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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition

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    Chapter 1 1-1 A Review of Basic Concepts (Optional) 1.1 a. High school GPA is a number usually between 0.0 and 4.0. Therefore, it is quantitative. b. Country of citizenship: USA, Japan, etc. is qualitative. c. The scores on the SAT's are numbers between 200 and 800. Therefore, it is quantitative. d. Gender is either male or female. Therefore, it is qualitative. e. Parent's income is a number: $25,000, $45,000, etc. Therefore, it is quantitative. f. Age is a number: 17, 18, etc. Therefore, it is quantitative. 1.2 a. The experimental units are the new automobiles. The model name, manufacturer, type of transmission, engine size, number of cylinders, estimated city miles/gallon, and estimated highway miles/gallon are measured on each automobile. b. Model name, manufacturer, and type of transmission are qualitative. None of these is measured on a numerical scale. Engine size, number of cylinders, estimated city miles/gallon, and estimated highway miles/gallon are all quantitative. Each of these variables is measured on a numerical scale. 1.3 Both the variables current position and type of organization are qualitative. The variable years of experience is quantitative because it is measured on a numerical scale. 1.4 The experimental units are the operational satellites currently in orbit around Earth. The variables country of operator/owner, primary use, and class of orbit are all qualitative because none are measured on a numerical scale. The variables longitudinal position, apogee, launch mass, usable electric power, and expected lifetime are all quantitative variables. All of these variables are measured on a numerical scale. 1.5 a. Species of sea buckthorn is a qualitative variable. b. Altitude of collection location is a quantitative variable. c. Total flavonoid content in berries is a quantitative variable. 1.6 Gender and level of education are both qualitative since neither is measured on a numerical scale. Age, income, job satisfaction score, and Machiavellian rating score are all quantitative since they can be measured on a numerical scale. 1.7 a. The population of interest is all decision makers. The sample set is 155 volunteer students. Variables measured were the emotional state and whether to repair a very old car (yes or no). b. Subjects in the guilty-state group are less likely to repair an old car. Chapter 1

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    1-2 A Review of Basic Concepts 1.8 a. The data would represent the population. These data are all of the data that are of interest to the researchers. b. If the 80 jamming attacks are actually a sample, then the population would be all jamming attacks by the U.S. military over the past several years. 1.9 a. The experimental units are the participants in the study. b. The variables of interest are the price of the engagement ring and the level of appreciation. The price of the ring is quantitative, while the level of appreciation is qualitative. c. The population of interest is average American engaged couples. d. The sample of 33 respondents is probably not representative of the population. Only engaged couples who used a popular website for engaged couples were used. Those who used this website were probably not representative of all average American engaged couples. 1.10 a. The sample is the set of 505 teenagers selected at random from all U.S. teenagers. b. The population from which the sample was selected is the set of all teenagers in the U.S. c. Since the sample was a random sample, it should be representative of the population. d. The variable of interest is the topics that teenagers most want to discuss with their parents. e. The inference is expressed as a percent of the population that want to discuss particular topics with their parents. f. The “margin of error” is the measure of reliability. This margin of error measures the uncertainty of the inference. 1.11 a. The population of interest is all young women who recently participated in a STEM program. b. The sample is the 159 young women who were recruited to complete an online survey. c. We would infer that 27% of all young women who recently participated in a STEM program felt that participation in the STEM program increased their interest in science. 1.12 a. The population of interest is the Machiavellian traits in accountants. b. The sample is 198 accounting alumni of a large southwestern university. c. The Machiavellian behavior is not necessary to achieve success in the accounting profession. d. Non-response could bias the results by not including potential other important information that could direct the researcher to a conclusion.

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    Chapter 1 1-3 1.13 a. A relative frequency table is: Rhino Species Frequency Relative Frequency African Black 5,000 0.1745 African White 20,000 0.6978 (Asian) Sumatran 100 0.0035 (Asian) Javan 60 0.0021 (Asian) Greater One-Horned 3,500 0.1221 Total 28,660 1.000 b. Using MINITAB, the relative frequency bar graph is: Greater 1-horned Javan Sumatran White Black 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Species Rel Freq Chart of Rel Freq c. The proportion of African rhinos is 0.1745 0.6978 0.8723.   The proportion of Asian rhinos is 0.0035 0.0021 0.1221 0.1277.    d. Using MINITAB, the pie chart for these proportions is: African Asian Category 12.8% Asian 87.2% African Pie Chart of Species

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    1-4 A Review of Basic Concepts 1.14 a. From the pie chart, 76.0% of the sample have a cable/satellite subscription at home. The proportion would be 0.76. This can be found by computing the relative frequency or 1,521/ 2, 001 0.76.  b. Using MINITAB, the pie chart is: Current Subscriber Cord Cutter Category 16.5% Cord Cutter 83.5% Current Subscriber Pie Chart of Type 1.15 Using MINITAB, the side-by-side bar graphs are: Dive Percent Right Middle Left 80 60 40 20 0 Right Middle Left 80 60 40 20 0 Ahead Behind Tied Chart of Dive Panel variable: Situation; Percent within all data. From the graphs, it appears that if the team is either tied or ahead, the goal-keepers tend to dive either right or left with equal probability, with very few diving in the middle. However, if the team is behind, then the majority of goal-keepers tend to dive right (71%). 1.16 a. 196 0.3889 504  is the proportion of ice melt ponds that had landfast ice. b. Yes, since 88 0.1746 504  is approximately 17%. c. The multiyear ice type appears to be significantly different from the first-year ice melt.

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    Chapter 1 1-5 1.17 a. Private Public Category 53.8% 46.2% Pie Chart of WellClass b. Bedrock Unconsoli Category 9.9% 90.1% Pie Chart of Aquifer c. MTBE-Detect Percent Detect Below Limit 70 60 50 40 30 20 10 0 Chart of MTBE-Detect Percent within all data.

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    1-6 A Review of Basic Concepts d. MTBE-Detect Percent Detect Below Limit 80 70 60 50 40 30 20 10 0 Detect Below Limit Private Public Chart of MTBE-Detect Panel variable: WellClass; Percent within all data. Public wells (40%); Private wells (21%). 1.18 a. The estimated percentage of aftershocks measuring between 1.5 and 2.5 on the Richter scale is approximately 68%. b. The estimated percentage of aftershocks measuring greater than 3.0 on the Richter scale is approximately 12%. c. The data are skewed right. 1.19 a. The graph is a frequency histogram. b. The quantitative variable summarized in the graph is the fup/fumic ratio. c. The proportion of ratios greater than 1 is 8 5 1 14 0.034. 416 416     d. The proportion of ratios less than 0.4 is 181 108 289 .695. 416 416    1.20 a. Using MINITAB, the frequency histogram is: 255 225 195 165 135 105 75 45 15 -15 -45 25 20 15 10 5 0 RDER Percent Histogram of RDER b. From the graph, it appears that about 0.18 of the RDER values are between 75 and 105.

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    Chapter 1 1-7 c. From the graph, it appears that about 0.10 of the RDER values are below 15. 1.21 The tem-and-leaf display with the leaves for the honey dosage group bolded. Stem-and-leaf of TotalScore N = 105 1 0 0 4 1 000 4 2 7 3 000 16 4 000000000 20 5 0000 28 6 00000000 41 7 0000000000000 52 8 00000000000 (13) 9 0000000000000 40 10 0000000000 30 11 000000 24 12 0000000000000 11 13 0000 7 14 0 6 15 00000 1 16 0 Leaf Unit = 0.1 Yes. Most of the scores for the honey dosage tend to be higher than the other treatments. 1.22 a. Using MINITAB, the frequency histogram is: 10.4 10.0 9.6 9.2 8.8 8.4 8.0 70 60 50 40 30 20 10 0 OLD Percent Histogram of OLD

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    1-8 A Review of Basic Concepts b. Using MINITAB, the stem-and-leaf display is: Stem-and-leaf of VOLTAGE LOCATION_OLD = 1 N = 30 Leaf Unit = 0.10 1 8 0 1 8 1 8 3 8 77 4 8 8 4 9 4 9 5 9 5 7 9 77 (10) 9 8888889999 13 10 000000111 4 10 222 1 10 5 The stem-and-leaf is more informative since the actual values of the old location can be found. The histogram is useful if shape and spread of the data is what is needed, but the actual data points are absorbed in the graph. c. Using MINITAB, the frequency histogram is: 10.00 9.75 9.50 9.25 9.00 8.75 8.50 6 5 4 3 2 1 0 NEW Frequency Histogram of NEW d. Side-by-side graphs are: 10.4 10.0 9.6 9.2 8.8 8.4 8.0 70 60 50 40 30 20 10 0 10.4 10.0 9.6 9.2 8.8 8.4 8.0 OLD Percent NEW Histogram of OLD, NEW

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    Chapter 1 1-9 The old process appears to be better than the new process. For the old process, only about 0.13 of the observations are less than 9.2. For the new process, about 0.3 of the observations are below 9.2. 1.23 a. Using MINITAB, the stem-and-leaf display is: Stem-and-leaf of Score N = 194 1 6 1 1 6 1 7 3 7 88 5 8 24 24 8 6667777788889999999 83 9 00000000001111112222222222222223333333333344444444444444444 (84) 9 5555555555555555555566666666666666666777777777777777777777778888888889999999999+ 27 10 000000000000000000000000000 Leaf Unit = 1 b. Of the 194 observations, 189 have acceptable standard of sanitation scores. The proportion is 189 0.974. 194  c. The score of 78 is highlighted in bold below. Stem-and-leaf of Score N = 194 1 6 1 1 6 1 7 3 7 8 8 5 8 24 24 8 6667777788889999999 83 9 00000000001111112222222222222223333333333344444444444444444 (84) 9 5555555555555555555566666666666666666777777777777777777777778888888889999999999+ 27 10 000000000000000000000000000 Leaf Unit = 1

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    1-10 A Review of Basic Concepts d. Using MINITAB, the histogram of the data is: 96 90 84 78 72 66 60 20 15 10 5 0 Score Percent Histogram of Score e. The proportion of ships with acceptable sanitation scores is about 0.97. 1.24 Using MINITAB, the histogram is: 525 450 375 300 225 150 75 0 60 50 40 30 20 10 0 INTTIME Frequency Histogram of INTTIME The data are skewed right. Answers may vary on whether the phishing attack against the organization was an “inside job.” 1.25 a. Using MINITAB, the histograms of the number of deaths is: 1000 800 600 400 200 0 40 30 20 10 0 Deaths Percent Histogram of Deaths b. The interval containing the largest proportion of estimates is 0-50. Almost half of the estimates fall in this interval.

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    Chapter 1 1-11 1.26 a. The sample mean is: 1 2 3 1 5 6 2 4 1 2 4 2 9 42 3.231 13 13 y y n                  b. The sample variance is:   2 2 2 2 42 202 13 5.526 1 13 1 y y n s n          The standard deviation is: 5.526 2.351 s   c. Using Tchebysheff’s Theorem, at least 75% of the observations will fall within 2 standard deviations of the mean. This interval is:     2 3.231 2 2.351 3.231 4.702 1.471,7.933 y s        At least 75% of all shaft graves will contain between 0 and 7 sword shafts. 1.27 a. 2.12; y  The average magnitude for the aftershocks is 2.12. b. Range = 6.7. The difference between the largest and smallest magnitude is 6.7. c. 0.66; s  About 95% of the magnitudes fall in the interval     2 2.1197 2 0.6636 2.1197 1.3272 0.79,3.44 y s       d.  = mean;  = Standard deviation 1.28 a. The mean RDER score is 78.1885. On average, subjects make 78.19 more error under the irrelevant background speech than under silence. b. From the histogram in Exercise 1.20, it appears that the data are approximately mound- shaped. By the rule of thumb, approximately 95% of the observations will fall within 2 standard deviations of the mean. c. We would expect approximately 95% of the observations to fall within the following interval:     2 78.19 2 63.24 78.19 126.48 48.29, 204.67 y s        1.29 a. Using MINITAB, the descriptive statistics are: Statistics Variable N Mean StDev Minimum Maximum Score 194 94.474 4.897 61.000 100.000 94.474, y  4.897 s  b.     2 94.474 2 4.897 94.474 9.794 84.680,104.268 y s      

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    1-12 A Review of Basic Concepts c. The percentage of scores that fall in the interval is   189 100% 97.42%. 194  This number is agrees with the rule of thumb which says that approximately 95% of the observations will fall within 2 standard deviations of the mean. 1.30 a. The average score for Energy Star is 4.44. The average score is close to 5 meaning the average score is close to ‘very familiar’. b. The ecolabel that had the most variation in the numerical responses is Audubon International because it has the largest standard deviation. c. The interval would be     2 4.44 2 0.82 4.44 1.64 2.80,6.08 y s       1.31 a. 293 353 2 30 z     A score of 293 would be 2 standard deviations below the mean. 413 353 2 30 z    A score of 413 would be 2 standard deviations above the mean. Using Tchebysheff’s Theorem, at least 3/4 of the observations will be within 2 standard deviations of the mean. b. For a mound-shaped, symmetric distribution, approximately 95% of the observations will be within 2 standard deviations of the mean, using the rule of thumb. c. 134 184 2 25 z     A score of 134 would be 2 standard deviations below the mean. 234 184 2 25 z    A score of 234 would be 2 standard deviations above the mean. Using Tchebysheff’s Rule, at least 3/4 of the observations will be within 2 standard deviations of the mean. d. For a mound-shaped, symmetric distribution, approximately 95% of the observations will be within 2 standard deviations of the mean, using the rule of thumb. 1.32 We will find intervals within 2 standard deviations for each group. Group T:     2 10.5 2 7.6 10.5 15.2 4.7, 25.7 y s        Group V:     2 3.9 2 7.5 3.9 15.0 11.1,18.9 y s        Group C:     2 1.4 2 7.5 1.4 15.0 13.6,16.4 y s        Since the only interval that contains 22.5 is the interval for Group T, the patient is most likely to have come from Group T. 1.33 a. Using Table 1, Appendix D:       1 1 1 0 0 1 0.3413 0.3413 0.6826 P z P z P z             

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    Chapter 1 1-13 b.       1.96 1.96 1.96 0 0 1.96 0.4750 0.4750 0.9500 P z P z P z              c. Since 1.645 is half way between 1.64 and 1.65 in the table, we will use halfway between the corresponding areas. Halfway between the 2 areas is 0.4495 0.4505 0.9 0.4500 2 2          1.645 1.645 1.645 0 0 1.645 0.4500 0.4500 0.9000 P z P z P z              d.       3 3 3 0 0 3 0.4987 0.4987 0.9974 P z P z P z              1.34 a. The z -score for 2   - is ( ) 2 2. z     - - = = - The z -score for 2   + is ( ) 2 2. z     + - = = ( ) ( ) 2 2 2 2 P y P z     - £ £ + = - £ £ ( ) ( ) 2 0 0 2 P z P z = - £ £ + £ £ Using Table 1 in Appendix D, ( ) 2 0 0.4772 P z - £ £ = and ( ) 0 2 0 .4772. P z £ £ = So ( ) 2 2 0 .4772 0 .4772 0.9544 P y     - £ £ + = + = b. The z -score for 108 y = is 108 100 1. 8 y z   - - = = = ( ) ( ) 108 1 P y P z ³ = ³ Using Table 1 of Appendix D, we find ( ) 0 1 0.3413. P z £ £ = so ( ) 1 0.5 0.3413 0.1587 P z ³ = - = c. The z -score for 92 y = is 92 100 1. 8 y z   - - = = = - ( ) ( ) 92 1 P y P z £ = £ - Using Table 1 of Appendix D, we find ( ) 1 0 0.3413, P z - £ £ = so ( ) 1 0.5 0.3413 0.1587 P z £ - = - = d. The z -score for 92 y = is 92 100 1. 8 y z   - - = = = -

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    1-14 A Review of Basic Concepts The z -score for 116 y = is 116 100 2. 8 y z   - - = = = ( ) ( ) 92 116 1 2 P y P z £ £ = - £ £ Using Table 1 of Appendix D, ( ) 1 0 0.3413 P z - £ £ = and ( ) 0 2 0.4772. P z £ £ = So ( ) ( ) 92 116 1 2 P y P z £ £ = - £ £ 0.3413 0.4772 0.8185. = + = e. The z -score for 92 y = is 92 100 1. 8 y z   - - = = = - The z -score for 96 y = is 96 100 0.5. 8 y z   - - = = = - ( ) ( ) 92 96 1 .5 P y P z £ £ = - £ £ - Using Table 1 of Appendix D, ( ) 1 0 0.3413 P z - £ £ = and ( ) 0.5 0 0.1915 P z - £ £ = . So ( ) 92 96 P y £ £ ( ) 1 0.5 0.3413 0.1915 0.1498. P z = - £ £ - = - = f. The z -score for 76 y = is 76 100 3. 8 y z   - - = = = - The z -score for 124 y = is 124 100 3. 8 y z   - - = = = ( ) ( ) 76 124 3 3 P y P z £ £ = - £ £ Using Table 1 of Appendix D, ( ) 3 0 0.4987 P z - £ £ = and ( ) 0 3 0.4987. P z £ £ = So ( ) 76 124 P y £ £ = ( ) 3 3 0.4987 0.4987 0.9974. P z - £ £ = + = 1.35 a. Using Table 1, Appendix D:         4 4.44 4 0.54 0.54 0 0 0.82 P y P z P z P z P z                    0.2054 0.5 0.7054    b.     2 4.44 4 4.44 2 4 2.98 0.54 0.82 0.82 P y P z P z                       2.98 0 0.54 0 0.4986 0.2054 0.2932 P z P z            c.         1 4.44 1 4.20 0 4.20 0 0.5 0.5 0 0.82 P y P z P z P z P z                       Since the probability of observing a value of 1 or less is so small, it would be extremely unlikely that the ecolabel shown was Energy Star .

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    Chapter 1 1-15 1.36 a. Using Table 1, Appendix D:         400 353 400 1.57 0 1.57 0 30 P y P z P z P z P z                  0.4418 0.5 .9418    b.         100 184 100 3.36 3.36 0 0 25 P y P z P z P z P z                    0.5 0.5 1.0    1.37 a. Using Table 1, Appendix D:         60 59 60 0.2 0 0 0.2 5 P y P z P z P z P z                  0.5 0.0793 0.4207    b.         60 43 60 3.4 0 0 3.4 0.5 0.5 0 5 P y P z P z P z P z                     1.38 We want to find y 1 and y 2 such that   1 2 0.90. P y y y    First, we must find z 1 and z 2 such that   1 2 0.90. P z z z    By symmetry, we know that     1 2 0 0 0.90 / 2 0.4500. P z z P z z        Using Table 1, Appendix D, 1 1.645 z   and 2 1.645. z    1 1 1 1 64 1.645 64 1.645 2.6 64 4.277 59.723 2.6 y y z y                 2 2 2 2 64 1.645 64 1.645 2.6 64 4.277 68.277 2.6 y y z y              Thus, the interval would be   59.723,68.277 . 1.39 For Tunnel Face ,       1 1.2 1 1.25 0.5 1.25 0 0.5 0.3944 0.1056 0.16 P y P z P z P z                      For Tunnel Walls ,       1 1.4 1 2.00 0.5 2.00 0 0.5 0.4772 0.0228 0.20 P y P z P z P z                      For Tunnel Crown ,       1 2.1 1 1.57 0.5 1.57 0 0.5 0.4418 0.0582 0.70 P y P z P z P z                      The probability of failing for Tunnel Face is larger than the probabilities of failure for the other two areas. Thus, Tunnel Face is more likely to result in failure.

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