Please do the ones that haven't been answered

8HEOCS6-E Question Help V at According to the Bureau of Labor Statistics, the average hourly wage in the United States was $26.84 in April 2018. To conrm this wage, a random sample of 39 hourly workers was selected during the month. The average wage for this sample was $25.14. Assume the standard deviation of wages for the country is $4.25. Complete parts a and b. a. Are the results of this sample consistent with the claim made by the Bureau of Labor Statistics using a 95% condence interval? The condence interval is $ 23.81 s x S $ 26.47 . The Bureau's claim is not between the lower limit and the upper limit of the condence interval, so the results of the sample are not consistent with the claim. (Round to two decimal places as needed.) I). What is the margin of error for this sample? The margin of error is $ , (Round to two decimal places as needed.) @ 8HEOC57-E Question Help V a A sports league introduced a video referee to review certain critical plays. However, the system has not been without its controversies. One particular unfortunate incident led to an outcry among fans. Nevertheless. a poll after this event found that 62% of fans were still in favor of the continued use of the video referee. Suppose the size of this sample was 600 people. Complete parts a and b. 3. Estimate the true proportion of fans in favor of keeping the video referee using a 95% condence interval. The oonfidenoe interval has a lower limit of 0.581 and an upper limit of 0.659 . (Round to three decimal places as needed.) b. What is the margin of error for this sample? The margin of error is . (Round to three decimal places as needed.) @ 8.EOC.58-E Question Help V 0 According to a government agency. 53.8% of mothers in a certain region were breast-feeding their nine-monthpld babies in 2016. Suppose a health insurance company would like to estimate the proportion of mothers in the region who breast-fed their children at nine months in 2018. A random sample of 240 mothers with nine-month-old babies was selected, 145 of which were breast-feeding. Complete parts a through c. a. Construct a 90% condence interval to estimate the actual proportion of mothers with nine-month-old babies that were breast-feeding in 2018. The condence interval has a lower limit of 0.552 and an upper limit of 0.656 . (Round to three decimal places as needed.) b. What is the margin of error for this sample? The margin of error is t (Round to three decimal places as needed) x) 8.EOC.59-E Question Help The following data show the costs charged by a tax preparation service for a random sample of 15 tax returns. Complete parts a through c. $130 $115 $125 $205 $95 $125 $95 $200 $145 $75 $155 $115 $185 $155 $65 a. Using a 95% confidence interval, estimate the average cost of the service for preparing a tax return for a customer. The 95% confidence interval is from $ 108.86 to $ 155.80 . (Round to two decimal places as needed. Use ascending order.) b. What is the margin of error for this sample? The margin of error is $. (Round to two decimal places as needed.) 8.EOC.64-E Question Helpv a A governmental department would like to estimate the average weekly wages for adults with a margin of error equal to $18, Determine the sample sizes needed to construct a condence interval for this estimate using the following condence levels. Assume the population standard deviation for the weekly wage is $160. a. 90% h. 98% c. 99% d. Explain why the condence level affects the sample size needed to obtain a particular margin of error. a. The sample size needed is 214 . (Round up to the nearest integer.) b. The new sample size needed would be 428 . (Round up to the nearest integer.) c. The new sample size needed would be 525 . (Round up to the nearest integer.) d. To be more condent that the interval contains the true population parameter given a constant margin of error, a sample size would be needed, since the value of iv will as the condence level increases. x 8.EOC.66-E Question Help Your school claims that the average student graduates with debt of $1,750. To validate this claim, you create a random sample of 20 students during graduation and ask each student to anonymously report the amount of their student loan debt. Complete parts a through c. $0 $0 $3 $300 $1,000 $0 $1,900 $2,250 $2,900 $1,900 $2,200 $2,200 $3, 100 $2,800 $0 $2, 100 $350 $2,350 $2,850 a. Construct a 95% confidence interval with these data to estimate the average student loan debt of students graduating at your school. The 95% confidence interval is from $ 848.95 to $ 1971.35 . (Round to two decimal places as needed. Use ascending order.) b. Do these results validate the claim of your school? Since $1,750 is contained within the 95% confidence interval, it can be said with 95% confidence that the sample validates the school's claim. c. What is the margin of error for this sample? The margin of error is $. (Round to two decimal places as needed.) 8.EOC.69'E Question Helpv at A government bureau would like to estimate the average annual contributions to charity by the members of Generation Y (people born in the United States in the years 1981 through 1991). The following data show the annual contributions to charity for a random sample of 18 individuals in this age group. Complete pans a and b. $20 $750 $390 $160 $195 $515 D1 $270 $420 $445 $192 $460 $330 $60 $595 $405 $160 $655 $220 a. Construct a 90% condence interval with these data to estimate the average annual contributions to charity for this age group. The 90% condence interval is from $ 263.06 to $ 430.50 . (Round to two decimal places as needed Use ascending order.) b. What is the margin of error for this sample? The margin of error is $ . (Round to two decimal places as needed.) 8HE0C70-E Question HelpV a According to recent data by the American Marketing Association, 76.5% of people in the 18-24 age group used social media in 2017. You want to nd out what the social media proportion is among your classmates who are all between 18 and 24 years old. Suppose you obtain a random sample of 231 students and nd that 207 students in this sample are active social media users. Complete parts a through c below. a. Construct a 90% confidence interval to estimate the actual proportion of social media users. A 90% condence interval to estimate the actual proportion has a lower limit of 0.863 and an upper limit of 0.929 . (Round to three decimal places as needed.) I). What is the margin of error for this sample? The margin of error is . (Round to three decimal places as needed.) 8.EOC.71-E Question Help At its peak after the 2007-2008 recession, credit card delinquency rates in a certain region reached 10.02%. Since its peak, delinquency rates have declined to a new low of 5.41% in the second quarter of 2015. You are charged with checking the most recent credit card delinquency rates in the region. Using a random sample of 449 credit card customers, you found that 28 were at least 30 days overdue on their monthly payments. Using a 90% confidence interval, can you conclude that the proportion of delinquent card holders changed since its lowest point in 2015? The 90% confidence interval to estimate the actual most recent proportion of delinquent card holders has a lower limit of 0.044 and an upper limit of 0.081 . (Round to three decimal places as needed.) Based on the confidence interval, it be concluded that this proportion has changed since 2015, because the confidence interval includes zero. confidence interval does not include zero. the margin of error is greater than zero. confidence interval does not include the proportion from 2015. confidence interval includes the proportion from 2015. the margin of error is less than zero. the margin of error is less than the proportion from 2015. the margin of error is greater than the proportion from 2015