Download 1. A – The strongest linear relationship is –0.9. Correlation must

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STAT200(Wisener)–Exam3–PracticeExamSolutions
1. A–Thestrongestlinearrelationshipis–0.9.Correlationmusthaveavalue
between–1and+1soanswerchoiceEisnotanoption.Todeterminethe
strongestlinearrelationship,youwantthevalueclosesttotheabsolutevalue
of1.
2. C–Testoftwoindependentmeansbecausewearedealingwithtwodifferent
gendersandtheamountofmoneyspentontextbookseachyearisa
quantitativevariable.
3. B–Staticscomefromsamples,andparameterscomefrompopulations.
4. C–Ho:p=0.71
Ha:p>0.71.Sincewearedealingwithproportions,we
usethepopulationparameter,p.Weknowitisagreaterthanalternative
hypothesistestbecausethedoctorwantstoseeifthenewmethodismore
effectivethanthecurrentmethod.
5. A–Testofonemeanbecausewearelookingfortheaveragevalueofasingle
quantitativevariable.
6. F–Chi-squaretestbecausetheresearcheristestingtoseeifthereisa
relationshipbetweentwoqualitativevariables.
7. B–Yourscoreonexam1isusedtopredictyourscoreonexam2.The
predictorvariablewillbethevariablelistedinthepredictorcolumnofthe
Minitaboutputundertherowwiththeinformationabouttheconstant.
8. B–Positivesquarerootof0.439.YouaregivenR-Sq(r2)intheregression
output.Youareaskedtofindcorrelation(r).Thismeansyouneedtotakethe
squarerootofr2.Youknowtheansweristhepositivesquarerootbecause
thecoefficientforthepredictorvariable(Exam1)isapositivenumber.This
meansthattheregressionlinehasapositiveslope.Whenaregressionline
hasapositiveslope,correlation(r)willbepositive.
9. A–Exam2=45.1+0.469(Exam1).Theconstantvalue(y-intercept)inthe
equationisthecoefficientforthe“constant”rowintheregressionoutput.
Theslopeoftheregressionequationisthecoefficientforthepredictor
variable(Exam1).
10. C–43.9%becauseitisthevalueofR-Sq.
11. B–Ho:B1=0
Ha:B1≠0
12. E–Sincetheteststatisticis12.46andthep-valueis0.000yourejectHoand
concludethatExam1isasignificantlinearpredictorofExam2.
13. C–0.100to0.200.Sincethesamplesizeis5,youknowthatdegreesof
freedomis4.Youcanfindthat1.988isbetween1.533and2.132inthedf=4
row.Youcanseefromthetoprowofthetablethatateststatisticof1.988
willhaveap-valuebetween0.100and0.050foraone-sidedtest.However,
theproblemtellsyouthatthetestisanotequaltestwhichmeansitisa
two-sidedtest.Sinceitisatwosidedtest,youneedtodoublethep-values.
14. E–Regressionbecausethedoctorwantstofindtherelationshipbetween
twoquantitativevariables.
15. B–Testofpairedmeansbecauseyouarecomparingestimatesoftwolawn
servicefirmstocutthesamelawns.
16. D–18.17to37becausethisisQ3tothemaximum,whichisonly25%ofthe
valuesinafivenumbersummary.
17. B–Thenewweightlossmethodhasasuccessratethatisgreaterthan60%.
Youknowthatitisagreaterthantestbecausethedoctorwanttoseeifthe
methodismoreeffective.Youcantellthatthenewweightlossmethoddoes
haveagreatersuccessratethanthecurrentmethodbecausethep-valueis
lessthan0.05.
18. F–One-wayAnalysisofVariance(ANOVA)becausetheproblemis
comparingsalary,aquantitativevariable,acrossfourdifferentgroups.
19. B–0.050to0.100.Sincethesamplesizeis5,youknowthatdegreesof
freedomis4.Youcanfindthat1.988isbetween1.533and2.132inthedf=4
row.Youcanseefromthetoprowofthetablethatateststatisticof1.988
willhaveap-valuebetween0.100and0.050foraone-sidedtest.
20. C–Thegroupofpeoplefromwhichthesamplewastaken.
21. A–Themeanwouldbegreaterthanthemedian.
22. B–Parameterbecausewearetalkingabouttheentirepopulationofout-ofstatePennStatestudents.
23. C–Itshouldberepresentativeofthepopulation.
24. C–Thelargerstandarddeviationwasnotmorethantwicethesmaller
standarddeviation.Youneedtocomparethetwostandarddeviationsinthe
outputwhendeterminingifyoucanusepooledvariances.
25. A–Ho:μ1–μ2=0
Ha:μ1–μ2≠0
26. C–4.604.Youknowthatdegreesoffreedomisfourbecausethesamplesize
isfive.Soyouneedtolookupthemultipliervalueinthe99%columninthe
rowforfourdegreesoffreedom.
27. D–Testoftwoproportionsbecausewearecomparingtwocategorical
variables.
28. A–Statisticbecauseonlyasubsetofthepopulation(allPennStatestudents)
wasusedinthestudy.
29. C–4
dfFactor(betweengroups)=k–1
3=k–1
k=4
30. E–19
dfTotal=N–1
18=N–1
N=19
31. C–RejectHoandconcludethatnotallofthemeansareequalbecausepvalueis0.001.
32. A–Thereisadifferencebetweenthemeansofgroup1andgroup3because
theconfidenceintervaldoesnotcontain0.
33. D–Thereisarelationshipinthepopulationbetweengenderandheart
attacksbecausethealternativehypothesisalwaysstatesthereisa
relationshipwhilethenullhypothesisstatesthereisnotarelationship.
34. D–85/420
Risk=(Numberincategory)/(Totalnumberingroup)=85/420
35. B–(150/380)/(85/420)
Relativerisk=(Riskincategory1)/(Riskincategory2)
Theproblemtellsyoutocomparemalerisktofemalerisksomaleriskwill
becategory1andfemaleriskwillbecategory2.
Malerisk=150/380
Femalerisk=85/420
Relativerisk=(150/380)/(85/420)
36. A–Statisticallysignificantinthepopulationbecausethep-valueislessthan
0.05.
37. C–150/380becausetheprobabilityofhavingaheartattackisthesame
thingastheriskofhavingaheartattack.
Malerisk=150/380
38. B–Estimatethepopulationparameter
39. C–Negative0.9becauseyouarelookingforthecorrelationwithanabsolute
valueclosestto1.
40. A–Positive0.1becauseyouarelookingforthecorrelationwithanabsolute
valueclosestto0.
41. B–82and4becausethemeanwillincreasebythefourpointsaddedto
everyone’sscore;however,thestandarddeviationwillnotbeaffected.
42. B–Negative1
𝑧=
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑𝑣𝑎𝑙𝑢𝑒 − 𝑀𝑒𝑎𝑛 74 − 78
=
= −1
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
4
43. D–90
z=
3=
Observedvalue − Mean
Standarddeviation
𝑥 − 78
4
x=90
44. C–70to86because95%ofvalueswillfallwithin2standarddeviationsof
themean.
45. C–Inastatisticalstudy,asampleshouldberepresentativeofthepopulation.
46. C–Orange,orange,orangebecauseorangeisthemostlikelyeventtooccur
eachtimethediceisrolled.
47. C–Theyareusedtoestimatetheaccuracyofthesamplestatistic.
48. D–18.17to37
49. B–Statisticallysignificantinthepopulationbecausethep-valueislessthan
0.05.
50. B–Testofpairmeans
51. C–Testoftwoindependentmeans
52. D–Testoftwoproportions
53. A–Testofonemean
54. G–Chi-squaretestbecausetheresearcheriscomparingtwocategorical
variables.
55. G–One-wayAnalysisofVariance(ANOVA)becausewearecomparingthe
meansalaryacrossmorethantwogroups.
56. B–92%ofallPennStatethinkthatmarijuanashouldbedecriminalizedis
thepopulationproportion.
57. A–56/1626
58. B–(70/650)/(56/1626)Theproblemtellsuswewant“thosewhotookthe
placebocomparedtothosewhotookthemalariapills.”Soweneedtotake
theriskoftheplacebogroupdividedbytheriskofthemalariapillsgroup.
59. A–Peopletakingtheplaceboare3.13timesmorelikelytogetmalariathan
thosewhoreceivedthemalariapills.
60. D–Thereisnorelationshipinthepopulationbetweenthetreatmentand
whetherornotsomeonegetsmalaria.Thenullhypothesisalwaysstates
thereisnorelationshipinthepopulation.
61. B–Thereisarelationshipinthepopulationbetweenthetreatmentand
whetherornotsomeonegetsmalaria.Thealternativehypothesisalways
statesthereisarelationshipinthepopulation.
62. B–Inthepopulation,thereisastatisticallysignificantrelationshipbetween
thetreatmentandgettingmalariabecausethep-valueislessthan0.05.
63. B–Ho:p=0.5
Ha:p>0.5
64. C–Ho:μ=10,000
Ha:μ≠10,000
65. A–Testofonemean.ThesamplestatisticwouldbetheaverageACTverbal
scoreforincomingfreshmen.Youwouldgetthescorefromeachfreshmen
sampledandtakeanaverage.Theresultwouldbeonemean.Thepartabout
comparingthescoretoascoreof26isincludedtomaketheproblem
“tricky.”
66. B–Testofpairmeans.Youarecomparingthedifferenceinperson
freshmen’smathandverbalscore.
67. C–Testoftwoindependentmeans.Highschoolstudentsandcollege
studentsareindependentgroups.
68. D–Testoftwoproportions.Youarecomparingpercentagesformalesand
females.
69. A–Comparingtheaverageamountoftarforfivebrandsofcigarettes.You
arecomparingonemeanacrossfivecategoricalvariables.
70. E–ComparingtestscoresofstudentsinaMondaylabtostudentsina
Wednesdaylab.StudentsinMondayandWednesdaylabsareindependent
groups.
71. D–Comparingstudentgenderandwhethertheyattendedacollegein-state
orout-of-state.Youarecomparingtwocategoricalvariables.
72. A–92%ofPennStatestudentssampledthinkmarijuanashouldbe
decriminalized.Statisticsaretheresultofasample.
73. B–0.050to0.100.Thesamplesizeis5sodf=4.Lookingacrossthedf=4
row,2.01fallsbetween0.050and0.100.Thistableisgiving“right-tail
probabilities”whichmeanstheareatotherightofthepositiveteststatistic.
Theareatotherightoftheteststatisticcorrespondstoagreaterthantest.
However,inthisproblemyouweretoldtheteststatisticwasnegative2.01
andthealternativehypothesiswasalessthantest.Becauseofthesymmetry
ofthet-distribution,weknowtheareatotheleftofnegative2.01isthesame
areatotherightofpositive2.01.
74. C–0.100to0.200.Thisproblemtellsusthealternativehypothesisisanot
equaltest.Thus,weneedtodoubletheanswerfromthepreviousproblem.
75. A–Standarddeviationmeasuresthevariationinanindividualsample;
Standarderroristheestimatedstandarddeviationinasampling
distribution.Thismeansthatitmeasuresthevariabilityamongthe
distributionofpossiblevaluesofastatistic.
76. C–Thenullhypothesis:μ1–μ2=0isrejectedbecausewewouldbe
concludingthatμ1–μ2=0isdifferencefrom0.
77. B–Theyhavethesamevaluesforthestandarderrorintheteststatistic.
Lookingattheequationsbelow,itisclearthattherewillbedifferentvalues
forstandarderror.
IndependentSamples(Two-sampletprocedure)
𝒕 = 𝑺𝒂𝒎𝒑𝒍𝒆𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄– 𝑵𝒖𝒍𝒍𝒗𝒂𝒍𝒖𝒆
(𝒙𝟏 − 𝒙𝟐 ) − 𝟎
=
𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅𝒆𝒓𝒓𝒐𝒓
𝟐
𝟐
𝒔𝟏
𝒔
+ 𝟐
𝒏𝟏
𝒏𝟐
DependentSamples(Pairedtprocedure)
𝑺𝒂𝒎𝒑𝒍𝒆𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄– 𝑵𝒖𝒍𝒍𝒗𝒂𝒍𝒖𝒆
𝒅 − 𝟎
𝒕 = = 𝒔
𝒅
𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅𝒆𝒓𝒓𝒐𝒓
𝒏
AtfirstitwouldalsoseemthatanswerAwouldbefalsebecausetheylookto
havedifferencevaluesinthenumeratoraswell.However,youneedto
rememberthat 𝒅 = 𝒙𝟏 − 𝒙𝟐 78. D–Twooddswillbedifferent.Ifthechi-squaredstatisticisequalto0,the
actualcountwillequaltheexpectedcountforeachcell.Thismeansitisnot
possibleforthetwooddstobedifferent.
79. C–Themultiplelinearregressionequationusestheconstantplusthe
coefficienttimeseachexplanatoryvariable.
80. A–Thepredictedfinalscorewillincreasebytheslopeforeach1point
increaseinthemidtermscore.
81. B–Adummyvariablewilleitherbeequalto1or0.Sincetheslopeis-1.158,
thepredictedfinalscorewilldecreaseby1.158whenthevariableispresent.
82. B–Quizaverageisasignificantlinearpredictoroffinalexamscorebecause
ithasap-valuelessthan0.05.Thisonlyholdstruewhenallexplanatory
variablesarepresentinthemodel.
83. D–DummyGenderhasap-valuegreaterthan0.05,soitisnotasignificant
linearpredictorwhencombinedwiththeotherexplanatoryvariablesinthis
model.
84. C–IntheANOVAtable,weseethatthep-valueislessthan0.05.Thattellsus
thatatleastoneoftheslopesinthemodeldiffersfrom0,orisasignificant
linearpredictoroffinalexamscore.Itdoesnottellusthatalloftheslopes
differfrom0.
85. C–R-squared=38.5%
86. B–Odds=(Numberincategory/Numbernotincategory)
Oddsforwomen=(Female“yes”/Female“no”)=196/387
87. D–Oddsratio=(Oddsforcategory1/Oddsforcategory2)
Oddsforwomen=(Female“yes”/Female“no”)=196/387
Oddsformen=(Male“yes”/Male“no”)=196/387
Oddsratioforwomencomparedtomen=(Oddsforwomen/Oddsformen)
=(196/387)/(196/387)