9 Chi squarenew2Word文档下载推荐.docx

9 Chi squarenew2Word文档下载推荐.docx

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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

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

2.058<

6.251;

notinC.R.,Donotrejectthenullhypothesis.

Theprincipal’sclaimisprobablytrue.