SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
Chapter 2 M
Simple Comparative Experiments
M M
Solutions
2.1. TheMMinitabMoutputMforMaMrandomMsampleMofMdataMisMshownMbelow.M SomeMofMtheMquantitiesMare
Mmissing. MCompute Mthe Mvalues MofMthe Mmissing Mquantities.
Variable N Mean SEMMean Std.MDev Sum
.
Y 15 ? 0.159 ? 399.851
MeanM=M26.657 Std.MDev.M=M0.616
2.2. TheMMinitabMoutputMforMaMrandomMsampleMofMdataMisMshownMbelow.M SomeMofMtheMquantitiesMare
Mmissing. MCompute Mthe Mvalues MofMthe Mmissing Mquantities.
Variable N Mean SEMMea Std.MDev. Variance Minimum Maximum
n
Y 11 19.96 ? 3.12 ? 15.94 27.16
SEMMean M=M0.941 VarianceM=M9.73
2.3. ConsiderM theMMinitabM outputMshownM below.
One-SampleMZ
TestMofMmuM=M30MvsMnotM=M30
TheMassumedMstandardMdeviationM=M1.2
N Mean SEMMea 95%MCI Z P
n
16 31.2000 0.3000 (30.6120,M 31.7880 ? ?
)
(a) FillMinMtheMmissingMvaluesMinMtheMoutput.M WhatMconclusionMwouldMyouMdraw?
ZM=M4 PM=M0.00006
(b) IsMthisMaMone-sidedMorMtwo-sidedMtest?
Two-sided.
(c) UseMtheMoutputMandMtheMnormalMtableMtoMfindMaM99MpercentMCIMonMtheMmean.
CIM =M 30.42725,M31.97275
(d) WhatM isM theM P-valueM ifMtheM alternativeM hypothesisM isMH1:M µM >M 30?
P-valueM=M 0.00003
2-1
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.4. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM>Mµ0.MCalcualte MtheMP-
valueMforMtheMfollowing MobservedMvaluesMofMtheMtestMstatistic:
(a)MMMMMM Z0M=M2.45 P-
valueM=M 0.00714M(b)MMMMMZ0M=M-
1.53MMMMMP-
valueM=M0.93699M(c)MMMMMM Z0M=M2.15
P-
valueM=M 0.01578M(d)MMMMM Z0M=M1.95
P-
valueM=M 0.02559M(e)MZ0M=M-0.25M P-
valueM=M0.59871
2.5. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM≠Mµ0.MCalcualte MtheMP-
valueMforMtheMfollowing MobservedMvaluesMofMtheMtestMstatistic:
(a)MMMMMM Z0M=M2.25 P-
valueM=M 0.02445M(b)MMMMM Z0M=M1.55
P-
valueM=M 0.12114M(c)MMMMMM Z0M=M2.10
P-
valueM=M 0.03573M(d)MMMMM Z0M=M1.95
P-
valueM=M 0.05118M(e)MZ0M=M-0.10M P-
valueM=M0.92034
2.6. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM≠Mµ0MwithMaMsample MsizeMofMnM=M9.MC
alculate MboundsMonMtheMP-valueMforMtheMfollowingMobservedMvaluesMofMtheMtest Mstatistic:
(a) t0M=M2.48 TableMP-valueM=M0.02,M0.05 CumputerM P-valueM =M 0.03811
(b) t0M=M-3.95 TableMP-valueM=M0.002,M0.005 CumputerM P-valueM =M 0.00424
(c) t0M=M2.69 TableMP-valueM=M0.02,M0.05 CumputerM P-valueM =M 0.02750
(d) t0M=M1.88 TableMP-valueM=M0.05,M0.10 CumputerM P-valueM =M 0.09691
(e) t0M=M-1.25 TableMP-valueM=M0.20,M0.50 CumputerM P-valueM =M 0.24663
2-2
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.7. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM>Mµ0MwithMaMsample MsizeMofMnM=M13.M
Calculate MboundsMonMtheMP-valueMforMtheMfollowingMobservedMvaluesMofMthe MtestMstatistic:
(a) t0M=M2.35 TableMP-valueM=M0.01,M0.025 CumputerM P-valueM =M 0.01836
(b) t0M=M3.55 TableMP-valueM =M 0.001,M 0.0025 CumputerM P-valueM =M 0.00200
(c) t0M=M2.00 TableMP-valueM=M0.025,M0.005 CumputerM P-valueM =M 0.03433
(d) t0M=M1.55 TableMP-valueM=M0.05,M0.10 CumputerM P-valueM =M 0.07355
2-3
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.8. ConsiderM theMMinitabM outputMshownM below.
One-SampleMT:M Y
TestMofMmuM=M25MvsM>M25
Variable N Mean Std.MDev SEMMean 95%MLowerMBoun T P
. d
Y 12 25.6818 ? 0.3360 ? ? 0.034
(a) HowM manyMdegreesM ofM freedomM areM thereMonM theM t-testM statistic?
11
(b) FillMinMtheMmissingMinformation.
Std.MDev.M=M1.1639 95%MLowerMBoundM=M2.0292
2.9. ConsiderM theMMinitabM outputMshownM below.
Two-SampleMT-
TestMandMCI:M Y1,MY2
Two-sampleMTMforMY1MvsMY2
N Mean Std.MDev SEMMea
. n
Y1 20 50.19 1.71 0.38
Y2 20 52.52 2.48 0.55
DifferenceM=MmuM(X1)M–
Mmu M(X2) MEstimateMfor Mdifferen
ce:M -2.33341
95%MCIMforMdifference:M (-3.69547,M-0.97135)
T-TestMofMdiffenceM=M0M(vsMnotM=M)M:MT-
ValueM=M-3.47MP-ValueM=M0.01M DFM=M 38
BothMuseMPooledMStd.MDev.M=M2.1277
(a) CanMtheMnullMhypothesisMbeMrejectedMatMtheM0.05Mlevel?M Why?
Yes,M theMP-ValueMofM0.001M isMmuchM lessM thanM 0.05.
(b) IsMthisMaMone-sidedMorMtwo-sidedMtest?
Two-sided.
(c) IfMtheMhypothesis MhadMbeenMH0:Mµ1M-Mµ2M=M2MversusMH1:Mµ1M-
Mµ2M≠ M2 Mwould Myou Mreject MtheMnull Mhypothesis Mat Mthe M0.05 Mlevel?
Yes.
(d) IfMtheMhypothesisMhadMbeenMH0:Mµ1M-Mµ2M=M2MversusMH1:Mµ1M-
Mµ2M<M2 Mwould Myou Mreject MtheMnull Mhypothesis MatM theM 0.05M level?M CanM youM answerM thisM question
M withoutM doingM anyMadditionalM calculations?M Why?
Yes.M NoM additional M calculations M areM requiredM because M theM test M becomes M moreM significantM w
ithM the MchangeMfromM-2.33341MtoM-4.33341.
2-4
,M NY
Chapter 2 M
Simple Comparative Experiments
M M
Solutions
2.1. TheMMinitabMoutputMforMaMrandomMsampleMofMdataMisMshownMbelow.M SomeMofMtheMquantitiesMare
Mmissing. MCompute Mthe Mvalues MofMthe Mmissing Mquantities.
Variable N Mean SEMMean Std.MDev Sum
.
Y 15 ? 0.159 ? 399.851
MeanM=M26.657 Std.MDev.M=M0.616
2.2. TheMMinitabMoutputMforMaMrandomMsampleMofMdataMisMshownMbelow.M SomeMofMtheMquantitiesMare
Mmissing. MCompute Mthe Mvalues MofMthe Mmissing Mquantities.
Variable N Mean SEMMea Std.MDev. Variance Minimum Maximum
n
Y 11 19.96 ? 3.12 ? 15.94 27.16
SEMMean M=M0.941 VarianceM=M9.73
2.3. ConsiderM theMMinitabM outputMshownM below.
One-SampleMZ
TestMofMmuM=M30MvsMnotM=M30
TheMassumedMstandardMdeviationM=M1.2
N Mean SEMMea 95%MCI Z P
n
16 31.2000 0.3000 (30.6120,M 31.7880 ? ?
)
(a) FillMinMtheMmissingMvaluesMinMtheMoutput.M WhatMconclusionMwouldMyouMdraw?
ZM=M4 PM=M0.00006
(b) IsMthisMaMone-sidedMorMtwo-sidedMtest?
Two-sided.
(c) UseMtheMoutputMandMtheMnormalMtableMtoMfindMaM99MpercentMCIMonMtheMmean.
CIM =M 30.42725,M31.97275
(d) WhatM isM theM P-valueM ifMtheM alternativeM hypothesisM isMH1:M µM >M 30?
P-valueM=M 0.00003
2-1
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.4. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM>Mµ0.MCalcualte MtheMP-
valueMforMtheMfollowing MobservedMvaluesMofMtheMtestMstatistic:
(a)MMMMMM Z0M=M2.45 P-
valueM=M 0.00714M(b)MMMMMZ0M=M-
1.53MMMMMP-
valueM=M0.93699M(c)MMMMMM Z0M=M2.15
P-
valueM=M 0.01578M(d)MMMMM Z0M=M1.95
P-
valueM=M 0.02559M(e)MZ0M=M-0.25M P-
valueM=M0.59871
2.5. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM≠Mµ0.MCalcualte MtheMP-
valueMforMtheMfollowing MobservedMvaluesMofMtheMtestMstatistic:
(a)MMMMMM Z0M=M2.25 P-
valueM=M 0.02445M(b)MMMMM Z0M=M1.55
P-
valueM=M 0.12114M(c)MMMMMM Z0M=M2.10
P-
valueM=M 0.03573M(d)MMMMM Z0M=M1.95
P-
valueM=M 0.05118M(e)MZ0M=M-0.10M P-
valueM=M0.92034
2.6. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM≠Mµ0MwithMaMsample MsizeMofMnM=M9.MC
alculate MboundsMonMtheMP-valueMforMtheMfollowingMobservedMvaluesMofMtheMtest Mstatistic:
(a) t0M=M2.48 TableMP-valueM=M0.02,M0.05 CumputerM P-valueM =M 0.03811
(b) t0M=M-3.95 TableMP-valueM=M0.002,M0.005 CumputerM P-valueM =M 0.00424
(c) t0M=M2.69 TableMP-valueM=M0.02,M0.05 CumputerM P-valueM =M 0.02750
(d) t0M=M1.88 TableMP-valueM=M0.05,M0.10 CumputerM P-valueM =M 0.09691
(e) t0M=M-1.25 TableMP-valueM=M0.20,M0.50 CumputerM P-valueM =M 0.24663
2-2
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.7. SupposeMthatMweMareMtestingMH0:MµM=Mµ0MversusMH1:MµM>Mµ0MwithMaMsample MsizeMofMnM=M13.M
Calculate MboundsMonMtheMP-valueMforMtheMfollowingMobservedMvaluesMofMthe MtestMstatistic:
(a) t0M=M2.35 TableMP-valueM=M0.01,M0.025 CumputerM P-valueM =M 0.01836
(b) t0M=M3.55 TableMP-valueM =M 0.001,M 0.0025 CumputerM P-valueM =M 0.00200
(c) t0M=M2.00 TableMP-valueM=M0.025,M0.005 CumputerM P-valueM =M 0.03433
(d) t0M=M1.55 TableMP-valueM=M0.05,M0.10 CumputerM P-valueM =M 0.07355
2-3
, SolutionsM fromM Montgomery,M D.M C.M (2008)M DesignM andM AnalysisM ofM Experiments,M Wiley
,M NY
2.8. ConsiderM theMMinitabM outputMshownM below.
One-SampleMT:M Y
TestMofMmuM=M25MvsM>M25
Variable N Mean Std.MDev SEMMean 95%MLowerMBoun T P
. d
Y 12 25.6818 ? 0.3360 ? ? 0.034
(a) HowM manyMdegreesM ofM freedomM areM thereMonM theM t-testM statistic?
11
(b) FillMinMtheMmissingMinformation.
Std.MDev.M=M1.1639 95%MLowerMBoundM=M2.0292
2.9. ConsiderM theMMinitabM outputMshownM below.
Two-SampleMT-
TestMandMCI:M Y1,MY2
Two-sampleMTMforMY1MvsMY2
N Mean Std.MDev SEMMea
. n
Y1 20 50.19 1.71 0.38
Y2 20 52.52 2.48 0.55
DifferenceM=MmuM(X1)M–
Mmu M(X2) MEstimateMfor Mdifferen
ce:M -2.33341
95%MCIMforMdifference:M (-3.69547,M-0.97135)
T-TestMofMdiffenceM=M0M(vsMnotM=M)M:MT-
ValueM=M-3.47MP-ValueM=M0.01M DFM=M 38
BothMuseMPooledMStd.MDev.M=M2.1277
(a) CanMtheMnullMhypothesisMbeMrejectedMatMtheM0.05Mlevel?M Why?
Yes,M theMP-ValueMofM0.001M isMmuchM lessM thanM 0.05.
(b) IsMthisMaMone-sidedMorMtwo-sidedMtest?
Two-sided.
(c) IfMtheMhypothesis MhadMbeenMH0:Mµ1M-Mµ2M=M2MversusMH1:Mµ1M-
Mµ2M≠ M2 Mwould Myou Mreject MtheMnull Mhypothesis Mat Mthe M0.05 Mlevel?
Yes.
(d) IfMtheMhypothesisMhadMbeenMH0:Mµ1M-Mµ2M=M2MversusMH1:Mµ1M-
Mµ2M<M2 Mwould Myou Mreject MtheMnull Mhypothesis MatM theM 0.05M level?M CanM youM answerM thisM question
M withoutM doingM anyMadditionalM calculations?M Why?
Yes.M NoM additional M calculations M areM requiredM because M theM test M becomes M moreM significantM w
ithM the MchangeMfromM-2.33341MtoM-4.33341.
2-4
