Chapter 8 answers - Amazon S3

[Pages:6]Chapter 8 Answers:

Section 8.1:

1) A confidence interval will become narrower if the sample size is increased. 3) A confidence interval will become narrower if the confidence level is decreased.

5) There is a 95% chance that the interval 6353 km < ? < 6384 km contains the true

value of the mean diameter of the Earth. 7) The proportion of Americans who believe it is the government's responsibility for

health care is between 52% and 60%.

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 p^ 0.1693 and n = 34,641. np^ = 34641* 0.1693 5864.7 5

and nq^ = 34641*(1- 0.1693) 28776.3 5 . So, the sampling distribution for

p^ is a normal distribution.

c.) Find the sample statistic and confidence interval Sample Proportion:

x = 5865 n = 34641 p^ = x = 5865 0.1693

n 34641 Confidence Interval zC = 2.575

p^q^

0.1693(1- 0.1693)

E = zC

= 2.575 n

34641

0.0052

p^ - E < p < p^ + E

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%. 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 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^ 0.4598 and n = 631. np^ = 3482 * 0.4598 1601.02 5 and

nq^ = 3482 *(1- 0.4598) 1880.98 5 . So, the sampling distribution for p^

is a normal distribution. c.) Find the sample statistic and confidence interval

Sample Proportion:

x = 1601 n = 3482 p^ = x = 1601 0.4598

n 3482 Confidence Interval zC = 1.645

p^ q^

0.4598(1- 0.4598)

E = zC

= 1.645 n

0.0139 3482

p^ - E < p < p^ + E

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%. 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 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^ 0.6102 and n = 1039. np^ = 1039 * 0.6102 633.998 5 and

nq^ = 1039 *(1- 0.6102) 405.002 5 . So, the sampling distribution for p^ is

a normal distribution. c.) Find the sample statistic and confidence interval

Sample Proportion:

x = 634 n = 1039 p^ = x = 634 0.6102

n 1039 Confidence Interval zC = 2.33

p^q^

0.6102(1- 0.6102)

E = zC

= 2.33 n

0.0353 1039

p^ - E < p < p^ + E 0.5750 < p < 0.6454 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%.

Section 8.3:

1) a.) State the random variable and the parameter in words. x = CO2 emissions in 2010

? = 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.

Co1l2lection 1

Histogram

Collection 1

Normal Quantile Plot

10 2

8

1

6

0

4

-1

2

-2

0

2

4

6

8

10

0

2

4

6

8

10

12

Emission

Emission

Normal Quantile = 0.337Emission - 1.16

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

tc = 2.797

s

2.96828

E = tc

= 2.797 n

1.6605 metric tons per capita 25

1.7944 < ? < 5.1152 metric tons per capita

d.) Statistical Interpretation: There is a 99% chance that the interval

1.7944 < ? < 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.

3) a.) State the random variable and the parameter in words.

x = amount of mercury in bass fish in Florida

? = 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

tc = 1.675

s

0.341036

E = tc

= 1.675 n

0.07847 mg/kg 53

0.44872 < ? < 0.60562 mg/kg

d.) Statistical Interpretation: There is a 90% chance that the interval

0.44872 < ? < 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.

5) a.) State the random variable and the parameter in words.

x = pulse rate after running for 1 minute of a female who drinks alcohol

? = 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.

Colle1c2tion 1 10 8 6 4 2

Histogram

Collection 1 2 1 0 -1 -2

Normal Quantile Plot

0 40 80 120 160 200 Pulse_rate

0 40 80 120 160 Pulse_rate

Normal Quantile = 0.0298Pulse_rate - 3

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

tc = 2.056

s

33.5609

E = tc

= 2.056 n

13.279 beats/min 27

87.2423 < ? < 113.795 beats/min

d.) Statistical Interpretation: There is a 95% chance that the interval

87.2423 < ? < 113.795 beats/mincontains 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.

7) a.) State the random variable and the parameter in words. x = percentage of women receiving prenatal care per country in 2009

? = 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

tc = 1.679

s

8.47562

E = tc

= 1.679 n

2.08% 47

88.8747% < ? < 93.0253%

d.) Statistical Interpretation: There is a 90% chance that the interval

88.8747% < ? < 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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