﻿
431
t-score
(
x
1
x
2
) (m
1
m
2
)
r
(S
1
)
2
n
1
+
(S
2
)
2
n
2
(10.2)
where:
 s
1
ands
2
,thesamplestandarddeviations,areestimatesofs
1
ands
2
,respectively.
 s
1
ands
2
aretheunknownpopulationstandarddeviations.
x
1
and
x
2
arethesamplemeans.m
1
andm
2
arethepopulationmeans.
Thedegreesoffreedom(df)isasomewhatcomplicatedcalculation.However,acomputerorcalculatorcal-
culatesiteasily.Thedfsarenotalwaysawholenumber.Theteststatisticcalculatedaboveisapproximated
bythestudent’s-tdistributionwithdfsasfollows:
Degreesoffreedom
df =
(s
1
)
2
n
1
+
(s
2
)
2
n
2
2
1
n
1
1
(s
1
)
2
n
1
2
+
1
n
2
1
(s
2
)
2
n
2
2
(10.3)
Whenbothsamplesizesn
1
andn
2
areﬁveorlarger,thestudent’s-tapproximationisverygood.Noticethat
thesamplevariancess
1
2
ands
2
2
arenotpooled.(Ifthequestioncomesup,donotpoolthevariances.)
NOTE
: Itisnotnecessarytocomputethisbyhand.Acalculatororcomputereasilycomputesit.
Example10.1:Independentgroups
Theaverageamountoftimeboysandgirlsages7through11spendplayingsportseachdayis
believedtobethesame. Anexperimentisdone,dataiscollected,resultinginthetablebelow.
Bothpopulationshaveanormaldistribution.
SampleSize
AverageNumberof
HoursPlayingSports
PerDay
SampleStandard
Deviation
Girls
9
2hours
p
0.75
Boys
16
3.2hours
1.00
Table10.1
Problem
day?Testatthe5%levelofsigniﬁcance.
Solution
Thepopulationstandarddeviationsarenotknown.Letgbethesubscriptforgirlsandbbethe
subscriptforboys. Then,m
g
isthepopulationmeanforgirlsandm
b
isthepopulationmeanfor
boys.Thisisatestoftwoindependentgroups,twopopulationmeans.
Randomvariable:
X
g
X
b
=differenceinthesamplemeanamountoftimegirlsandboysplay
sportseachday.
H
o
:m
g
=m
b
m
g
m
b
=0
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432
CHAPTER10. HYPOTHESISTESTING:TWOMEANS,PAIREDDATA,TWO
PROPORTIONS
H
a
:m
g
6=m
b
m
g
m
b
6=0
Thewords"thesame"tellyouH
o
hasan"=".SincetherearenootherwordstoindicateH
a
,then
assume"isdifferent."Thisisatwo-tailedtest.
Distributionforthetest: Uset
df
wheredf iscalculatedusingthedf formulaforindependent
groups,twopopulationmeans.Usingacalculator,dfisapproximately18.8462. Donotpoolthe
variances.
Calculatethep-valueusingastudent’s-tdistribution:p-value=0.0054
Graph:
Figure10.1
s
g
=
p
0.75
s
b
=1
So,
x
g
x
b
=2 3.2= 1.2
Halfthep-valueisbelow-1.2andhalfisabove1.2.
o
.
Thismeansyourejectm
g
=m
b
.Themeansaredifferent.
Conclusion: Atthe5%levelofsigniﬁcance,thesampledatashowthereissufﬁcientevidenceto
concludethatthemeannumberofhoursthatgirlsandboysaged7through11playsportsperday
isdifferent(meannumberofhoursboysaged7through11playsportsperdayisgreaterthanthe
meannumberofhoursplayedbygirlsORthemeannumberofhoursgirlsaged7through11play
sportsperdayisgreaterthanthemeannumberofhoursplayedbyboys).
NOTE
:TI-83+andTI-84:Press
.Arrowoverto
andpress
. Arrowover
toStatsandpress
.Arrowdownandenter
fortheﬁrstsamplemean,
p
0.75 forSx1,
forn1,
forthesecondsamplemean,
forSx2,and
forn2. Arrowdowntom1: andarrow
to
m2.Press
.ArrowdowntoPooled:and
.Press
.Arrowdownto
andpress
.Thep-valueisp=0.0054,thedfsareapproximately18.8462,andthe
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433
Example10.2
ageis3.5mathclasseswithastandarddeviationof1mathclass.Thecommunitygroupbelieves
populationshaveanormaldistribution. Testata1%signiﬁcance e level. . Answer r thefollowing
questions.
Problem1
(Solutiononp.466.)
Isthisatestoftwomeansortwoproportions?
Problem2
(Solutiononp.466.)
Arethepopulationsstandarddeviationsknownorunknown?
Problem3
(Solutiononp.466.)
Whichdistributiondoyouusetoperformthetest?
Problem4
(Solutiononp.466.)
Whatistherandomvariable?
Problem5
(Solutiononp.466.)
Whatarethenullandalternatehypothesis?
Problem6
(Solutiononp.466.)
Isthistestright,left,ortwotailed?
Problem7
(Solutiononp.466.)
Whatisthep-value?
Problem8
(Solutiononp.466.)
Doyourejectornotrejectthenullhypothesis?
Conclusion:
Atthe1%levelofsigniﬁcance,fromthesampledata,thereisnotsufﬁcientevidencetoconclude
10.3ComparingTwoIndependentPopulationMeanswithKnownPop-
ulationStandardDeviations
3
Eventhoughthissituationisnotlikely(knowingthepopulationstandarddeviationsisnot likely), the
followingexampleillustrateshypothesistestingforindependentmeans,knownpopulationstandardde-
viations. Thesamplingdistributionforthedifferencebetweenthemeansisnormalandbothpopulations
mustbenormal.Therandomvariableis
X
1
X
2
.Thenormaldistributionhasthefollowingformat:
Normaldistribution
X
1
X
2
N
2
4
u
1
u
2
,
s
(s
1
)
2
n
1
+
(s
2
)
2
n
2
3
5
(10.4)
3
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434
CHAPTER10. HYPOTHESISTESTING:TWOMEANS,PAIREDDATA,TWO
PROPORTIONS
Thestandarddeviationis:
s
(s
1
)
2
n
1
+
(s
2
)
2
n
2
(10.5)
Theteststatistic(z-score)is:
z=
(
x
1
x
2
) (m
1
m
2
)
r
(s
1
)
2
n
1
+
(s
2
)
2
n
2
(10.6)
Example10.3
independentgroups,populationstandarddeviationsknown: Themeanlastingtimeof2com-
petingﬂoorwaxesistobecompared.Twentyﬂoorsarerandomlyassignedtotesteachwax.Both
populationshaveanormaldistribution.Thefollowingtableistheresult.
Wax
SampleMeanNumberofMonthsFloorWaxLast
PopulationStandardDeviation
1
3
0.33
2
2.9
0.36
Table10.2
Problem
Doesthedataindicatethatwax1ismoreeffectivethanwax2?Testata5%levelofsigniﬁcance.
Solution
Thisisatestoftwoindependentgroups,twopopulationmeans,populationstandarddeviations
known.
RandomVariable:
X
1
X
2
=differenceinthemeannumberofmonthsthecompetingﬂoorwaxes
last.
H
o
:m
1
m
2
H
a
:m
1
>m
2
Thewords"ismoreeffective"saysthatwax1lastslongerthanwax2,ontheaverage."Longer"
isa”>”symbolandgoesintoH
a
.Therefore,thisisaright-tailedtest.
Distributionforthetest: Thepopulationstandarddeviationsareknownsothedistributionis
normal.Usingtheformulaabove,thedistributionis:
X
1
X
2
N
0,
q
0.33
2
20
+
0.36
2
20
Sincem
1
m
2
thenm
1
m
2
0andthemeanforthenormaldistributionis0.
Calculatethep-valueusingthenormaldistribution:p-value=0.1799
Graph:
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435
Figure10.2
x
1
x
2
=3 2.9=0.1
Compareaandthep-value:a=0.05andp-value=0.1799.Therefore,ap-value.
o
.
Conclusion:Atthe5%levelofsigniﬁcance,fromthesampledata,thereisnotsufﬁcientevidence
toconcludethatthemeantimewax1lastsislonger(wax1ismoreeffective)thanthemeantime
wax2lasts.
NOTE
:TI-83+andTI-84:Press
.Arrowoverto
andpress
. Arrowover
to
andpress
.Arrowdownandenter
forsigma1,
forsigma2,
fortheﬁrst
samplemean,
forn1,
forthesecondsamplemean,and
forn2.Arrowdowntom1:and
arrowto>m2.Press
.Arrowdownto
andpress
.Thep-valueisp=0.1799
do
.
10.4ComparingTwoIndependentPopulationProportions
4
1. Thetwoindependentsamplesaresimplerandomsamplesthatareindependent.
2. Thenumberofsuccessesisatleastﬁveand d thenumberoffailuresisatleastﬁveforeachofthe
samples.
Comparingtwoproportions,likecomparingtwomeans,iscommon. Iftwoestimatedproportionsare
A
P
B
proportions.
Thedifferenceoftwoproportionsfollowsanapproximatenormaldistribution.Generally,thenullhypoth-
esisstatesthatthetwoproportionsarethesame.Thatis,H
o
:p
A
=p
B
.Toconductthetest,weuseapooled
proportion,p
c
.
4
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436
CHAPTER10. HYPOTHESISTESTING:TWOMEANS,PAIREDDATA,TWO
PROPORTIONS
Thepooledproportioniscalculatedasfollows:
p
c
=
x
A
+x
B
n
A
+n
B
(10.7)
Thedistributionforthedifferencesis:
P
A
P
B
N
"
0,
s
p
c
(1 p
c
)
1
n
A
+
1
n
B
#
(10.8)
Theteststatistic(z-score)is:
z=
(p
A
p
B
)
(p
A
p
B
)
r
p
c
(1 p
c
)
1
n
A
+
1
n
B
(10.9)
Example10.4:Twopopulationproportions
Testata1%levelofsigniﬁcance.
10.4.1Determiningthesolution
Thisisatestof2populationproportions.
Problem
(Solutiononp.466.)
Howdoyouknow?
LetAandBbethesubscriptsformedicationAandmedicationB.Thenp
A
andp
B
arethedesired
populationproportions.
RandomVariable:
P
A
P
B
medicationAandmedicationB.
H
o
:p
A
=p
B
p
A
p
B
=0
H
a
:p
A
6=p
B
p
A
p
B
6=0
Distributionforthetest:Sincethisisatestoftwobinomialpopulationproportions,thedistribu-
tionisnormal:
p
c
=
x
A
+x
B
n
A
+n
B
=
20+12
200+200
=0.08 1 p
c
=0.92
Therefore, P
A
P
B
N
0,
r
(0.08)(0.92)
1
200
+
1
200
P
A
P
B
followsanapproximatenormaldistribution.
Calculatethep-valueusingthenormaldistribution:p-value=0.1404.
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437
EstimatedproportionforgroupA: p
A
=
x
A
n
A
=
20
200
=0.1
EstimatedproportionforgroupB: p
B
=
x
B
n
B
=
12
200
=0.06
Graph:
Figure10.3
P
A
P
B
=0.1 0.06=0.04.
Halfthep-valueisbelow-0.04andhalfisabove0.04.
Compareaandthep-value:a=0.01andthep-value=0.1404.ap-value.
o
.
Conclusion:Ata1%levelofsigniﬁcance,fromthesampledata,thereisnotsufﬁcientevidenceto
minutestomedicationAandmedicationB.
NOTE
:TI-83+andTI-84:Press
.Arrowoverto
andpress
.Arrowdown
andenter
forx1,
forn1,
forx2,and
forn2. Arrowdownto
: andarrowto
. Press
.Arrowdownto
andpress
.Thep-valueis=0.1404
do
.
10.5MatchedorPairedSamples
5
1. Simplerandomsamplingisused.
2. Samplesizesareoftensmall.
3. Twomeasurements(samples)aredrawnfromthesamepairofindividualsorobjects.
4. Differencesarecalculatedfromthematchedorpairedsamples.
5. Thedifferencesformthesamplethatisusedforthehypothesistest.
5
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438
CHAPTER10. HYPOTHESISTESTING:TWOMEANS,PAIREDDATA,TWO
PROPORTIONS
6. Thematchedpairshavedifferencesthateithercomefromapopulationthatisnormalorthenumberof
differencesissufﬁcientlylargesothedistributionofthesamplemeanofdifferencesisapproximately
normal.
Inahypothesistestformatchedorpairedsamples,subjectsarematchedinpairsanddifferencesarecal-
culated. Thedifferencesarethedata. . Thepopulationmeanforthedifferences,m
d
, isthentestedusing
aStudent-ttestforasingle populationmeanwith1degreesoffreedomwherenisthe numberof
differences.
Theteststatistic(t-score)is:
t=
x
d
m
d
s
d
p
n
(10.10)
Example10.5:Matchedorpairedsamples
Astudywasconductedtoinvestigatetheeffectivenessofhypnotisminreducingpain. Results
forrandomlyselectedsubjectsareshowninthetable.The"before"valueismatchedtoan"after"
valueandthedifferencesarecalculated.Thedifferenceshaveanormaldistribution.
Subject:
A
B
C
D
E
F
G
H
Before
6.6
6.5
9.0
10.3
11.3
8.1
6.3
11.6
After
6.8
2.4
7.4
8.5
8.1
6.1
3.4
2.0
Table10.3
Problem
Arethesensorymeasurements,onaverage,lowerafterhypnotism?Testata5%signiﬁcancelevel.
Solution
Corresponding"before"and"after"valuesformmatchedpairs.(Calculate"sfter"-"before").
AfterData
BeforeData
Difference
6.8
6.6
0.2
2.4
6.5
-4.1
7.4
9
-1.6
8.5
10.3
-1.8
8.1
11.3
-3.2
6.1
8.1
-2
3.4
6.3
-2.9
2
11.6
-9.6
Table10.4
Thedataforthetestarethedifferences:{0.2,-4.1,-1.6,-1.8,-3.2,-2,-2.9,-9.6}
The samplemeanand sample standarddeviationofthedifferencesare:
x
d
3.13and
s
d
=2.91Verifythesevalues.
Letm
d
bethepopulationmeanforthedifferences.Weusethesubscriptdtodenote"differences."
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439
RandomVariable:
X
d
=themeandifferenceofthesensorymeasurements
H
o
:m
d
0
(10.11)
Thereisnoimprovement.(m
d
isthepopulationmeanofthedifferences.)
H
a
:m
d
<0
(10.12)
Thereisimprovement. Thescoreshouldbelowerafterhypnotismsothedifferenceoughttobe
negativetoindicateimprovement.
Distributionforthetest: Thedistributionisastudent-twithdf =117. Uset
7
.
(Noticethatthetestisforasinglepopulationmean.)
Calculatethep-valueusingtheStudent-tdistribution:p-value=0.0095
Graph:
Figure10.4
X
d
istherandomvariableforthedifferences.
Thesamplemeanandsamplestandarddeviationofthedifferencesare:
x
d
3.13
s
d
=2.91
Compareaandthep-value:a=0.05andp-value=0.0095.a>p-value.
o
.
Thismeansthatm
d
<0andthereisimprovement.
Conclusion:Ata5%levelofsigniﬁcance,fromthesampledata,thereissufﬁcientevidencetocon-
cludethatthesensorymeasurements,onaverage,arelowerafterhypnotism.Hypnotismappears
tobeeffectiveinreducingpain.
NOTE
(after-before)andputthedifferencesintoalistoryoucanputtheafterdataintoaﬁrstlistand
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440
CHAPTER10. HYPOTHESISTESTING:TWOMEANS,PAIREDDATA,TWO
PROPORTIONS
thebeforedataintoasecondlist.Thengotoathirdlistandarrowuptothename. Enter1stlist
name-2ndlistname. Thecalculatorwilldothesubtractionandyouwillhavethedifferencesin
thethirdlist.
NOTE
: TI-83+andTI-84: : Useyourlistofdifferencesasthedata. . Press
andarrowoverto
.Press
.Arrowoverto
andpress
.Arrowdownandenter
form
0
,the
nameofthelistwhereyouputthedata,and
forFreq:.Arrowdowntom: andarrowoverto<
m
0
. Press
.Arrowdownto
andpress
.Thep-valueis0.0094andthetest
statisticis-3.04. Dotheseinstructionsagainexceptarrowto
). Press
.
Example10.6
Acollegefootballcoachwasinterestedinwhetherthecollege’sstrengthdevelopmentclassin-
creased hisplayers’maximumlift (inpounds)onthe benchpressexercise. . He e asked4ofhis
playerstoparticipateinastudy. Theamountofweighttheycouldeachliftwasrecordedbefore
theytookthestrengthdevelopmentclass.Aftercompletingtheclass,theamountofweightthey
couldeachliftwasagainmeasured.Thedataareasfollows:
Weight(inpounds)
Player1
Player2
Player3
Player4
Amountofweightedliftedpriortotheclass
205
241
338
368
Amountofweightliftedaftertheclass
295
252
330
360
Table10.5
Thecoach wantstoknow if thestrengthdevelopmentclassmakeshisplayers stronger, on
average.
Problem
(Solutiononp.466.)
Recordthedifferencesdata.Calculatethedifferencesbysubtractingtheamountofweightlifted
priortotheclassfromtheweightliftedaftercompletingtheclass.Thedataforthedifferencesare:
{90,11,-8,-8}.Thedifferenceshaveanormaldistribution.
Usingthedifferencesdata,calculatethesamplemeanandthesamplestandarddeviation.
x
d
=21.3
s
d
=46.7
Usingthedifferencedata,thisbecomesatestofasingle__________(ﬁllintheblank).
Deﬁnetherandomvariable:
X
d
=meandifferenceinthemaximumliftperplayer.
Thedistributionforthehypothesistestist
3
.
H
o
:m
d
0
H
a
:m
d
>0
Graph:
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