Follow-up Tests and Two-Way ANOVA

41 Pages·2013·3.29 MB·English

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CHAPTER 26 More about Analysis of
Variance: Follow-up Tests
and Two-Way ANOVA
A
nalysisofvariance(ANOVA)isastatisticalmethodforcomparingthemeans IN THIS CHAPTER
ofseveralpopulationsbasedonindependentrandomsamples,orthemean WE COVER...
responses to several treatments in a randomized comparative experiment.
Whenwecomparejusttwomeans,weusethetwo-samplet proceduresdescribed ■ Beyondone-wayANOVA
in Chapter 18. ANOVA allows comparison of any number of means. The basic ■ Follow-upanalysis:Tukeypairwise
multiplecomparisons
formofANOVAisone-wayANOVA,whichtreatsthemeansbeingcomparedas
■ Follow-upanalysis:contrasts*
meanresponsestodifferentlevelsofasinglevariable.Forexample,inChapter24
■ Two-wayANOVA:conditions,
weusedone-wayANOVAtocomparethemeanweightsofadultmaleWistarrats
maineffects,andinteraction
fedoneofthreetypesofdiets.Figure26.1showstheMinitabANOVAoutputfor
■ Inferencefortwo-wayANOVA
thesedata(displayedinTable24.1,page598).
■ Somedetailsoftwo-wayANOVA*
Beyondone-wayANOVA
You should recall or review the big ideas of one-way ANOVA from Chapter 24. one-wayANOVA
One-wayANOVAcomparesthemeansμ ,μ ,...,μ ofk populationsbasedon
1 2 k
samplesofsizesn ,n ,...,n fromthesepopulations.
1 2 k
26-1
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26-2 CHAPTER26 ■ MoreaboutAnalysisofVariance:Follow-upTestsandTwo-WayANOVA
■ Usingseparatetwo-samplet procedurestocomparemanypairsofmeansis
abadidea,becausewedon’thavea P-valueoraconfidencelevelforthe
completesetofcomparisonstogether.Thisistheproblemofmultiple
multiplecomparisons comparisons.
■ One-wayANOVAgivesasingletestforthenullhypothesisthatallthe
populationmeansarethesameagainstthealternativehypothesisthatnotall
arethesame(H simplyisnottrue).
0
■ ANOVAworksbycomparinghowfarapartthesamplemeansarerelativeto
thevariationamongindividualobservationsinthesamesample.Thetest
ANOVAF statistic statisticistheANOVA F statistic
variationamongthesamplemeans
F =
variationamongindividualsinthesamesample
F distribution The P-valuecomesfroman F distribution.
ANOVAconditions ■ Therequiredconditions for ANOVAareindependentrandomsamplesfrom
eachofthek populations(orarandomizedcomparativeexperimentwithk
treatments),Normaldistributionsfortheresponsevariableineachpopulation,
andacommonstandarddeviationσ inallpopulations.Fortunately,ANOVA
inferenceisquiterobustagainstmoderateviolationsoftheNormalityand
commonstandarddeviationconditions.
■ Inbasicstatisticalpractice,wecombinethe F testwithdescriptivedata
analysistochecktheconditionsforANOVAandtoseewhichmeansappear
todifferandbyhowmuch.
Examples 24.1 and 24.2 (pages 597 and 600) showed all the steps required for a
one-way ANOVA. This chapter moves beyond basic one-way ANOVA in two
directions.
Minitab
Session
One-way ANOVA: Chow, Restricted, Extended
Source DF SS MS F P
Factor 2 63400 31700 10.71 0.000
Error 47 139174 2961
Total 49 202573
S = 54.42 R-Sq = 31.30% R-Sq(adj) = 28.37%
Individual 95% CIs for Mean
Based on Pooled StDev
Level N Mean StDev
Chow 19 605.63 49.64
Restricted 16 657.31 50.68
Extended 15 691.13 63.41
595 630 665 700
FIGURE 26.1 MinitabANOVA Pooled StDev = 54.42
outputfortheratweightdataof
Examples24.1and24.2.
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■ Follow-upanalysis:Tukeypairwisemultiplecomparisons 26-3
Follow-up analysis. The ANOVA F test in Figure 26.1 tells us only that the
populationmeansarenotthesame.Wewouldliketosaywhichmeansdifferandby
howmuch.Forexample,dothedataallowustosaythatthe“extendeddiet”pop-
ulationdoeshaveahighermeanweightthanthe“chowdiet”andthe“restricted
diet”populationsofadultmaleWistarlabrats?Thisisafollow-upanalysistothe F
testthatgoesbeyonddataanalysistoconfidenceintervalsandtestsofsignificance
forspecificcomparisonsofmeans.
Two-way ANOVA. One-way ANOVA compares mean responses for several
levelsofjustoneexplanatoryvariable.InExamples24.1and24.2,thatvariableis
“the type of diet provided.” Suppose that we have data on two explanatory vari-
ables,say,thetypeofdietprovidedandwhetherthelabratsarephysicallyactive.
Therearenow6groupsformedbycombinationsofdiettypeandphysicalactivity,
asfollows:
Variable2
Active Inactive
Chow Group1 Group2
Variable1 Restricted Group3 Group4
Extended Group5 Group6
One-way ANOVA will still tell us if there is evidence that mean body weight
inthese6experimentalgroupsdiffers.Butwewantmore:Doesdiettypematter?
Doesphysicalactivitymatter?Anddothesetwovariablesinteract?Thatis,doesthe
effectofdiettypechangewhenthelabratsarephysicallyactive?Perhapsphysical
activityreducesthecravingforcafeteriafood,sothatdiettypehaslesseffectwhen
theratsareactivethanwhentheyareinactive.Toanswerthesequestionswemust
extend ANOVA to take into account the fact that the 6 groups are formed from
twoexplanatoryvariables.Thisistwo-wayANOVA.
Wewilldiscussfollow-upanalysisinANOVAfirst,andthentwo-wayANOVA.
Fortunately,thedistinctionbetweenone-wayandtwo-waydoesn’taffectthefollow-
up methods we will present. So once you have mastered these methods in the
one-waysetting,youcanapplythemimmediatelytotwo-wayproblems.
Follow-upanalysis:Tukeypairwisemultiplecomparisons
InExamples24.1and24.2wesawthatthereisgoodevidencethatthemeanbody
weight of adult male Wistarratsis not the same when they are assignedto a diet
consisting of chow only, chow plus restricted access to cafeteria food, and chow
plusextendedaccesstocafeteriafood.1 ThesamplemeansinFigure26.1suggest
that(aswemightexpect)themeanbodyweightishighestinratsgivenextended
accesstocafeteriafoodandlowestinratsgivenchowonly.
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26-4 CHAPTER26 ■ MoreaboutAnalysisofVariance:Follow-upTestsandTwo-WayANOVA
EXAMPLE 26.1 Comparinggroups:individualtprocedures
Let’s use A, B, and C to refer to the chow, restricted, and extended groups, respec-
tively.Howmuchhigheristhemeanbodyweightofratsgivenrestrictedaccesstocafe-
teria food than that of rats given chow only? A 95% confidence interval comparing
GroupsAandBanswersthisquestion.BecausetheconditionsforANOVArequirethat
thepopulationstandarddeviationbethesameinallthreepopulationsofrats,wecan
useaversionofthetwo-samplet confidenceintervalthatalsoassumesequalstandard
deviations.
CristianCiobanu/Alamy TheMinitaboutputinFigure26.1givesthepooledstandarddeviation(firstdefined
inchapters18and24,pages453and616)ass = 54.42grams(g).Thisisanestimate
p
of the common standard deviation σ based on all three samples. It has 47 degrees of
freedom,thedegreesoffreedomfor“Error”intheANOVAtable.Thestandarderrorfor
thedifferenceinsamplemeansx −x is(page453)
A B
(cid:2)
1 1
s +
p
n n
A B
A95%confidenceintervalforμ −μ wouldthereforebe
A B
(cid:2)
1 1
(x −x )±t∗s +
A B p
n n
A B
usingt∗ = 2.012fromtechnology(orapproximatelyt∗ = 2.021fromTableCfordf=
40,conservatively,sincethereisnorowfordf=47inTableC).
pairwisedifference However,wereallywanttoestimateallthreepairwisedifferencesamongthepop-
ulationmeans,
μ −μ μ −μ μ −μ
A B A C B C
Three95%confidenceintervalswillnotgiveus95%confidencethatallthreesimultaneously
capturetheirtrueparametervalues.Thisisthefamiliarproblemofmultiplecomparisons
thatwediscussedinChapters22and24.■
In general, we want to give confidence intervals for all pairwise differences
overallconfidence amongthepopulationmeansμ1,μ2,...,μk ofkpopulations.Wewantanoverall
confidence level of (say) 95%. That is, in very many uses of the method, all the
intervals will simultaneously capture the true differences 95% of the time. To do
this,takethenumberofcomparisonsintoaccountbyreplacingthet criticalvalue
∗
t in Example 26.1 with another critical value based on the distribution of the
differencebetweenthelargestandsmallestofasetofksamplemeans.Wewillcall
∗ ∗
thiscriticalvaluem ,formultiplecomparisons.Valuesofm nolongercomefrom
a t table. They depend on the number of populations we are comparing and on
thetotalnumberofobservationsinthesamples,aswellasontheconfidencelevel
we want. Software is very helpful for practical use. This method is named after
its inventor, John Tukey (1915–2000), the same man who developed the ideas
∗
of modern data analysis. A short table of m values for a 95% confidence level
(TableG)isprovidedforconvenienceattheendofthischapter.
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■ Follow-upanalysis:Tukeypairwisemultiplecomparisons 26-5
TUKEY PAIRWISE MULTIPLE COMPARISONS
IntheANOVAsetting,wehaveindependentSRSsofsizen fromeachofk
i
populationshavingNormaldistributionswithmeansμ andacommon
i
standarddeviationσ.Tukey simultaneous confidence intervalsforall
pairwisedifferencesμ −μ amongthepopulationmeanshavetheform
i j
(cid:2)
1 1
(x −x )±m∗s +
i j p
n n
i j
Herex isthesamplemeanoftheithsampleands isthepooledestimateof
i p
σ.Thecriticalvaluem∗ dependsontheconfidencelevelC,thenumberof
populationsk,andthetotalnumberofobservations N.
Ifallsamplesarethesamesize,theTukeysimultaneousconfidenceintervals
provideanoveralllevelC ofconfidencethatalltheintervalssimultaneously
capturethetruepairwisedifferences.Ifthesamplesdifferinsize,thetrue
confidencelevelisatleastaslargeasC.Thatis,theconclusionsarethen
conservative.
Tocarryoutsimultaneous testsofthehypotheses
H : μ = μ
0 i j
H : μ =(cid:3) μ
a i j
forallpairsofpopulationmeans,reject H foranypairwhoseconfidence
0
intervaldoesnotcontain0.Thesetestshaveoverallsignificancelevelnoless
than1−C.Thatis,1−C istheprobabilitythat,whenallthepopulation
meansareequal,anyofthetestsincorrectlyrejectsitsnullhypothesis.
Minitab
Session
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 98.05%
Chow subtracted from:
Lower Center Upper
Restricted 7.03 51.68 96.33
Extended 40.05 85.50 130.95
−60 0 60 120
Restricted subtracted from:
Lower Center Upper FIGURE 26.2 Additional
Extended −13.47 33.82 81.12
MinitabANOVAoutputshowing
−60 0 60 120 Tukeypairwisemultiplecomparisons
fortheratweightdata,for
Example26.2.
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26-6 CHAPTER26 ■ MoreaboutAnalysisofVariance:Follow-upTestsandTwo-WayANOVA
EXAMPLE 26.2 Ratsandacafeteria-stylediet:multipleintervals
Figure26.2containsmoreMinitaboutputfortheANOVAcomparingthemeanbody
weightsin3experimentalgroupsoflabrats.WeaskedforTukeymultiplecomparisons
withanoverallerrorrateof5%.Thatis,theoverallconfidencelevelforthethreein-
tervalstogetheris95%.
The format of the Minitab output takes some study. Be sure you can see that the
Tukeyconfidenceintervalsare
7.03to 96.33 for μ −μ
B A
40.05to130.95 for μ −μ
C A
−13.47to 81.12 for μ −μ
C B
Ifyoudonothaveaccesstotechnology,theseintervalscaneasilybecomputedbyhand.
Let’sseehowtoobtainthefirstinterval,forμ −μ .TableGattheendofthischapter
A B
gives values of m∗ when using an overall confidence level of 95% and various combi-
nationsof Nandk.Startbyfindingtherightcombinationofk comparisons(toprow)
and N−k degreesoffreedom(leftmargin).Inourexample,k = 3and N−k = 47,
so m∗ = 2.434, approximately (based on a conservative df = 40, since df = 47 is not
available).Theintervalforμ −μ istherefore
B A
(cid:2) (cid:2)
1 1 1 1
(x −x )±m∗s + =(657.31−605.63)±(2.434)(54.42) +
B A p
n n 16 19
B A
=51.68±44.94
=6.74to96.62
Noticethatthevalueofm∗ weusehereislargerthanthevalueoft∗ inExample26.1.
This is the price we pay for having 95% confidence not just in one interval but in all
threesimultaneously.■
EXAMPLE 26.3 Ratsandacafeteria-stylediet:multipletests
TheANOVAnullhypothesisisthatallpopulationmeansareequal,
H:μ =μ =μ
0 A B C
WeknowfromtheoutputinFigure26.1thattheANOVAF testrejectsthishypothesis
(F = 10.71, P < 0.0005).Sowehavegoodevidencethatsomepairsofmeansarenot
the same. Which pairs? Look at the simultaneous 95% confidence intervals in Exam-
ple26.2.Whichoftheseintervalsdonotcontain0?Ifanintervaldoesnotcontain0,
werejectthehypothesisthatthispairofpopulationmeansareequal.
Theconclusionsare
Wecanreject H:μ =μ
0 B A
Wecanreject H:μ =μ
0 C A
Wecannotreject H:μ =μ
0 C B
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■ Follow-upanalysis:Tukeypairwisemultiplecomparisons 26-7
Thatis,
Wedohaveenoughevidencetoconcludethatμ =(cid:3) μ
A B
Wedohaveenoughevidencetoconcludethatμ =(cid:3) μ
A C
Wedonothaveenoughevidencetoconcludethatμ =(cid:3) μ
B C
This Tukey simultaneous test of three null hypotheses has the property that when all
three hypotheses are true, there is only a 5% probability that any of the three tests
wronglyrejectsitshypothesis.■
Recall what a test at a fixed significance level such as 5% tells us: either we
dohaveenoughevidencetorejectthenullhypothesis,orthedatadonotgiveenough
evidencetoallowrejection.
Thestudyfoundevidencethatratsonachow-onlydietdiffersignificantlyin
bodyweightfromratsgivenrestrictedaccessandfromratsgivenextendedaccess
to a cafeteria-style diet. However, the study did not find evidence that restricted
andextendedaccesstocafeteriafoodresultinratswithsignificantlydifferentbody
weights.Thatis, x = 605.63and x = 657.31arefarenoughaparttoconclude
A B
thatthepopulationmeansμ andμ differ, x = 605.63and x = 691.13are
A B A C
far enough apart to conclude that the population means μ and μ differ, but
A C
x = 657.31isnotfarenoughfrom x = 691.13toruleoutthepossibilitythat
B C
thepopulationmeansμ andμ mightbethesame.
B C
NoticethattheTukeymethoddoesnotgiveaP-valueforthethreeteststaken
together. Rather, we have a set of “reject” or “fail to reject” conclusions with an
overallsignificancelevelthatwefixedinadvance,5%inthisexample.Thereare
several other multiple-comparisons procedures that produce simultaneous confi-
dence intervals with an overall confidence level or simultaneous tests with an
overall probability of any false rejection. The Tukey procedures are arguably the
most useful.2 If you can interpret results from Tukey, you can understand output
fromothermultiple-comparisonsprocedures.
APPLY YOUR KNOWLEDGE
26.1 Caffeineandsugar. Exercise 24.26 (page 624) describes a double-blind random-
izedexperimentthatassignedhealthyundergraduatestudentstodrinkoneoffour
beveragesafterfastingovernight:water,waterwith75mgofcaffeine,waterwith
75gofglucose,andwaterwith75mgofcaffeineand75gofglucose.Thesubjects
performed a cognitive task, and their reaction times in the task are summarized
below(SEMisthestandarderrorofthemean):3
Beverage n x SEM
Water 18 389.35 18.50
Waterandcaffeine 18 320.16 17.98
Waterandglucose 18 318.16 17.04
Water,caffeine,andglucose 18 336.44 14.02
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26-8 CHAPTER26 ■ MoreaboutAnalysisofVariance:Follow-upTestsandTwo-WayANOVA
AnANOVAF testgivesasignificant P-valueof0.0134,withMSE=5186.0358.
(a)Becauseallfourgroupsarethesamesize,themarginoferroristhesamefor
all6pairwisecomparisons.ObtainthismarginoferrorusingTableGon
page26-41.FindtheTukeysimultaneous95%confidenceintervalsforall
pairwisecomparisonsofpopulationmeans.
(b)Explaininsimplelanguagewhat“95%confidence”meansforthese
intervals.
(c)Whichpairsofmeansdiffersignificantlyattheoverall5%significance
level?
26.2 Loggingintherainforest. Exercise 24.3 (page 604) describes a study comparing
forest plots in Borneo that had never been logged (Group 1) with similar plots
nearbythathadbeenlogged1yearearlier(Group2)and8yearsearlier(Group3).
Thethreegroupscanbeconsideredtobeindependentrandomsamples.Thedata
appearinTable24.2(page604);thevariableTreesisthenumberoftreesinaplot.4
Theone-wayANOVAshowninFigure24.4comparedthemeancountsoftreesin
the3typesofforestplotsandwasstatisticallysignificant,with P =0.0002.Italso
gaveMSE=27.3574.
(a)FindtheTukeysimultaneous95%confidenceintervalsforallpairwise
comparisonsofpopulationmeans.UsesoftwareorTableGonpage26-41.
(b)Explaininsimplelanguagewhat“95%confidence”meansforthese
intervals.
(c)Whichpairsofmeansdiffersignificantlyattheoverall5%significancelevel?
26.3 Whichcolorattractsbeetlesbest? Example24.4(page611)presentsdataonthe
numbers of cereal leaf beetles trapped by boards of four different colors.5 Yellow
boards appear most effective. ANOVA gives very strong evidence (P < 0.0005,
MSE=32.167)thatthecolorsdifferintheirabilitytoattractbeetles.
(a)Howmanypairwisecomparisonsaretherewhenwecomparefourcolors?
(b)UsesoftwareorTableGonpage26-41toobtaintheTukeysimultaneous95%
confidenceintervalsforallpairwisecomparisonsofpopulationmeans.Which
pairsofcolorsaresignificantlydifferentwhenwerequireasignificancelevelof
5%forallcomparisonsasagroup?
26.4 Dogs,friends,andstress. InExercise24.4(page605)youexaminedtheeffectof
petsinstressfulsituationsfromtheEESEEstory“StressamongPetsandFriends.”
The ANOVA F test had a very small P-value, giving good reason to conclude
thatmeanheartratesunderstressdodifferdependingonwhetherapet,afriend,
ornooneispresent.Table24.3(page606)displaysthesubjects’meanheartrate
duringastressfultask.Wewanttoknowwhetherthemeansforthetwotreatments
(pet,friend)differsignificantlyfromeachotherandfromthemeanforthecontrol
group.
(a)Whatarethecorrespondingthreenullhypotheses?
(b)Wewanttobe95%confidentthatwedon’twronglyrejectanyofthethree
nullhypotheses.Tukeypairwisecomparisonscangiveconclusionsthat
meetthiscondition.Whataretheconclusions?UsesoftwareorTableGon
page26-41.TheMinitaboutputforthesedatashowninFigure24.5
(page606)indicates“PooledStDev=9.208.”
■
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■ Follow-upanalysis:contrasts 26-9
Follow-upanalysis:contrasts*
Multiple-comparisons methods give conclusions about all comparisons in some
class with a measure of confidence that applies to all the comparisons taken to-
gether.Forexample,Tukey’smethodgivesconclusionsaboutallpairwisecompar-
isons among a set of population means. These methods are most useful when we
didnothaveanyspecificcomparisoninmindbeforeweproducedthedata.
Multiple-comparisonsproceduressometimesgivetestsorconfidenceintervals
for comparisons that don’t interest us. And they may leave out comparisons that
dointerestus.Ifwehavespecificquestionsinmindbeforeweproducedata,itismore
efficienttoplanananalysisthatasksthesespecificquestions.
EXAMPLE 26.4 Whichcolorattractsbeetlesbest?
Whatcolorshouldweuseonstickyboardsplacedinafieldofoatstoattractcerealleaf
beetles? Example 24.4 (page 611) gives data from an experiment in which 24 boards
(6 each of blue, green, white, and yellow) were placed at random locations in a field.
ANOVAshowsthattherearesignificantdifferencesamongthemeannumbersofbeetles
trapped by these colors. We might follow ANOVA with Tukey pairwise comparisons
(Exercise26.3).
But in fact we have specific questions in mind: We suspect that warm colors are
generallymoreattractivethancoldcolors.Thatis,beforeanydataaregathered,wesuspect
thatblueandwhiteboardswillhavesimilarproperties,thatgreenandyellowboardswill
givesimilarresults,andalsothattheaveragebeetlecountforgreenandyellowwillbe HoltStudiosInternational/Alamy
greater than the average count for blue and white. Therefore, we want to test three
hypotheses:
Hypothesis1 Hypothesis2 Hypothesis3
H:μ =μ H:μ =μ H:(μ +μ )/2=(μ +μ )/2
0 B W 0 G Y 0 Y G B W
H:μ =(cid:3) μ H:μ =(cid:3) μ H:(μ +μ )/2>(μ +μ )/2
a B W a G Y a Y G B W
Twoofthesehypothesesinvolvepairwisecomparisons.Thethirddoesnotandalsohas
aone-sidedalternative.■
We can ask questions about population means by specifying contrasts among
themeans.
CONTRASTS
IntheANOVAsettingcomparingthemeansμ ,μ ,...,μ ofk
1 2 k
populations,apopulation contrastisacombinationofthemeans
L = c μ +c μ +···+c μ
1 1 2 2 k k
withnumericalcoefficientsthataddto0,c +c +···+c = 0.
1 2 k
*Thismaterialisoptional.
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26-10 CHAPTER26 ■ MoreaboutAnalysisofVariance:Follow-upTestsandTwo-WayANOVA
EXAMPLE 26.5 Attractingbeetles:contrasts
WecanrestatethethreehypothesesinExample26.4intermsofthreecontrasts:
L =(1)(μ )+(0)(μ )+(−1)(μ )+(0)(μ )
1 B G W Y
L =(0)(μ )+(1)(μ )+(0)(μ )+(−1)(μ )
2 B G W Y
L =(−1/2)(μ )+(1/2)(μ )+(−1/2)(μ )+(1/2)(μ )
3 B G W Y
Checkthatthefourcoefficientsineachlinedoaddto0.Intermsofthesecontrasts,the
hypothesesbecome
Hypothesis1 Hypothesis2 Hypothesis3
H: L =0 H: L =0 H: L =0
0 1 0 2 0 3
H: L =(cid:3) 0 H: L =(cid:3) 0 H: L >0
a 1 a 2 a 3 ■
Somestatisticalsoftwarewilltesthypothesesandgiveconfidenceintervalsfor
anycontrastsyouspecify.Becauseothersoftwarelacksthiscapability,here’show
toproceedbyhand,usinginformationfrombasicANOVAoutput.
Toestimateapopulationcontrast
L = c μ +c μ +···+c μ
1 1 2 2 k k
samplecontrast usethecorrespondingsample contrast
^L = c x +c x +···+c x
1 1 2 2 k k
^
Thesamplecontrast L hasstandarderror(estimatedstandarddeviation)
(cid:3)
(cid:4)
(cid:4)(cid:5)c2 c2 c2
SE = s 1 + 2 +···+ k
^L p n n n
1 2 k
INFERENCE ABOUT A POPULATION CONTRAST
IntheANOVAsetting,alevelC confidence intervalforapopulation
contrastis
^L ±t∗SE
^L
^ ∗
where L isthecorrespondingsamplecontrastandt isacriticalvaluefrom
thet distributionwiththedegreesoffreedomforerrorintheANOVA.
Totest H : L = 0,usethet statistic
0
^
L
t =
SE
^L
withthesamedegreesoffreedom.
For one-way ANOVA, the degrees of freedom for error are N−k, where N
isthetotalnumberofobservationsandk isthenumberofpopulationscompared