9 Chi squarenew2.docx
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9 Chi squarenew2.docx
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9Chisquarenew2
TheChi-SquareDistribution,
:
ChiSquarevariablecannotbenegative
ChiSquaredistributionsarepositivelyskewed
1.Findthecriticalchi-squaredvaluefor15degreesoffreedomwhen
andthetestisonetailedright.
2.Findthecriticalchi-squaredvaluefor10degreesoffreedomwhen
andthetestisonetailedleft
3.Findthecriticalchi-squaredvaluesfor22degreesoffreedomwhen
andatwo-tailedtestisconducted.
AssumptionsfortheUseofChi-SquareTest
1.Thesamplemustberandomlyselectedfromthepopulation.
2.Thepopulationmustbenormallydistributedforthevariableunderstudy.
3.Theobservationsmustbeindependentofeachother.
Question:
1.AninstructorwishestoseewhetherthevariationinIQofthe23studentsinherclassislessthanthevarianceofthepopulation.Thevarianceoftheclassis198.Testtheclaimthatthevariationoftheclassislessthanthepopulationvariance(
)at
.
2.Amedicalresearcherbelievesthatthestandarddeviationofthetemperaturesofnewborninfantsisgreaterthan
.Asampleof15infantswasfoundtohaveastandarddeviationof
.At
doestheevidencesupporttheresearcher’sbelief?
3.Acigarettemanufacturerwishestotesttheclaimthatthevarianceofthenicotinecontentofthecigaretteshiscompanymanufacturesisequalto0.644milligram.Asampleof20cigaretteshasastandarddeviationof1gram.At
testthemanufacturer’sclaim.
FormulafortheConfidenceIntervalofaVariance&StandardDeviation
Theformulafortheconfidenceintervalofavarianceif
Theformulafortheconfidenceintervalofastandarddeviationis
Thedegreesoffreedomforbothformulasidd.f.=n-1
4.Findthe95%confidenceintervalofthevarianceandstandarddeviationforthevarianceinthepreviousexample.
Acigarettemanufacturerwishestotesttheclaimthatthevarianceofthenicotinecontentofthecigaretteshiscompanymanufacturesisequalto0.644milligram.Asampleof20cigaretteshasastandarddeviationof1gram.At
testthemanufacturer’sclaim.
Onecanbe95%confidentthatthetruevarianceforthenicotinecontentisbetween0.578and2.133.
Andstandarddeviationis:
Wecanbe95%confidentthatthetruestandarddeviationisbetween0.760and1.46.
FormulafortheChi-SquareGoodness-of-FitTest
Withdegreesoffreedomequaltothenumberofcategoriesminus1,andwhere
O=observedfrequency
E=expectedfrequency
TheChi-SquareGoodness-of-Fittest
Step1:
Statethehypotheses.
Step2:
Findthecriticalvalue.
Step3:
Computethetestvalue,followingthesesteps.
a.Makeatable,asshown,andplacetheobservedvaluesinthefirstcolumn(O)
b.Placetheexpectedvaluesinthesecondcolumn(E).
c.Subtracttheexpectedvaluesfromtheobservedvalues,andplacetheanswerinthethirdcolumn(O-E).
d.Squaretheanswersinthethirdcolumn,andplacethesquaresinthefourthcolumn
.
e.Divideeachvalueinthefourthcolumnbyitscorrespondingvalueinthesecondcolumn
.
f.Findthesumofthenumbersinthelastcolumntogetthetestvalue.
Step4:
Makeadecision.
Step5:
Summarizetheresults.
5.Supposeamarketanalystwishestoseewhetherconsumershaveanypreferenceamongfiveflavorsofanewsoda.Asampleof100peopleprovidedthefollowingdata:
CherryStrawberryOrangelimeGrape
3228161410
Iftherewasnopreference,onewouldexpectthateachflavorwouldbeselectedwithequalfrequencywhichis100/5=20.
Thefrequenciesfromsamplearecalledobservedfrequencies.
Thefrequenciesthoughtarecalledexpectedfrequencies.
FrequencyCherryStrawberryOrangeLimeGrape
Observed3228161410
Expected2020202020
1.
Consumersshownopreferencefortheflavor.
Consumersshowapreferencefortheflavor.
2.C.V.
c.v.=9.488
3.TestValue:
18.0
Decision:
18>9.488;Rejectthenullhypothesis.
Conclusion:
Thereissufficientevidencetosupportthatsomeflavorsaremorepopularthanothers.
6.AhighschoolprincipalbelievesthatwhentheSATtestsaregiven,thegroupconsistsof10%freshmen,20%sophomores,40%juniors,and30%seniors.Thegroupthatjusttookthetestconsistedof12freshmen,18sophomores,45juniors,and25seniors.At
testtheprincipal’sconjecture.
7.
Thegroupconsistsof10%freshmen,20%sophomores,40%juniors,and30%seniors.
Thedistributionisnotthesameasstatedinthenullhypothesis.
8.C.V.
c.v.=6.251
9.TestValue:
2.058
Decision:
2.058<6.251;notinC.R.,Donotrejectthenullhypothesis.
Conclusion:
Theprincipal’sclaimisprobablytrue.
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