Fiddle Faddle« & Screaming Yellow Zonkers«, przepisy, Przepisy Pizza Hut KFC MC DONALDS, Przepisy po angielsku
Fatburger«, przepisy, Przepisy Pizza Hut KFC MC DONALDS, Przepisy po angielsku
Food 20 adjectives, ANGIELSKI, !!!!!!!!!!pomoce, słownictwo, food
Frommer's Naples and the Amalfi Coast day BY Day, Travel Guides- Przewodniki (thanx angielski i stuff)
Frommer s Sicily Day By Day, Travel Guides- Przewodniki (thanx angielski i stuff)
Fuyumi Ono - Twelve Kingdoms 01 - Shadow of the Moon a Sea of Shadows, Angielskie [EN](4)(2)
Farago&Zwijnenberg (eds) - Compelling Visuality ~ The work of art in and out of history, sztuka i nie tylko po angielsku
Forex Study Book For Successful Foreign Exchange Dealing, gielda walutowa, Angielskie
Fabulous Creatures Mythical Monsters and Animal Power Symbols-A Handbook, Angielski
Functioning in the Real World Precalculus - Chapter 06, Angielskie [EN](4)(2)
Fundamentals of Statistics - Chapter12, Angielskie [EN](4)(2)
[ Pobierz całość w formacie PDF ]12
Additional
InferentialProcedures
CHAPTER
Outline
12.1
Goodness-of-FitTest
12.2
TestsforIndependenceandtheHomogeneityof
Proportions
12.3
TestingtheSignificanceoftheLeast-Squares
RegressionModel
12.4
ConfidenceandPredictionIntervals
"
ChapterReview
"
CaseStudy:FeelingLucky?Well,AreYou?
(OnCD)
DECISIONS
Aretherebenefitstoattendingcollege?Ifso,whatarethey?SeetheDecisions
projectonpage573.
PuttingItAllTogether
InChapters9–11,weintroducedstatisticalmethodsthat
canbeusedtotesthypothesesregardingapopulation
parametersuchasor
p
.Thischaptercanbeconsidered
intwoparts.Thefirstpart,Sections12.1and12.2,intro-
ducesinferenceusingthechi-squaredistribution.
Often,ratherthanbeinginterestedintesting a
hypothesesregardingaparameterofaprobabilitydistri-
bution,weareinterestedintestinghypothesesregarding
theentireprobabilitydistribution.Forexample,wemight
wishtotestwhetherthedistributionofcolorsinabagof
plainM&Mcandiesis13%brown,14%yellow,13%red,
20%orange,24%blue,and16%green.Weintroduce
methodsfortestinghypothesessuchasthisinSection12.1.
InSection12.2,wediscussamethodthatcanbeused
todeterminewhethertwovariablesareindependentbased
onasample.WeconcludeSection12.2byintroducingtests
forhomogeneity.Thisprocedureisusedtocompare
proportionsfromtwoormorepopulations.Itisanexten-
sionofthetwo-sample
z
-testforproportionsdiscussedin
Section11.3.
Thesecondpartofthischapterextendsinferential
statisticstotheleast-squaresregressionline.Sections
12.3and12.4introduceinferentialmethodsthatcanbe
usedontheleast-squaresregressionline.
InChapter4,wepresentedmethodsfordescribing
therelationbetweentwovariables-bivariatedata.
InSection12.3,weusethemethodsofhypothesis
testingfirstpresentedinChapter10totestwhethera
linearrelationexistsbetweentwoquantitativevariables.
InSection12.4,weconstructconfidenceintervals
aboutthepredictedvalueoftheleast-squaresregression
line.
550
Section12.1Goodness-of-FitTest
551
12.1
Goodness-of-FitTest
PreparingforThisSection
Beforegettingstarted,reviewthefollowing:
•
Expectedvalue(Section6.1,pp.291–292)
•
Meanofabinomialrandomvariable(Section6.2,
pp.305–306)
•
Mutuallyexclusive(Section5.2,pp.238–241)
Objective
Performagoodness-of-fittest
PerformaGoodness-of-FitTest
Inthissection,wepresentaprocedurethatcanbeusedtotesthypothesesre-
gardingaprobabilitydistribution.Forexample,wemightwanttotestwhether
thedistributionofplainM&Mcandiesinabagis13%brown,14%yellow,13%
red,20%orange,24%blue,and16%green.Orwemightwanttotestifthe
numberofhitsaplayergetsinhisnextfourat-batsfollowsabinomialdistribu-
tionwith and
Weusethesymbol chi-square,pronounced“kigh-square”(torhyme
with“sky-square”),torepresentvaluesofthechi-squaredistribution.Wecan
findcriticalvaluesofthechi-squaredistributioninTableVIinAppendixAof
thetext.BeforediscussinghowtoreadTableVI,weintroducecharacteristicsof
thechi-squaredistribution.
=
0.298.
Figure1
CharacteristicsoftheChi-SquareDistribution
1.
Itisnotsymmetric.
2.
Theshapeofthechi-squaredistributiondependsonthedegreesof
freedom,justlikeStudent’s
t
-distribution.
3.
Asthenumberofdegreesoffreedomincreases,thechi-squaredistribu-
tionbecomesmoresymmetric,asillustratedinFigure1.
4.
Thevaluesof arenonnegative.Thatis,thevaluesof aregreater
thanorequalto0.
with2degreesoffreedom
with5degreesoffreedom
with10degreesoffreedom
0.2
with15degreesoffreedom
with30degreesoffreedom
0.1
0
1 0
2 0
3 0
4 0
5 0
TableVIisstructuredsimilarlytoTableVforthe
t
-distribution.Theleftcol-
umnrepresentsthedegreesoffreedom,andthetoprowrepresentsthearea
underthechi-squaredistributiontotherightofthecriticalvalue.Weusetheno-
tation todenotethecritical suchthattheareaunderthechi-square
distributiontotherightof is
-value
EXAMPLE1
FindingCriticalValuesfortheChi-SquareDistribution
Problem
:
Findthecriticalvaluesthatseparatethemiddle90%ofthechi-square
distributionfromthe5%areaineachtail,assuming15degreesoffreedom.
Approach
:
Weperformthefollowingstepstoobtainthecriticalvalues.
Step1
Figure2
Area
$
0.05
Area
$
0.05
:
Drawachi-squaredistributionwiththecriticalvaluesandareaslabeled.
Step2
:
UseTableVItofindthecriticalvalues.
Area
$
0.90
Solution
Step1
:
Figure2showsthechi-squaredistributionwith15degreesoffreedom
andtheunknowncriticalvalueslabeled.Theareatotherightof is0.05.The
areatotherightof
Step2
:
Figure3showsapartialrepresentationofTableVI.Therowcontaining
15degreesoffreedomisboxed.Thecolumnscorrespondingtoanareatothe
x
0.95
x
0.05
x
2
2
x
0.05
2
0.95
is0.95.
552
Chapter12AdditionalInferentialProcedures
rightof0.95and0.05arealsoboxed.Thecriticalvaluesare and
x
0.95
=
7.261
x
0.05
=
24.996.
Figure3
AreatotheRightoftheCriticalValue
Degreesof
Freedom
0.9950.990.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005
1
— — 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879
2
0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597
3
0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838
12
3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.299
13
3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.819
14
4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.319
15
4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.801
16
5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.267
17
5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.718
18
6.365 7.215 8.231 9.288 10.265 25.200 28.601 31.595 24.265 36.456
InstudyingTableVI,wenoticethatthedegreesoffreedomarenumbered1
to30inclusive,then Ifthenumberofdegreesoffreedomis
notfoundinthetable,wefollowthepracticeofchoosingthedegreesoffree-
domclosesttothatdesired.Ifthedegreesoffreedomareexactlybetweentwo
values,findthemeanofthevalues.Forexample,tofindthecriticalvaluecorre-
spondingto75degreesoffreedom,computethemeanofthecriticalvaluescor-
respondingto70and80degreesoffreedom.
40,50,60,
Á
,100.
Nowthatweunderstandthechi-squaredistribution,wecandiscussthe
goodness-
of-fit
test.
Definition
A
goodness-of-fittest
isaninferentialprocedureusedtodetermine
whetherafrequencydistributionfollowsaclaimeddistribution.
Asanexample,wemightwanttotestwhetheradieisfair.Thiswouldmean
theprobabilityofeachoutcomeiswhenadieiscast.Weexpressthisas
H
0
:
p
Here’sanotherexample:AccordingtotheU.S.CensusBureau,in2000,
19.0%ofthepopulationoftheUnitedStatesresidedintheNortheast;22.9%
residedintheMidwest,35.6%residedintheSouth,and22.5%residedinthe
West.WemightwanttotestwhetherthedistributionofU.S.residentsisthe
sametodayasitwasin2000.Remember,thenullhypothesisisastatementof
“nochange,”soforthistest,thenullhypothesisis
ThedistributionofresidentsintheUnitedStatesisthesametodayasit
wasin2000.
0
:
Theideabehindtestingthesetypesofhypothesesistocomparetheactual
numberofobservationsforeachcategoryofdatawiththenumberofobserva-
tionswewouldexpectifthenullhypothesisweretrue.Ifasignificantdifference
betweentheobservedcountsandexpectedcountsexists,wehaveevidence
againstthenullhypothesis.
Themethodforobtainingtheexpectedcountsisanextensionoftheex-
pectedvalueofabinomialrandomvariable.Recallthatthemean(andthere-
foreexpectedvalue)ofabinomialrandomvariablewith
n
independenttrials
andprobabilityofsuccess,
p
,isgivenby
E
= np
.
Section12.1Goodness-of-FitTest
553
ExpectedCounts
Supposethereare
n
independenttrialsofanexperimentwith mutu-
allyexclusivepossibleoutcomes.Let representtheprobabilityofobserv-
ingthefirstoutcomeand representtheexpectedcountofthefirst
outcome,representtheprobabilityofobservingthesecondoutcomeand
representtheexpectedcountofthesecondoutcome,andsoon.Theex-
pectedcountsforeachpossibleoutcomearegivenby
InOtherWords
Theexpectedcountforeachcategory
isthenumberoftrialsoftheexperiment
timestheprobabilityofsuccessinthe
category.
= np
i
for
i
=
1,2,
Á
,
k
EXAMPLE2
FindingExpectedCounts
Problem
:
Anurbaneconomistwishestodeterminewhetherthedistribution
ofresidentsintheUnitedStatesisthesametodayasitwasin2000.Thatyear,
19.0%ofthepopulationoftheUnitedStatesresidedintheNortheast,22.9%
residedintheMidwest,35.6%residedintheSouth,and22.5%residedinthe
West(basedondataobtainedfromtheCensusBureau).Iftheeconomist
randomlyselects1500householdsintheUnitedStates,computetheexpected
numberofhouseholdsineachregion,assumingthatthedistributionofhouse-
holdsdidnotchangefrom2000.
Approach
Step1
:
Determinetheprobabilitiesforeachoutcome.
Step2
:
Thereare trials(the1500householdssurveyed)oftheexper-
iment.Weexpect ofthehouseholdssurveyedtoresideintheNorth-
east, ofthehouseholdstoresideintheMidwest,andsoon.
=
1500
np
northeast
np
midwest
HistoricalNote
Solution
Step1
:
Theprobabilitiesaretherelativefrequenciesfromthe2000distribution:
and
Step2
:
TheexpectedcountsforeachlocationwithintheUnitedStatesareas
follows:
Thegoodness-of-fittestwasinvented
byKarlPearson(thePearsonof
correlationcoefficientfame).
Pearsonbelievedthatstatistics
shouldbedonebydeterminingthe
distributionofarandomvariable.
Suchadeterminationcouldbemade
onlybylookingatlargenumbersof
data.ThisphilosophycausedPearson
to“buttheads”withRonaldFisher,
becauseFisherbelievedinanalyzing
smallsamples.
p
northeast
=
0.190,
p
midwest
=
0.229,
p
south
=
0.356,
p
west
=
0.225.
ExpectedcountofNortheast:
np
northeast
=
1500
1
0.190
2
=
285
ExpectedcountofMidwest:
np
midwest
=
1500
1
0.229
2
=
343.5
ExpectedcountofSouth:
np
south
=
1500
1
0.356
2
=
534
ExpectedcountofWest:
np
west
=
1500
1
0.225
2
=
337.5
Ofthe1500householdssurveyed,theeconomistexpects285householdsinthe
Northeast,343.5householdsintheMidwest,534householdsintheSouth,and
337.5householdsintheWestifthedistributionofresidentsoftheUnitedStates
isthesametodayasitwasin2000.
NowWorkProblem5.
Totestahypothesis,wecomparetheobservedcountswiththeexpected
counts.Iftheobservedcountsaresignificantlydifferentfromtheexpectedcounts,
wehaveevidenceagainstthenullhypothesis.Toperformthistest,weneedatest
statisticandsamplingdistribution.
554
Chapter12AdditionalInferentialProcedures
TestStatisticforGoodness-of-FitTests
Let representtheobservedcountsofcategory representtheexpect-
edcountsofcategory
i,k
representthenumberofcategories,and
n
repre-
sentthenumberofindependenttrialsofanexperiment.Thentheformula
CAUTION
Goodness-of-fittestsareused
totesthypothesesregardingthe
distributionofavariablebasedon
asinglepopulation.Ifyouwishto
comparetwoormorepopulations,you
mustusethetestsforhomogeneity
presentedinSection12.2.
i
,
E
i =
1,2,
Á
,
k
approximatelyfollowsthechi-squaredistributionwith degreesof
freedom,providedthat
1.
allexpectedfrequenciesaregreaterthanorequalto1(all )and
2.
nomorethan20%oftheexpectedfrequenciesarelessthan5.
Note:
for
= np
i
=
1,2,
Á
,
k
FromExample2,therewere categories(Northeast,Midwest,South,and
West).FortheNortheast,theexpectedfrequency,
E
,is295.
Nowthatweknowthedistributionofgoodness-of-fittests,wecanpresenta
methodfortestinghypothesesregardingthedistributionofarandomvariable.
TheGoodness-of-FitTest
Totesthypothesesregardingadistribution,wecanusethestepsthatfollow.
Step1
:
Determinethenullandalternativehypotheses:
Therandomvariablefollowsacertaindistribution.
Therandomvariabledoesnotfollowacertaindistribution.
Step2
:
Decideonalevelofsignificance,dependingontheseriousnessof
makingaTypeIerror.
Step3
:
(a)
Calculatetheexpectedcountsforeachofthe
k
categories.Theexpect-
edcountsare for where
n
isthenumberoftri-
alsand istheprobabilityofthe
i
thcategory,assumingthatthenull
hypothesisistrue.
(b)
Verifythattherequirementsforthegoodness-of-fittestaresatisfied.
1.
Allexpectedcountsaregreaterthanorequalto1(all ).
2.
Nomorethan20%oftheexpectedcountsarelessthan5.
(c)
Computethe
teststatistic:
= np
i
=
1,2,
Á
,
k
CAUTION
IfrequirementsinStep3(b)are
notsatisfied,oneoptionisto
combinetwoofthelow-frequency
categoriesintoasinglecategory.
Note:
istheobservedcountforthe
i
thcategory.
ClassicalApproach
P-ValueApproach
Step4
:
UseTableVItoobtainanapproximate
P
-valuebydeterminingtheareaunderthechi-square
distributionwith degreesoffreedomtothe
rightoftheteststatistic.SeeFigure5.
Step4
:
Determinethecriticalvalue.Allgoodness-of-fit
testsareright-tailedtests,sothecriticalvalueis with
degreesoffreedom.SeeFigure4.
(
continuedonnextpage
)
[ Pobierz całość w formacie PDF ]
