Chapter 7 answers - Amazon S3

Chapter 7 Answers:

Section 7.1:

1) Random variable: x = number of defects from scratches Population parameter: p = proportion of defects from scratches Ho : p = 0.11 H1 : p > 0.11

3) Random variable: x = CO2 emissions in 2010

Population parameter: ? = mean CO2 emissions in 2010 Ho : ? = 4.87 metric tons per capita H1 : ? < 4.87 metric tons per capita

5) 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.

7) 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.

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 Ho : p = 0.11 H1 : p > 0.11

= 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 5 and

nq = 34641*(1- 0.11) = 30830.59 5 . So, the sampling distribution for p^

is a normal distribution. d.) Find the sample statistic, test statistic, and p-value

Sample Proportion:

x = 5865 n = 34641 0.1693

Test Statistic:

35.274 p-value: p-value = e.) Conclusion

> 35.274 0

Reject Ho . 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. 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

Ho : p = 0.23

H1 : p > 0.23

= 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 5 and

nq = 3482 *(1- 0.23) = 2681.14 5 . So, the sampling distribution for p^ is a

normal distribution. d.) Find the sample statistic, test statistic, and p-value

Sample Proportion:

x = 1601 n = 3482

0.4598

Test Statistic:

32.222 p-value: p-value = e.) Conclusion

> 32.222 0

Since the p-value < 0.05, reject Ho . f.) Interpretation

There is enough evidence to show that the proportion of complaints due to identity theft in Arkansas is more than 23%. 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

Ho : p = 0.81

H1 : p < 0.81 = 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 5 and

nq = 1039 *(1- 0.81) = 197.41 5 . So, the sampling distribution for p^ is a

normal distribution. d.) Find the sample statistic, test statistic, and p-value

Sample Proportion:

x = 634 n = 1039

0.6102

Test Statistic:

-16.417

p-value: p-value = < -16.417 0 e.) Conclusion Since the p-value < 0.01, reject Ho . 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.

Section 7.3:

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

? = mean CO2 emissions in 2010

b.) State the null and alternative hypotheses and the level of significance

Ho : ? = 4.87 metric tons per capita

H1 : ? < 4.87 metric tons per capita = 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.

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

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:

-2.384 p-value: p-value = < -2.384 0.0127 e.) Conclusion Since the p-value > 0.01, fail to reject Ho .

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.

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 the null and alternative hypotheses and the level of significance

Ho : ? = 1 mg/kg

H1 : ? > 1 mg/kg = 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:

-10.0935 p-value: p-value = > -10.0935 1 e.) Conclusion Since the p-value > 0.10, fail to reject Ho . 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. 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 the null and alternative hypotheses and the level of significance

Ho : ? = 97 beats/min

H1 : ? > 97 beats/min = 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

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