebraska. A previous study had $1700 listed as
the sample standard deviation. You want to have a
90% confidence interval with a margin of error of
$250. How large a sample do you need?
a) 178
b) 126
c) 91
d) 12
e) Cannot be determined
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8.19 Suppose that you want to estimate the entry
level salaries of high school teachers for the state
of Nebraska. A previous study had $1700 listed as
the sample standard deviation. You want to have a
90% confidence interval with a margin of error of
$250. How large a sample do you need?
a) 178
b) 126
c) 91
d) 12
e) Cannot be determined
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8.20 Suppose that you are a writer for a university
newspaper and due to time constraints you conduct a
survey of only 20 randomly selected students. You ask
them, “Do you plan to watch the homecoming
parade?” and 19 of them say they plan to watch.
Create a 95% confidence interval for the proportion of
students that plan on watching the parade.
a) (0.85, 1)
b) (.74, 1)
c) (.76, .99)
d) Cannot be determined
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8.20 Suppose that you are a writer for a university
newspaper and due to time constraints you conduct a
survey of only 20 randomly selected students. You ask
them, “Do you plan to watch the homecoming
parade?” and 19 of them say they plan to watch.
Create a 95% confidence interval for the proportion of
students that plan on watching the parade.
a) (0.85, 1)
b) (.74, 1)
c) (.76, .99)
d) Cannot be determined
Copyright © 2013 Pearson Education, Inc.
8.21 The margin of error of a confidence interval
estimates the error…
a) caused by bad sampling techniques.
b) caused by measurement error.
c) caused by not controlling lurking variables.
d) caused by using a sample rather than the whole
population.
e) all of the above.
Copyright © 2013 Pearson Education, Inc.
8.21 The margin of error of a confidence interval
estimates the error…
a) caused by bad sampling techniques.
b) caused by measurement error.
c) caused by not controlling lurking variables.
d) caused by using a sample rather than the whole
population.
e) all of the above.
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8.22 The bootstrap method is a method that
constructs a confidence interval by…
a) repeatedly sampling from the population.
b) repeatedly sampling from the sample.
c) repeatedly sampling from the sampling
distribution.
d) none of the above.
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8.22 The bootstrap method is a method that
constructs a confidence interval by…
a) repeatedly sampling from the population.
b) repeatedly sampling from the sample.
c) repeatedly sampling from the sampling
distribution.
d) none of the above.
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8.23 Why is the bootstrap method used?
a) Because it is easier than traditional methods.
b) Because it can be used when the formula for
the confidence interval cannot easily be found
mathematically.
c) Because it uses additional sampling – this
makes the confidence interval much better than
traditional methods.
d) Because it allows us to use 10,000 samples
rather than just 1 sample that is used in
traditional methods.
Copyright © 2013 Pearson Education, Inc.
8.23 Why is the bootstrap method used?
a) Because it is easier than traditional methods.
b) Because it can be used when the formula for
the confidence interval cannot easily be found
mathematically.
c) Because it uses additional sampling – this
makes the confidence interval much better than
traditional methods.
d) Because it allows us to use 10,000 samples
rather than just 1 sample that is used in
traditional methods.
Copyright © 2013 Pearson Education, Inc.
8.24 To make a 90% confidence interval for the
population median using the bootstrap method you first
randomly select 20,000 separate samples of size 8 from
the original data and then you compute the medians
from each of the new samples. What is the next step?
a) Find the average and standard deviation of the
20,000 medians. Then, compute a traditional 95%
s
x

t
confidence interval (
) with those values.
n
b) Find the 2.5th and 97.5th percentile of the medians,
this is your confidence interval.
c) Find the 5th and 95th percentiles of the medians, this
is your confidence interval.
Copyright © 2013 Pearson Education, Inc.
8.24 To make a 90% confidence interval for the
population median using the bootstrap method you first
randomly select 20,000 separate samples of size 8 from
the original data and then you compute the medians
from each of the new samples. What is the next step?
a) Find the average and standard deviation of the
20,000 medians. Then, compute a traditional 95%
s
x

t
confidence interval (
) with those values.
n
b) Find the 2.5th and 97.5th percentile of the medians,
this is your confidence interval.
c) Find the 5th and 95th percentiles of the medians, this
is your confidence interval.
Copyright © 2013 Pearson Education, Inc.