1. Find the critical value z Subscript czc necessary to form a confidence interval at the level of confidence shown below. c=0.87 z Subscript czc= nothing (Round to two decimal places as needed.) 2. Use the values on the number line to find the sampling error. 24 25 26 27 x=26.08 =24.49 The sampling error is nothing. 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 czc 90% 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=51 A 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 a sample must she select if she desires to be 99% confident that the true mean is within 2 ounces of 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 the population favors decriminalizing marijuana possession. How large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 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 error The sample mean is nothing. 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 (in centimeters) of a sample of seedlings. The estimated margin of error is nothing. The sample mean is Enter 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 many subjects are needed to estimate the HDL cholesterol within 44 points with 99% confidence assuming sigma equals 18.6 question mark =18.6? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required? A 99% confidence level requires nothing subjects. (Round up to the nearest whole number as needed.) A 95% confidence level requires nothing 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=1919 LOADING... Click the icon to view the t-distribution table. The margin of error is nothing. (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=25 LOADING... Click the icon to view the t-distribution table. The margin of error is nothing. (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 the standard 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 the margin 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 the confidence interval. B. With 80% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval. C. It can be said that 80% of people have a commute time between the bounds of the confidence interval. D. If a large sample of people are taken approximately 80% of them will have commute times between 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 standard deviation was $15.10. Using the standard normal distribution with the appropriate calculations for a standard deviation that is known, assume the population is normally distributed, find the margin 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). Compare the results. The margin of error of is nothing. (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 than the confidence interval found using the student's t-distribution. B. The confidence interval found using the standard normal distribution has smaller lower and upper confidence interval limits. C. The confidence interval found using the standard normal distribution is narrower thannarrower than 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 as the 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) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 11.90 29.80 27.10 16.51 11.40 8.81 5.30 20.46 14.90 33.67 30.91 14.86 11.40 15.35 9.72 19.80 14.86 8.09 22.49 18.30 (a) Find the sample mean. The sample mean is nothing. (Round to two decimal places as needed.) (b) Find the sample standard deviation. The sample standard deviation is nothing. (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 in economics, are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. 4450.914450.91 4596.174596.17 4366.424366.42 4455.284455.28 4151.744151.74 3727.093727.09 4283.424283.42 4527.954527.95 4407.934407.93 3946.263946.26 4023.834023.83 4221.124221.12 (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 interval for 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 the standard 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 random sample, unknown, e. cannot standard normal distribution or n less than 30n<30, not select correct choice below and, if necessary, fill in any answer boxes to complete your choice. a. 90% confidence interval .( , ). (round two decimal places as needed.) b. neither can be used construct interval. interpret results. choose below. said that of institutions have an rate between bounds large sample taken approximately them will c. with confidence, population mean click answer. 19. let p proportion for following condition. find point estimates q. 871 children surveyed, 117 plan join armed forces future. estimate p, caretp, nothing. three q, q caretq, enter each boxes. 20. given margin error proportion. (0.605,0.633) e =(Type integer decimal.) caretpequals =nothing (type 21. survey 655 males ages 18-64, 396 say they gone dentist past year. 95% intervals results compare widths intervals. convenient, technology is.( both 18-64 who year endpoints which wider? answer(s). 22. 8000 women, 5431 change their nail polish once week. 99% women ( 23. researcher wishes estimate, adults think president country control price gasoline. her must accurate within 33% true a) no preliminary available. minimum size needed. b) needed, using prior study found 38% respondents c) from parts (a) (b). what needed assuming information available? nothing up nearest whole number (b) gasoline? (c) how do compare? having has effect on raises reduces 24. table right shows 2592 a, 1119 b, 1051 c were asked believe climate poses threat world. (a), (b), (c). does pose world? yes ci 3030% (0.277,0.3230.277,0.323) b 5757% (0.532,0.6080.532,0.608) 2929% (0.254,0.3260.254,0.326) determine whether reasonably possible equal explain reasoning. yes, overlap. no, 25. separate samples 400 east, south, midwest, west traffic congestion serious problem community. east south midwest 37% 28% 54% problem. is. 1. critical value z subscript czc necessary form at level shown look tables, 1.13 2. values line sampling error. 24 25 26 27 x =26.08 >