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Chapter 7: Sampling and Sampling Distributions 1 Chapter 7 Sampling and Sampling Distributions LEARNING OBJECTIVES The two main objectives for Chapter 7 are to give you an appreciation for the proper application of sampling techniques and an understanding of the sampling distributions of two statistics, thereby enabling you to: 1. Contrast sampling to census and differentiate among different methods of sampling, which include simple, stratified, systematic, and cluster random sampling; and convenience, judgment, quota, and snowball nonrandom sampling, by assessing the advantages associated with each 2. Describe the distribution of a sample’s mean using the central limit theorem, correcting for a finite population if necessary 3. Describe the distribution of a sample’s proportion using the z formula for sample proportions SOLUTIONS TO PROBLEMS IN CHAPTER 7 7.1 a) i. A union membership list for the company. ii. A list of all employees of the company. b) i. White pages of the telephone directory for Utica, New York. ii. Utility company list of all customers. c) i. Airline company list of phone and mail purchasers of tickets from the airline during the past six months. ii. A list of frequent flyer club members for the airline. d) i. List of boat manufacturer's employees. ii. List of members of a boat owners association. e) i. Cable company telephone directory. ii. Membership list of cable management association. Chapter 7: Sampling and Sampling Distributions 2 7.4 a) Size of motel (rooms), age of motel, geographic location. b) Gender, age, education, social class, ethnicity. c) Size of operation (number of bottled drinks per month), number of employees, number of different types of drinks bottled at that location, geographic location. d) Size of operation (sq.ft.), geographic location, age of facility, type of process used. 7.5 a) Under 21 years of age, 21 to 39 years of age, 40 to 55 years of age, over 55 years of age. b) Under $1,000,000 sales per year, $1,000,000 to $4,999,999 sales per year, $5,000,000 to $19,999,999 sales per year, $20,000,000 to $49,000,000 per year, $50,000,000 to $99,999,999 per year, over $100,000,000 per year. c) Less than 2,000 sq. ft., 2,000 to 4,999 sq. ft., 5,000 to 9,999 sq. ft., over 10,000 sq. ft. d) East, southeast, midwest, south, southwest, west, northwest. e) Government worker, teacher, lawyer, physician, engineer, business person, police officer, fire fighter, computer worker. f) Manufacturing, finance, communications, health care, retailing, chemical, transportation. 7.6 n = N/k = 100,000/200 = 500 7.7 N = nk = 75(11) = 825 7.8 k = N/n = 3,500/175 = 20 Start between 0 and 20. The human resource department probably has a list of company employees which can be used for the frame. Also, there might be a company phone directory available. Chapter 7: Sampling and Sampling Distributions 3 7.9 a) i. Counties ii. Metropolitan areas b) i. States (beside which the oil wells lie) ii. Companies that own the wells c) i. States ii. Counties 7.10 Go to the district attorney's office and observe the apparent activity of various attorneys at work. Select some who are very busy and some who seem to be less active. Select some men and some women. Select some who appear to be older and some who are younger. Select attorneys with different ethnic backgrounds. 7.11 Go to a conference where some of the Fortune 500 executives attend. Approach those executives who appear to be friendly and approachable. 7.12 Suppose 40% of the sample is to be people who presently own a personal computer and 60% with people who do not. Go to a computer show at the city's conference center and start interviewing people. Suppose you get enough people who own personal computers but not enough interviews with those who do not. Go to a mall and start interviewing people. Screen out personal computer owners. Interview non personal computer owners until you meet the 60% quota. 7.13 µ = 50, = 10, n = 64 a) P( x > 52): x 52 50 z = = 1.6 10 n 64 from Table A.5, Prob. = .4452 P( x > 52) = .5000 - .4452 = .0548 b) P( x < 51): x 51 50 z = = 0.80 10 n 64 from Table A.5 prob. = .2881 P( x < 51) = .5000 + .2881 = .7881 Chapter 7: Sampling and Sampling Distributions 4 c) P( x < 47): x 47 50 z = = -2.40 10 n 64 from Table A.5 prob. = .4918 P( x < 47) = .5000 - .4918 = .0082 d) P(48.5 < x < 52.4): x 48.5 50 z = = -1.20 10 n 64 from Table A.5 prob. = .3849 x 52.4 50 z = = 1.92 10 n 64 from Table A.5 prob. = .4726 P(48.5 < x < 52.4) = .3849 + .4726 = .8575 e) P(50.6 < x < 51.3): x 50.6 50 z = = 0.48 10 n 64 from Table A.5, prob. = .1844 x 51.3 50 z = 1.04 10 n 64 from Table A.5, prob. = .3508 P(50.6 < x < 51.3) = .3508 - .1844 = .1644 Chapter 7: Sampling and Sampling Distributions 5 7.14 µ = 23.45 = 3.8 a) n = 10, P( x > 22): x 22 23.45 z = = -1.21 3.8 n 10 from Table A.5, prob. = .3869 P( x > 22) = .3869 + .5000 = .8869 b) n = 4, P( x > 26): x 26 23.45 z = = 1.34 3.8 n 4 from Table A.5, prob. = .4099 P( x > 26) = .5000 - .4099 = .0901 7.15 n = 36 µ = 278 P( x < 280) = .86 .3600 of the area lies between x = 280 and µ = 278. This probability is associated with z = 1.08 from Table A.5. Solving for : x z = n 280 278 1.08 = 36 1.08 = 2 6 12 = = 11.11 1.08 7.16 n = 81 = 12 P( x > 300) = .18 Chapter 7: Sampling and Sampling Distributions 6 .5000 - .1800 = .3200 and from Table A.5, z.3200 = 0.92 Solving for µ: x z = n 300 0.92 = 12 81 12 0.92 = 300 - 9 1.2267 = 300 - µ = 300 - 1.2267 = 298.77 7.17 a) N = 1,000 n = 60 µ = 75 =6 P( x < 76.5): x 76.5 75 z = = 2.00 N n 6 1000 60 n N 1 60 1000 1 from Table A.5, prob. = .4772 P( x < 76.5) = .4772 + .5000 = .9772 b) N = 90 n = 36 µ = 108 = 3.46 P(107 < x < 107.7): x 107 108 z = = -2.23 N n 3.46 90 36 n N 1 36 90 1 from Table A.5, prob. = .4871 x 107.7 108 z = = -0.67 N n 3.46 90 36 n N 1 36 90 1 Chapter 7: Sampling and Sampling Distributions 7 from Table A.5, prob. = .2486 P(107 < x < 107.7) = .4871 - .2486 = .2385 c) N = 250 n = 100 µ = 35.6 = 4.89 P( x > 36): x 36 35.6 z = = 1.05 N n 4.89 250 100 n N 1 100 250 1 from Table A.5, prob. = .3531 P( x > 36) = .5000 - .3531 = .1469 d) N = 5000 n = 60 µ = 125 = 13.4 P( x < 123): x 123 125 z = = -1.16 N n 13.4 5000 60 n N 1 60 5000 1 from Table A.5, prob. = .3770 P( x < 123) = .5000 - .3770 = .1230 7.18 µ = 99.9 = 30 n = 38 a) P( x < 90): x 90 99.9 z = = -2. 03 30 n 38 from table A.5, area = .4788 P( x < 90) = .5000 - .4788 = .0212 b) P(98 < x < 105): x 105 99.9 z = = 1.05 30 n 38 Chapter 7: Sampling and Sampling Distributions 8 from table A.5, area = .3531 x 98 99.9 z = = -0.39 30 n 38 from table A.5, area = .1517 P(98 < x < 105) = .3531 + .1517 = .5048 c) P( x < 112): x 112 99.9 z = = 2.49 30 n 38 from table A.5, area = .4936 P( x < 112) = .5000 + .4936 = .9936 d) P(93 < x < 96): x 93 99.9 z = = -1.42 30 n 38 from table A.5, area = .4222 x 96 99.9 z = = -0.80 30 n 38 from table A.5, area = .2881 P(93 < x < 96) = .4222 - .2881 = .1341 7.19 N = 1500 n = 100 µ = 177,000 = 8,500 P( X > $185,000): Chapter 7: Sampling and Sampling Distributions 9 X 185,000 177,000 z = = 9.74 N n 8,500 1500 100 n N 1 100 1500 1 from Table A.5, prob. = .5000 P( X > $185,000) = .5000 - .5000 = .0000 7.20 µ = $65.12 = $21.45 n = 45 P( x > x 0 ) = .2300 Prob. x lies between x 0 and µ = .5000 - .2300 = .2700 from Table A.5, z.2700 = 0.74 Solving for x 0 : x0 z = n x 0 65.12 0.74 = 21.45 45 2.366 = x 0 - 65.12 and x 0 = 65.12 + 2.366 = 67.486 7.21 µ = 50.4 = 11.8 n = 42 a) P( x > 52): x 52 50.4 z = = 0.88 11.8 n 42 from Table A.5, the area for z = 0.88 is .3106 P( x > 52) = .5000 - .3106 = .1894 b) P( x < 47.5): x 47.5 50.4 z = = -1.59 11.8 n 42 from Table A.5, the area for z = -1.59 is .4441 Chapter 7: Sampling and Sampling Distributions 10 P( x < 47.5) = .5000 - .4441 = .0559 c) P( x < 40): x 40 50.4 z = = -5.71 11.8 n 42 from Table A.5, the area for z = -5.71 is .5000 P( x < 40) = .5000 - .5000 = .0000 d) 71% of the values are greater than 49. Therefore, 21% are between the sample mean of 49 and the population mean, µ = 50.4. The z value associated with the 21% of the area is -0.55 z.21 = -0.55 x z = n 49 50.4 -0.55 = 42 = 16.4964 7.22 p = .25 a) n = 110 ˆ P( p < .21): p p ˆ .21 .25 z = = -0.97 pq (. 25 )(. 75 ) n 110 from Table A.5, prob. = .3340 ˆ P( p < .21) = .5000 - .3340 = .1660 b) n = 33 ˆ P( p > .24): p p ˆ .24 .25 z = = -0.13 pq (. 25 )(. 75 ) n 33 Chapter 7: Sampling and Sampling Distributions 11 from Table A.5, prob. = .0517 ˆ P( p > .24) = .5000 + .0517 = .5517 c) n = 59 ˆ P(.24 < p < .27): p p ˆ .24 .25 z = = -0.18 pq (. 25 )(. 75 ) n 59 from Table A.5, prob. = .0714 pP ˆ .27 .25 z = = 0.35 pq (. 25 )(. 75 ) n 59 from Table A.5, prob. = .1368 ˆ P(.24 < p < .27) = .0714 + .1368 = .2082 d) n = 80 ˆ P( p > .30): p p ˆ .30 .25 z = = 1.03 pq (. 25 )(. 75 ) n 80 from Table A.5, prob. = .3485 ˆ P( p > .30) = .5000 - .3485 = .1515 e) n = 800 ˆ P( p > .30): p p ˆ .30 .25 z = = 3.27 pq (. 25 )(. 75 ) n 800 from Table A.5, prob. = .4995 ˆ P( p > .30) = .5000 - .4995 = .0005 7.23 p = .58 n = 660 ˆ a) P( p > .60): Chapter 7: Sampling and Sampling Distributions 12 p p ˆ .60 .58 z = = 1.04 pq (. 58 )(. 42 ) n 660 from table A.5, area = .3508 ˆ P( p > .60) = .5000 - .3508 = .1492 ˆ b) P(.55 < p < .65): p p ˆ .65 .58 z = = 3.64 pq (. 58 )(. 42 ) n 660 from table A.5, area = .4998 p p ˆ .55 .58 z = = 1.56 pq (. 58 )(. 42 ) n 660 from table A.5, area = .4406 ˆ P(.55 < p < .65) = .4998 + .4406 = .9404 ˆ c) P( p > .57): p p ˆ .57 .58 z = = -0.52 pq (. 58 )(. 42 ) n 660 from table A.5, area = .1985 ˆ P( p > .57) = .1985 + .5000 = .6985 ˆ d) P(.53 < p < .56): p p ˆ .56 .58 p p ˆ .53 .58 z = = -1.04 z = = -2.60 pq (. 58 )(. 42 ) pq (. 58 )(. 42 ) n 660 n 660 from table A.5, area for z = -1.04 is .3508 from table A.5, area for z = -2.60 is .4953 ˆ P(.53 < p < .56) = .4953 - .3508 = .1445 Chapter 7: Sampling and Sampling Distributions 13 ˆ e) P( p < .48): p p ˆ .48 .58 z = = -5.21 pq (. 58 )(. 42 ) n 660 from table A.5, area = .5000 ˆ P( p < .48) = .5000 - .5000 = .0000 7.24 p = .40 ˆ P( p > .35) = .8000 ˆ P(.35 < p < .40) = .8000 - .5000 = .3000 from Table A.5, z.3000 = -0.84 Solving for n: p p ˆ z = pq n .35 .40 .05 -0.84 = = (. 40 )(. 60 ) .24 n n 0.84 .24 n .05 8.23 = n n = 67.73 68 7.25 p = .28 n = 140 ˆ ˆ P( p < p0 ) = .3000 ˆ ˆ P( p < p0 < .28) = .5000 - .3000 = .2000 from Table A.5, z.2000 = -0.52 ˆ Solving for p0 : p0 p ˆ z = pq n Chapter 7: Sampling and Sampling Distributions 14 p0 .28 ˆ -0.52 = (. 28 )(. 72 ) 140 ˆ -.02 = p0 - .28 ˆ p0 = .28 - .02 = .26 7.26 P(x > 150): n = 600 p = .21 x = 150 150 ˆ p = = .25 600 p p ˆ .25 .21 z = = 2.41 pq (. 21)(. 79 ) n 600 from table A.5, area = .4920 P(x > 150) = .5000 - .4920 = .0080 7.27 p = .48 n = 200 a) P(x < 90): 90 ˆ p = = .45 200 p p ˆ .45 .48 z = = -0.85 pq (. 48 )(. 52 ) n 200 from Table A.5, the area for z = -0.85 is .3023 P(x < 90) = .5000 - .3023 = .1977 b) P(x > 100): 100 ˆ p = = .50 200 p p ˆ .50 .48 z = = 0.57 pq (. 48 )(. 52 ) n 200 from Table A.5, the area for z = 0.57 is .2157 Chapter 7: Sampling and Sampling Distributions 15 P(x > 100) = .5000 - .2157 = .2843 c) P(x > 80): 80 ˆ p = = .40 200 p p ˆ .40 .48 z = = -2.26 pq (. 48 )(. 52 ) n 200 from Table A.5, the area for z = -2.26 is .4881 P(x > 80) = .5000 + .4881 = .9881 7.28 p = .19 n = 950 ˆ a) P( p > .25): p p ˆ .25 .19 z = = 4.71 pq (. 19 )(. 89 ) n 950 from Table A.5, area = .5000 ˆ P( p > .25) = .5000 - .5000 = .0000 ˆ b) P(.15 < p < .20): p p ˆ .15 .19 z = = -3.14 pq (. 19 )(. 81) n 950 p p ˆ .20 .19 z = = 0.79 pq (. 19 )(. 89 ) n 950 from Table A.5, area for z = -3.14 is .4992 from Table A.5, area for z = 0.79 is .2852 ˆ P(.15 < p < .20) = .4992 + .2852 = .7844 c) P(133 < x < 171): Chapter 7: Sampling and Sampling Distributions 16 133 171 ˆ p1 = = .14 ˆ p2 = = .18 950 950 ˆ P(.14 < p < .18): p p ˆ .14 .19 p p ˆ .18 .19 z = = -3.93 z = = -0.79 pq (. 19 )(. 81) pq (. 19 )(. 81) n 950 n 950 from Table A.5, the area for z = -3.93 is .49997 the area for z = -0.79 is .2852 P(133 < x < 171) = .49997 - .2852 = .21477 7.29 µ = 76, = 14 a) n = 35, P( x > 79): x 79 76 z = = 1.27 14 n 35 from table A.5, area = .3980 P( x > 79) = .5000 - .3980 = .1020 b) n = 140, P(74 < x < 77): x 74 76 x 77 76 z = = -1.69 z = = 0.85 14 14 n 140 n 140 from table A.5, area for z = -1.69 is .4545 from table A.5, area for 0.85 is .3023 P(74 < x < 77) = .4545 + .3023 = .7568 Chapter 7: Sampling and Sampling Distributions 17 c) n = 219, P( x < 76.5): x 76.5 76 z = = 0.53 14 n 219 from table A.5, area = .2019 P( x < 76.5) = .5000 + .2019 = .7019 7.30 p = .46 a) n = 60 ˆ P(.41 < p < .53): p p ˆ .53 .46 z = = 1.09 pq (. 46 )(. 54 ) n 60 from table A.5, area = .3621 p p ˆ .41 .46 z = = -0.78 pq (. 46 )(. 54 ) n 60 from table A.5, area = .2823 ˆ P(.41 < p < .53) = .3621 + .2823 = .6444 b) n = 458 ˆ P( p < .40): p p .40 .46 z = = -2.58 p q (.46)(.54) n 458 from table A.5, area = .4951 P( p < .40) = .5000 - .4951 = .0049 Chapter 7: Sampling and Sampling Distributions 18 c) n = 1350 ˆ P( p > .49): p p ˆ .49 .46 z = = 2.21 pq (. 46 )(. 54 ) n 1350 from table A.5, area = .4864 ˆ P( p > .49) = .5000 - .4864 = .0136 7.31 Under 18 250(.22) = 55 18 – 25 250(.18) = 45 26 – 50 250(.36) = 90 51 – 65 250(.10) = 25 over 65 250(.14) = 35 n = 250 7.32 p = .55 n = 600 x = 298 x 298 ˆ p = = .497 n 600 ˆ P( p < .497): p p ˆ .497 .55 z = = -2.61 pq (. 55 )(. 45 ) n 600 from Table A.5, Prob. = .4955 ˆ P( p < .497) = .5000 - .4955 = .0045 No, the probability of obtaining these sample results by chance from a population that supports the candidate with 55% of the vote is extremely low (.0045). This is such an unlikely chance sample result that it would cause the researcher to probably reject her claim of 55% of the vote. Chapter 7: Sampling and Sampling Distributions 19 7.33 a) Roster of production employees secured from the human resources department of the company. b) Alpha/Beta store records kept at the headquarters of their California division or merged files of store records from regional offices across the state. c) Membership list of Maine lobster catchers association. 7.34 µ = $ 17,755 = $ 650 n = 30 N = 120 P( x < 17,500): 17,500 17,755 z = = -2.47 650 120 30 30 120 1 from Table A.5, the area for z = -2.47 is .4932 P( x < 17,500) = .5000 - .4932 = .0068 7.35 Number the employees from 0001 to 1250. Randomly sample from the random number table until 60 different usable numbers are obtained. You cannot use numbers from 1251 to 9999. 7.36 µ = $125 n = 32 x = $110 2 = $525 P( x > $110): x 110 125 z = = -3.70 525 n 32 from Table A.5, Prob.= .5000 P( x > $110) = .5000 + .5000 = 1.0000 Chapter 7: Sampling and Sampling Distributions 20 P( x > $135): x 135 125 z = = 2.47 525 n 32 from Table A.5, Prob.= .4932 P( x > $135) = .5000 - .4932 = .0068 P($120 < x < $130): x 120 125 z = = -1.23 525 n 32 x 130 125 z = = 1.23 525 n 32 from Table A.5, Prob.= .3907 P($120 < x < $130) = .3907 + .3907 = .7814 7.37 n = 1100 a) x > 810, p = .73 x 810 ˆ p = n 1100 p p ˆ .7364 .73 z = = 0.48 pq (. 73)(. 27 ) n 1100 from table A.5, area = .1844 P(x > 810) = .5000 - .1844 = .3156 b) x < 1030, p = .96, x 1030 ˆ p = = .9364 n 1100 Chapter 7: Sampling and Sampling Distributions 21 p p ˆ .9364 .96 z = = -3.99 pq (. 96 )(. 04 ) n 1100 from table A.5, area = .49997 P(x < 1030) = .5000 - .49997 = .00003 c) p = .85 ˆ P(.82 < p < .84): p p ˆ .82 .85 z = = -2.79 pq (. 85 )(. 15 ) n 1100 from table A.5, area = .4974 p p ˆ .84 .85 z = = -0.93 pq (. 85 )(. 15 ) n 1100 from table A.5, area = .3238 ˆ P(.82 < p < .84) = .4974 - .3238 = .1736 7.38 1) The managers from some of the companies you are interested in studying do not belong to the American Managers Association. 2) The membership list of the American Managers Association is not up-to-date. 3) You are not interested in studying managers from some of the companies belonging to the American Management Association. 4) The wrong questions are asked. 5) The manager incorrectly interprets a question. 6) The assistant accidentally marks the wrong answer. 7) The wrong statistical test is used to analyze the data. 8) An error is made in statistical calculations. 9) The statistical results are misinterpreted. 7.39 Divide the factories into geographic regions and select a few factories to represent those regional areas of the country. Take a random sample of employees from each selected factory. Do the same for distribution centers and retail outlets. Divide the United States into regions of areas. Select a few areas. Take a random sample from each of the selected area distribution centers and retail outlets. 7.40 N = 12,080 n = 300 Chapter 7: Sampling and Sampling Distributions 22 k = N/n = 12,080/300 = 40.27 Select every 40th outlet to assure n > 300 outlets. Use a table of random numbers to select a value between 0 and 40 as a starting point. 7.41 p = .54 n = 565 a) P(x > 339): x 339 ˆ p = = .60 n 565 p p ˆ .60 .54 z = = 2.86 pq (. 54 )(. 46 ) n 565 from Table A.5, the area for z = 2.86 is .4979 P(x > 339) = .5000 - .4979 = .0021 b) P(x > 288): x 288 ˆ p = = .5097 n 565 p p ˆ .5097 .54 z = = -1.45 pq (. 54 )(. 46 ) n 565 from Table A.5, the area for z = -1.45 is .4265 P(x > 288) = .5000 + .4265 = .9265 ˆ c) P( p < .50): p p ˆ .50 .54 z = = -1.91 pq (. 54 )(. 46 ) n 565 from Table A.5, the area for z = -1.91 is .4719 ˆ P( p < .50) = .5000 - .4719 = .0281 7.42 µ = $550 n = 50 = $100 Chapter 7: Sampling and Sampling Distributions 23 P( x < $530): x 530 550 z = = -1.41 100 n 50 from Table A.5, Prob.=.4207 P(x < $530) = .5000 - .4207 = .0793 7.43 µ = 56.8 n = 51 = 12.3 a) P( x > 60): x 60 56.8 z = = 1.86 12.3 n 51 from Table A.5, Prob. = .4686 P( x > 60) = .5000 - .4686 = .0314 b) P( x > 58): x 58 56.8 z = = 0.70 12.3 n 51 from Table A.5, Prob.= .2580 P( x > 58) = .5000 - .2580 = .2420 c) P(56 < x < 57): x 56 56.8 x 57 56.8 z = = -0.46 z = = 0.12 12.3 12.3 n 51 n 51 from Table A.5, Prob. for z = -0.46 is .1772 from Table A.5, Prob. for z = 0.12 is .0478 P(56 < x < 57) = .1772 + .0478 = .2250 Chapter 7: Sampling and Sampling Distributions 24 d) P( x < 55): x 55 56.8 z = = -1.05 12.3 n 51 from Table A.5, Prob.= .3531 P( x < 55) = .5000 - .3531 = .1469 e) P( x < 50): x 50 56.8 z = = -3.95 12.3 n 51 from Table A.5, Prob.= .5000 P( x < 50) = .5000 - .5000 = .0000 7.45 p = .73 n = 300 a) P(210 < x < 234): x 210 x 234 ˆ p1 = = .70 ˆ p2 = = .78 n 300 n 300 p p ˆ .70 .73 z = = -1.17 pq (. 73)(. 27 ) n 300 p p ˆ .78 .73 z = = 1.95 pq (. 73)(. 27 ) n 300 from Table A.5, the area for z = -1.17 is .3790 the area for z = 1.95 is .4744 P(210 < x < 234) = .3790 + .4744 = .8534 ˆ b) P( p > .78): p p ˆ .78 .73 z = = 1.95 pq (. 73)(. 27 ) n 300 from Table A.5, the area for z = 1.95 is .4744 Chapter 7: Sampling and Sampling Distributions 25 ˆ P( p > .78) = .5000 - .4744 = .0256 c) p = .73 n = 800 ˆ P( p > .78): p p ˆ .78 .73 z = = 3.19 pq (. 73)(. 27 ) n 800 from Table A.5, the area for z = 3.19 is .4993 ˆ P( p > .78) = .5000 - .4993 = .0007 7.46 n = 140 P(x > 35): 35 ˆ p = = .25 p = .22 140 p p ˆ .25 .22 z = = 0.86 pq (. 22 )(. 78 ) n 140 from Table A.5, the area for z = 0.86 is .3051 P(x > 35) = .5000 - .3051 = .1949 P(x < 21): 21 ˆ p = = .15 140 p p ˆ .15 .22 z = = -2.00 pq (. 22 )(. 78 ) n 140 from Table A.5, the area for z = 2.00 is .4772 P(x < 21) = .5000 - .4772 = .0228 n = 300 p = .20 ˆ P(.18 < p < .25): p p ˆ .18 .20 z = = -0.87 pq (. 20 )(. 80 ) n 300 from Table A.5, the area for z = -0.87 is .3078 Chapter 7: Sampling and Sampling Distributions 26 p p ˆ .25 .20 z = = 2.17 pq (. 20 )(. 80 ) n 300 from Table A.5, the area for z = 2.17 is .4850 ˆ P(.18 < p < .25) = .3078 + .4850 = .7928 7.47 By taking a sample, there is potential for obtaining more detailed information. More time can be spent with each employee. Probing questions can be asked. There is more time for trust to be built between employee and interviewer resulting in the potential for more honest, open answers. With a census, data is usually more general and easier to analyze because it is in a more standard format. Decision-makers are sometimes more comfortable with a census because everyone is included and there is no sampling error. A census appears to be a better political device because the CEO can claim that everyone in the company has had input. 7.48 p = .75 n = 150 x = 120 ˆ P( p > .80): p p ˆ .80 .75 z = = 1.41 pq (. 75 )(. 25 ) n 150 from Table A.5, the area for z = 1.41 is .4207 ˆ P( p > .80) = .5000 - .4207 = .0793 7.49 a) Switzerland: n = 40 µ = $ 30.67 =$3 P(30 < x < 31): x 31 30.67 z 0.70 3 n 40 x 30 30.67 z 1.41 3 n 40 from Table A.5, the area for z = 0.70 is .2580 the area for z = -1.41 is .4207 Chapter 7: Sampling and Sampling Distributions 27 P(30 < x < 31) = .2580 + .4207 = .6787 b) Japan: n = 35 µ = $ 20.20 = $3 P( x > 21): x 21 20.20 z 1.58 3 n 35 from Table A.5, the area for z = 1.58 is .4429 P( x > 21) = .5000 - .4429 = .0571 c) U.S.: n = 50 µ = $ 23.82 =$3 P( x < 22.75): x 22.75 23.82 z 2.52 3 n 50 from Table A.5, the area for z = -2.52 is .4941 P( x < 22.75) = .5000 - .4941 = .0059 7.50 a) Age, Ethnicity, Religion, Geographic Region, Occupation, Urban-Suburban-Rural, Party Affiliation, Gender b) Age, Ethnicity, Gender, Geographic Region, Economic Class c) Age, Ethnicity, Gender, Economic Class, Education d) Age, Ethnicity, Gender, Economic Class, Geographic Location 7.51 µ = $281 n = 65 = $47 P( x > $273): x 273 281 z = = -1.37 47 n 65 from Table A.5 the area for z = -1.37 is .4147 P( x > $273) = .5000 + .4147 = .9147