Chapter 8 Estimation (S)
1. In September a biological research team caught, weighed, and released a random sample of 54 chipmunks in Rocky Mountain National Park. The mean of the sample weights was x = 8.7 oz with sample standard deviation s = 1.4 oz. Find a 90% confidence interval for the mean September weight of all chipmunks in Rocky Mountain National Park.
(S)
2. A sociologist would like to estimate the total amount of federal, state and city taxes paid by a family of four with income level $50,000 per year. A random sample of 100 families produced sample mean x = $4,258 with known standard deviation σ = $1,098. Find a 95% confidence interval for the population mean.
(S)
3. Using a standard normal distribution table, find the critical value, zc , for P(!z0.97 < z < z0.97 ) = 0.97.
(S)
4. A random sample of 60 employees of a large corporation had a sample mean x = 32.5 accumulated vacation days with known standard deviation σ = 18.5 days. Find a 90% confidence level for the population mean number of accumulated vacation days.
(S)
5. A college career center conducted a study of skill levels required for jobs. A random sample of 200 job listings showed 120 required a high level of technical skill. Find a 90% confidence interval for the population proportion of jobs requiring a high level of technical skill.
(M) 6. For σ = 1.1, n = 120, and z0.95 = 1.96, find the maximal error tolerance, E. A. 0.018
B. 5.08
C. 0.197
D. 0.026
E. 0.279
(S)
7. An air pollution index (from 1 to 25 with 25 being the most serious pollution) takes into account the amount of carbon dioxide in the air. A random sample of 42 readings taken one hour before sunset had a mean index reading of 9.7 with standard deviation 3.2. Find a 90% confidence interval for the population mean index reading.
(S)
8. A random sample of 200 businesses showed that 120 of them ban smoking on their premises. Find a 95% confidence interval for the population proportion of all businesses than ban smoking on their premises.
(M) 9. Given that P(–0.227 < x – µ < 0.227) = 0.90 and n = 150, find σ. A. 4.57 B. 03 C. 1.66 D. 1.54 E. 1.69
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(S) 10. The national program planner for the Girl Scouts wants to estimate the average number of girls per troop. A random sample of 81 troops shows sample mean x = 14.8 girls with standard deviation 9.0 girls. Find a 99% confidence interval for the population mean number of girls per troop. (S) 11. A random sample of 50 calls initiated on cellular car phones had a mean duration of 3.5 minutes with known standard deviation 1.2 minutes. Find a 99% confidence interval for the population mean duration of telephone calls initiated on cellular car phones. (S) 12. Given σ = 2.4, c = 0.85, n = 110, and x = 9.65, compute the confidence interval for µ. (S) 13. The director of the city summer recreation program would like to estimate the proportion of children in the summer learn to swim program who do not know how to swim. A random sample of 80 registration records from last year showed 36 children who did not know how to swim. It is reasonable to assume that the population for this year has the same characteristics as the population in last year’s program. Find a 90% confidence interval for the proportion of children entering the program who do not know how to swim. (S) 14. Karl would like to estimate the proportion of eligible voters is a large metropolitan district who are registered to vote. A random sample of 175 eligible voters showed 110 of them were registered. (a) Find a 95% confidence interval for the population proportion of registered voters. (b) Find the margin of error for this estimate. (S) 15. A candy store sells packages of saltwater taffy. Because the size and weight of the individually wrapped taffy pieces varies, the number of pieces per pound is not consistent. The storekeeper counted a sample of 50 one-pound packages and found that x = 13.5 pieces and s = 1.8 pieces. The storekeeper wants to label the packages with a range that represents a 95% confidence interval. Compute this range. (S) 16. Roger would like to estimate the proportion of registered voters in his city who voted in the last general election. A random sample of 225 registered voters showed 98 actually voted. (a) Find a 90% confidence interval for the population proportion of registered voters who voted in the last general election. (b) Find the margin of error for this estimate. (S) 17. The postmaster in a large city found that 56 packages out of a random sample of 400 packages had insufficient postage. Find a 99% confidence interval for the population proportion of packages with insufficient postage.
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(M) 18. Given the confidence interval, 6.71 < µ < 7.23, and that σ = 1.15 and n = 75, find the confidence level, c. A. 0.99 B. 0.95 C. 0.90 D. 0.85 E. 0.80 (S) 19. A random sample of 48 fish taken from a popular fishing lake showed 15 to be yellow perch. Let p represent the population proportion of yellow perch in the lake. (a) Find a point estimate for p. (b) Find an 85% confidence interval for p. (S) 20. A biologist has found the average weight of 12 randomly selected mud turtles to be 8.7 lb. with standard deviation 3.6 lb. Find a 90% confidence interval for the population mean weight of all such turtles. (M) 21. Given the confidence interval, 11.29 < µ < 12.01, and that n = 90 and c = 0.90, find the sample standard deviation, σ. A. 2.08 B. 0.48 C. 0.72 D. 1.44 E. 4.15 (S) 22. The Aloha Taxi Company of Honolulu Hawaii wants to estimate the mean life of tires on its cabs. A random sample of 15 tires from the cabs had a mean life of 18,280 mi. with standard deviation 1,310 mi. Find a 90% confidence interval for the population mean life of the tires. (S) 23. A park ranger has timed the mating calls of 22 bull elk. The mean duration of these calls is 14.6 sec. with standard deviation 2.8 sec. Find a 95% confidence interval for the population mean duration of mating calls of the bull elk. (M) 24. Use a student’s t distribution table to find the critical value tc for a 0.95 confidence level for a t distribution with sample size n = 9. A. 1.860 B. 2.896 C. 3.355 D. 2.306 E. 2.365 (S) 25. The weights of grapefruit follow a normal distribution. A random sample of 12 new hybrid grapefruit has a mean weight of 1.7 lb. with sample standard deviation 0.24 lb. Find a 95% confidence interval for the population mean weight of the hybrid grapefruit. (S) 26. What is the average cost of a 30-second ad on prime time TV? A random sample of 12 such ads had an average cost of 122 thousand dollars with sample standard deviation 15 thousand dollars. Find a 95% confidence interval for the cost of the ads. (M) 27. For t0.98 = 3.747, we use a student’s t distribution table to find n. A. 3
B. 4
C. 5
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D. 6
E. 7
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(S) 28. A random sample of 15 novels making a national best seller list was selected. The average length of time these novels stayed on the best seller list was 5.2 weeks with standard deviation 2.3 weeks. Find a 99% confidence interval for the average number of weeks a book is on this best seller list. (S) 29. The Tender Turkey Farm hatches turkey eggs. A random sample of 100 eggs hatched 84 chicks. Let p be the proportion of eggs from Tender Turkey Farm that hatch. (a) Find a point estimate for p. (b) Find a 90% confidence interval for p. (S) 30. A random sample of ten jumbo burgers from a local diner had the following weights (in ounces): 12.1 12.4 11.8 12.6 11.9 11.5 12.2 11.3 11.7 12.3 If µ is the mean weight of all jumbo burgers served at the diner, find a 99% confidence interval for µ. (S) 31. Scott wants to estimate the cost of a night at a Los Angeles hotel. He called a random sample of 8 hotels and got a sample mean cost for one night of $145 with sample standard deviation $40. (a) Find an 85% confidence interval for the population mean cost. (b) What size sample does he need to say with 90% confidence that the sample mean is within $10 of the population mean? (S) 32. A biology student is doing a study of the nesting habits of the American Robin. She checks a random sample of 6 nests and finds a sample mean x = 4.9 eggs with sample standard deviation s = 1.4. Find a 90% confidence interval for the population mean number of eggs. (M) 33. The 95% confidence interval for a certain set of random sample data is 4.36 < µ < 4.68. The sample size is 14. Find s. A. 0.55 B. 0.28 C. 0.31 D. 0.54 E. 0.27 (S) 34. Country Boy Seed Company wants to estimate the yield from their corn seed on prime Iowa farm land. A random sample of 20 one acre plots gave an average yield of 170 bushels per acre with standard deviation 6.4 bushels per acre. Find a 99% confidence interval for the true mean yield per acre. (S) 35. The heights of a random sample of six fifth-graders are the following (in inches): 49.4 59.5 50.8 55.0 60.1 53.3 Compute a 90% confidence interval for all fifth-graders from this data, letting µ represent the average height of all fifth-graders.
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(S) 36. An official of the National Parks Service would like to determine what proportion of the visitors to Yellowstone National Park would be willing to pay an additional $25 for a national park pass. A random sample of 275 visitors were surveyed. 90 of them said they would be willing to pay the additional amount. Let p be the population proportion of visitors who would be willing to pay $25 more. (a) Find a point estimate for p. (b) Find a 95% confidence interval for p. (c) Express the results of (a) and (b) using the margin of error. (S) 37. A preliminary study of the fall lobster catch showed that for 35 lobsters selected at random the mean weight was 2.9 lb with standard deviation 0.6 lb. (a) Find a 90% confidence interval for the population mean weight of the fall lobsters. (b) How many more lobsters should be included in the sample if we want to say with 90% confidence that the population mean weight of the fall lobster catch is within 0.1 lb of the sample mean? (M) 38. A set of random sample data has s = 3.4 and x = 21.8. Compute the maximal error of estimate if the degrees of freedom are 11 in number and the confidence level is 85%. A. 1.59 B. 1.51 C. 1.58 D. 1.52 E. 1.13 (S) 39. A member of the House of Representatives wants to determine the proportion of voters in her district who favor a flat income tax. A random sample of 200 voters in her district showed 89 in favor. Let p represent the proportion of voters who favor a flat income tax. (a) Find a point estimate for p. (b) Find a 95% confidence interval for p. (c) Does the data indicate whether a majority of the voters oppose the tax? Explain. (S) 40. A member of the state legislature wants to determine what fraction of the voters in his district favor a proposal refunding excess state sales tax revenues to the taxpayers. A random sample of 145 voters in his district show 97 in favor. Let p be the population proportion in favor of the proposal. (a) Find a 95% confidence interval for p. (b) What is the margin of error for this estimate? (c) What size sample does he need in order to state with 95% confidence that the sample estimate is within 0.05 of p?
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(M) 41. The weights of eight two-year-olds were selected at random from a hospital’s records. They are the following (in pounds): 29.0 27.4 35.1 31.2 30.6 33.8 26.3 29.7 Let µ be the mean weights of all the two-year-olds’ weights listed in the hospital’s records, and find a 98% confidence interval for µ. A. 27.2 < µ < 33.6 B. 27.0 < µ < 33.8 C. 28.3 < µ < 32.5 D. 27.9 < µ < 32.9 E. 27.4 < µ < 33.8 (S) 42. The Director of a Museum would like to know what fraction of the museum associates make purchases through the gift shop catalogue. (a) If no preliminary study is done, how large a sample must be taken if the director is to say with 90% confidence that the sample estimate is within 2% of the population proportion? (b) A preliminary study showed that out of 60 associates, 12 have used the gift shop catalogue. What size sample does the director need in order to say with 90% confidence that the sample estimate is within 2% of the population proportion? (M) 43. A random sample of 80 mature coast redwoods was selected, and the heights were measured. The mean height was x = 281 feet with s = 24 feet. Compute a 99% confidence interval if µ represents the mean height of all mature coast redwoods. A. 275 < µ < 287 B. 280 < µ < 282 C. 274 < µ < 288 D. 279 < µ < 283 E. 277 < µ < 285 (S) 44. The marketing director for a cereal company would like to know what proportion of households that receive free samples of the cereal with their newspapers later purchase the cereal. A random sample of 175 households showed that 45 purchased the cereal after receiving the free sample. (a) Find a point estimate for p. (b) Find a 95% confidence interval for p. (c) Express the results of (a) and (b) using the margin of error. (S) 45. The time it takes broken bones of a given size and type of break to heal is normally distributed. A random sample of 17 simple arm breaks occurring in people in their twenties took an average of 53 days to heal with standard deviation 5 days. Find a 90% confidence interval for the population mean healing time for all broken bones of this type.
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(M) 46. Five hundred members of a health club out of a total of 10,000 are selected at random and asked if they are satisfied or unsatisfied with the club’s facilities. Out of the 500, 375 said they were satisfied. Compute the point estimate for the probability that any of the 10,000 members selected at random will answer that they are satisfied with the club’s facilities. A. pˆ = 0.75 B. pˆ = 0.0375 C. pˆ = 1.33 D. pˆ = 0.0875 E. pˆ = 0.25 (S) 47. The past few years a nursery has kept records of blue spruce trees they have replanted. Over this period it was found that 421 out of 518 blue spruce trees survived replanting. Let p be the population proportion of blue spruce trees that survive replanting. (a) Find a point estimate for p. (b) Find a 99% confidence interval for p. Give a brief interpretation of the meaning of your confidence interval. (c) Assuming that the number of trees, r, which survive replanting is a binomial random variable, what additional conditions must be satisfied to insure the accuracy of your confidence interval? (S) 48. Why do people enroll in fitness programs? A random sample of 500 people were surveyed. 335 of them reported that an important reason for enrolling was to prevent heart disease. (a) Find a point estimate for p. (b) Find a 90% confidence interval for p. (M) 49. Compute ! = A. 0.917
pˆ qˆ n given pˆ = 0.6 and n = 35.
B. 0.007
C. 0.993
D. 0.083
E. 0.041
(S) 50. In a study regarding public opinion about American involvement in foreign affairs, a random sample of 40 people showed that 25 people favored increasing US involvement in foreign affairs. How many more people should be included in a the sample for the study to state with 90% confidence that the sample proportion is within 0.05 of the population proportion? (S) 51. A seed company advertises that the mean time from planting to harvest for a new variety of zucchini seeds is 50 days. The standard deviation is estimated to be 10.3 days. If we wanted to verify this claim how large a sample would we need in order to state with 95% confidence that our sample mean differs from the population mean by no more than 3 days? (M) 52. Compute the maximal error tolerance of the error of estimate pˆ ! p for a confidence level of 99%, given pˆ = 0.7 and n = 60. A. 0.85
B. 0.15
C. 0.009
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D. 0.991
E. 0.07
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(S) 53. A legislative aide is doing research on the duration of calls initiated on cellular car phones. A preliminary study showed that the standard deviation of these calls is about 1.8 minutes. How large a sample will be needed for the aide to be able to state with 99% confidence that the sample mean duration time for the calls differs from the population mean by no more than 0.25 minute? (S) 54. A doctor is testing a new sinus medication. A random sample of subjects will be given the medication and the number who are cured will be counted. How large a sample does he need to be able to state with 95% confidence that the sample proportion cured is within 0.03 of the actual population proportion? (S) 55. If n = 70 and r = 21, compute a 98% confidence interval for p. (S) 56. A random sample of 40 cups of coffee from a vending machine had a sample mean volume of coffee dispensed equal to 7.1 oz with sample standard deviation s = 0.3 oz. Find a 90% confidence interval for the mean amount of coffee dispensed per cup. (S) 57. The Junetag Company makes washing machines. In one phase of production it is necessary to estimate the drying time of enamel paint on metal sheeting. Analysis of a random sample of 420 sheets shows that the average drying time is 56 hr with standard deviation 12 hr. Find a 99% confidence interval for the population mean drying time. (M) 58. For the confidence interval 0.64 < p < 0.76, given that n = 100, determine the confidence level, c. A. 0.85 B. 0.84 C. 0.83 D. 0.82 E. 0.81 (S) 59. A random sample of 100 felony trials in a large city in the Midwest shows the mean waiting time between arrest and trial is 173 days with a known standard deviation of 28 days. Find a 99% confidence interval for the mean time interval between arrest and trial. (S) 60. A local C. J. Nickel department store took a random sample of 49 charge accounts and found that the average balance due was $63.19 with standard deviation $13.50. Find a 95% confidence interval for the average balance due at this store. (M) 61. For the confidence interval 0.83 < p < 0.91, given that the confidence level is 0.95, determine n. A. 50 B. 6 C. 272 D. 11 E. 68 (S) 62. Pro Computer Services has computer monitoring of all outgoing and incoming telephone calls to record the duration of technical consulting calls. A random sample of 40 calls has sample mean x = 15.95 min with standard deviation s = 5.43 min. Find a 90% confidence interval for the population mean duration of technical consulting calls at Pro Computer Services.
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(S) 63. From a survey of 18 patients, symptoms of a new flu virus have been determined to have a mean duration of 9.7 days with standard deviation 4.8 days. Find a 95% confidence interval for the population mean duration of these flu symptoms. (S) 64. A random sample of 212 El Cheapo brand mechanical pencils showed that 53 were defective. Let p represent the proportion of mechanical pencils that are defective to all mechanical pencils produced by El Cheapo. (a) What is the point estimate for p? (b) Find a 95% confidence interval for p. (c) The confidence interval was computed using a normal approximation. Was this justified? (S) 65. A random sample of four boxes of Purr Lots cat food had weights (in ounces) of 12.7, 11.6, 11.2, and 12.4. Find a 99% confidence interval for the mean weight of all such boxes of Purr Lots cat food. (S) 66. The average cholesterol level for a random sample of 26 adult women from Dairyhaven Wisconsin is 263 units (mg. per dl. of blood). The sample standard deviation is 43 units. Find a 99% confidence interval for the mean cholesterol level of all adult women in Dairyhaven Wisconsin. (M) 67. Compute an 85% confidence interval for the following situation: Out of 160 basketballs purchased, 16 have leaks. Let p represent the proportion of basketballs with leaks to all basketballs produced by the manufacturer. A. 0.07 < p < 0.13 B. 0.06 < p < 0.14 C. 0.09 < p < 0.11 D. 0.08 < p < 0.12 E. 0.05 < p < 0.15 (S) 68. In wine making, acidity of the grape is a crucial factor. A pH range from 3.1 to 3.6 is considered very acceptable. A random sample of 12 bunches of grapes was taken from a California vineyard. The sample mean acidity was 3.38 with sample standard deviation 0.20. Find a 99% confidence interval for the mean acidity of the entire harvest of grapes from this vineyard. (S) 69. Certain rivers in Labrador are known to have phenomenally large brook trout. A biologist using catch-and-release tagging of trout obtained the following information about the annual weight gain of seven trout (in pounds). The results are: 0.73 1.05 0.81 1.12 0.92 1.17 0.98. Find a 90% confidence interval for the mean annual weight gain of all brook trout from these rivers. (M) 70. For a particular research project, the maximum error of estimate must be no larger than 0.10 unit, with a confidence level of 99%. A preliminary sample study produced a sample standard deviation of s = 2.5 units. How large must the sample taken be to ensure the above requirements for the research project? A. 64 B. 65 C. 4160 D. 4161 E. 2081
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(S) 71. Shoplifting has been a problem in a large men’s clothing store. Using special security measures to monitor shoplifting, it was found that there were attempts to shoplift the following dollar values of items of merchandise during the last nine weeks: $356 $285 $310 $375 $290 $325 $331 $342 $335. (a) Find a 90% confidence interval for the population mean shoplifting loss per week. (b) Estimate the mean shoplifting losses for a four-week month. (S) 72. Air Connect airlines found that 88 out of a random sample of 121 passengers purchased round-trip tickets. Let p be the proportion of all Air Connect passengers who purchase round-trip tickets. (a) Find a point estimate for p. (b) Find a 95% confidence interval for p. (c) Express the results of parts (a) and (b) using the margin of error. (S) 73. A company with 2200 employees is planning to alter their health insurance plan. A random sample of 330 employees, when asked whether or not they are satisfied with their current coverage, resulted in 132 replying with “satisfied” and 198 replying with “unsatisfied.” Let p represent the proportion of employees of the entire company that are unsatisfied with their coverage. (a) What is the point estimate for p? (b) Find a 99% confidence interval for p. (c) The confidence interval was computed using a normal approximation. Was this justified? (S) 74. An automobile manufacturer used a random sample of 50 cars of a certain model to estimate the mileage this model car gets in highway driving. The sample standard deviation was found to be 5.7 mpg. How large a sample do we need to state with 90% confidence that the sample mean mileage is within 1 mpg of the population mean for all cars of this model. (S) 75. Jim is doing a research project in political science to determine the proportion of registered voters, p, in his district who favor capital punishment. (a) If no preliminary sample is taken to estimate p, how large a sample is necessary if he wants to state with 95% confidence that the margin of error is 2%? (b) Jim did a preliminary study in which he found that in a random sample of 932 registered voters 136 favored capital punishment. How many more voters should be included in the sample if the margin if error is to be 2%? (M) 76. A biologist is studying varieties of fungi in a state park. A preliminary study found that p was approximately 0.42, where p represents the proportion of the dominant variety of fungi to all varieties occurring in the park. For a 95% confidence level, how large a sample should be taken to ensure that the point estimate for p will be in error either way by less than 0.02? A. 1194 B. 24 C. 2340 D. 47 E. 570 Copyright © Houghton Mifflin Company. All rights reserved.
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(S) 77. Union officials want to estimate the percent, p, of workers in the Big Bend Metalworks who favor a strike. The union wants to have 90% confidence that its estimate for p is within 2.5% of the true proportion of workers who favor a strike. (a) If no accurate estimate for p is available, how large a sample of workers is necessary? (b) A preliminary sample of 100 workers gave r/n = 0.35. How many more workers should be included in this sample? (S) 78. A random sample of 40 Salt Lake City teachers showed the standard deviation of teaching experience to be 5.3 years. How many more teachers should be included in the sample to have 95% confidence that the sample mean number of years teaching experience is within six months of the population mean? (M) 79. A car company is investigating whether or not to issue a recall of a model of car that has shown signs of having a defective emissions system. If no preliminary study is made to estimate p, the probability that the emissions system is defective, how large a sample should the investigator use to obtain a 98% confidence level that the point estimate for p will be in error either way by less than 0.03? A. 6032 B. 1509 C. 6033 D. 1508 E. 20 (S) 80. Aerodynamic engineers want to know the mean shear force on the wings of an FX5000 fighter jet as the jet comes out of a 3,000 ft. vertical dive. A random sample of FX5000 jets flown by Air Force fighter pilots was used for 37 test flights. The sample standard deviation of wing shear force was s = 948 foot-pounds. The engineers want to have 95% confidence that the sample mean is within 250 footpounds of the population mean. How many more test flights should be made? (S) 81. A prospective medical student wants to know if there is a difference in acceptance rates to medical school for liberal arts majors as compared to science majors. A random sample of 150 science majors who applied to medical schools last year showed 50 acceptances. An independent random sample of 75 liberal arts majors who applied to medical school last year had 19 acceptances. (a) Find an 85% confidence interval for the difference in the two proportions. (b) Does the data indicate a difference in acceptance rates? Explain. (S) 82. Rocky mows lawns for a living. For a random sample of 40 6000 sq. ft suburban lawns the sample mean time to mow the lawn and trim is 75 minutes with standard deviation 10 minutes. Find an 80% confidence interval for the mean time it takes Rocky to mow all such lawns.
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(S) 83. Dawn is comparing the fat content of Healthy Snack trail mix with that of Overland trail bars. A random sample of 12 28-gram samples of Healthy Snack trail mix had mean fat content 6.8 grams with standard deviation 1.2 grams. An independent random sample of 10 28-gram samples of Overland trail bars had mean fat content 4.2 grams with sample standard deviation 1.6 grams. (a) Find a 90% confidence interval for the difference of the two population means. (b) Does there appear to be a difference? Explain. (S) 84. x1 has a normal distribution with mean µ1 = 21.1 and standard deviation !1 = 3.1. x2 has a normal distribution with µ2 = 13.4 and ! 2 = 2.2. Independent random samples of size n1 = 90 was taken from x1 and n2 = 110 from x2 . (a) Does the variable x1 ! x2 have a normal distribution? (b) What is the mean of x1 ! x2 ? (c) What is the standard deviation of x1 ! x2 ? (S) 85. A naturalist is studying groups of Kingfishers nesting along two stretches of a river. On the first stretch of the river which passes by several industrial plants she found 6 Kingfisher nests with sample mean 5.4 eggs per nest and sample standard deviation 1.2 eggs. On the second stretch of the river which passes through a rural area she found 8 Kingfisher nests with sample mean 6.5 eggs per nest and sample standard deviation 0.8 eggs. (a) Find an 80% confidence interval for the difference in population mean numbers of eggs per nest. (b) Does it seem likely that there is a difference in the population means? Explain. (S) 86. A business finance class is studying start-up costs for various kinds of small businesses. A random sample of 9 sports equipment and apparel stores had sample mean start-up costs (in thousands of dollars) of 91.0 with sample standard deviation 33.92. An independent random sample of 10 travel agencies had sample mean startup costs of 61.5 with sample standard deviation 34.03. (a) Find a 90% confidence interval for the difference in population mean start-up costs for the two types of businesses. (b) Does it seem likely that there is a difference? Explain. (S) 87. A random sample of 100 art books showed a sample average price $40.52 with a known sample standard deviation of $15.30. Find a 90% confidence interval for the population mean price of art books.
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(S) 88. Start-up costs (in thousands of dollars) for a random sample of 13 print shops had sample mean 113.08 with sample standard deviation 41.83. An independent random sample of 12 pet stores had sample mean start-up costs 52.75 with sample standard deviation 41.5. (a) Find a 95% confidence interval for the difference in population means. (b) Does there appear to be a difference in mean start-up costs for print shops as compared to pet shops? Explain. (M) 89. In 1980, a random ample of n1 = 87 gasoline stations was chosen, which showed that the mean price per gallon of gasoline was x1 = 98.9 cents with a sample standard deviation of s1 = 8.4 cents. In 2000, a random sample of n2 = 95 stations was chosen, which gave x2 = 137.9 cents and s2 = 11.6 cents. Compute a 95% confidence interval for µ1 ! µ2 , the difference of gas price means. A. 38.1 < µ1 ! µ2 < 39.9
B. 36.1 < µ1 ! µ2 < 41.9
C. –41.9 < µ1 ! µ2 < –36.1
D. –39.9 < µ1 ! µ2 < –38.1
E. –44.7 < µ1 ! µ2 < –33.3 (S) 90. Membership in the 1000 Pound Club is for ocean fishermen who have caught a fish weighing 1000 lb or more. A random sample of 9 catches of Atlantic Blue Marlin had sample weight 1093.9 lb with sample standard deviation 125.1 lb. An independent random sample of 7 Pacific Blue Marlin catches had sample mean weight of 1114 lb with sample standard deviation 125.6 lb. (a) Construct a 95% confidence interval for the difference in population mean weights. (b) Does the data indicate that there is a difference in the mean weights for the two populations? Explain. (S) 91. A conservationist is interested in the level of support for a water diversion project that would direct water from a rural agricultural area to a rapidly growing urban district. A random sample of 200 people from the rural area showed 40 people in favor of the project. An independent random sample of 300 people in the urban area showed 180 people in favor of the project. (a) Find a 99% confidence interval for the difference of the two proportions. (b) Does your data indicate a difference in the level of support among people in the two regions? Explain. (S) 92. Cedaredge Public Library keeps daily circulation records. A random sample of 64 circulation records for the past year showed mean daily circulation to be 150 books with a known standard deviation of 25.5 books. An independent random sample of 64 circulation records for two years ago had sample mean daily circulation 75 books with a known standard deviation of 10.2 books. Find a 90% confidence interval for the difference of population means.
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(M) 93. A random sample of a certain breed of cat of size n1 = 120 was taken, which showed that the mean litter size was x1 = 7.5 kittens with a standard deviation of s1 = 1.2 kittens. Also, a random sample of a certain breed of dog of size n2 = 135 was taken, which showed that the mean litter size was x2 = 4.6 puppies with a standard deviation of s2 = 1.1 puppies. Compute a 98% confidence interval for µ1 ! µ2 , the difference in litter sizes. A. 2.8 < µ1 ! µ2 < 3.0
B. –3.0 < µ1 ! µ2 < –2.8
C. 2.5 < µ1 ! µ2 < 3.3
D. 2.6 < µ1 ! µ2 < 3.2
E. –3.2 < µ1 ! µ2 < –2.6 (S) 94. During the fall semester math lab attendance was recorded daily. A random sample of 10 records showed sample mean attendance 10.2 students with standard deviation 2.4 students. During the spring semester attendance was also recorded daily. An independent random sample of 10 records for spring semester showed sample mean 15.9 students with sample standard deviation 1.5 students. Find a 90% confidence interval for the difference of the population means. (S) 95. Stellar Supermarket has several stores in different locations. They have recently started offering customers the option of using charge cards to pay their grocery bills. At location A, of a random sample of 50 customers 15 customers said that they regularly used charge cards to pay for their groceries. At location B an independent random sample of 50 customers showed 30 who regularly use credit cards to pay for their groceries. Find a 90% confidence interval for the difference of the two population proportions. (S) 96. Marybelle has developed a new way to teach problem-solving strategies in sixth grade arithmetic. She used her method on a random sample of 16 sixth graders. An independent random sample of 12 sixth graders was taught problem-solving by a traditional method. All students were given the same test. Marybelle’s students had an average score of 82 (out of a possible 100 points) with standard deviation 8.5. The other group had a class average of 75 with standard deviation 10. Find a 95% confidence interval for the difference of population means. (M) 97. Calculate the pooled variance of s12 and s2 2 values for two independent random samples with s1 = 1.5, n1 = 111 and s2 = 1.4, n2 =143. A. –0.1
B. 1.4
C. 17.7
D. 0.02
E. 2.1
(S) 98. A random sample of 100 families showed that 65 had attended at least one major league baseball game last year. An independent random sample of 100 families who had children in a recreational baseball league showed 90 had attended at least one major league baseball game last year. Find a 99% confidence interval for the difference of the two population proportions.
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Chapter 8 Estimation
225
(S) 99. A random sample of 8 boxes of Happy Harvest cereal had sample mean weight 16.1 oz with sample standard deviation 2.2 oz. An independent random sample of 6 boxes of Glorious Grains cereal had mean weight 15.8 oz and standard deviation 1.5 oz. Find a 99% confidence interval for the difference of the population means. (M)100. Use a standard normal distribution table to compute the critical value for an 89% level of confidence. A. 1.59 B. 1.60 C. 1.61 D. 1.62 E. 1.63 (S) 101. A controversial issue is a state legislature is whether excess state revenue from income taxes should be refunded to the taxpayer or whether it should be spent for other purposes such as education and highway repair and construction. A random sample of 200 registered Republicans showed 178 of them in favor of taxpayer refunds. An independent random sample of 200 registered Democrats showed 125 in favor of taxpayer refunds. (a) Find a 95% confidence interval for the difference in population proportions. (b) Does your data indicate that there may be a difference in the level of support for taxpayer refunds between the two parties? Explain. (S) 102. A random sample of 225 books in the fields of sociology and economics had sample mean price $50.24 with sample standard deviation $8.50. An independent random sample of 250 books in the fields of philosophy and psychology had a sample mean price of $44.71 with sample standard deviation $8.75. (a) Find a 90% confidence interval for the difference in population means. (b) Does your data indicate a difference in the population means? Explain. (M)103. Compute the approximate standard deviation, !ˆ , for the difference of two proportions from binomial probability distributions, pˆ1 ! pˆ 2 , where pˆ1 = 0.6, pˆ 2 = 0.5, n1 = 256, and n2 = 183. A. 0.048
B. 0.002
C. 0.00001
D. 0.000003
E. 0.024
(S) 104. A sociologist wants to study the level of support for a proposed system of graduated driving privileges for young automobile drivers. A random sample of 1000 licensed drivers 25 years of age or younger was surveyed. 375 of them said that they were in favor of the proposed system. An independent random sample of 1000 drivers older than 25 were also surveyed. 550 of them expressed support for the proposed system. (a) Find a 90% confidence interval for the difference of population proportions. (b) Does the data indicate a difference in population proportions? Explain. (S) 105. The question of whether Quebec should separate from the rest of Canada has been controversial for Canadians. A random sample of 1000 residents of Eastern Quebec showed 625 favored separation. An independent random sample of 1000 residents of Western Quebec showed 415 in favor of separation. Find a 90% confidence interval for the difference of population proportion
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226
Test Item File Understandable Statistics, 8th Edition
(M)106. In January 2001, a random sample of registered voters in a city was asked whether or not they approved of the job the mayor was doing. Three hundred thirty-six out of the 600 polled said they approved. In July 2001, another random sample of registered voters in the city was asked the same question. This time, 210 out of the 500 polled said they approved. Compute a 95% confidence interval for p1 ! p2 , the difference of proportions of registered voters who approved of the mayor’s performance relative to all registered voters. Let p1 represent the proportion from January and p2 from July. A. 0.11 < p1 ! p2 < 0.17
B. 0.139 < p1 ! p2 < 0.141
C. 0.08 < p1 ! p2 < 0.20
D. 0.138 < p1 ! p2 < 0.142
E. 0.12 < p1 ! p2 < 0.16 (S) 107. An agricultural agent wants to compare the corn yield in bushels per acre for farms in the Midwest in 1996 and 1995. A random sample of 200 farms showed that a mean corn yield in 1996 of 121.1 bushels per acre with a known standard deviation of 17.22 bushels per acre. An independent random sample of 200 farms showed a mean corn yield for 1995 of 108.9 bushels per acre with a known standard deviation of 11.89 bushels per acre. (a) Find a 95% confidence interval for the difference in the population means for 1996 and 1995. (b) Does your data indicate a difference in population means? Explain. (S) 108. A sociologist wants to study the level of support for a proposed restructuring of the social security system. A random sample of 1000 adults who are employed full-time shows 520 in favor of the new system. An independent random sample of 1000 retired people shows 320 in favor. (a) Find a 90% confidence interval for the difference in population proportions. (b) Does the data indicate a difference in population proportions? Explain.
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