MATH 227 CP 8 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find the indicated critical z value. 1) Find z /2 for = 0.07. 1) 2) Find the value of z /2 that corresponds to a confidence level of 89.48%. 2) 3) Find the critical value z /2 that corresponds to a 91% confidence level. 3) 4) Find the critical value z /2 that corresponds to a 98% confidence level. 4) Express the confidence interval using the indicated format. 5) Express the confidence interval 0.491 ± 0.057 in the form of p - E < p < p + E. 5) 6) Express the confidence interval (0.432, 0.52) in the form of p ± E. 6) Solve the problem. 7) The following confidence interval is obtained for a population proportion, p: 0.753 < p < 0.797. Use these confidence interval limits to find the point estimate, p. 8) The following confidence interval is obtained for a population proportion, p: (0.399, 0.437). Use these confidence interval limits to find the margin of error, E. 7) 8) Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 9) 90% confidence; n = 300, x = 140 9) 10) 99% confidence; n = 5900, x = 1770 10) 11) In a random sample of 192 college students, 129 had part-time jobs. Find the margin of error for the 95% confidence interval used to estimate the population proportion. 11) 12) In a clinical test with 1600 subjects, 800 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion. 12) 13) 95% confidence; n = 2388, x = 1672 13) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 14) n = 130, x = 69; 90% confidence 14) 15) n = 109, x = 65; 88% confidence 15) 16) n = 195, x = 162; 95% confidence 16) 17) n = 85, x = 49; 98% confidence 17) 1
Use the given data to find the minimum sample size required to estimate the population proportion. 18) Margin of error: 0.012; confidence level: 93%; p and q unknown 18) 19) Margin of error: 0.005; confidence level: 97%; p and q unknown 19) 20) Margin of error: 0.07; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 92%. 20) 21) Margin of error: 0.007; confidence level: 99%; from a prior study, p is estimated by 0.255. 21) Solve the problem. Round the point estimate to the nearest thousandth. 22) Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 898 people, 46 people had hearing aids. 22) 23) 50 people are selected randomly from a certain population and it is found that 12 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall? 23) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 24) Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for 24) the proportion of all such items that are defective. 25) Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. 25) 26) Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke. 26) 27) Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community. 27) 28) Of 260 employees selected randomly from one company, 18.46% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool. 28) Solve the problem. 29) In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error is no more than 4 percentage points. 29) Do one of the following, as appropriate: (a) Find the critical value z /2, (b) find the critical value t /2, (c) state that neither the normal nor the t distribution applies. 30) 90%; n = 10; is unknown; population appears to be normally distributed. 30) 2
31) 99%; n = 17; is unknown; population appears to be normally distributed. 31) 32) 91%; n = 45; is known; population appears to be very skewed. 32) Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution. 33) n = 10, x = 8.1, s = 4.8, 95% confidence 33) 34) A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers. 34) 35) The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.4 15.8 15.4 15.1 15.8 15.9 15.8 15.7 Construct a 98% confidence interval for the mean amount of juice in all such bottles. 35) 36) Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students. 36) 37) A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s = 17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs. 37) Use the given information to find the minimum sample size required to estimate an unknown population mean µ. 38) Margin of error: $126, confidence level: 99%, = $512 38) 39) How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confidence that the sample mean is within $2 of the population mean, and the population standard deviation is known to be $60. 39) 40) How many commuters must be randomly selected to estimate the mean driving time of Chicago commuters? We want 90% confidence that the sample mean is within 4 minutes of the population mean, and the population standard deviation is known to be 12 minutes. 40) 41) How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98% confidence that the sample mean is within $400 of the population mean, and the population standard deviation is known to be $1400. 41) Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean. 42) A random sample of 187 full-grown lobsters had a mean weight of 19 ounces and a 42) standard deviation of 3.3 ounces. Construct a 98% confidence interval for the population mean µ. 43) Test scores: n = 104, x = 95.3, = 6.5; 99% confidence 43) 3
44) A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 90% confidence interval for the mean score, µ, of all students taking the test? 44) 45) 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.3 pounds and a standard deviation of 2.4 pounds. What is the 95% confidence interval for the true mean weight, µ, of all packages received by the parcel service? 45) Solve the problem. 46) Find the chi-square value 2 L of 98 percent. corresponding to a sample size of 4 and a confidence level 46) 47) Find the critical value 2 R percent. corresponding to a sample size of 6 and a confidence level of 95 47) 48) Find the critical value 2 R 99 percent. corresponding to a sample size of 19 and a confidence level of 48) Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. 49) Weights of men: 90% confidence; n = 14, x = 161.5 lb, s = 13.7 lb 49) 50) College students' annual earnings: 98% confidence; n = 9, x = $3361, s = $865 50) 51) The mean replacement time for a random sample of 20 washing machines is 10.9 years and the standard deviation is 2.7 years. Construct a 99% confidence interval for the standard deviation,, of the replacement times of all washing machines of this type. 51) 52) A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the standard deviation,, of the scores of all subjects. 52) Find the appropriate minimum sample size. 53) You want to be 95% confident that the sample variance is within 40% of the population variance. 53) 54) You want to be 99% confident that the sample standard deviation s is within 5% of the population standard deviation. 54) 4
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. 55) The football coach randomly selected ten players and timed how long each player took to 55) perform a certain drill. The times (in minutes) were: 9 7 15 6 15 12 8 6 14 5 Find a 95% confidence interval for the population standard deviation. 56) The daily intakes of milk (in ounces) for ten randomly selected people were: 23.3 28.4 10.5 16.4 26.4 18.1 20.4 17.3 27.4 13.2 Find a 99% confidence interval for the population standard deviation. 56) 57) The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.2 15.1 15.9 15.5 15.6 15.1 15.8 15.0 Find a 98% confidence interval for the population standard deviation. 57) Provide an appropriate response. 58) The confidence interval, 17.12 < 2 < 78.88, for the population variance is based on the following sample statistics: n = 25, x = 42.6 and s = 5.7. What is the degree of confidence? 59) The confidence interval, 3.56 < < 5.16, for the population standard deviation is based on the following sample statistics: n = 41, x = 30.8, and s = 4.2. What is the degree of confidence? 58) 59) 5
Answer Key Testname: MATH227CP8 1) 1.81 2) 1.62 3) 1.70 4) 2.33 5) 0.434 < p < 0.548 6) 0.476 ± 0.044 7) 0.775 8) 0.019 9) 0.0474 10) 0.0154 11) 0.0664 12) 0.0322 13) 0.0184 14) 0.459 < p < 0.603 15) 0.523 < p < 0.669 16) 0.778 < p < 0.883 17) 0.451 < p < 0.701 18) 5688 19) 47,089 20) 58 21) 25,708 22) 0.051 23) 0.24 24) 0.0118 < p < 0.0576 25) 0.582 < p < 0.744 26) 11.6% < p < 28.4% 27) 0.0323 < p < 0.0782 28) 14.5% < p < 22.4% 29) 658 30) t /2 = 1.833 31) t /2 = 2.921 32) z /2 = 1.70 33) 4.67 < µ < 11.53 34) $453.59 < µ < $874.69 35) 15.33 oz < µ < 15.89 oz 36) 91.68 < µ < 98.32 37) 173.8 mg < µ < 196.2 mg 38) 110 39) 3458 40) 25 41) 67 42) 18 oz < µ < 20 oz 43) 93.7 < µ < 96.9 44) 28.4 < µ < 30.6 45) 9.5 lb < µ < 11.1 lb 46) 0.115 47) 12.833 48) 37.156 49) 10.4 lb < < 20.3 lb 6
Answer Key Testname: MATH227CP8 50) $546 < < $1907 51) 1.9 yr < < 4.5 yr 52) 16.9 < < 29.3 53) 57 54) 1336 55) 2.7 min < < 7.2 min 56) 3.78 oz < < 13.95 oz 57) 0.21 oz < < 0.82 oz 58) 99% 59) 90% 7