Please solve this.
The data shown below represent the repair cost for a lowimpact collision in a simple random sample of mini and micro-vehicles. Complete parts (a) through (d) below. $3119 $1007 $713 $635 $799 Q $1?80 $3380 $203?r $2605 $1360 a Click the icon to View the table of critical correlation coefcient values for normal probability plots. (a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. Choose the correct answer below. O A O B. O c O D ., 2 '5?- ., 2 '9' m 2 s ., 2 "its 1 s 1 5 1 Q Q Q Q g 0 g 0 g 0 g 0 13' -' ". E - T5 _ 1'5 , ' 3-1 '3': 3-1 cr- 3-1 -I:."- 3-1 .3; m . 7 . 7 m . 7 . _7 0 2000 4000 0 2000 4000 0 2000 4000 0 2000 4000 Repair Cost (5) Repair Cost ($3 Repair Cost (5] Repair Cost (3) Since the correlation between the expected z-scores and the observed data, , rvl the critical value , it 'Yl reasonable to conclude that the data come from a population that is normally distributed. (Round to three decimal places as needed ) The data shown below represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles. Complete parts (a) through (d) below $3119 $1007 $713 $635 $799 $1780 $3380 $2037 $2605 $1360 Click the icon to view the table of critical correlation coefficient values for normal probability plots. (b) Draw a boxplot to check for outliers. Choose the correct answer below. O A. O B. O c. O D. 2000 4000 2000 4000 o 2000 4000 O 2000 4000 Does the boxplot suggest that there are outliers? O A. Yes, there is at least one point that is greater than the third quartile or less than the first quartile. O B. Yes, there is at least one point that is outside of the 1.5(IQR) boundaries. O C. No, there are no points that are outside of the 1.5(IQR) boundaries. O D. No, there are no points that are greater than the third quartile or less than the first quartile.The data shown below represent the repair cost for a low-impact collision in a simple random sample of mini and micro-vehicles. Complete parts (a) through (d) below. $3119 $100? $713 $635 $799 Q $1?80 $3380 $203? $2605 $1360 5 Click the icon to view the table of critical correlation coefcient values for normal probability plots. O B. Yes, there is at least one point that is outside of the 1.5(IQR) boundaries. 0 C. No, there are no points that are outside of the 1.5(IQR) boundaries. 0 D. No, there are no points that are greater than the third quartile or less than the rst quartile (c) Construct and interpret a 95% condence interval for population mean cost of repair. Select the correct choice below and ll in the answer boxes to complete your choice. (Round to the nearest dollar. Use ascending order.) O A- There is a 95% probability that the mean cost of repair is between $ and $ O B- We are 95% confident that the mean cost of repair lS between $ and $ (d) Suppose you obtain a simple random sample of size n =10 of a specic type of minivehicle that was in a lowimpact collision and determine the cost of repair. Do you think a 95% condence interval would be wider or narrower? Explain. O A. Wider, because there is more variability in the data because variability in the repair cost of the car has been added. 0 B. Narrower, because there is less variability in the data because any variability caused by the different types of vehicles has been removed. 0 C. Narrower, because taking a second random sample will always lead to a narrower condence interval. 0 D. Wider, because taking a second random sample will always lead to a wider condence interval A The accompanying data represent the total travel tax (in dollars} for a 37day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 6924 7929 67.03 83.93 82.96 882610199 99.54EI-_ (3) Determine a point estimate for the population mean travel tax. Apoint estimate for the population mean travel tax is $ (Round to two decimal places as needed.) (b) Construct and interpret a 99% condence interval for the mean tax paid for a threeday business trip Select the correct choice below and ll in the answer boxes to complete your choice. (Round to two decimal places as needed.) O A- One can be % condent that all cities have a travel tax between \".5 and $ O B- The travel tax is between $ and $ for % of all cities. O C. There is a % probability that the mean travel tax for all Cities is be ween $ and $ O 9- One can be % condent that the mean travel tax for all cities is between $ and $ (c) What would you recommend to a researcher who wants to increase the preciston of the interval. but does not have access to additional data? The researcher could "'I' the 3'" A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 114, and the sample standard deviation, s, is found to be 11. (a) Construct a 95% confidence interval about p if the sample size, n, is 23. (b) Construct a 95% confidence interval about p if the sample size, n, is 14. (c) Construct a 90% confidence interval about p if the sample size, n, is 23. (d) Should the confidence intervals in parts (a)-(c) have been computed if the population had not been normally distributed? (a) Construct a 95% confidence interval about p if the sample size, n, is 23. Lower bound: ; Upper bound: (Round to one decimal place as needed.) (b) Construct a 95% confidence interval about p if the sample size, n, is 14. Lower bound: ; Upper bound: (Round to one decimal place as needed.) How does decreasing the sample size affect the margin of error, E? O A. As the sample size decreases, the margin of error increases. O B. As the sample size decreases, the margin of error stays the same. O C. As the sample size decreases, the margin of error decreases. (c) Construct a 90% confidence interval about p if the sample size, n, is 23. Lower bound: ; Upper bound: (Round to one decimal place as needed.)The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (b) According to a certain survey, adults spend 2.45 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching television on a weekday" is 1.93 hours. If a random sample of 60 adults is obtained, describe the sampling distribution of x, the mean amount of time spent watching television on a weekday. is approximately normal with us = 2.45 and o- = 1.93 (Round to six decimal places as needed.) (c) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is . (Round to four decimal places as needed.) (d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.08 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.08 hours or less from a population whose mean is presumed to be 2.45 hours. The likelihood is . (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) O A. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect a sample mean of 2.08 or more in about of the samples. O B. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect a sample mean of exactly 2.08 in about of the samples. O C. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect aThe amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (c) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is (Round to four decimal places as needed.) (d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.08 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.08 hours or less from a population whose mean is presumed to be 2.45 hours. The likelihood is (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) O A. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect a sample mean of 2.08 or more in about of the samples. B. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect a sample mean of exactly 2.08 in about of the samples. O C. If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.45 hours is obtained, we would expect a sample mean of 2.08 or less in about of the samples. Based on the result obtained, do you think avid Internet users watch less television? O No O Yes
