Chapter 7 Solutions:
Section 7.1:
1.)
2.)
3.)
4.)
5.)
6.)
7.)
8.)
Random variable: x = number of defects from scratches
Population parameter: p = proportion of defects from scratches
H o : p = 0.11
H1 : p > 0.11
Random variable: x = CO2 emissions in 2010
Population parameter: m = mean CO2 emissions in 2010
H o : m = 4.87 metric tons per capita
H1 : m < 4.87 metric tons per capita
Type I error: saying that the proportion of defects from scratches is higher than
11%, when the proportion is actually 11%. One consequence of this error is that
the company may waste time and energy looking at their manufacturing process
to see why there are more scratches than other defects when there really aren’t
more scratches.
Type II error: saying that the proportion of defects from scratches is 11%, when in
reality it is higher than 11%. One consequence of this error is that the company
will not look to see if they can change their manufacturing process to create less
scratches. This will mean there will be more lenses that are thrown away due to a
defect that is fixable.
The best alpha level in this case would be 5%, since both errors seem to have
similar consequences.
Type I error: saying that the mean CO2 emission in 2010 is less than 4.87 metric
tons per capita (2004 value) when the emissions haven’t changed. One
consequence of this error is that governments would think that what they are
doing to curb emissions hasn’t actually worked.
Type II error: saying that the mean CO2 emission in 2010 is 4.87 metric tons per
capita (2004 value) when the emissions are actually less. One consequence of this
error is the governments would think that what they are doing isn’t helping and
might abandon a method that is working.
The best alpha in this case would be 5%, since both errors have consequences that
are equally dire.
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Section 7.2:
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 the null and alternative hypotheses and the level of significance
H o : p = 0.11
H1 : p > 0.11
a = 0.01
c.) 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 p = 0.11 and n = 34,641. np = 34641*0.11 = 3810.51 ³ 5 and
nq = 34641* (1- 0.11) = 30830.59 ³ 5 . So, the sampling distribution for p̂ is
a normal distribution.
d.) Find the sample statistic, test statistic, and p-value
Sample Proportion:
x = 5865
n = 34641
x 5865
=
» 0.1693
n 34641
Test Statistic:
p̂ - p
0.1693 - 0.11
z=
=
» 35.274
pq
0.11(1- 0.11)
n
34641
p-value:
p-value = P ( z > 35.274 ) = normalcdf ( 35.274,1E99,0,1) » 0
or from technology
z » 35.262
p - value » 0
e.) Conclusion
Since the p-value < 0.01, reject H o .
f.) Interpretation
There is enough evidence to show that the proportion of defective lenses from
scratches is higher than 11%, the proportion from other causes.
p̂ =
2.)
3.)
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 the null and alternative hypotheses and the level of significance
H o : p = 0.23
H1 : p > 0.23
a = 0.05
c.) 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 p = 0.23 and n = 3482. np = 3482 *0.23 = 800.86 ³ 5 and
nq = 3482 * (1- 0.23) = 2681.14 ³ 5 . So, the sampling distribution for p̂ is a
normal distribution.
d.) Find the sample statistic, test statistic, and p-value
Sample Proportion:
x = 1601
n = 3482
x 1601
=
» 0.4598
n 3482
Test Statistic:
p̂ - p
0.4598 - 0.23
z=
=
» 32.222
pq
0.23(1- 0.23)
n
3482
p-value:
p-value = P ( z > 32.222 ) = normalcdf ( 32.222,1E99,0,1) » 0
or from technology
z » 32.221
p - value » 0
e.) Conclusion
Since the p-value < 0.05, reject H o .
f.) Interpretation
There is enough evidence to show that the proportion of complaints due to
identity theft in Arkansas is more than 23%.
p̂ =
4.)
5.)
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 the null and alternative hypotheses and the level of significance
H o : p = 0.81
H1 : p < 0.81
a = 0.01
c.) 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 p = 0.81 and n = 1029. np = 1039 *0.81 = 841.59 ³ 5 and
nq = 1039 * (1- 0.81) = 197.41 ³ 5 . So, the sampling distribution for p̂ is a
normal distribution.
d.) Find the sample statistic, test statistic, and p-value
Sample Proportion:
x = 634
n = 1039
x 634
=
» 0.6102
n 1039
Test Statistic:
p̂ - p
0.6102 - 0.81
z=
=
» -16.417
pq
0.81(1- 0.81)
n
1039
p-value:
p-value = P ( z < -16.417 ) = normalcdf ( -1E99,-16.417,0,1) » 0
or from technology
z » -16.416
p - value » 0
e.) Conclusion
Since the p-value < 0.01, reject H o .
f.) Interpretation
There is enough evidence to show that the proportion of American adults in
2013 who believe that there is a conspiracy in the death of President Kennedy
is less than 81%, the value in 2001.
p̂ =
6.)
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Section 7.3:
1.)
a.) State the random variable and the parameter in words.
x = CO2 emissions in 2010
m = mean CO2 emissions in 2010
b.) State the null and alternative hypotheses and the level of significance
H o : m = 4.87 metric tons per capita
H1 : m < 4.87 metric tons per capita
a = 0.01
c.) 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.
d.) Find the sample statistic, test statistic, and p-value
Sample mean and standard deviation:
x » 3.4548 metric tons per capita
s » 2.96828 metric tons per capita
n = 25
Test Statistic:
3.4548 - 4.87
t=
» -2.384
2.96828 25
p-value:
p-value = P ( t < -2.384 ) = tcdf ( -1E99,-2.384,24 ) » 0.0127
or from technology
t » -2.384
p - value » 0.013
e.) Conclusion
Since the p-value > 0.01, fail to reject H o .
f.) Interpretation
There is not enough evidence to show that the mean CO2 emissions in 2010 is
less than 4.87 metric tons per capita, the value in 2004.
2.)
3.)
4.)
5.)
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 the null and alternative hypotheses and the level of significance
H o : m = 1 mg/kg
H1 : m > 1 mg/kg
a = 0.10
c.) 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.
d.) Find the sample statistic, test statistic, and p-value
Sample mean and standard deviation:
x » 0.52717 mg/kg
s » 0.341036 mg/kg
n = 53
Test Statistic:
0.52717 -1
t=
» -10.0935
0.341036 53
p-value:
p-value = P ( t > -10.0935 ) = tcdf ( -10.0935,1E99,52 ) » 1
or from technology
t » -10.09
p - value » 1
e.) Conclusion
Since the p-value > 0.10, fail to reject H o .
f.) Interpretation
There is not enough evidence to show that the mean amount of mercury in
bass fish in Florida is more than the allowable amount of 1.0 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 the null and alternative hypotheses and the level of significance
H o : m = 97 beats/min
H1 : m > 97 beats/min
a = 0.05
c.) 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.
d.) Find the sample statistic, test statistic, and p-value
Sample mean and standard deviation:
x » 100.519 beats/min
s » 33.5609 beats/min
n = 27
Test Statistic:
100.519 - 97
t=
» 0.5448
33.5609 27
p-value:
p-value = P ( t > 0.5448 ) = tcdf ( 0.5448,1E99,26 ) » 0.2953
or from technology
t » 0.5448
6.)
7.)
p - value » 0.3
e.) Conclusion
Since the p-value > 0.05, fail to reject H o .
f.) Interpretation
There is not enough evidence to show that the mean pulse rate after running
for 1 minute of a female who drinks alcohol is more than 97 beats per minute,
the mean pulse rate after running for 1 minute for females who do not drink
alcohol.
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 the null and alternative hypotheses and the level of significance
8.)
H o : m = 80.1%
H1 : m > 80.1%
a = 0.05
c.) 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.
d.) Find the sample statistic, test statistic, and p-value
Sample mean and standard deviation:
x » 90.95%
s » 8.47562%
n = 47
Test Statistic:
90.95 - 80.1
t=
» 8.7762
8.47562 47
p-value:
p-value = P ( t > 8.7762 ) = tcdf ( 8.7762,1E99, 46 ) » 1.090 ´10 -11
or from technology
t » 8.776
p - value < 0.0001
e.) Conclusion
Since the p-value < 0.05, reject H o .
f.) Interpretation
There is enough evidence to show that the mean percentage of women
receiving prenatal care in 2009 is higher than 80.1%, the value in 1999.
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