SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Part 3 Estimation Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. 1) Explain the difference between descriptive and inferential statistics. 1) 2) Define margin of error. Explain the relation between the confidence interval and the error estimate. Suppose a confidence interval is 9.65 < μ < 11.35. Find the sample mean x and the error estimate E. 2) 3) Define a point estimate. What is the best point estimate for μ? 3) 4) Under what circumstances can you replace σ with s in the formula E = z α/2 σ n. 4) 5) How do you determine whether to use the z or t distribution in computing the margin of error, E = z α/2 σn or E = t α/2 s n? 5) 6) When determining the sample size needed to achieve a particular error estimate you need to know σ. What are the two methods of estimating σ if σ is unknown? 6) 7) What assumption about the parent population is needed to use the t distribution to compute the margin of error? 7) 8) Under what three conditions is it appropriate to use the t distribution in place of the standard normal distribution? 8) 9) Interpret the following 95% confidence interval for mean weekly salaries of shift managers at Guiseppeʹs Pizza and Pasta. 325.80 < μ < 472.30 9) 10) Identify the correct distribution (z, t, or neither) for each of the following. 10) 11) What is the best point estimate for the population proportion? Explain why that point estimate is best. 11) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the indicated critical z value. 12) Find the critical value zα/2 that corresponds to a 98% confidence level. A) 2.33 B) 1.75 C) 2.05 D) 2.575 12) 1

13) Find the critical value zα/2 that corresponds to a 91% confidence level. A) 1.34 B) 1.645 C) 1.75 D) 1.70 14) Find the critical value zα/2 that corresponds to a 90% confidence level. A) 2.33 B) 1.28 C) 1.75 D) 1.645 13) 14) Determine whether the given conditions justify using the margin of error E = z α/2 σ/ n when finding a confidence interval estimate of the population mean μ. 15) The sample size is n = 4, σ = 12.7, and the original population is normally distributed. 15) A) No B) Yes 16) The sample size is n = 2 and σ is not known. A) Yes B) No 16) 17) The sample size is n = 200 and σ = 19. A) No B) Yes 17) 18) The sample size is n = 9, σ is not known, and the original population is normally distributed. A) Yes B) No 18) Use the confidence level and sample data to find the margin of error E. Round your answer to the same number of decimal places as the sample mean unless otherwise noted. 19) Weights of eggs: 95% confidence; n = 53, x = 1.34 oz, σ = 0.58 oz A) 0.13 oz B) 0.16 oz C) 0.02 oz D) 0.36 oz 19) 20) Replacement times for washing machines: 90% confidence; n = 45, x = 11.9 years, σ = 2.0 years A) 2.9 yr B) 0.1 yr C) 0.4 yr D) 0.5 yr 20) 21) The duration of telephone calls directed by a local telephone company: σ = 3.6 minutes, n = 560, 90% confidence. Round your answer to the nearest thousandth. A) 0.250 min B) 0.006 min C) 0.092 min D) 0.011 min 21) 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. 22) Test scores: n = 92, x = 90.6, σ = 8.9; 99% confidence A) 88.8 < μ < 92.4 B) 88.4 < μ < 92.8 C) 89.1 < μ < 92.1 D) 88.2 < μ < 93.0 22) 23) Test scores: n = 75, x = 46.1, σ = 5.8; 98% confidence A) 44.4 < μ < 47.8 B) 44.5 < μ < 47.7 C) 45.0 < μ < 47.2 D) 44.8 < μ < 47.4 23) 24) A random sample of 130 full-grown lobsters had a mean weight of 21 ounces and a standard deviation of 3.0 ounces. Construct a 98% confidence interval for the population mean μ. A) 19 oz < μ < 21 oz B) 20 oz < μ < 22 oz C) 21 oz < μ < 23 oz D) 20 oz < μ < 23 oz 24) 2

25) A laboratory tested 83 chicken eggs and found that the mean amount of cholesterol was 233 milligrams with σ = 12.9 milligrams. Construct a 95% confidence interval for the true mean cholesterol content, μ, of all such eggs. A) 229 mg < μ < 235 mg B) 229 mg < μ < 236 mg C) 230 mg < μ < 236 mg D) 231 mg < μ < 237 mg 25) 26) 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 17.0 pounds and a standard deviation of 3.3 pounds. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service? A) 15.9 lb < μ < 18.1 lb B) 15.6 lb < μ < 18.4 lb C) 16.1 lb < μ < 17.9 lb D) 15.7 lb < μ < 18.3 lb 26) 27) A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 203 milligrams with σ = 15.0 milligrams. Construct a 95% confidence interval for the true mean cholesterol content, μ, of all such eggs. A) 199 mg < μ < 206 mg B) 199 mg < μ < 205 mg C) 201 mg < μ < 207 mg D) 200 mg < μ < 206 mg 27) 28) A random sample of 158 full-grown lobsters had a mean weight of 16 ounces and a standard deviation of 3.5 ounces. Construct a 98% confidence interval for the population mean μ. A) 15 oz < μ < 18 oz B) 16 oz < μ < 18 oz C) 14 oz < μ < 16 oz D) 15 oz < μ < 17 oz 28) 29) A group of 66 randomly selected students have a mean score of 22.4 with a standard deviation of 2.8 on a placement test. What is the 90% confidence interval for the mean score, μ, of all students taking the test? A) 21.8 < μ < 23.0 B) 21.7 < μ < 23.1 C) 21.5 < μ < 23.3 D) 21.6 < μ < 23.2 29) Do one of the following, as appropriate: (a) Find the critical value zα/2, (b) find the critical value tα/2, (c) state that neither the normal nor the t distribution applies. 30) 90%; n = 10; σ is unknown; population appears to be normally distributed. A) zα/2 = 2.262 B) tα/2 = 1.833 C) tα/2 = 1.812 D) zα/2 = 1.383 30) 31) 95%; n = 11; σ is known; population appears to be very skewed. A) zα/2 = 1.96 B) tα/2 = 2.228 C) zα/2 = 1.812 D) Neither the normal nor the t distribution applies. 31) Assume that a sample is used to estimate a population mean μ. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation. _ 32) 95% confidence; n = 91; x = 16, s = 9.1 A) 1.63 B) 4.10 C) 1.90 D) 1.71 32) _ 33) 99% confidence; n = 201; x = 276; s = 75 A) 12.4 B) 16.0 C) 13.8 D) 10.5 33) 3

_ 34) 95% confidence; n = 12; x = 76.4; s = 5.9 A) 3.75 B) 3.38 C) 2.81 D) 4.50 34) 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. 35) A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 225 35) milligrams with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs. A) 216.9 mg < μ < 233.1 mg B) 214.9 mg < μ < 235.1 mg C) 215.0 mg < μ < 235.0 mg D) 215.1 mg < μ < 234.9 mg 36) Thirty randomly selected students took the calculus final. If the sample mean was 83 and the standard deviation was 13.5, construct a 99% confidence interval for the mean score of all students. A) 76.23 < μ < 89.77 B) 78.81 < μ < 87.19 C) 76.21 < μ < 89.79 D) 76.93 < μ < 89.07 36) 37) A sociologist develops a test to measure attitudes towards 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 mean score of all such subjects. A) 74.6 < μ < 77.8 B) 64.2 < μ < 88.2 C) 69.2 < μ < 83.2 D) 67.7 < μ < 84.7 37) 38) 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. A) $453.59 < μ < $874.69 B) $493.71 < μ < $834.57 C) $492.52 < μ < $835.76 D) $455.65 < μ < $872.63 38) 39) The principal randomly selected six students to take an aptitude test. Their scores were: 76.5 85.2 77.9 83.6 71.9 88.6 Determine a 90% confidence interval for the mean score for all students. A) 85.74 < μ < 75.49 B) 75.49 < μ < 85.74 C) 75.39 < μ < 85.84 D) 85.84 < μ < 75.39 39) 40) The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.2 15.5 15.9 15.5 15.0 15.7 15.0 15.7 Construct a 98% confidence interval for the mean amount of juice in all such bottles. A) 15.00 oz < μ < 15.87 oz B) 15.87 oz < μ < 15.00 oz C) 15.77 oz < μ < 15.10 oz D) 15.10 oz < μ < 15.77 oz 40) 41) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 7.2 10.5 9.9 8.2 11.0 7.3 6.7 11.0 10.8 12.4 Determine a 95% confidence interval for the mean time for all players. A) 10.85 min < μ < 8.15 min B) 8.15 min < μ < 10.85 min C) 10.95 min < μ < 8.05 min D) 8.05 min < μ < 10.95 min 41) 4

Express the confidence interval using the indicated format. 42) Express the confidence interval 0.047 < p < 0.507 in the form of p^ ± E. A) 0.277-0.23 B) 0.277 ± 0.5 C) 0.277 ± 0.23 D) 0.23 ± 0.5 42) Solve the problem. 43) The following confidence interval is obtained for a population proportion, p: (0.505, 0.545). Use these confidence interval limits to find the point estimate, p^. A) 0.527 B) 0.525 C) 0.545 D) 0.505 43) 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. 44) 95% confidence; n = 320, x = 60 44) A) 0.0428 B) 0.0514 C) 0.0385 D) 0.0449 45) In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the 95% confidence interval used to estimate the population proportion. A) 0.0649 B) 0.00266 C) 0.126 D) 0.0721 45) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 46) n = 125, x = 72; 90% confidence 46) A) 0.506 < p < 0.646 B) 0.503 < p < 0.649 C) 0.502 < p < 0.650 D) 0.507 < p < 0.645 47) A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. A) 0.438 < p < 0.505 B) 0.444 < p < 0.500 C) 0.471 < p < 0.472 D) 0.435 < p < 0.508 47) 48) A study involves 669 randomly selected deaths, with 31 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents. A) 2.54% < p < 6.73% B) 3.04% < p < 6.23% C) 3.29% < p < 5.97% D) 2.74% < p < 6.53% 48) 49) Of 140 adults selected randomly from one town, 35 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke. A) 15.6% < p < 34.4% B) 17.8% < p < 32.2% C) 16.5% < p < 33.5% D) 19.0% < p < 31.0% 49) 50) Of 234 employees selected randomly from one company, 12.82% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool. A) 8.54% < p < 17.1% B) 7.18% < p < 18.5% C) 9.23% < p < 16.4% D) 7.73% < p < 17.9% 50) 51) Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective. A) 0.0118 < p < 0.0576 B) 0.0110 < p < 0.0584 C) 0.0154 < p < 0.0540 D) 0.0345 < p < 0.0349 51) 5

52) A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate. A) 0.301 < p < 0.445 B) 0.316 < p < 0.430 C) 0.304 < p < 0.442 D) 0.308 < p < 0.438 52) 53) When 319 college students are randomly selected and surveyed, it is found that 120 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. A) 0.323 < p < 0.429 B) 0.332 < p < 0.421 C) 0.306 < p < 0.446 D) 0.313 < p < 0.439 53) 54) Of 88 adults selected randomly from one town, 69 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. A) 0.671 < p < 0.897 B) 0.698 < p < 0.870 C) 0.682 < p < 0.886 D) 0.712 < p < 0.856 54) 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. 55) In a clinical test with 3300 subjects, 660 showed improvement from the treatment. Find the margin 55) of error for the 99% confidence interval used to estimate the population proportion. A) 0.0156 B) 0.0136 C) 0.0102 D) 0.0179 56) 90% confidence; n = 480, x = 120 A) 0.0387 B) 0.0325 C) 0.0406 D) 0.0348 56) 57) 99% confidence; n = 6500, x = 1950 A) 0.0128 B) 0.00833 C) 0.0146 D) 0.0111 57) Solve the problem. 58) The following confidence interval is obtained for a population proportion, p: 0.537 < p < 0.563. Use these confidence interval limits to find the point estimate, p^. A) 0.550 B) 0.537 C) 0.555 D) 0.545 59) The following confidence interval is obtained for a population proportion, p: 0.689 < p < 0.723. Use these confidence interval limits to find the margin of error, E. A) 0.034 B) 0.017 C) 0.706 D) 0.018 58) 59) Express the confidence interval using the indicated format. 60) Express the confidence interval 0.66 < p < 0.8 in the form of p^ ± E. A) 0.66 ± 0.07 B) 0.73 ± 0.14 C) 0.66 ± 0.14 D) 0.73 ± 0.07 60) 61) Express the confidence interval (0.668, 0.822) in the form of p^ ± E. A) 0.745 ± 0.077 B) 0.668 ± 0.154 C) 0.668 ± 0.077 D) 0.745 ± 0.154 61) 62) Express the confidence interval 0.457 ± 0.044 in the form of p^ - E < p < p^ + E. A) 0.413 < p < 0.501 B) 0.457 < p < 0.501 C) 0.435 < p < 0.479 D) 0.413 < p < 0.457 62) 6

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 63) Of 82 adults selected randomly from one town, 69 have health insurance. Find a 90% confidence 63) interval for the true proportion of all adults in the town who have health insurance. A) 0.762 < p < 0.921 B) 0.747 < p < 0.935 C) 0.738 < p < 0.945 D) 0.775 < p < 0.908 64) Of 367 randomly selected medical students, 30 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. A) 0.0449 < p < 0.119 B) 0.0537 < p < 0.110 C) 0.0582 < p < 0.105 D) 0.0484 < p < 0.115 64) 65) Of 87 adults selected randomly from one town, 63 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. A) 0.645 < p < 0.803 B) 0.630 < p < 0.818 C) 0.612 < p < 0.836 D) 0.601 < p < 0.848 65) 66) Of 253 employees selected randomly from one company, 18.97% of them commute by carpooling. Construct a 90% confidence interval for the true percentage of all employees of the company who carpool. A) 13.2% < p < 24.7% B) 12.6% < p < 25.3% C) 14.1% < p < 23.8% D) 14.9% < p < 23.0% 66) Use the given information to find the minimum sample size required to estimate an unknown population mean μ. 67) How many women must be randomly selected to estimate the mean weight of women in one age 67) group. We want 90% confidence that the sample mean is within 3.7 lb of the population mean, and the population standard deviation is known to be 28 lb. A) 153 B) 221 C) 155 D) 156 68) 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 $5 of the population mean, and the population standard deviation is known to be $63. A) 430 B) 610 C) 537 D) 862 68) 69) How many business students must be randomly selected to estimate the mean monthly earnings of business students at one college? We want 95% confidence that the sample mean is within $128 of the population mean, and the population standard deviation is known to be $536. A) 47 B) 68 C) 59 D) 95 69) 70) How many commuters must be randomly selected to estimate the mean driving time of Chicago commuters? We want 98% confidence that the sample mean is within 4 minutes of the population mean, and the population standard deviation is known to be 12 minutes. A) 35 B) 49 C) 25 D) 60 70) 71) How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 99% confidence that the sample mean is within $200 of the population mean, and the population standard deviation is known to be $1100. A) 117 B) 82 C) 165 D) 201 71) 7

Use the given data to find the minimum sample size required to estimate the population proportion. 72) Margin of error: 0.004; confidence level: 95%; p^ and q^ unknown A) 50,024 B) 60,148 C) 60,018 D) 60,025 72) 73) Margin of error: 0.01; confidence level: 95%; from a prior study, p^ is estimated by the decimal equivalent of 52%. A) 9589 B) 8630 C) 19,976 D) 16,551 73) Solve the problem. 74) A newspaper article about the results of a poll states: ʺIn theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States.ʺ Find the sample size suggested by this statement. A) 385 B) 544 C) 27 D) 664 74) 75) 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. A) 658 B) 317 C) 501 D) 232 75) Use the given data to find the minimum sample size required to estimate the population proportion. 76) Margin of error: 0.05; confidence level: 99%; from a prior study, p^ is estimated by 0.15. A) 17 B) 339 C) 407 D) 196 76) 77) Margin of error: 0.04; confidence level: 95%; from a prior study, p^ is estimated by the decimal equivalent of 89%. A) 708 B) 236 C) 9 D) 209 77) 78) Margin of error: 0.015; confidence level: 96%; p^ and q^ unknown A) 6669 B) 4519 C) 4670 D) 3667 78) 79) Margin of error: 0.04; confidence level: 94%; p^ and q^ unknown A) 587 B) 486 C) 553 D) 572 79) Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. 80) Independent samples from two different populations yield the following data. x1 = 236, x2 = 905, s1 = 88, s2 = 13. The sample size is 381 for both samples. Find the 85% confidence interval for μ1 - μ2. A) -676 < μ1 - μ2 < -662 B) -683 < μ1 - μ2 < -655 C) -670 < μ1 - μ2 < -668 D) -677 < μ1 - μ2 < -661 80) 8

81) Two types of flares are tested and their burning times are recorded. The summary statistics are given below. 81) Brand X Brand Y n = 35 n = 40 x = 19.4 min x = 15.1 min s = 1.4 min s = 0.8 min Construct a 95% confidence interval for the differences between the mean burning time of the brand X flare and the mean burning time of the brand Y flare. A) 3.8 min < μx - μy < 4.8 min B) 3.2 min < μx - μy < 5.4 min C) 3.5 min < μx - μy < 5.1 min D) 3.6 min < μx - μy < 5.0 min Construct the indicated confidence interval for the difference between population proportions p1 - p2. Assume that the samples are independent and that they have been randomly selected. 82) x1 = 22, n1 = 38 and x2 = 31, n2 = 52; Construct a 90% confidence interval for the difference 82) between population proportions p1 - p2. A) 0.406 < p1 - p2 < 0.752 B) -0.190 < p1 - p2 < 0.156 C) 0.373 < p1 - p2 < 0.785 D) 0.785 < p1 - p2 < 0.373 83) A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product. Construct a 99% confidence interval for the difference between the two population proportions. A) 0.0247 < p1 - p2 < 0.0286 B) -0.0443 < p1 - p2 < 0.0976 C) -0.0177 < p1 - p2 < 0.1243 D) -0.0034 < p1 - p2 < 0.0566 83) 84) In a random sample of 300 women, 49% favored stricter gun control legislation. In a random sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the population proportions p1 - p2. A) 0.126 < p1 - p2 < 0.294 B) 0.122 < p1 - p2 < 0.298 C) 0.099 < p1 - p2 < 0.321 D) 0.110 < p1 - p2 < 0.310 84) 85) In a random sample of 500 people aged 20-24, 22% were smokers. In a random sample of 450 people aged 25-29, 14% were smokers. Construct a 95% confidence interval for the difference between the population proportions p1 - p2. A) 0.025 < p1 - p2 < 0.135 B) 0.035 < p1 - p2 < 0.125 C) 0.032 < p1 - p2 < 0.128 D) 0.048 < p1 - p2 < 0.112 85) 9