=0.5-0.35
=a
a=1.04（fromZtables）
NBlowest15%thereforeSUBTRACTFROMTHEMEAN!
Thereforex=80-（1.04*4）
=80–4.16
=75.84ml
iii）Ifarandomsampleof100bottlesarechosenfortesting,whatistheprobabilitythattheaveragecontentsis75mls?
（NBTYPOSHOULDHAVEBEEN79）
X=contentsofthebottleoflotionX~N（80,42）x=79andn=100
Z=（75-80）/（4/√100）
=5/0.4
=12.5
p~0.0000almostimpossible!
Ifyouwroteprobabilityof1–nomarksdeducted.
C）Writeyouranswertothefollowingtwoquestions.Youmayuseadiagram&words（4Marks）
i）Whichmeasureofcentraltendencyismosteffectedbythedegreeofskewnessinafrequencydistribution?
Why?
Themeanbecauseitisthemosteffectedbyoutlyingorextremevaluesduetobeingcalculatedbysummingallvaluesanddividingbythenumberofvaluesieeveryvaluewhetheranoutlierornearthemiddlehasthesameweight.
ii）Whichmeasureofcentraltendencyisleasteffectedbythedegreeofskewnessinafrequencydistribution?
Why?
Themedianbecauseitisthemiddlescorewhenthedataarerankedinorderfromlowesttohighestvalue,iethe50thpercentileandisthereforeleasteffectedbyoutlyingvaluesorskews.
Question2:
A）Asoccerteamplayssocceronarosterof0,1or2daysaweek.Theyplay0days20%ofthetime,theyplay1day50%ofthetimeandtheyplay2days30%ofthetime.
（7marks）
i）Constructaprobabilitydistributionfunctiontableforthisseries.
X
P（x）
0
0.20
1
0.50
2
0.30
TOTAL
1
ii）Whatistheexpectedvalueofthedaysoftheweekthesoccerteamplays?
E（X）=0*P（X=0）+1*P（X=1）+2*P（X=2）
E（X）=0*0.2+1*0.5+2*0.3
=0+0.5+0.6
=1.1
iii）Whatistheexpectedvarianceofthisseries?
Var（X）=Σx2p（x）-μ2=Σ（x2-μ2）p（x）
E（V）=（0-1.1）2*（.2）+（1-1.1）2*.5+（2-1.1）2*.3
=0.242+0.005+0.243
=0.490
B）SusanisanexpertOlympicbasketballplayerandshehasimprovedforthepast10yearsandisnowasgoodasshewilleverbe.Hersuccessratehasbeenmeasuredat80%.Shetakes6shotsongoalintheBeijingOlympicfinalforherteam.（7marks）
i）WritedowntheprobabilityforeachvalueofX（thenumberofsuccesses）usingtheBernoullitables
FORMULAIS:
ButnoneedtocalculateascanuseBERNOULLITABLESprovidedforp=0.80andN=6
P（X=0）=0.000
P（X=1）=0.002
P（X=2）=0.015
P（X=3）=0.082
P（X=4）=0.246
P（X=5）=0.393
P（X=6）=0.262
ii）WhatistheprobabilityofSusanshootingatleast5goalsoutof6?
=P（5<=X<=6）
=P（5）+P（6）
=0.393+0.262
=0.655
Or65.5%
iii）Whatistheprobabilityofatmost2successesinhershootingagoal?
=P（0）+P
（1）+P
（2）
=0.000+0.002+0.015
=0.017
Or1.7%
C）Answerthefollowingquestionsinyourownwords–youmayusediagramsandwords.
（6marks）
i）DefineandexplainthedifferencebetweenaOneTailedTestandaTwoTailedTest?
Aonetailedtestistherejectionareaforahypothesisinvolvingonedirectionandnormallyincludestermssuchasgreaterthanorlessthanandistestedbycriticalvaluesofhalfthenormalcurve.
Therejectionregionstellswherewewillrejectanullhypothesis.
Atwotailedtestistherejectionregionforahypothesistestthatdoesnotspecifyadirectioneg“isnotequalto”.Atwotailedtestisastrictertestinthatthecriticalvalueishalvedforeachdirection,egifthecriticalvalueisthe.05%level,thetwotailedtestmustbeassessedfortheα=0.025criticalregion.
ii）DefineandexplainthedifferencebetweenaTypeIErrorandaTypeIIError
ATypeIErrorismadebyrejectinganullhypothesiswhenitistrue,itistheprobabilityofaphaαItoccurswhenyouarelenientintherejectionofthenullhypothesisegusep=.010.
ATypeIIErrorismadebynotrejectinganullhypothesiswhenitisfalse,itistheprobabilityofbetaβ.Itoccurswhenyouaretoostrictintherejectionofthenullhypothesisegusep=0.01.
REALITY
Decision
TRUE
FALSE
Donotreject
OK
Type2
Reject
Type1
OK
Wecannotreducetheprobabilityofonetypeoferrorwithoutincreasingtheprobabilityoftheother.Asthesignificanceleveldecreasesegapproaches0.01itismorelikelythatyouwillmakeaTypeIIErrorwhileaTypeIErrorismorelikelytooccurasporαlevelincreasesegto0.10.
NB:
nomarksdeductedforusingstrictandlenientin
termsofthealternativehypothesis
Question3:
A）Afinancialadvisingcompanywasconcernedaboutwhethertheirlevelsofcustomerservicewereconsistentacrossbranches.Asatisfactionratingwasobtainedfortwobranchesfromindependentsamples.Thedataareshownbelow.（10marks）
SAMPLE1
SAMPLE2
42
41
69
38
48
46
37
49
42
37
18
36
64
42
71
18
=391
=307
Performahypothesistestforthisproblem,listingoutall7stepsofhypothesistesting.
NBusingmeanandsdformula:
OR
1.Data:
Sample1:
N=8=48.88s=18.18s2=330.41
Sample2:
N=8=38.38s=9.36s2=87.70
2.H0:
Sample1mean=Sample2meanH1:
Sample1mean≠2mean
3.AsN<30,usethetdistributionandthetwosamplet-test
4.Chooseapvalueof0.05or0.01torejectnullhypothesis.With14degreesoffreedom（8+8-2）
Thecriticalvalueoftis2.145at0.05and2.648at0.01levelsattwotailedlevel
5.tstatistic:
D=X1–X2~（md,sd2）
t=1.45
1.Asthecalculatedvalueoftislessthanthecriticalvalue,wedonotrejectthenullhypothesis.
7.Concludethatthesatisfactionratingisnotsignificantlydifferentbetweenthetwobranches.
B）Aneconomistwantstochoosebetweentherateofreturnfortwodifferentproductsandhewantstotellhisbosswhetherthesearelowerthantheexpectedrateofreturnof$95.Hewantstobesureattheα=0.05.Stateanullandalternativehypothesistotestandselecttheappropriateteststatisticsandstatewhetherthisisaonetailedortwotailedtest.（5marks）
Rateofreturn
Average
$90.76
StandardDeviation
$19.32
N
200
H0:
μ=95
H1:
μ<95
AlthoughσisunknownasN>30wecanuseZdistribution（Nbhassameformulaast）
z=-4.24/1.37
z=-3.10
Onetailedhypothesisselectedastheproblemistoinvestigatewhethertherateofreturnis
lowerthanexpected
p=0.5-0.4990
=0.001
AsZ<-1.65ieoutsidethecriticalboundary,rejectnullhypothesisandconcludethattherateofreturnissignificantlylowerthanexpectedrateof$95
C）Beforeconductingmanystatisticaltestsitisimportanttotestwhethertheassumptionofhomogeneityofvariancesismet.Setupandtestahypothesisforthefollowingdata.
（5
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