Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition Answers to Sample Chapter Tests All calculations performed using Minitab 15 or a TI-83 Plus. Answers given as provided by the technology. CHAPTER 1, FORM A 1. 2. 3. 4. 5. 6. 7. (a) Length of time to earn a bachelor’s degree (b) Quantitative (c) All Colorado residents who earned (or will earn) a bachelor’s degree Explanations will vary. (a) Nominal (b) Nominal (c) Ratio (d) Interval (e) Nominal (f) Ordinal (g) Ordinal (a) Census (b) Experiment (c) Sampling (d) Simulation Answers will vary. The outcomes are the number of dots on the face, 1 through 6. Consider single digits in the random number table. Select a starting place at random. Record the first five digits you encounter that are between (and including) 1 and 6. The first five outcomes are 3, 6, 1, 5, and 6. (a) Systematic (b) Cluster (c) Stratified (d) Convenience (e) Simple random (a) Parameter (b) Parameter (c) Statistic 3. 4. 5. 6. 7. CHAPTER 1, FORM C 1. 2. 3. CHAPTER 1, FORM B 1. 2. (a) Observed book purchase (mystery or not mystery) (b) Qualitative (c) All current customers of the bookstore Explanations will vary. (a) Nominal (b) Ratio (c) Interval (d) Ordinal (e) Ratio (a) Census (b) Sampling (c) Simulation (d) Experiment Answers will vary. Assign each of the 736 employees a number between 1 and 736. Select a starting place in the random number table at random. Use groups of three digits. Use the first 30 distinct groups of three digits that correspond to employees numbers. For this problem, use employees 622, 413, 055, 401, and 334. (a) Simple random (b) Stratified (c) Cluster (d) Convenience (e) Systematic (a) Statistic (b) Parameter (c) Statistic 4. 5. 6. (b) A. B. C. D. E. A. B. C. D. (e) (d) A. B. C. D. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. (b) (a) (d) (c) (a) (b) (c) (a) (d) (e) (d) (a) (c) Part III: Answers to Sample Chapter Tests E. (b) 4. Time Series Plot of Air Quality Index 5.5 CHAPTER 2, FORM A 5.0 1. Air Quality Index 4.5 Pareto Chart of Majors at Hendrix College 700 600 4.0 3.5 3.0 2.5 Number of Students 500 2.0 400 1.5 Sunday Monday Tuesday 300 Wednesday Day Thursday Friday Saturday 200 100 0 5. Business Administration Social Science Natural Science Major Humanities Philosophy Circle Graph of Noise Preference Background 8.0% 2. (a) The class width is 6 Quiet 38.0% Frequency and Relative Frequency Table Stereo 34.0% Class Rel. Class Cum. Boundaries Freq. Frequency Midpoint Freq. 2.5 8.5 8.5 14.5 14.5 20.5 20.5 26.5 26.5 32.5 10 9 6 2 3 0.3333 0.3000 0.2000 0.0667 0.1000 5.5 11.5 17.5 23.5 29.5 10 19 25 27 30 TV 20.0% 6. Circle Graph of Age of Shoppers Men 60+ 20.0% (b) Women under 60 35.0% Relative Frequency Histogram of Student Visits 35 Relative Frequency 30 25 Men under 60 25.0% 20 15 Women 60+ 20.0% 10 5 0 2.5 8.5 14.5 20.5 Weekly Student Visits 26.5 32.5 7. Dotplot of Student Visits 3. Minitab output: Stem-and-leaf of Dollars N = 20 Leaf Unit = 1.0 0 1 2 3 4 5 5789 25679 124 357 269 17 4 8 12 16 20 Student Visits 24 28 32 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition Relative Frequency Histogram of Incomplete Grades 35 30 Relative Frequency 8. 9. The dotplot and the histogram show the same information, except the dotplot retains the original values in the data set. Eleven motorists were speeding. Minitab output: Stem-and-leaf of Age of Runner N = 22 Leaf Unit = 1.0 25 20 15 10 5 0 1 2 3 4 5 8 089 2445889 12568 0237 05 0 3. The distribution is fairly symmetrical with a center near 30. 1. 33.5 48.5 63.5 Number of Incomplete Grades 78.5 93.5 Minitab output: Stem-and-leaf of Study Abroad Students N = 20 Leaf Unit = 1.0 2 3 4 5 6 CHAPTER 2, FORM B 18.5 169 3468 2457 012689 348 Pareto Chart of Sales 1400 4. Time Series Plot of Price of Gold 1000 325 800 320 600 315 Price of Gold ($) Number of Books Sold 1200 400 200 0 Nonfiction Fiction Children's Electronic Media Genre 310 305 300 295 290 1 2. (a) The class width is 15 2 3 4 5 6 7 Week 8 9 10 Frequency and Relative Frequency Table Class Rel. Class Cum. Boundaries Freq. Frequency Midpoint Freq. 18.5 33.5 33.5 48.5 48.5 63.5 63.5 78.5 78.5 93.5 3 7 8 4 2 0.1250 0.2917 0.3333 0.1667 0.0833 26 41 56 71 86 5. Circle Graph of Chosen Major Undecided 8.0% 3 10 18 22 24 Social Science 16.0% Engineering 18.0% Science 8.0% Computers 28.0% 6. (b) Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Humanities 22.0% 11 12 Part III: Answers to Sample Chapter Tests 8. 9. Circle Graph of Bus Riders Junior 12.0% Senior 2.0% (a) (c) CHAPTER 3, FORM A 1. Sophomore 25.0% Freshman 61.0% 2. 3. 7. (c) s 2 81.67 Dotplot of Incomplete Grade 4. 5. 20 30 40 50 60 Incomplete Grade 70 x = 16.61; median = 13; mode = 12 $42.19 (thousand); $35.76 (thousand); The trimmed mean because it does not include the extreme value. Note that all the salaries are below $42.19 (thousand) except one. (a) Range = 25 (b) x = 33 80 (d) (a) (b) (a) 90 s = 9.04 CV = 20.9% $6.04 to $14.72 Low value = 15; Q1 = 23; median = 33; Q3 = 49; High value = 72 (b) Boxplot of Age of Skier 80 70 8. 9. Age of Skier 60 The dotplot and the histogram show the same information, except the dotplot retains the original values in the data set. 6 days Minitab output: Stem-and-leaf of Diameter N = 16 Leaf Unit = 0.10 0 1 2 3 4 9 278 345568 01677 1 The distribution is fairly symmetrical with a center near 2.6 mm. (c) (b) (e) (a) (d) (e) (b) 40 30 20 10 6. 7. 8. (c) Interquartile range = 26 Weighted average = 85.65 163.31 lb 79% below; 21% above CHAPTER 3, FORM B 1. x = 10.55; median = 11; mode = 11 2. 37.8 inches; 31.63 inches; The trimmed mean because it does not include the extreme value. Note that all but three values are below 37.8 inches. (a) Range = 14 (b) x = 12.17 CHAPTER 2, FORM C 1. 2. 3. 4. 5. 6. 7. 50 3. (c) s 2 25.37 4. (d) s = 5.04 (a) CV = 26.5% (b) 15.3 to 49.7 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition 5. (a) Low value = 27; Q1 = 35.25; median = 42.5; Q3 = 57.5; High value = 68 (b) 2. (b) x 167; y 53; x2 6069; y 2 789; xy 2103 Boxplot of Faculty Age 70 (c) b = 0.678; a = 12.0; yˆ 0.678x 12.0 60 Age y 10.6 (a) x 33.4; 50 3. 40 4. 5. 30 (d) See problem 1. r = 0.996; r2 = 0.992; 99.2% of the variation in y can be explained by the least squares line using x as the predicting variable. For x = 32 inches, y = 9.696 pounds. Scatter Diagram of Sales ($) versus TV Ads ($) 900 800 Sales ($ Thousands) 6. 7. 8. (c) Interquartile range = 22.25 Weighted average = 84.75 $44.40 19% above; 81% below CHAPTER 3, FORM C 1. 2. 3. 4. 5. 6. 7. 8. (b) (d) A. B. C. D. A. B. (b) (a) (d) (e) 600 500 400 300 200 100 0 (e) (a) (c) (d) (b) (a) 6. 10 20 30 40 50 60 TV Ads ($ Thousands) 70 80 90 The linear correlation coefficient appears to be positive since as x increases, y tends to increase. (a) x = 35.2; y = 358 (b) x 211; y 2150; x2 12143; y 2 1124798; xy 114639 (c) b = 8.26; a = 67.7; yˆ 8.26 x 67.7 7. CHAPTER 4, FORM A 1. 8. (d) See problem 5. r = 0.954; r2 = 0.910; 91.0% of the variation in y can be explained by the least squares line using x as the predicting variable. For x = $37, y = $373.32 (in thousands). Scatter Diagram of Pounds versus Inches for Northern Pike 20 15 Pounds 700 10 5 0 20 25 30 35 40 45 Inches The linear correlation coefficient appears to be positive since as x increases, y tends to increase. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Part III: Answers to Sample Chapter Tests 7. CHAPTER 4, FORM B 1. 8. r = 0.994; r2 = 0.988; 98.8% of the variation in y can be explained by the least squares line using x as the predicting variable. For x = $35, y = $100.40. Scatter Diagram of Revenue versus Salary 26 CHAPTER 4, FORM C Revenue ($ Billions) 24 22 20 18 16 14 12 10 1.0 2. 1.5 Salary ($ Millions) 2.0 2.5 The linear correlation coefficient appears to be positive since as x increases, y tends to increase. (a) x 1.38; y 17.8 (b) 4. 5. Scatter Diagram of Price ($) versus Cost ($) 175 Price ($) 150 125 100 75 50 20 30 40 50 Cost ($) 60 2. 3. 4. 5. 6. 200 6. 1. x 6.9; y 89; x 11.03; y 2 1703; xy 134.6 (d) See problem 1. r = 0.880; r2 = 0.775; 77.5% of the variation in y can be explained by the least squares line using x as the predicting variable. For x = $1.5 million, y = $18.735 billion. 70 (a) (d) (b) (a) (c) (e) (d) (c) CHAPTER 5, FORM A 2 (c) b = 7.81; a = 7.02; yˆ 7.81x 7.02 3. 1. 2. 3. 4. 5. 6. 7. 8. (a) Relative frequency; 19/317 = 0.0599 or 5.99% (b) 1 0.0599 = 0.9401 = 94.01% (c) Defective, not defective; yes 1/3 (a) 1/12 (b) 1/12 (c) 1/4 (a) 21/144 = 0.149 (b) 21/132 = 0.159 P(Approval on written and interview) = P(written)P(interview | written) = (0.63)(0.85) = 0.5355 = 53.55% (a) 35/360 (b) 21/360 (c) 5/149 (d) 149/360 (e) 48/149 (f) 31/360 (g) No; P(referred) = 149/360 ≠ P(referred | satisfied) = 59/138. The linear correlation coefficient appears to be positive since as x increases, y tends to increase. (a) x = 44.83; y = 126.3 (b) x 269; y 758; x2 13659; y 2 107100; xy 38216 (c) b = 2.6471; a = 7.653; yˆ 2.65x 7.65 (d) See problem 5. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition 7. There are 8 packages to display. 7. He can walk 6 different routes. 8. 9. P16,5 = 524,160 C24,4 = 10,626 CHAPTER 5, FORM C 8. 9. P40,4 = 2,193,360 C11, 3 = 165 CHAPTER 5, FORM B 1. 2. 3. 4. 5. 6. (a) P(woman) = 4/10 = 0.4 = 40% (b) P(man) = 6/10 = 0.6 = 60% (c) P(President) = 1/10 = 0.1 = 10% 2/3 (a) 1/6 (b) 1/6 (c) 1/9 (a) (6/17)(8/17) = 0.166 = 16.6% (b) (6/17)(8/16) = 0.176 = 17.6% P(female and computer science major) = P(female)P(computer science major | female) = (0.64)(0.12) = 0.0768 = 7.68% (a) 180/417 (b) 65/115 (c) 98/417 (d) 61/417 (e) 61/101 (f) 65/180 (g) No; P(neutral) = 166/417 ≠ P(neutral | freshman) = 61/101 1. 2. 3. 4. 5. 6. 7. 8. 9. A. B. (a) A. B. C. A. B. (b) A. B. C. D. E. F. (d) (e) (a) (d) (b) (c) (e) (b) (a) (d) (e) (c) (a) (d) (c) (b) CHAPTER 6, FORM A 1. (a) P(0) = 0.038; P(1) = 0.300; P(2) = 0.263; P(3) = 0.171; P(4) = 0.117; P(5) = 0.058; P(6) = 0.033; P(7) = 0.021 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Part III: Answers to Sample Chapter Tests (b) (b) Probability Distribution of x 0.25 Probability Distribution of x 0.20 0.30 Probability Probability 0.25 0.20 0.15 0.10 0.15 0.05 0.10 0.00 0.05 1 2 3 4 5 6 7 8 9 10 x 0.00 0 1 2 3 4 5 6 7 x (c) 0.641 (d) 0.101 (e) = 4.40 (c) P(2 x 5) = 0.609 (d) P(x < 3) = 0.601 (e) = 2.44 (f) σ = 1.828 2. 3. 2. 3. (f) = 1.574 P(4) = 0.122 (a) P(11) = 0.0088 (b) P(r 8) = 0.050 (c) P(r 4) = 0.515 (b) P(r 9) = 0.200 4. 5. (c) P(r 3) = 0.012 4. 5. (f) = 2.08 P(3) = 0.251 (a) P(15) = 0.000 (to three digits) (d) P(5 r 10) = 0.484 51 (a) (d) P(9 r 11) = 0.200 6 (a) Distribution Plot Binomial, n=7, p=0.55 0.30 0.25 Probability Distribution Plot Binomial, n=5, p=0.75 0.4 0.20 0.15 0.10 Probability 0.3 0.05 0.00 0.2 0 1 2 3 4 5 6 7 8 r 0.1 0.0 0 1 2 3 4 5 (b) = 3.85 6 (c) = 1.316 r 6. 6. (a) = 1.2 (b) = 3.75 (b) = 0.980 (c) = 0.968 (a) µ = 1.05 (c) Yes, since 5 > µ + 2.5 = 1.2 + (2.5)(0.980) = 3.65. (b) = 0.945 (c) Yes, since 4 > µ + 2.5 = 1.05 + (2.5)(0.945) = 3.4125. CHAPTER 6, FORM C 1. CHAPTER 6, FORM B 1. (a) P(1) = 0.066; P(2) = 0.092; P(3) = 0.202; P(4) = 0.224; P(5) = 0.184; P(6) = 0.079; P(7) = 0.053; P(8) = 0.044; P(9) = 0.035; P(10) = 0.022 2. 3. A. B. C. D. (e) A. (c) (e) (b) (d) (b) Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition 4. 5. 6. B. C. D. (b) A. B. (c) (a) (d) (c) CHAPTER 7, FORM B (e) (c) 2. 3. (a) The tails of a normal curve must get closer and closer to the x-axis. (b) A normal curve must be symmetrical. 68% 1.82 ≈ 2 babies 4. (a) z 0.33 1. (b) 0.86 z 0.57 CHAPTER 7, FORM A (c) z 1.33 1. (d) x 2.6 2. 3. (a) A normal curve is bell-shaped with one peak. This curve has two peaks. (b) A normal curve never crosses the horizontal axis. 99.7% 0.91 ≈ 1 boy 4. (a) z 1.8 (e) 6.8 x 11 5. (b) 1.2 z 0.8 (c) z 2.2 6. (d) x 3.1 (e) 4.1 x 4.6 5. 6. (f) x 4.35 (a) No, look at z values to compare different populations. (b) For Joel, z = 1.51. For John, z = 2.77. Relative to the district, John is a better salesman. (a) P(x 5) = 0.2198 (b) P(x 10) = 0.0668 7. 8. 9. 7. 8. (c) P(5 x 10) = 0.7134 29.90 minutes or 30 minutes (a) Here, np = (57)(.87) > 5 and nq = (57)(.13) > 5. Yes, the assumptions are satisfied. (b) P(r ≤ 47) ≈ 0.2053 (c) P(47 ≤ r ≤ 55) ≈ 0.8781 9. (f) x 13.1 (a) Generator II is hotter; see the z values. (b) z = 0.72 for Generator I z = 2.75 for Generator II Generator II could be near a melt down since it is hotter. (a) P(x < 6) = 0.0912 (b) P(x > 7) = 0.0228 (c) P(6 x 7) = 0.8860 3 years (a) Here, np = (88)(.71) > 5 and nq = (88)(.29) > 5. Yes, the assumptions are satisfied. (b) P(r ≥ 60) ≈ 0.6791 (c) P(60 ≤ r ≤ 70) ≈ 0.7283 (a) P 5.2 x = 0.9664 (b) P x 7.1 = 0.0916 (c) P 5.2 x 7.1 = 0.8748 10. (a) P 8 x = 0.9845 (b) P x 9 = 0.2949 (c) P 8 x 9 = 0.6896 (a) P x 3 = 0.3036 (b) P x 3.5 = 0.2203 (c) P 3.1 x 3.3 = 0.2029 10. (a) P x 23 = 0.0472 (b) P x 28 = 0.1129 (c) P 23 x 28 = 0.8399 CHAPTER 7, FORM C 1. 2. 3. 4. 5. (c) (b) (e) A. B. A. B. C. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. (d) (a) (d) (b) (a) Part III: Answers to Sample Chapter Tests 6. A. B. C. 7. (c) 8. A. B. 9. A. B. C. D. 10. A. B. C. 11. A. B. C. (c) (e) (b) 3. 4. (a) (b) (c) (b) (a) (a) (b) (d) (c) (a) (d) (e) 5. 6. 7. 2. 3. 4. 5. 6. 7. (a) p̂ = 41/56 or 0.732 (b) (c) (d) 20 (a) (b) 0.62 to 0.85 np > 5 and nq > 5; yes 246 more 271 167 CHAPTER 8, FORM C 1. 2. CHAPTER 8, FORM A 1. (b) The distribution is mound-shaped symmetrical and is known. (c) $120 to $156 10.90 to 18.50 pages 89.0 to 97.4 minutes; We are 99% confident that the interval 89.0 to 97.4 contains the population mean repair time at Computer Depot. 27.70 to 31.90 inches We are 95% confident that the interval 27.70 to 31.90 contains the population mean Blue Spruce circumference. (a) $33.04 to $36.36 (b) The distribution is mound-shaped symmetrical and is unknown. (c) $826.00 to $909.00 27.98 to 38.78 minutes 10.05 to 13.75 inches; We are 99% confident that the interval 10.05 to 13.75 contains the population mean rainbow trout length. 3. 4. 5. 6. 7. (b) A. B. (a) (b) A. B. C. (d) A. B. (c) (e) (d) (a) (e) (a) (d) (a) p̂ = 59/78 0.756 (b) 0.68 to 0.84 (c) np > 5 and nq > 5; yes (d) 122 more 147 (a) 1692 (b) 715 CHAPTER 8, FORM B 1. 2. 1.88 to 2.92; We are 95% confident that the interval 1.88 to 2.92 contains the population mean number of calls received each night at the O’Sullivan household. (a) $6.00 to $7.80 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition For all hypothesis testing questions in chapters 9 – 11, the P-values will be given as determined by a TI-84 Plus. Sketches of the sampling distributions were generated using Minitab 15. As such, there may be slight rounding discrepancies in the P-values between the technologies. (e) Do not reject H0. At the 5% level of significance, there is not enough evidence to CHAPTER 9, FORM A conclude that the waiting time has 1. (a) 0.01 decreased. 3. (a) 0.01 H0: = 24,819 H0: p = 0.75 H1: > 24,819 H1: p < 0.75 Right-tailed Left-tailed (b) Standard normal; z = 3.60 (b) Standard normal; z = 2.81 (c) P-value = 0.0024 (c) P-value = 0.00016 Distribution Plot Normal, Mean=0, StDev=1 Distribution Plot Normal, Mean=0, StDev=1 0.4 0.4 0.3 Density Density 0.3 0.2 0.2 0.1 0.00248 0.1 0.0 0.000159 0.0 2. 0 z (d) z0 = 2.33 (e) Reject H0. At the 1% level of significance, there is evidence that the population mean daily sales has increased. (a) 0.05 H0: = 18.3 H1: < 18.3 Left-tailed (b) Student’s t; d.f. = 4; t = 1.154 (c) P-value = 0.1564 -2.81 0 4. (d) z0 = 2.33 (e) Reject H0. At the 1% level of significance, there is evidence that the proportion of rainbow trout is less than 75%. (a) 0.05 H0: = 71 H1: 71 Two-tailed (b) Standard normal; z = 6.236 (c) P-value ≈ 0.0000 Distribution Plot Distribution Plot T, df=4 Normal, Mean=0, StDev=1 0.4 0.4 0.3 0.3 Density Density z 3.6 0.2 0.1 0.2 0.1 0.156 0.0 2.2445E-10 -1.154 0 t (d) t0 = -2.132 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. 0.0 -6.24 2.2445E-10 z 0 6.236 Part III: Answers to Sample Chapter Tests 5. (d) z0 = 1.96 (e) Reject H0. At the 5% level of significance, there is evidence that the population mean height is different from 71 inches. (a) 0.05 H0: = 30 (e) Do not reject H0. At the 5% level of significance, there is not enough evidence to conclude that the proportion is less than 60%. CHAPTER 9, FORM B H1: > 30 Right-tailed (b) Student’s t; d.f. = 13; t = 0.364 (c) P-value = 0.361 1. (a) 0.01 H0: = 5.3 H1: < 5.3 Left-tailed (b) Standard normal; z = 3.28 (c) P-value = 0.00053 Distribution Plot T, df=13 Distribution Plot Normal, Mean=0, StDev=1 0.4 0.4 0.3 Density Density 0.3 0.2 0.1 0.361 0.1 0.000519 0.0 t 0.364 0.0 (d) t0 = 1.771 (e) Do not reject H0. At the 5% level of significance, there is not enough evidence to conclude that the delivery time is greater than 30 minutes. (a) 0.05 H0: p = 0.6 H1: p < 0.6 Left-tailed (b) Standard normal; z = 1.14 (c) P-value = 0.1269 2. -3.28 z 0 (d) z0 = 2.33 (e) Reject H0. At the 1% level of significance, there is evidence that the average recovery time is less as an outpatient. (a) 0.05 H0: = 4.19 H1: > 4.19 Right-tailed (b) Student’s t; d.f. = 15; t = 3.2 (c) P-value = 0.0030 Distribution Plot T, df=15 0.4 Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.3 Density 6. 0.2 0.2 0.2 0.1 0.00298 0.1 0.0 0.127 0.0 (d) z0 = 1.645 0 t -1.14 3.2 0 z (d) t0 = 1.753 (e) Reject H0. At the 5% level of significance, there is evidence that the average yield of Arizona municipal bonds is higher than the national average yield. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition 3. (a) 0.01 Right-tailed (b) Student’s t; d.f. = 15; t = 2.38 (c) P-value = 0.0155 H0: p = 0.37 H1: p 0.37 Two-tailed (b) Standard normal; z = 0.467 (c) P-value = 0.6408 Distribution Plot T, df=15 0.4 Distribution Plot 0.3 Normal, Mean=0, StDev=1 Density 0.4 0.2 0.3 Density 0.1 0.2 0.0155 0.0 0.1 0.320 0.0 2.38 t 0.320 -0.467 0 0.467 z 4. 0 (d) z0 = 2.576 (e) Do not reject H0. At the 1% level of significance, there is no evidence to conclude that the proportion of freshmen on the Dean’s List is different than 37%. (a) 0.05 6. (d) t0 = 1.753 (e) Reject H0. At the 5% level of significance, there is evidence that the average age is greater than 30 years. (a) 0.05 H0: p = 0.27 H1: p > 0.27 Right-tailed (b) Standard normal; z = 2.746 (c) P-value = 0.0030 H0: = 2.8 H1: 2.8 Two-tailed Distribution Plot Normal, Mean=0, StDev=1 (b) Standard normal; z = 2.598 (c) P-value = 0.0094 0.4 Density 0.3 0.2 Distribution Plot Normal, Mean=0, StDev=1 0.1 0.4 0.00302 0.0 Density 2.746 0.2 (d) z0 = 1.645 (e) Reject H0. At the 5% level of significance, there is sufficient evidence to conclude that the proportion is greater than 27%. 0.1 0.00469 0.0 -2.598 0.00469 0 z 2.60 CHAPTER 9, FORM C 5. 0 z 0.3 (d) z0 = 1.96 (e) Reject H0. At the 5% level of significance, there is evidence that the population mean time is different from 2.8 hours. (a) 0.05 H0: = 30 H1: > 30 1. 2. A. B. C. D. E. A. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. (b) (b) (b) (d) (b) (e) Part III: Answers to Sample Chapter Tests 4. 5. 6. (a) (b) (d) (b) (d) (e) (c) (a) (a) (b) (c) (e) (c) (a) (c) (b) (d) (c) (a) (c) (e) (d) (a) (a) Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.115 0.0 5. z 1.2 (d) z0 = 1.96 (e) Do not reject H0. At the 5% level of significantly, there is not enough evidence to conclude that there is difference in population proportions of success between the two different sales pitches. (a) 0.05 (b) Student’s t; t = 2.263 (c) P-value = 0.0365 Distribution Plot T, df=5 0.4 0.3 8.29 to 1.71 with d.f. = 15.14 using a TI-83; Since the interval contains numbers that are all negative it, appears that the population mean duration of the first drug is less than that of the second drug. 3. 0.189 to 0.130; Since the interval contains both positive and negative values, there is no evidence of a difference in the population proportions of successful portfolios managed by Kendra compared to those managed by Lisa. (a) 0.05 0.2 0.1 (a) 0.61 to 2.79 (b) Since the interval contains all positive numbers, it seems that battery I has a larger population mean. H 0: p 1 = p 2 H 1: p 1 p 2 Two-tailed (b) Standard normal; z = 1.20 (c) P-value = 0.2304 0 H1: d < 0 left-tailed 2. 4. 0.115 -1.2 H0: d = 0 CHAPTER 10, FORM A 1. 0.2 0.1 Density 3. B. C. D. E. A. B. C. D. E. A. B. C. D. E. A. B. C. D. E. A. B. C. D. E. 0.0 6. 0.0365 -2.263 t 0 (d) t0 = 2.015 (e) Reject H0. At the 5% level of significance, there is evidence to conclude that the new process increases the mean number of items processed per shift. (a) 0.01 H0: 1 = 2 H1: 1 > 2 Right-tailed (b) Student’s t; d.f. = 8; t = 1.480 (c) P-value = 0.0886 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition Distribution Plot T, df=8 0.4 Density 0.3 0.2 3. 0.1 0.0886 0.0 7. 0 X 1.48 (d) t0 = 2.896 (e) Do not reject H0. At the 1% level of significance, there is not sufficient evidence to show that the population mean profit per employee in computer stores is higher than those for building supply stores. (a) 0.05 4. (b) n1p1, n1q1, n2p2, n2q2 are all greater than 5; yes (c) The interval contains all positive values and shows that at the 90% confidence level, the population proportion of defects is greater on shift I. (a) 15.49 to 46.51 (b) The interval contains values that are all positive. At the 90% confidence level, the population mean stopping distance of the old tread design is greater than that for the new. (a) 0.05 H0: p1 = p2 H1: p1 < p2 Left-tailed (b) Standard normal; z = 0.894 (c) P-value = 0.1858 H0: 1 = 2 H1: 1 2 Two-tailed Distribution Plot Normal, Mean=0, StDev=1 (b) Standard normal; z = 1.52 (c) P-value = 0.1292 0.4 Density 0.3 Distribution Plot Normal, Mean=0, StDev=1 0.2 0.4 0.1 0.186 Density 0.3 0.0 0.2 -0.894 0 z 0.1 0.0643 0.0 0.0643 -1.52 0 z 1.52 (d) z0 = 1.96 (e) Do not reject H0. At the 5% level of significance, there is not enough evidence to say that the mean weight of tomatoes grown organically is different than that of other tomatoes. CHAPTER 10, FORM B 1. 2. (a) -$103.60 to $461.60 with d.f. = 29.29 using a TI-83. (b) Since the interval contains both positive and negative values, it does not appear that the population mean daily sales of the printers differ. (a) 0.018 to 0.093 5. (d) z0 = 1.645 (e) Do not reject H0. At the 5% level of significance, there is not enough evidence to say that the proportion of seniors is higher than the proportion of juniors. (a) 0.05 H0: d = 0 H1: d < 0 Left-tailed (b) Student’s t; d.f. = 4; t = 2.566 (c) P-value = 0.0311 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Part III: Answers to Sample Chapter Tests Distribution Plot Normal, Mean=0, StDev=1 Distribution Plot 0.4 T, df=4 0.4 0.3 Density Density 0.3 0.2 0.2 0.1 0.000200 0.1 0.0 0.0 6. 0.0311 -2.566 (d) z0 = 1.645 (e) Reject H0. At a 5% level of significance, the evidence indicates that the average egg production of range free chickens is higher. CHAPTER 10, FORM C 1. 2. 3. 4. H1: 1 > 2 Right-tailed (b) Student’s t; d.f. = 8; t = 1.744 (c) P value = 0.0597 Distribution Plot T, df=8 0.4 5. 0.3 Density 3.54 t (d) t0 = 2.132 (e) Reject H0. At the 5% level of significance, there is evidence that the average time for runners at higher elevation is longer. (a) 0.01 0.2 0.1 0.0597 0.0 z 0 H0: 1 = 2 0 t 7. 0 1.744 (d) t0 = 2.896 (e) Do not reject H0. At the 1% level of significance, there is not enough evidence to conclude that the average mileage for the Pacer is greater. (a) 0.05 6. 7. H0: 1 = 2 H1: 1 > 2 Right-tailed (b) Standard normal; z = 3.54 (c) P-value = 0.0002 (b) (e) (d) A. B. C. D. E. A. B. C. D. E. A. B. C. D. E. A. B. C. D. E. (e) (b) (c) (d) (a) (d) (c) (c) (e) (a) (b) (b) (b) (a) (a) (c) (e) (c) (e) (b) CHAPTER 11, FORM A 1. Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition Scatter Diagram of Pounds versus Inches for Northern Pike (c) P-value = 0.0046 20 Distribution Plot Chi-Square, df=4 0.20 0.15 10 Density Pounds 15 0.10 5 0.05 0.00464 0 20 25 30 35 40 45 0.00 Inches 2. The linear correlation coefficient appears to be positive since as x increases, y tends to increase. (a) x 33.4; y 10.6 (b) x 167; y 53; x2 6069; y 2 789; xy 2103 (c) b = 0.678; a = 12.0; yˆ 0.678x 12.0 4. 5. (d) See problem 1. r = 0.996; r2 = 0.992; 99.2% of the variation in y can be explained by the least squares line using x as the predicting variable. Se = 0.757 For x = 32 inches, y = 9.65 pounds 6. 7.70 y 11.61 pounds 7. = 0.01 H0: = 0 H1: 0 3. 9. = 0.05 H0: = 0 H1: > 0 t = 19.31; d.f. = 3; P-value = 0.00015 Reject H0; At the 5% level of significance, there is sufficient evidence that the slope is positive. Chi-square goodness of fit (a) = 0.05 H0: The distribution of ages of college students is the same in the Western Association as in the nation. H1: The distribution of ages of college students is different in the Western Association than it is in the nation. (b) Chi-Square; d.f. = 4; 2 = 15.03 15.03 X (d) 02 9.49 (e) Reject H0. At the 5% level of significance, there is evidence that the distribution of ages is different. 10. Chi-square test of independence (a) = 0.05 H0: Teacher evaluations are independent of midterm grades H1: Teacher evaluations are not independent of midterm grades 2 (b) Chi-Square; d.f. = 4; 12.36 (c) P-value = 0.0149 Distribution Plot Chi-Square, df=4 0.20 0.15 Density t = 19.31; d.f. = 3; P-value = 0.000304 Reject H0; There is evidence that the population correlation coefficient is not equal to 0 at the 1% significance level. 8. 0 0.10 0.05 0.00 0.0149 0 x 12.36 (d) 02 9.49 (e) Reject H0. There is evidence to say that at the 5% level of significance teacher evaluations are not independent of midterm grades. 11. Chi-square test for (a) = 0.01 H0: = 4.1 H1: > 4.1 (b) Chi-Square; d.f. = 24; 2 = 34.28 (c) P-value = 0.0798 Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Part III: Answers to Sample Chapter Tests Distribution Plot Chi-Square, df=24 5. For a CEO salary of $1.5 million, the model predicts an annual company revenue of $18.74 billion. 6. $11.00 y $26.43 billion 7. = 0.01 H0: = 0 H1: 0 0.06 0.05 Density 0.04 0.03 0.02 0.01 0.0798 0.00 x t = 3.21; d.f. = 3; P-value = 0.0490 Do not reject H0; There is not enough evidence that the population correlation coefficient is not equal to 0 at the 1% significance level. 34.28 (d) 02 42.98 (e) Do not reject H0. At the 1% significance level, there is not enough evidence to reject = 4.1. 8. 9. = 0.05 H0: = 0 H1: > 0 t = 3.21; d.f. = 3; P-value = 0.0245 Reject H0; At the 5% level of significance, there is sufficient evidence that the slope is positive. Chi-square goodness of fit (a) = 0.05 H0: The distribution of fish fits the initial stocking distribution H1: The distribution of fish after six years does not fit the initial stocking distribution CHAPTER 11, FORM B 1. (b) Chi-Square; d.f. = 4; 2 = 19.0 Scatter Diagram of Revenue versus Salary (c) P-value = 0.00079 26 Distribution Plot Chi-Square, df=4 22 0.20 20 18 0.15 16 Density Revenue ($ Billions) 24 14 0.10 12 0.05 10 1.0 2. 1.5 Salary ($ Millions) 2.0 2.5 The linear correlation coefficient appears to be positive since as x increases, y tends to increase. (a) x 1.38; y 17.8 (b) x 6.9; y 89; x2 11.03; y 2 1703; xy 134.6 (c) b = 7.81; a = 7.02; yˆ 7.81x 7.02 3. 4. 0.000786 0.00 (d) See problem 1. r = 0.880; r2 = 0.775; 77.5% of the variation in y can be explained by the least squares line using x as the predicting variable. Se = 2.988 0 x 19 (d) 02 9.49 (e) Reject H0. At the 5% level of significance, there is enough evidence to conclude that the fish distribution has changed. 10. Chi-square test of independence (a) = 0.01 H0: The choice of college major is independent of grade average H1: The choice of college major is not independent of college average Copyright © Houghton Mifflin Harcourt Company. All rights reserved. Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition 6. 7. 8. 9. (c) (b) (c) A. B. C. D. 10. A. B. C. D. 11. A. B. 2 (b) Chi-Square; d.f. = 4; = 2.64 (c) P-value = 0.6198 Distribution Plot Chi-Square, df=4 0.20 Density 0.15 0.10 0.05 0.620 0.00 0 2.64 x (d) 02 13.28 (e) Do not reject H0. At the 1% level of significance, there is not enough evidence to conclude that college major is not independent of grade average. 11. Chi-square test for (a) = 0.01 H 0: = 6 H 1: ≠ 6 (b) Chi-Square; d.f. = 21; 2 = 30.24 (c) P-value = 0.1745 Distribution Plot Chi-Square, df=21 0.07 0.06 Density 0.05 0.04 0.03 0.02 0.01 0.00 0.0873 0 z 30.24 2 41.40 (d) L2 8.03; H (e) Do not reject H0. At the 1% significance level, there is not enough evidence to reject = 60. CHAPTER 11, FORM C 1. 2. 3. 4. 5. (a) (d) (b) (a) (a) Copyright © Houghton Mifflin Harcourt Company. All rights reserved. (c) (e) (d) (a) (d) (c) (a) (b) (e) (b)