MATH 123-Find the critical value z Subscript czc necessary

1. Find the critical value z Subscript czc necessary to form a confidence interval at the level ofconfidence shown below.c=0.87z Subscript czc=nothing(Round to two decimal places as needed.) 2. Use the values on the number line to find the sampling error.24252627x=26.08 ?=24.49The sampling error isnothing. 3. Find the margin of error for the given values of c,?, and n.c=0.900.90,?=3.73.7,n=4949 Click the icon to view a table of common critical values.E= (Round to three decimal places as needed.)Enter your answer in the answer box. Table of Common Critical Values Level of Confidence z Subscript czc90% 1.645 95% 1.96 99% 2.575 4. Construct the confidence interval for the population mean ?.c=0.95, x=6.5, ?=0.7, and n=51A 95% confidence interval for ? is,). (Round to two decimal places as needed.)Enter your answer in each of the answer boxes.5. A nurse at a local hospital is interested in estimating the birth weight of infants. How large asample must she select if she desires to be 99% confident that the true mean is within 2 ouncesof the sample mean? The standard deviation of the birth weights is known to be 5 ounces. Enter your answer in the answer box. 6. A private opinion poll is conducted for a politician to determine what proportion of thepopulation favors decriminalizing marijuana possession. How large a sample is needed in orderto be 95% confident that the sample proportion will not differ from the true proportion by morethan 5%?Enter your answer in the answer box.7. Use the confidence interval to find the margin of error and the sample mean.(0.462,0.560)The margin of errorThe sample mean isnothing.Enter your answer in each of the answer boxes.8. Use the confidence interval to find the estimated margin of error. Then find the sample mean.A biologist reports a confidence interval of (1.7,3.1) when estimating the mean height (incentimeters) of a sample of seedlings.The estimated margin of error isnothing. The sample mean isEnter your answer in each of the answer boxes.9. A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How manysubjects are needed to estimate the HDL cholesterol within 44 points with 99% confidenceassuming sigma equals 18.6 question mark ?=18.6? Suppose the doctor would be content with95% confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requiresnothing subjects.(Round up to the nearest whole number as needed.)A 95% confidence level requiresnothing subjects.(Round up to the nearest whole number as needed.)How does the decrease in confidence affect the sample size required?A.The lower the confidence level the smaller the sample size.B.The sample size is the same for all levels of confidence.C.The lower the confidence level the larger the sample size.Click to select your answer(s).10. Find the critical value tc for the confidence level c=0.9999 and sample size n=10.LOADING… Click the icon to view the t-distribution table.tc=nothing (Round to the nearest thousandth as needed.)Enter your answer in the answer box.11. Find the margin of error for the given values of c, s, and n.c=0.98, s=6, n=1919LOADING… Click the icon to view the t-distribution table.The margin of error isnothing. (Round to one decimal place as needed.)Enter your answer in the answer box.12. Find the margin of error for the given values of c, s, and n.c=0.9999, s=3.3, n=25LOADING… Click the icon to view the t-distribution table.The margin of error isnothing. (Round to one decimal place as needed.)Enter your answer in the answer box.13. Use the given confidence interval to find the margin of error and the sample mean.(5.795.79,9.679.67)The sample mean is(Type an integer or a decimal.)Enter your answer in each of the answer boxes. 14. In a random sample of 25 people, the mean commute time to work was 30.9 minutes and thestandard deviation was 7.1 minutes. Assume the population is normally distributed and use a tdistribution to construct a 80% confidence interval for the population mean ?. What is themargin of error of ?? Interpret the results.The confidence interval for the population mean ? is.( , ).(Round to one decimal place as needed.)The margin of error of ? is(Round to one decimal place as needed.)Interpret the results.A.With 80% confidence, it can be said that the commute time is between the bounds of theconfidence interval.B.With 80% confidence, it can be said that the population mean commute time is between thebounds of the confidence interval.C.It can be said that 80% of people have a commute time between the bounds of the confidenceinterval.D.If a large sample of people are taken approximately 80% of them will have commute timesbetween the bounds of the confidence interval.Click to select your answer(s).15. In a random sample of 13 microwave ovens, the mean repair cost was $90.00 and the standarddeviation was $15.10. Using the standard normal distribution with the appropriate calculationsfor a standard deviation that is known, assume the population is normally distributed, find themargin of error and construct a 98% confidence interval for the population mean ?. A 98%confidence interval using the t-distribution was left parenthesis 78.8, 101.2 (78.8,101.2)….pare the results.The margin of error of ? isnothing.(Round to two decimal places as needed.)A 98% confidence interval for ? is .( , ).(Round to one decimal place as needed.)16. Compare the results. Choose the correct answer below.A.The confidence interval found using the standard normal distribution is wider thanwider thanthe confidence interval found using the student’s t-distribution.B.The confidence interval found using the standard normal distribution has smaller lower andupper confidence interval limits.C.The confidence interval found using the standard normal distribution is narrower thannarrowerthan the confidence interval found using the student’s t-distribution. D.The confidence interval found using the standard normal distribution is the same asthe same asthe confidence interval found using the student’s t-distribution.Click to select your answer(s).The following data represent the concentration of dissolved organic carbon (mg/L) collectedfrom 20 samples of organic soil. Assume that the population is normally distributed. Completeparts (a) through (c) on the right.11.9029.8027.1016.5111.408.815.3020.4614.9033.6730.9114.8611.4015.359.7219.8014.868.0922.4918.30(a) Find the sample mean.The sample mean isnothing.(Round to two decimal places as needed.)(b) Find the sample standard deviation.The sample standard deviation isnothing.(Round to two decimal places as needed.)(c) Construct a 98% confidence interval for the population mean ?.The 98% confidence interval for the population mean ? is.( , ).(Round to two decimal places as needed.)Enter your answer in each of the answer boxes.17. The monthly incomes for 12 randomly selected people, each with a bachelor’s degree ineconomics, are shown on the right. Complete parts (a) through (c) below.Assume the population is normally distributed. 4450.914450.914596.174596.174366.424366.424455.284455.284151.744151.743727.093727.094283.424283.424527.954527.954407.934407.933946.263946.264023.834023.834221.124221.12(a) Find the sample mean.x=nothing (Round to one decimal place as needed.)(b) Find the sample standard deviation.s=nothing (Round to one decimal place as needed.)(c) Construct a 90% confidence interval for the population mean ?.A 90% confidence interval for the population mean is .( , ).(Round to one decimal place as needed.)Enter your answer in each of the answer boxes.18. Use the standard normal distribution or the t-distribution to construct a 90% confidence intervalfor the population mean. Justify your decision. If neither distribution can be used, explain why.Interpret the results.In a random sample of 24 mortgage institutions, the mean interest rate was 3.49% and thestandard deviation was 0.41%. Assume the interest rates are normally distributed.Which distribution should be used to construct the confidence interval?A.Use a t-distribution because the interest rates are normally distributed and ? is known.B.Use a normal distribution because the interest rates are normally distributed and ? is known.C.Use a normal distribution because nless than<30 and the interest rates are normally distributed.D.Use a t-distribution because it is a random sample, ? is unknown, and the interest rates arenormally distributed.E.Cannot use the standard normal distribution or the t-distribution because ? is unknown, n lessthan 30n<30, and the interest rates are not normally distributed.Select the correct choice below and, if necessary, fill in any answer boxes to complete yourchoice.A.The 90% confidence interval is .( , ).(Round to two decimal places as needed.) B.Neither distribution can be used to construct the confidence interval.Interpret the results. Choose the correct answer below.A.It can be said that 90% of institutions have an interest rate between the bounds of theconfidence interval.B.If a large sample of institutions are taken approximately 90% of them will have an interest ratebetween the bounds of the confidence interval.C.With 90% confidence, it can be said that the population mean interest rate is between thebounds of the confidence interval.D.Neither distribution can be used to construct the confidence interval.Click to select your answer.19. Let p be the population proportion for the following condition. Find the point estimates for pand q.Of 871 children surveyed, 117 plan to join the armed forces in the future.The point estimate for p, p with caretp, isnothing.(Round to three decimal places as needed.)The point estimate for q, q with caretq, is(Round to three decimal places as needed.)Enter your answer in each of the answer boxes.20. Use the given confidence interval to find the margin of error and the sample proportion.(0.605,0.633)E=(Type an integer or a decimal.)p with caretpequals=nothing (Type an integer or a decimal.)Enter your answer in each of the answer boxes.21. In a survey of 655 males ages 18-64, 396 say they have gone to the dentist in the past year.Construct 90% and 95% confidence intervals for the population proportion. Interpret the resultsand compare the widths of the confidence intervals. If convenient, use technology to constructthe confidence intervals.The 90% confidence interval for the population proportion p is.( , ).(Round to three decimal places as needed.)The 95% confidence interval for the population proportion p is.( , ).(Round to three decimal places as needed.)Interpret your results of both confidence intervals.A. With the given confidence, it can be said that the sample proportion of males ages 18-64 whosay they have gone to the dentist in the past year is between the endpoints of the givenconfidence interval.B.With the given confidence, it can be said that the population proportion of males ages 18-64who say they have gone to the dentist in the past year is between the endpoints of the givenconfidence interval.C.With the given confidence, it can be said that the population proportion of males ages 18-64who say they have gone to the dentist in the past year is not between the endpoints of the givenconfidence interval.Which interval is wider?The 90% confidence intervalThe 95% confidence intervalClick to select your answer(s).22. In a survey of 8000 women, 5431 say they change their nail polish once a week. Construct a 99%confidence interval for the population proportion of women who change their nail polish once aweek.A 99% confidence interval for the population proportion is ( , ).(Round to three decimal places as needed.)Enter your answer in each of the answer boxes.23. A researcher wishes to estimate, with 99% confidence, the population proportion of adults whothink the president of their country can control the price of gasoline. Her estimate must beaccurate within 33% of the true proportion.a) No preliminary estimate is available. Find the minimum sample size needed.b) Find the minimum sample size needed, using a prior study that found that 38% of therespondents said they think their president can control the price of gasoline.c) Compare the results from parts (a) and (b).(a) What is the minimum sample size needed assuming that no prior information is available?n=nothing (Round up to the nearest whole number as needed.)(b) What is the minimum sample size needed using a prior study that found that 38% of therespondents said they think their president can control the price of gasoline?n=nothing (Round up to the nearest whole number as needed.)(c) How do the results from (a) and (b) compare?A.Having an estimate of the population proportion has no effect on the minimum sample sizeneeded.B.Having an estimate of the population proportion raises the minimum sample size needed.C. Having an estimate of the population proportion reduces the minimum sample size needed.Click to select your answer(s).24. The table to the right shows the results of a survey in which 2592 adults from Country A, 1119adults from Country B, and 1051 adults from Country C were asked if they believe climatechange poses a large threat to the world. Complete parts (a), (b), and (c).Does climate change pose a large threat to the world?CountryYes99% CIA3030%(0.277,0.3230.277,0.323)B5757%(0.532,0.6080.532,0.608)C2929%(0.254,0.3260.254,0.326)(a) Determine whether it is reasonably possible that the proportion of adults from Country A andthe proportion of adults from Country B are equal and explain your reasoning. Choose thecorrect answer below.A.Yes, because the two confidence intervals do not overlap.B.Yes, because the two confidence intervals overlap.C.No, because the two confidence intervals overlap.D.No, because the two confidence intervals do not overlap.(b) Determine whether it is reasonably possible that the proportion of adults from Country B andthe proportion of adults from Country C are equal and explain your reasoning. Choose thecorrect answer below.A.No, because the two confidence intervals do not overlap.B.Yes, because the two confidence intervals do not overlap.C.Yes, because the two confidence intervals overlap.D.No, because the two confidence intervals overlap.(c) Determine whether it is reasonably possible that the proportion of adults from Country A andthe proportion of adults from Country C are equal and explain your reasoning. Choose thecorrect answer below.A. Yes, because the two confidence intervals do not overlap.B.No, because the two confidence intervals do not overlap.C.No, because the two confidence intervals overlap.D.Yes, because the two confidence intervals overlap.Click to select your answer.25. The table shows the results of a survey in which separate samples of 400 adults each from theEast, South, Midwest, and West were asked if traffic congestion is a serious problem in theircommunity. Complete parts (a) and (b).EastSouthMidwestWest37%33%28%54%(a) Construct a 95% confidence interval for the proportion of adults from the Midwest who saytraffic congestion is a serious problem.The 95% confidence interval for the proportion of adults from the Midwest who say trafficcongestion is a serious problem is. ( , ).(Round to three decimal places as needed.)(b) Construct a 95% confidence interval for the proportion of adults from the East who say trafficcongestion is a serious problem.The 95% confidence interval for the proportion of adults from the East who say traffic congestionis a serious problem is .( , ).(Round to three decimal places as needed.)Enter your answer in each of the answer boxes.