=totalmarks
Marks
1a
Ho:p_Ottawa=
1
H1:p_Ottawa<>
824 TAB]
SolutiontoFXdraft
60 =n
465 =k
IgnorepossiblyfiniteN
p_hat=
0.5643203883
1
Teststat=
(d8e3)/sqrt(e3*(1e3)/c5)=
Thisisclearlyimprobablerejectthenullhypothesis
Theproportioncontinuinghaschanged(increased).
1
1
0.51
0.51
1b
H0:p_Ottawap_Hearst=
H1:p_Ottawap_Hearst>
3.12 tailarea=
0.00091
0.02 )/SE=
2.20292
0.02
0.02
Ignorepossiblyfinitepopulationsizes
Hearst
Ottawa
n
460
824
k
221
465
p_hat
0.48043 0.56432
1
SE_diff=
sqrt((d32*(1d32)/d30)+(e32*(1e32)/e30)
=
0.0290004031
Teststat=
(p_hat_Ottawap_hat_Hearst
0.0138001481
At1%level,cannotreject,sosupportthetrustee'sclaim.
OR:Below5%sowerejectnullandDONOTsupport.
1
1
1c
1965
1999
average
73.4
81.2
sd
8.1
6.6
n
473
383
856
30967.92
16639.92
47607.84 55.746885246
ThisisactuallyacomparisonofMEANS
1
H0:mu_99mu_65=
H1:mu_99mu_65>
6
6
SincetheSdsroughlysimilar,wewillusethepooledvariancetest.
var_pooled= ((e42*e42*(f421)+e43*e43*(f431))/(f42+f432)
55.746885246
s_pooled=
7.4663836793
1
(D34D33E39)/(D45*SQRT(1/F33+1/F34))
3.507157556
RejectnullH>gradeinflationofmorethan6%canbeclaimed
1
Teststat=
1
11
==================================================================
2a
Teststat:Chisquarebasedonmodelofproportions.
Mark
A
B
C
Total
freq
33
15
10
58
4
2
1
7
propn
0.5714285714 0.2857143 0.14285714
Expected
33.142857143 16.571429 8.28571429
1
OE
0.142857143 1.571429 1.71428571
(OE)^2/E
0.0006157635 0.1490148 0.3546798 0.50431034
1
1
1
Chisquaretailarea=
Thereforecannotrejectprof'sclaim.
HO:marksfithypothesizeddistribution
H1:donotfit
ConcludeProf'sclaimisreasonable.
1
0.77712414 orcriticalvalue=
p(A)=.57,p(B)=.286,p(C)=.143
2b
H0:Thereisnoessentialdifferencebetweentheprofs'marksdistributions
(MarksandProfareindependent)
H1:Thereisadifference(MarksandProfareassociated)
Usechisquare.Computeexpectedsbasedonindependenceofmarkandprof.
ChiSquare
Test:
novice,
midlife,
oldfogey
Expected
counts
are
printed
below
observed counts
ChiSquare
contributions are
printed
below
expected counts
1
novice
F
DB
2
A
1
midlife
11
15
1.067
oldfogey Total
15
19
15
15
0
1.067
45
2
30
30
0
30
30
0
30
30
0
3
19
15
1.067
15
15
0
11
15
1.067
45
60
60
60
180
Total
90
ChiSq
=
4.267,
DF
=
4,
PValue
=
Donotrejecthypothesisthatdistributionsessentiallythesame.
ConcludethatMarksandProfareindependentorthatthereisnodiffbetweenthedistributions.
10
==================================================================
3a
mean=
188.96 sd=
21.2
P(my<U)=.95where
2
U<
188.96 +
t(13,.05)*
21.2 /sqrt(14)
U<
188.96 +
1.771 *
5.6659383
1
U<
198.99437678
1
1
1
3b
0.371
ThestemandleafdiagramdoesNOTshowsymmetryormoundshape,
thoughthesampleisquitesmall.
Ifwecannotassumenearnormalityinthepopulation,thentheWilcoxon
testmaybeappropriateifthepopulationissymmetric.Thisisnotapparent
fromthestemandleaf(butdoesshowupinthefulldata,interestingly)
IfwetrusttheWilcoxon,thenmedianisquitelikelybelow$200,000.
1
3c
P(L<=mu_billings<=U)=0.95,where
mean=
200.94 sd=
24.24
{L,U}=
200.94 +/
24.24*1.96/sqrt(325)
200.94 +/
2.6354
{
198.30459718 ,
203.5754 }
1
1
3d
Noassumptionssincedistributionissymmetricandsamplesizelarge,
soCentralLimitTheoremapplies.
Sincetheintervalcontainsthehypothesizedvalueof200,
itisconsistentwithBattleaxe'sclaim.
1
1
3e
Fromthefulldata,MsBattleaxewasright.
Problemissamplingfromadistributionthathaslongtails,where
wearegettingsomeoftheextremeelements.
sampling
longtailsorequivalent
1
1
3f
dropped
1
2
3
4
5
6
7
8
25
27
24
32
33
29
27
31
34
31
23
32
38
40
33
33
9
4
1
0
5
11
6
2
X_bar=
4.5
sd_diff
4.1748
Usingthedifference,whichhasasymmetricboxplot,wecandoa
hypothesistest
H0:difference(mu_Jmu_S)=0
Ha:difference>0
Teststat=T=(X_bar_diff0)/(SD_diff/sqrt(n))
=
3.0487836865
Comparetot(7,0.05)=
1.895
ThusrejectH0infavourofHa,soJansenbillsmore.
3g
dropped
Schmidtissuggestingwhatisessentiallyasigntest.Thatisheis
sayingthatifheandJansenhaveequalsales,thereisa.5chance
onewillhavehighersalesinagivenweek.Wethenneedtoknow
thechanceSchmidtwillexceedJansen1timein8atrandom.
P_binomial(K<=1,n=8,p=0.5)=
0.03515625
Clearly,atthe5%level,weREJECTthenullhypotheisthattheyhave
equalchanceofhigherbillings,andconcludeJansensellsmore.
11
==================================================================
4a
Assumptionofnormalerrorsmaybeviolated(histogramandNPPshowasymmetry,deviation
fromsightlineinNPP)
1
Equalvarianceassumptionmaybeviolatednarrowinginmiddleofresidsvs.fit.
1
1
1
1
1
1
1
1
1
1
1
4b
R_squared=1(resid.SS/totalSS)=1
=
4c
4d
Model3hasprecipitationNOTsignificant,andthefitisreallynobetterthanModel2.
FitofModel2muchbetterthanmodel1,allparameterssignificant,andthegraphs,
whilenotperfect,doindicateapproximatesymmetryand"nottoobad"NPPaswell
asareasonablyconstantspreadofresidualsvsfit(equivarianceassumption)
4f
sales=2014.83temp23.6rollrim
=
201
4.83 *
23.6 *
=
70.59 $
1
4g
27
0
PI=pointest+/t(df,0.025)*sqrt(SE_mu^2+s^2)
s=
16.68
pointest=
t(57,.025)~=
2
SE_mu=
ME=
34.5683844
PI=(
132.1016156 ,
201.238384 )
1
1
1
1
219319
H0:Allregressioncoeffsarezero.Ha:Atleastonenonzero.
UseFtest(ANOVAtable)andfindthatitishighlyunlikelyallcoefficientsarezero.
Atleastoneoftheregressioncoefficientsisnonzero,somodelis"significant".
4e
1
1
20018 /
0.90872656
H0:beta1=0,HA:beta1<>0
Usingeitherthettestonthecoefficientoftemperature,ortheFtestonthe"entire"
model(whichonlyhasonepredictortemp),wehaveaverysmallpvalue,sothatthe
nullhypothesisisrejected.
concludethedailytemperatureisausefulpredictor.
166.67
4.53
4h
Predictionisfor$70insales(hotday!).However,themaximumvalueoftempinthedatasetis
only16.54Celsius.Weareoutsidethedomainofthepredictorvariablesandshouldnottrustthis
prediction.
19
==================================================================
5a
1
linesegmentsininteractionsarenonparallel,suggestsinteracdtion
1
1
5b
Ho:nointeraction,Ha:Interaction.Usealpha=0.05
Teststat=F,pvalueisverysmall,thereforerejectH0>thereISinteraction.
5c
Estimatethedifferencebetweentheaveragefuelconsumptionforflightswhichare10minutesearly
withatailwindandforflightswhichare10minuteslatewithaheadwind.
earlywtailw. latewhw
6335.72 9262.42
6703.13 9896.97
6564.22 9475.43
6742.98 9754.93
1
Difference
avg
6586.5125 9597.4375
3010.925
averagefortimecategoryandsubtractingthegrandmeantoavoiddoublecounting.
Or:grandmean+windeffect+timeeffect.Thereisnocommenthereonsignificanceneeded.
5d
B1:MSE
35723
SE=sqrt(MSE*(1/4+1/4))=
Tstat(0.025,27)=
CIupper=
3285.17
CIlower=
2736.6815142
1
0.5
0.5
1
5e
2.052
133.65
274.24
Notethatthecriticalvaluehasnotbeenadjustedformultiplecomparisons.
Giventheexamquestion,anadjustmentiscalledfor.
UsemodelB1interactionissignificantandthemodelexplainsthefuelconsumption
quitewell(seeR_squared)
ModelB2isaonewaymodel,andis"significant",butdoesnotfitaswell.
1
1
ModelB3isatwowayadditivemodel,butdoesn'tfitwell.R_squared_adjustedverypoor.
Windnotsignificantinthisadditivemodel.
9
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