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

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 1 preview image

Chapter 11-1A Review of Basic Concepts(Optional)1.1a.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.1.2a.The experimental units are the new automobiles. The model name, manufacturer, type oftransmission, engine size, number of cylinders, estimated city miles/gallon, and estimatedhighway miles/gallon are measured on each automobile.b.Model name, manufacturer, and type of transmission are qualitative. None of these ismeasuredon a numerical scale. Engine size, number of cylinders, estimated citymiles/gallon, and estimated highway miles/gallon are all quantitative. Each of thesevariables is measured on a numerical scale.1.3Both the variables current position and type of organization are qualitative. The variable years ofexperience is quantitative because it is measured on a numerical scale.1.4The experimental units are the operational satellites currently in orbit around Earth. Thevariables country of operator/owner, primary use, and class of orbit are all qualitative becausenone are measured on a numerical scale. The variables longitudinal position, apogee, launchmass, usable electric power, and expected lifetime are all quantitative variables. All of thesevariables are measured on a numerical scale.1.5a.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.1.6Gender 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 theycan be measured on a numerical scale.1.7a.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 orno).b.Subjects in the guilty-state group are less likely to repair an old car.Chapter1

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1-2A Review of Basic Concepts1.8a.The data would represent the population. These data are all of the data that are of interest tothe researchers.b.If the 80 jamming attacks are actually a sample, then the population would be all jammingattacks by the U.S. military over the past several years.1.9a.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. Onlyengaged couples who used a popular website for engaged couples were used. Those whoused this website were probably not representative of all average American engagedcouples.1.10a.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 particulartopics with their parents.f.The “margin of error” is the measure of reliability. This margin of error measures theuncertainty of the inference.1.11a.The population of interest is all young women who recently participated in a STEMprogram.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 programfelt that participation in the STEM program increased their interest in science.1.12a.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 accountingprofession.d.Non-response could bias the results by not including potential other important informationthat could direct the researcher to a conclusion.

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 4 preview image

Chapter 11-31.13a.A relative frequency table is:Rhino SpeciesFrequencyRelativeFrequencyAfrican Black5,0000.1745African White20,0000.6978(Asian) Sumatran1000.0035(Asian) Javan600.0021(Asian) Greater One-Horned3,5000.1221Total28,6601.000b.Using MINITAB, the relative frequency bar graph is:Greater 1-hornedJavanSumatranWhiteBlack0.70.60.50.40.30.20.10.0SpeciesRel FreqChart of Rel Freqc.The proportion of African rhinos is0.17450.69780.8723.The proportion of Asianrhinos is0.00350.00210.12210.1277.d.Using MINITAB, the pie chart for these proportions is:AfricanAsianCategory12.8%Asian87.2%AfricanPie Chart of Species

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 5 preview image

1-4A Review of Basic Concepts1.14a.From the pie chart, 76.0% of the sample have a cable/satellite subscription at home. Theproportion would be 0.76. This can be found by computing the relative frequencyor1,521/ 2, 0010.76.b.Using MINITAB, the pie chart is:Current SubscriberCord CutterCategory16.5%Cord Cutter83.5%Current SubscriberPie Chart of Type1.15Using MINITAB, the side-by-side bar graphs are:DivePercentRightMiddleLeft806040200RightMiddleLeft806040200AheadBehindTiedChart of DivePanel 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 diveeither right or left with equal probability, with very few diving in the middle. However, if theteam is behind, then the majority of goal-keepers tend to dive right (71%).1.16a.1960.3889504is the proportion of ice melt ponds that had landfast ice.b.Yes, since880.1746504is approximately 17%.c.The multiyear ice type appears to be significantly different from the first-year ice melt.

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 6 preview image

Chapter 11-51.17a.PrivatePublicCategory53.8%46.2%Pie Chart of WellClassb.BedrockUnconsoliCategory9.9%90.1%Pie Chart of Aquiferc.MTBE-DetectPercentDetectBelow Limit706050403020100Chart of MTBE-DetectPercent within all data.

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1-6A Review of Basic Conceptsd.MTBE-DetectPercentDetectBelow Limit80706050403020100DetectBelow LimitPrivatePublicChart of MTBE-DetectPanel variable: WellClass;Percent within all data.Public wells (40%); Private wells (21%).1.18a.The estimated percentage of aftershocks measuring between 1.5 and 2.5 on the Richter scaleis approximately 68%.b.The estimated percentage of aftershocks measuring greater than 3.0 on the Richter scale isapproximately 12%.c.The data are skewed right.1.19a.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 is851140.034.416416d.The proportion of ratios less than 0.4 is181108289.695.4164161.20a.Using MINITAB, the frequency histogram is:255225195165135105754515-15-452520151050RDERPercentHistogram of RDERb.From the graph, it appears that about 0.18 of the RDER values are between 75 and 105.

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Chapter 11-7c.From the graph, it appears that about 0.10 of the RDER values are below 15.1.21The tem-and-leaf display with the leaves for the honey dosage group bolded.Stem-and-leaf of TotalScoreN = 105100410004273000164000000000205000028600000000417000000000000052800000000000(13)9000000000000040100000000000301100000024120000000000000111300007140615000001160Leaf Unit = 0.1Yes. Most of the scores for the honey dosage tend to be higher than the other treatments.1.22a.Using MINITAB, the frequency histogram is:10.410.09.69.28.88.48.0706050403020100OLDPercentHistogram of OLD

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1-8A Review of Basic Conceptsb.Using MINITAB, the stem-and-leaf display is:Stem-and-leaf of VOLTAGE LOCATION_OLD = 1 N = 30Leaf Unit = 0.101801818387748849495957977(10)9888888999913100000001114102221105The stem-and-leaf is more informative since the actual values of the old location can befound. The histogram is useful if shape and spread of the data is what is needed, but theactual data points are absorbed in the graph.c.Using MINITAB, the frequency histogram is:10.009.759.509.259.008.758.506543210NEWFrequencyHistogram of NEWd.Side-by-side graphs are:10.410.09.69.28.88.48.070605040302010010.410.09.69.28.88.48.0OLDPercentNEWHistogram of OLD, NEW

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Chapter 11-9The old process appears to be better than the new process. For the old process, only about0.13 of the observations are less than 9.2. For the new process, about 0.3 of theobservations are below 9.2.1.23a.Using MINITAB, the stem-and-leaf display is:Stem-and-leaf of ScoreN = 194161161737885824248666777778888999999983900000000001111112222222222222223333333333344444444444444444(84)95555555555555555555566666666666666666777777777777777777777778888888889999999999+2710000000000000000000000000000Leaf Unit = 1b.Of the 194 observations, 189 have acceptable standard of sanitation scores. The proportionis 1890.974.194c.The score of 78 is highlighted in bold below.Stem-and-leaf of ScoreN = 194161161737885824248666777778888999999983900000000001111112222222222222223333333333344444444444444444(84)95555555555555555555566666666666666666777777777777777777777778888888889999999999+2710000000000000000000000000000Leaf Unit = 1

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 11 preview image

1-10A Review of Basic Conceptsd.Using MINITAB, the histogram of the data is:9690847872666020151050ScorePercentHistogram of Scoree.The proportion of ships with acceptable sanitation scores is about 0.97.1.24Using MINITAB, the histogram is:5254503753002251507506050403020100INTTIMEFrequencyHistogram of INTTIMEThe data are skewed right. Answers may vary on whether the phishing attack against theorganization was an “inside job.”1.25a.Using MINITAB, the histograms of the number of deaths is:10008006004002000403020100DeathsPercentHistogram of Deathsb.The interval containing the largest proportion of estimates is 0-50. Almost half of theestimates fall in this interval.

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Chapter 11-111.26a.The sample mean is:1231562412429423.2311313yynb.The sample variance is:222242202135.5261131yynsnThe standard deviation is:5.5262.351sc.Using Tchebysheff’s Theorem, at least 75% of the observations will fall within 2 standarddeviations of the mean. This interval is:23.2312 2.3513.2314.7021.471,7.933ysAt least 75% of all shaft graves will contain between 0 and 7 sword shafts.1.27a.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22.11972 0.66362.11971.32720.79,3.44ysd.=mean;=Standard deviation1.28a.The mean RDER score is 78.1885. On average, subjects make 78.19 more error under theirrelevant 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 2standard deviations of the mean.c.We would expect approximately 95% of the observations to fall within the followinginterval:278.192 63.2478.19126.4848.29, 204.67ys1.29a.Using MINITAB, the descriptive statistics are:StatisticsVariableNMeanStDevMinimumMaximumScore19494.4744.89761.000100.00094.474,y4.897sb.294.4742 4.89794.4749.79484.680,104.268ys

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 13 preview image

1-12A Review of Basic Conceptsc.The percentage of scores that fall in the interval is189 100%97.42%.194This number isagrees with the rule of thumb which says that approximately 95% of the observations willfall within 2 standard deviations of the mean.1.30a.The average score for Energy Star is 4.44. The average score is close to 5 meaning theaverage score is close to ‘very familiar’.b.The ecolabel that had the most variation in the numerical responses is Audubon Internationalbecause it has the largest standard deviation.c.The interval would be24.442 0.824.441.642.80,6.08ys1.31a.293353230z A score of 293 would be 2 standard deviations below the mean.413353230z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 standarddeviations of the mean.b.For a mound-shaped, symmetric distribution, approximately 95% of the observations will bewithin 2 standard deviations of the mean, using the rule of thumb.c.134184225z A score of 134 would be 2 standard deviations below the mean.234184225z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 standarddeviations of the mean.d.For a mound-shaped, symmetric distribution, approximately 95% of the observations will bewithin 2 standard deviations of the mean, using the rule of thumb.1.32We will find intervals within 2 standard deviations for each group.Group T:210.52 7.610.515.24.7, 25.7ysGroup V:23.92 7.53.915.011.1,18.9ysGroup C:21.42 7.51.415.013.6,16.4ysSince the only interval that contains 22.5 is the interval for Group T, the patient is most likely tohave come from Group T.1.33a.Using Table 1, Appendix D:1110010.34130.34130.6826PzPzPz

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 14 preview image

Chapter 11-13b.1.961.961.96001.960.47500.47500.9500PzPzPzc.Since 1.645 is half way between 1.64 and 1.65 in the table, we will use halfway between thecorresponding areas. Halfway between the 2 areas is 0.44950.45050.90.4500221.6451.6451.645001.6450.45000.45000.9000PzPzPzd.3330030.49870.49870.9974PzPzPz1.34a.Thez-score for2-is()22.z--== -Thez-score for2+is()22.z+-==()()2222PyPz-££+=-££()()2002PzPz=-££+££Using Table 1 in Appendix D,()200.4772Pz-££=and()020 .4772.Pz££=So()220 .47720 .47720.9544Py-££+=+=b.Thez-score for108y=is1081001.8yz--===()()1081P yP z³=³Using Table 1 of Appendix D, we find()010.3413.Pz££=so()10.50.34130.1587P z³=-=c.Thez-score for92y=is921001.8yz--=== -()()921P yP z£=£ -Using Table 1 of Appendix D, we find()100.3413,Pz- ££=so()10.50.34130.1587P z£ -=-=d.Thez-score for92y=is921001.8yz--=== -

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1-14A Review of Basic ConceptsThez-score for116y=is1161002.8yz--===()()9211612PyPz££=- ££Using Table 1 of Appendix D,()100.3413Pz- ££=and()020.4772.Pz££=So()()9211612PyPz££=- ££0.34130.47720.8185.=+=e.Thez-score for92y=is921001.8yz--=== -Thez-score for96y=is961000.5.8yz--=== -()()92961.5PyPz££=- ££ -Using Table 1 of Appendix D,()100.3413Pz- ££=and()0.500.1915Pz-££=. So()9296Py££()10.50.34130.19150.1498.Pz=- ££ -=-=f.Thez-score for76y=is761003.8yz--=== -Thez-score for124y=is1241003.8yz--===()()7612433PyPz££=-££Using Table 1 of Appendix D,()300.4987Pz-££=and()030.4987.Pz£ £=So()76124Py££=()330.49870.49870.9974.Pz-££=+=1.35a.Using Table 1, Appendix D:44.4440.540.54000.82PyPzP zPzP z 0.20540.50.7054b.24.4444.44242.980.540.820.82PyPzPz 2.9800.5400.49860.20540.2932PzPzc.14.4414.2004.2000.50.500.82PyPzP zP zPz Since the probability of observing a value of 1 or less is so small, it would be extremelyunlikely that the ecolabel shown wasEnergy Star.

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Solution Manual for A Second Course in Statistics: Regression Analysis, 8th Edition - Page 16 preview image

Chapter 11-151.36a.Using Table 1, Appendix D:4003534001.5701.57030PyPzP zPzP z0.44180.5.9418b.1001841003.363.360025PyPzP zPzP z 0.50.51.01.37a.Using Table 1, Appendix D:6059600.2000.25PyPzP zP zPz0.50.07930.4207b.6043603.4003.40.50.505PyPzP zP zPz1.38We want to findy1andy2such that120.90.PyyyFirst, we must findz1andz2suchthat120.90.P zzzBy symmetry, we know that12000.90 / 20.4500.P zzPzzUsing Table 1, Appendix D,11.645z and21.645.z1111641.645641.645 2.6644.27759.7232.6yyzy2222641.645641.645 2.6644.27768.2772.6yyzyThus, the interval would be59.723,68.277 .1.39ForTunnel Face,11.211.250.51.2500.50.39440.10560.16PyPzP zPz ForTunnel Walls,11.412.000.52.0000.50.47720.02280.20PyPzP zPz ForTunnel Crown,12.111.570.51.5700.50.44180.05820.70PyPzP zPz The probability of failing forTunnel Faceis larger than the probabilities of failure for the othertwo areas. Thus,Tunnel Faceis more likely to result in failure.

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