481
observedvalues(thedata)intoafirstlistandtheexpectedvalues(thevaluesyouexpectifthe
nullhypothesisistrue)intoasecondlist. Press
and
.Enterthelistnames
fortheObservedlistandtheExpectedlist.Enterthedegreesoffreedomandpress
or
.Makesureyouclearanylistsbeforeyoustart.
11.5TestofIndependence
5
Testsofindependenceinvolveusingacontingencytableofobserved(data)values.Youfirstsawacontin-
gencytablewhenyoustudiedprobabilityintheProbabilityTopics(Section3.1)chapter.
Theteststatisticforatestofindependenceissimilartothatofagoodness-of-fittest:
S
(ij)
(O E)
2
E
(11.2)
where:
 O=observedvalues
 E=expectedvalues
 i=thenumberofrowsinthetable
 j=thenumberofcolumnsinthetable
Thereare jtermsoftheform
(O E)
2
E
.
Atestofindependencedetermineswhethertwofactorsareindependentornot. Youfirstencountered
thetermindependenceinChapter3.Asareview,considerthefollowingexample.
NOTE
:Theexpectedvalueforeachcellneedstobeatleast5inordertousethistest.
Example11.5
SupposeA=aspeedingviolationinthelastyearandB=acellphoneuserwhiledriving.IfAand
BareindependentthenP(AANDB)=P(A)P(B).AANDBistheeventthatadriverreceived
aspeedingviolationlastyearandisalsoacellphoneuserwhiledriving. Suppose,inastudyof
driverswhoreceivedspeedingviolationsinthelastyearandwhousescellphoneswhiledriving,
that755peopleweresurveyed. Outofthe755,70hadaspeedingviolationand685didnot;305
werecellphoneuserswhiledrivingand450werenot.
Lety=expectednumberofdriversthatuseacellphonewhiledrivingandreceivedspeeding
violations.
IfAandBareindependent,thenP(AANDB)=P(A)P(B).Bysubstitution,
y
755
=
70
755
305
755
Solvefory:y=
70305
755
=28.3
5
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482
CHAPTER11. THECHI-SQUAREDISTRIBUTION
About28peoplefromthesampleareexpectedtobecellphoneuserswhiledrivingandtoreceive
speedingviolations.
Inatestofindependence, westatethenullandalternatehypothesesinwords. . Sincethecon-
tingencytableconsistsoftwofactors,thenullhypothesisstatesthatthefactorsareindependent
andthealternatehypothesisstatesthattheyarenotindependent(dependent). Ifwedoatestof
independenceusingtheexampleabove,thenthenullhypothesisis:
H
o
: Beingacellphoneuserwhiledrivingandreceivingaspeedingviolationareindependent
events.
Ifthenullhypothesisweretrue,wewouldexpectabout28peopletobecellphoneuserswhile
drivingandtoreceiveaspeedingviolation.
Thetestofindependenceisalwaysright-tailedbecauseofthecalculationoftheteststatistic. If
theexpectedandobservedvaluesarenotclosetogether,thentheteststatisticisverylargeand
wayoutintherighttailofthechi-squarecurve,likegoodness-of-fit.
Thedegreesoffreedomforthetestofindependenceare:
df=(numberofcolumns-1)(numberofrows-1)
Thefollowingformulacalculatestheexpectednumber(E):
E=
(rowtotal)(columntotal)
totalnumbersurveyed
Example11.6
Inavolunteergroup,adults21andoldervolunteerfromonetoninehourseachweektospend
time withadisabledseniorcitizen. . The e programrecruitsamongcommunitycollegestudents,
four-yearcollegestudents,andnonstudents. Thefollowingtableisasampleoftheadultvolun-
teersandthenumberofhourstheyvolunteerperweek.
NumberofHoursWorkedPerWeekbyVolunteerType(Observed)
TypeofVolunteer
1-3Hours
4-6Hours
7-9Hours
RowTotal
CommunityCollegeStudents
111
96
48
255
Four-YearCollegeStudents
96
133
61
290
Nonstudents
91
150
53
294
ColumnTotal
298
379
162
839
Table11.9
:Thetablecontainsobserved(O)values(data).
Problem
Arethenumberofhoursvolunteeredindependentofthetypeofvolunteer?
Solution
Theobservedtableandthequestionattheendoftheproblem,"Arethenumberofhoursvol-
unteeredindependentofthetypeofvolunteer?"tellyouthisisatestofindependence. Thetwo
factorsarenumberofhoursvolunteeredandtypeofvolunteer.Thistestisalwaysright-tailed.
H
o
:Thenumberofhoursvolunteeredisindependentofthetypeofvolunteer.
H
a
:Thenumberofhoursvolunteeredisdependentonthetypeofvolunteer.
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483
Theexpectedtableis:
NumberofHoursWorkedPerWeekbyVolunteerType(Expected)
TypeofVolunteer
1-3Hours
4-6Hours
7-9Hours
CommunityCollegeStudents
90.57
115.19
49.24
Four-YearCollegeStudents
103.00
131.00
56.00
Nonstudents
104.42
132.81
56.77
Table11.10
:Thetablecontainsexpected(E)values(data).
Forexample,thecalculationfortheexpectedfrequencyforthetopleftcellis
E=
(rowtotal)(columntotal)
totalnumbersurveyed
=
255298
839
=90.57
Calculatetheteststatistic:c
2
=12.99
(calculatororcomputer)
Distributionforthetest:c
2
4
df=(3columns 1)(3rows 1)=(2)(2)=4
Graph:
Probabilitystatement:p-value=P
c
2
>12.99
=0.0113
Compareaandthep-value:Sincenoaisgiven,assumea=0.05.p-value=0.0113.a>p-value.
Makeadecision:Sincea>p-value,rejectH
o
.Thismeansthatthefactorsarenotindependent.
Conclusion: Ata5%levelofsignificance,fromthedata,thereissufficientevidencetoconclude
thatthenumberofhoursvolunteeredandthetypeofvolunteeraredependentononeanother.
Fortheaboveexample,iftherehadbeenanothertypeofvolunteer,teenagers,whatwouldthe
degreesoffreedombe?
NOTE
:Calculatorinstructionsfollow.
TI-83+andTI-84calculator: Pressthe
keyandarrowoverto
.Press
.Press
.EnterthetablevaluesbyrowfromExample11-6. Press
aftereach. Press
. Press
andarrow over to
.Arrow downto
c
. Press
. You
shouldsee
.Arrowdownto
. Press
.Thetest
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484
CHAPTER11. THECHI-SQUAREDISTRIBUTION
statisticis12.9909andthep-value=0.0113.Dotheprocedureasecondtimebutarrowdownto
insteadof
.
Example11.7
DeAnzaCollegeisinterestedintherelationshipbetweenanxietylevelandtheneedtosucceed
inschool. Arandomsampleof400studentstookatestthatmeasuredanxietylevelandneedto
succeedinschool. Thetableshowstheresults. DeAnzaCollegewantstoknowifanxietylevel
andneedtosucceedinschoolareindependentevents.
NeedtoSucceedinSchoolvs.AnxietyLevel
Need
to
Succeed in
School
High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
RowTotal
HighNeed
35
42
53
15
10
155
Medium
Need
18
48
63
33
31
193
LowNeed
4
5
11
15
17
52
Column To-
tal
57
95
127
63
58
400
Table11.11
Problem1
Howmanyhighanxietylevelstudentsareexpectedtohaveahighneedtosucceedinschool?
Solution
Thecolumntotalforahighanxietylevelis57.Therowtotalforhighneedtosucceedinschoolis
155.Thesamplesizeortotalsurveyedis400.
E=
(rowtotal)(columntotal)
totalsurveyed
=
15557
400
=22.09
Theexpectednumberofstudentswhohaveahighanxietylevelandahighneedtosucceedin
schoolisabout22.
Problem2
Ifthetwovariablesareindependent,howmanystudentsdoyouexpecttohavealowneedto
succeedinschoolandamed-lowlevelofanxiety?
Solution
Thecolumntotalforamed-lowanxietylevelis63. Therowtotalforalowneedtosucceedin
schoolis52.Thesamplesizeortotalsurveyedis400.
Problem3
a. E=
(rowtotal)(columntotal)
totalsurveyed
=
b. Theexpectednumberofstudentswhohaveamed-lowanxietylevelandalowneedtosucceed
inschoolisabout:
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485
11.6TheChi-SquareDistribution:TestforHomogeneity
6
TheGoodnessofFittestcanbeusedtodecidewhetherapopulationfitsagivendistribution,buttheGood-
nessofFittestwillnotsufficetocomparewhethertwopopulationsfollowthesameunknowndistribution.
Adifferenttest, calledtheTestforHomogeneity, canbe usedtomakeaconclusionaboutwhethertwo
populationshavethesamedistribution.Tocalculatetheteststatisticforatestforhomogeneity,followthe
sameprocedureaswiththeTestofIndependence.
NOTE
:Theexpectedvalueforeachcellneedstobeatleast5inordertousethistest.
Hypotheses
H
o
:Thedistributionsofthetwopopulationsarethesame.
H
a
:Thedistributionsofthetwopopulationsarenotthesame.
TestStatistic
Useac
2
teststatistic.Itiscomputedinthesamewayasthetestforindependence.
DegreesofFreedom(df)
df=numberofcolumns-1
Requirements
Allvaluesinthetablemustbegreaterthanorequalto5.
CommonUses
Comparingtwopopulations.Forexample:menversuswomen,beforevs.after,eastvs.west.Thevariable
iscategoricalwithmorethantwopossibleresponsevalues.
Example11.8
Domaleandfemalecollegestudentshavethesamedistributionoflivingconditions?Usealevelof
significanceof0.05.Supposethat250randomlyselectedmalecollegestudentsand300randomly
selectedfemalecollegestudentswereaskedabouttheirlivingconditions:Dormitory,Apartment,
WithParents,Other.Theresultsareshowninthetablebelow.
DistributionofLivingConditionsforCollegeMalesandCollegeFemales
Dormitory
Apartment
WithParents
Other
Males
72
84
49
45
Females
91
86
88
35
Table11.12
6
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486
CHAPTER11. THECHI-SQUAREDISTRIBUTION
Problem
Domaleandfemalecollegestudentshavethesamedistributionoflivingconditions?
Solution
H
o
:Thedistributionoflivingconditionsformalecollegestudentsisthesameasthedistribution
oflivingconditionsforfemalecollegestudents.
H
a
: The e distribution n of f living conditions for male college students is not the same as the
distributionoflivingconditionsforfemalecollegestudents.
DegreesofFreedom(df):
df=numberofcolumns-1=4-1=3
Distributionforthetest:c
2
3
Calculatetheteststatistic:c
2
=10.1287
(calculatororcomputer)
Probabilitystatement:p-value=P
c
2
>10.1287
=0.0175
TI-83+ and TI-84 calculator: : Press s the
keyand arrow over to
. Press
. Press
. Enter r the table e values by row. . Press
after each. . Press
.
Press
andarrowoverto
.Arrowdownto
c
. Press
.Youshouldsee
. Arrow down n to
. Press
. The test statistic
is10.1287andthep-value 0.0175. . Dotheprocedureasecondtimebutarrowdownto
insteadof
.
Compare and the e p-value: Since no is given, assume 0.05. p-value 0.0175.
a>p-value.
Makea decision: Since p-value, reject H
o
. This s means that the e distributions s are e not
thesame.
Conclusion: At a a 5% % level ofsignificance, from the data, , there e issufficient evidence to con-
cludethatthedistributionsoflivingconditionsformaleandfemalecollegestudentsarenotthe
same.
Notice that the conclusion n is only y that the e distributions s are not the same. . We e cannot use
theTestforHomogeneitytomakeanyconclusionsabouthowtheydiffer.
Example11.9
Bothbeforeandafterarecentearthquake,surveyswereconductedaskingvoterswhichofthe
threecandidatestheyplannedonvotingforintheupcomingcitycouncilelection.Hastherebeen
achange sincetheearthquake? ? Usealevelofsignificanceof0.05. Thetablebelowshowsthe
resultsofthesurvey.
Perez
Chung
Stevens
Before
167
128
135
After
214
197
225
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487
Table11.13
Problem
Hastherebeenachangeinthedistributionofvoterpreferencessincetheearthquake?
Solution
H
o
:Thedistributionofvoterpreferenceswasthesamebeforeandaftertheearthquake.
H
a
:Thedistributionofvoterpreferenceswasnotthesamebeforeandaftertheearthquake.
DegreesofFreedom(df):
df=numberofcolumns-1=3-1=2
Distributionforthetest:c2
2
Calculatetheteststatistic:c=3.2603
(calculatororcomputer)
Probabilitystatement:p-value=P
c
2
>3.2603
=0.1959
TI-83+ and TI-84 calculator: : Press s the
keyand arrow over to
. Press
. Press
. Enter r the table e values by row. . Press
after each. . Press
.
Press
andarrowoverto
.Arrowdownto
c
. Press
.Youshouldsee
.Arrowdownto
. Press
.Theteststatisticis
3.2603andthep-value=0.1959.Dotheprocedureasecondtimebutarrowdownto
instead
of
.
Compareaandthep-value:a=0.05andthep-value=0.1959.a<p-value.
Makeadecision:Sincea<p-value,donotrejectH
o
.
Conclusion: At a a 5% % level of f significance, from the e data, , there e is insufficient evidence to
conclude that t the e distribution of voter r preferences s was not the same e before and d after r the
earthquake.
**ContributedbyDr.LarryGreen
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488
CHAPTER11. THECHI-SQUAREDISTRIBUTION
11.7TheChi-Square Distribution: : Comparison n Summary ofthe Chi-
SquareTestsGoodness-of-Fit,IndependenceandHomogeneity
7
ComparisonSummaryoftheChi-SquareTests:Goodness-of-Fit,IndependenceandHomogeneity
Youhaveseenthec
2
teststatisticusedinthreedifferentcircumstances.Belowisasummarythatwillhelp
youdecidewhichc
2
testistheappropriateonetouse.
 Goodness-of-Fit:UsetheGoodness-of-FitTesttodecidewhetherapopulationwithunknowndistri-
bution"fits"aknowndistribution. Inthiscasetherewillbeasinglequalitativesurveyquestionora
singleoutcomeofanexperimentfromasinglepopulation. Goodness-of-Fitistypicallyusedtosee
ifthepopulationisuniform(alloutcomesoccurwithequalfrequency),thepopulationisnormal,or
thepopulationisthesameasanotherpopulationwithknowndistribution. Thenullandalternative
hypothesesare:
H
o
:Thepopulationfitsthegivendistribution.
H
a
:Thepopulationdoesnotfitthegivendistribution.
 Independence:UsetheTestforIndependencetodecidewhethertwovariables(factors)areindepen-
dentordependent. Inthiscasetherewillbetwoqualitativesurveyquestionsorexperimentsanda
contingencytablewillbeconstructed. Thegoalistoseeifthetwovariablesareunrelated(indepen-
dent)orrelated(dependent).Thenullandalternativehypothesesare:
H
o
:Thetwovariables(factors)areindependent.
H
a
:Thetwovariables(factors)aredependent.
 Homogeneity:UsetheTestforHomogeneitytodecideiftwopopulationswithunknowndistribution
havethesamedistributionaseachother.Inthiscasetherewillbeasinglequalitativesurveyquestion
orexperimentgiventotwodifferentpopulations.Thenullandalternativehypothesesare:
H
o
:Thetwopopulationsfollowthesamedistribution.
H
a
:Thetwopopulationshavedifferentdistributions.
**WithcontributionsbyDr.LarryGreen
11.8TestofaSingleVariance
8
A testofasinglevarianceassumesthatthe underlyingdistributionisnormal. Thenullandalternate
hypothesesarestatedintermsofthepopulationvariance(orpopulationstandarddeviation). Thetest
statisticis:
(n 1)s
2
s2
(11.3)
where:
 n=thetotalnumberofdata
 s
2
=samplevariance
 s
2
=populationvariance
Youmaythinkofsastherandomvariableinthistest.Thedegreesoffreedomaredf=1.
Atestofasinglevariancemayberight-tailed,left-tailed,ortwo-tailed.
7
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8
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489
The followingexample will show youhow to setup thenulland alternate hypotheses. . The e null and
alternatehypothesescontainstatementsaboutthepopulationvariance.
Example11.10
Mathinstructorsarenotonlyinterestedinhowtheirstudentsdoonexams,onaverage,buthow
theexamscoresvary. To o manyinstructors, thevariance(orstandarddeviation)maybemore
importantthantheaverage.
Supposeamathinstructorbelievesthatthestandarddeviationforhisfinalexamis5points.One
ofhisbeststudentsthinksotherwise.Thestudentclaimsthatthestandarddeviationismorethan
5points. Ifthestudent t were toconduct ahypothesistest, what wouldthe nullandalternate
hypothesesbe?
Solution
Eventhoughwearegiventhepopulationstandarddeviation,wecansetthetestupusingthe
populationvarianceasfollows.
 H
o
:s
2
=5
2
 H
a
:s
2
>5
2
Example11.11
Withindividuallinesatitsvariouswindows,apostofficefindsthatthestandarddeviationfor
normallydistributedwaitingtimesforcustomersonFridayafternoonis7.2minutes. Thepost
officeexperimentswithasinglemainwaitinglineandfindsthatforarandomsampleof25cus-
tomers,thewaitingtimesforcustomershaveastandarddeviationof3.5minutes.
Withasignificancelevelof5%, testtheclaimthatasinglelinecauseslowervariationamong
waitingtimes(shorterwaitingtimes)forcustomers.
Solution
Sincetheclaimisthatasinglelinecauseslowervariation,thisisatestofasinglevariance. The
parameteristhepopulationvariance,s
2
,orthepopulationstandarddeviation,s.
RandomVariable: Thesamplestandarddeviation,s,istherandomvariable. Lets=standard
deviationforthewaitingtimes.
 H
o
:s
2
=7.2
2
 H
a
:s
2
<7.2
2
Theword"lower"tellsyouthisisaleft-tailedtest.
Distributionforthetest:c
2
24
,where:
 n=thenumberofcustomerssampled
 df=n 1=25 1=24
Calculatetheteststatistic:
c
2
=
(n 1)s
2
s2
=
(25 1)3.5
2
7.2
2
=5.67
wheren=25,s=3.5,ands=7.2.
Graph:
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490
CHAPTER11. THECHI-SQUAREDISTRIBUTION
Probabilitystatement: p-value=P
c
2
5.67
0.000042
Compareaandthep-value:a=0.05
p-value=0.000042
a>p-value
Makeadecision:Sincea>p-value,rejectH
o
.
Thismeansthatyourejects=7.2
2
. Inotherwords,youdonotthinkthevariationinwaiting
timesis7.2minutes,butlower.
Conclusion: Ata5%levelofsignificance,fromthedata,thereissufficientevidencetoconclude
thatasinglelinecausesalowervariationamongthewaitingtimesorwithasingleline,thecus-
tomerwaitingtimesvarylessthan7.2minutes.
TI-83+andTI-84calculators:In
,use
c
.Thesyntaxis
for
theparameterlist.ForExample11-9,c
.Thep-value=0.000042.
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