Chapter 8 Solutions:
Section 8.1:
1.)
2.)
3.)
4.)
5.)
6.)
7.)
8.)
A confidence interval will become narrower if the sample size is increased.
A confidence interval will become narrower if the confidence level is decreased.
There is a 95% chance that the interval 6353 km < m < 6384 km contains the true
value of the mean diameter of the Earth.
The proportion of Americans who believe it is the government’s responsibility for
health care is between 52% and 60%.
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Section 8.2:
1.)
a.) State the random variable and the parameter in words.
x = number of defects from scratches
p = proportion of defects from scratches
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of defect types from 34,641 defective lenses was
taken. This assumption may not be met, since they collected the data in a
three-month time period. However, unless there was something special about
that time period, the sample is probably a representative sample. This
assumption is probably met.
ii. There are 34,641 defective lenses in this sample. The type of defect on one
lens should not affect the defect on the next one, unless there is something
wrong with a machine. There are only two outcomes, either the lens is
scratched or it isn’t. The chance that one defective lens is because of a scratch
is the same for all lenses, unless there is something wrong with the machine.
Thus the conditions for the binomial distribution are satisfied
iii. In this case p̂ » 0.1693 and n = 34,641. np̂ = 34641*0.1693 » 5864.7 ³ 5
and nq̂ = 34641* (1- 0.1693) » 28776.3 ³ 5 . So, the sampling distribution for
p̂ is a normal distribution.
c.) Find the sample statistic and confidence interval
Sample Proportion:
x = 5865
n = 34641
x 5865
=
» 0.1693
n 34641
Confidence Interval
zC = 2.575
p̂ =
E = zC
0.1693 (1- 0.1693)
p̂q̂
= 2.575
» 0.0052
n
34641
p̂ - E < p < p̂ + E
0.1693 - 0.0052 < p < 0.1693 + 0.0052
2.)
3.)
0.1641 < p < 0.1745
or from technology
0.1641 < p < 0.1745
d.) Statistical Interpretation: There is a 99% chance that the interval
0.1641 < p < 0.1745 contains the true proportion of defects that are from
scratches.
e.) Real World Interpretation: The proportion of defects that are from scratches is
between 16.4% and 17.5%.
a.) State the random variable and the parameter in words.
x = number of complaints from identity theft in Arkansas
p = proportion of complaints from identity theft in Arkansas
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the category of 3,482 complaints of identity theft
in Arkansas was taken. The study says that the complaints were out of all
complaints that year, but the year could have been chosen at random. This
assumption is may be met, but you can’t be sure.
ii. There are 3,482 complaints in this sample. The reason for the complaint does
not affect the next complaint. There are only two outcomes, either the
complaint was for identity theft or it wasn’t. The chance that one complaint
was for identity theft does not change. Thus the conditions for the binomial
distribution are satisfied
iii. In this case p̂ » 0.4598 and n = 631. np̂ = 3482 *0.4598 » 1601.02 ³ 5 and
nq̂ = 3482 * (1- 0.4598 ) » 1880.98 ³ 5 . So, the sampling distribution for p̂
is a normal distribution.
c.) Find the sample statistic and confidence interval
Sample Proportion:
x = 1601
n = 3482
x 1601
=
» 0.4598
n 3482
Confidence Interval
zC = 1.645
p̂ =
E = zC
0.4598 (1- 0.4598 )
p̂q̂
= 1.645
» 0.0139
n
3482
p̂ - E < p < p̂ + E
0.4598 - 0.0139 < p < 0.4598 + 0.0139
4.)
5.)
0.4459 < p < 0.4737
or from technology
0.4459 < p < 0.4737
d.) Statistical Interpretation: There is a 90% chance that the interval
0.4459 < p < 0.4737 contains the true proportion of complaints from identity
theft in Arkansas.
e.) Real World Interpretation: The proportion of complaints from identity theft in
Arkansas is between 44.6% and 47.4%.
a.) State the random variable and the parameter in words.
x = number of American adults in 2013 who believe that there was a conspiracy in
the death of President Kennedy
p = proportion of American adults in 2013 who believe that there was a
conspiracy in the death of President Kennedy
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the 1039 opinions of American adults about the
Kennedy assassination was taken in 2013. The study was conducted by the
Gallup poll so this assumption is probably true.
ii. There are 1039 opinions in this sample. The opinion of one American adult
doesn’t affect the opinion of the next one. There are only two outcomes, either
the American adult believes there was a conspiracy or they don’t. The chance
that one American believes there is a conspiracy does not change. Thus the
conditions for the binomial distribution are satisfied
iii. In this case p̂ » 0.6102 and n = 1039. np̂ = 1039 *0.6102 » 633.998 ³ 5 and
nq̂ = 1039 * (1- 0.6102 ) » 405.002 ³ 5 . So, the sampling distribution for p̂ is
a normal distribution.
c.) Find the sample statistic and confidence interval
Sample Proportion:
x = 634
n = 1039
x 634
=
» 0.6102
n 1039
Confidence Interval
zC = 2.33
p̂ =
E = zC
0.6102 (1- 0.6102 )
p̂q̂
= 2.33
» 0.0353
n
1039
p̂ - E < p < p̂ + E
0.6102 - 0.0353 < p < 0.6102 + 0.0352
0.5749 < p < 0.6455
or from technology
0.5750 < p < 0.6454
6.)
d.) Statistical Interpretation: There is a 98% chance that the interval
0.5750 < p < 0.6454 contains the true proportion of American adults in 2013
who believe that there was a conspiracy in the death of President Kennedy.
e.) Real World Interpretation: The proportion of American adults in 2013 who
believe that there was a conspiracy in the death of President Kennedy is
between 57.5% and 64.5%.
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Section 8.3:
1.)
a.) State the random variable and the parameter in words.
x = CO2 emissions in 2010
m = mean CO2 emissions in 2010
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the 25 CO2 emission was taken. The problem
mentioned that the sample was a random sample. So this requirement has been
met.
ii. The population of all CO2 emissions is normally distributed or the sample size
is 30 or more. The sample size is 25. The histogram looks skewed right, there
are no outliers, and the normal probability plot does not look linear. So this
requirement has not been met, so a larger sample might be in order.
c.) Find the sample statistic and confidence interval
Sample mean and standard deviation:
x » 3.4548 metric tons per capita
s » 2.96828 metric tons per capita
n = 25
Confidence Interval
t c = 2.797
s
2.96828
E = tc
= 2.797
» 1.6605 metric tons per capita
n
25
x-E<m<x+E
3.4548 -1.6605 < m < 3.4548 +1.6605
1.7943 < m < 5.1153 metric tons per capita
or from technology
1.7944 < m < 5.1152 metric tons per capita
2.)
3.)
d.) Statistical Interpretation: There is a 99% chance that the interval
1.7944 < m < 5.1152 metric tons per capita contains the true mean CO2
emissions in 2010.
e.) Real World Interpretation: The mean CO2 emissions in 2010 is between
1.7944 metric tons per capita and 5.1152 metric tons per capita.
a.) State the random variable and the parameter in words.
x = amount of mercury in bass fish in Florida
m = mean amount of mercury in bass fish in Florida
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the amount of mercury in bass fish in 53 lakes in
Florida. The problem doesn’t mention how the sample was taken. So this
requirement may not have been met.
ii. The population of the amount of mercury in bass fish in Florida is normally
distributed or the sample size is 30 or more. The sample size is 53. So this
requirement may be met.
c.) Find the sample statistic and confidence interval
Sample mean and standard deviation:
x » 0.52717 mg/kg
s » 0.341036 mg/kg
n = 53
Confidence Interval
t c = 1.675
s
0.341036
E = tc
= 1.675
» 0.07847 mg/kg
n
53
x-E<m<x+E
0.52717 - 0.07847 < m < 0.52717 + 0.07847
0.44870 < m < 0.60564 mg/kg
or from technology
0.44872 < m < 0.60562 mg/kg
4.)
5.)
d.) Statistical Interpretation: There is a 90% chance that the interval
0.44872 < m < 0.60562 mg/kg contains the true mean amount of mercury in
bass fish in Florida.
e.) Real World Interpretation: The mean amount of mercury in bass fish in
Florida is between 0.44872 mg/kg and 0.60562 mg/kg.
a.) State the random variable and the parameter in words.
x = pulse rate after running for 1 minute of a female who drinks alcohol
m = mean pulse rate after running for 1 minute of a female who drinks alcohol
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the pulse rates after running for 1 minute for a
female who drinks alcohol was taken. The problem doesn’t mention how the
sample was taken. So this requirement may not have been met.
ii. The population of the pulse rate after running for 1 minute of a female who
drinks alcohol is normally distributed or the sample size is 30 or more. The
sample size is 27. The histogram looks skewed right, but there are no outliers
and the normal probability plot does appear somewhat linear. So this
requirement may be met.
c.) Find the sample statistic and confidence interval
Sample mean and standard deviation:
x » 100.519 beats/min
s » 33.5609 beats/min
n = 27
Confidence Interval
t c = 2.056
s
33.5609
E = tc
= 2.056
» 13.279 beats/min
n
27
x-E<m<x+E
100.519 -13.279 < m < 100.519 + 13.279
87.240 < m < 113.798 beats/min
or from technology
87.2423 < m < 113.795 beats/min
6.)
7.)
d.) Statistical Interpretation: There is a 95% chance that the interval
87.2423 < m < 113.795 beats/min contains the true mean pulse rate after
running for 1 minute of a female who drinks alcohol.
e.) Real World Interpretation: The mean pulse rate after running for 1 minute of a
female who drinks alcohol is between 87.2 and 113.8 beats/min.
a.) State the random variable and the parameter in words.
x = percentage of women receiving prenatal care per country in 2009
m = mean percentage of women receiving prenatal care per country in 2009
b.) State and check the assumptions for a hypothesis test
i. A simple random sample of the percentage of women receiving prenatal care
in 2009 in 47 countries was taken. The problem doesn’t mention how the
sample was taken. So this requirement may not have been met.
ii. The population of the percentage of women receiving prenatal care per
country in 2009 is normally distributed or the sample is 30 or more. The
sample size is 47. So this requirement has been met.
c.) Find the sample statistic and confidence interval
Sample mean and standard deviation:
x » 90.95%
s » 8.47562%
n = 47
Confidence Interval
t c = 1.679
s
8.47562
E = tc
= 1.679
» 2.08%
n
47
x-E<m<x+E
90.95 - 2.08 < m < 90.95 + 2.08
88.87% < m < 93.03%
or from technology
88.8747% < m < 93.0253%
8.)
d.) Statistical Interpretation: There is a 90% chance that the interval
88.8747% < m < 93.0253% contains the true mean percentage of women
receiving prenatal care per country in 2009.
e.) Real World Interpretation: The mean percentage of women receiving prenatal
care per country in 2009 is between 88.87% and 93.03%.
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