Chapter 8 Answers: Section 8.1: Section 8.2:

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1 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 < µ < 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 5% and 60%. Section 8.: 1) a.) State the random variable and the parameter in words. x = number of defects from scratches p = proportion of defects from scratches 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 ˆp and n = 34,641. nˆp = 34641* and n ˆq = 34641* ( ) So, the sampling distribution for ˆp is a normal distribution. Sample Proportion: x = 5865 n = ˆp = x n = z C =.575 E = z C ˆp ˆq n ˆp E < p < ˆp + E =.575 ( )

2 < p < d.) Statistical Interpretation: There is a 99% chance that the interval < p < 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 i. A simple random sample of the category of 3,48 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 3,48 complaints in this sample. The reason for the complaint does not affect the nexomplaint. 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 nohange. Thus the conditions for the binomial distribution are satisfied iii. In this case ˆp and n = 631. nˆp = 348 * and n ˆq = 348 *( ) So, the sampling distribution for ˆp is a normal distribution. Sample Proportion: x = 1601 n = 348 ˆp = x n = z C = E = z C ˆp ˆq n = ( ) ˆp E < p < ˆp + E < p < d.) Statistical Interpretation: There is a 90% chance that the interval < p < 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 013 who believe that there was a conspiracy in the death of President Kennedy

3 p = proportion of American adults in 013 who believe that there was a conspiracy in the death of President Kennedy i. A simple random sample of the 1039 opinions of American adults about the Kennedy assassination was taken in 013. 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 nohange. Thus the conditions for the binomial distribution are satisfied iii. In this case ˆp and n = nˆp = 1039 * and n ˆq = 1039 *( ) So, the sampling distribution for ˆp is a normal distribution. Sample Proportion: x = 634 n = 1039 ˆp = x n = z C =.33 E = z C ˆp ˆq n =.33 ( ) ˆp E < p < ˆp + E < p < d.) Statistical Interpretation: There is a 98% chance that the interval < p < contains the true proportion of American adults in 013 who believe that there was a conspiracy in the death of President Kennedy. e.) Real World Interpretation: The proportion of American adults in 013 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 = CO emissions in 010 µ = mean CO emissions in 010 i. A simple random sample of the 5 CO emission was taken. The problem mentioned that the sample was a random sample. So this requirement has been met.

4 ii. The population of all CO emissions is normally distributed or the sample size is 30 or more. The sample size is 5. 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 Histogram Normal Quantile Plot x metric tons per capita s.9688 metric tons per capita n = 5 =.797 s.9688 E = = metric tons per capita n < µ < metric tons per capita d.) Statistical Interpretation: There is a 99% chance that the interval < µ < metric tons per capita contains the true mean CO emissions in 010. e.) Real World Interpretation: The mean CO emissions in 010 is between metric tons per capita and metric tons per capita. 3) a.) State the random variable and the parameter in words. x = amount of mercury in bass fish in Florida µ = mean amount of mercury in bass fish in Florida 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. x mg/kg s mg/kg n = Emission Emission Normal Quantile = 0.337Emission

5 = s E = = mg/kg n < µ < mg/kg d.) Statistical Interpretation: There is a 90% chance that the interval < µ < 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 mg/kg and 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 µ = mean pulse rate after running for 1 minute of a female who drinks alcohol 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 7. 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 Pulse_rate x beats/min s beats/min Histogram n = 7 =.056 s E = = beats/min n < µ < beats/min d.) Statistical Interpretation: There is a 95% chance that the interval < µ < beats/mincontains the true mean pulse rate after running for 1 minute of a female who drinks alcohol Normal Quantile Plot Pulse_rate Normal Quantile = 0.098Pulse_rate - 3

6 e.) Real World Interpretation: The mean pulse rate after running for 1 minute of a female who drinks alcohol is between 87. and beats/min. 7) a.) State the random variable and the parameter in words. x = percentage of women receiving prenatal care per country in 009 µ = mean percentage of women receiving prenatal care per country in 009 i. A simple random sample of the percentage of women receiving prenatal care in 009 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 009 is normally distributed or the sample is 30 or more. The sample size is 47. So this requirement has been met. x 90.95% s % n = 47 = s E = = % n % < µ < % d.) Statistical Interpretation: There is a 90% chance that the interval % < µ < % contains the true mean percentage of women receiving prenatal care per country in 009. e.) Real World Interpretation: The mean percentage of women receiving prenatal care per country in 009 is between 88.87% and 93.03%.