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1、Chapter12:ConsiderationsinDesigningExperiments12.1(SeeExample12.1)Letnσ131=()n=()90=33.75or34andn2=90−34=56.σ1+σ23+512.2(SeeEx.12.1).Ifn1=34andn2=56,then925σ=+=.7111Y1−Y23456Inordertoachievethissameboundwithequalsamplesizes,wemusthave925+=.7111nnThesolutionisn=47.8or48.Thus,itisnecessarytoh
2、aven1=n2=48sothatthesameamountofinformationisimplied.12.3Thelengthofa95%CIistwicethemarginoferror:9252(1.96)+,n1n2andthisisrequiredtobeequaltotwo.InEx.12.1,wefoundn1=(3/8)nandn1=(5/8)n,sosubstitutingthesevaluesintotheaboveandequatingittotwo,thesolutionisfoundtoben=245.9.Thus,n1=93andn2=154.
3、12.4(SimilartoEx.12.3)Here,theequationtosolveis9252(1.96)+=2.n1n1Thesolutionisn1=130.6or131,andthetotalsamplesizerequiredis131+131=262.12.5RefertoSection12.2.Thevarianceoftheslopeestimateisminimized(maximuminformation)whenSxxisaslargeaspossible.Thisoccurswhenthedataareasfarawayfromxaspossib
4、le.So,withn=6,threeratsshouldreceivex=2unitsandthreeratsshouldreceivex=5units.12.6Whenσisknown,a95%CIforβisgivenbyβˆ±zσ.1α/2SxxUnderthetwomethods,wecalculatethatSxx=13.5forMethod1andSxx=6.3forMethod2.Thus,Method2willproducethelongerinterval.Bycomputingtheratioof13.5themarginsoferrorforthetw
5、omethods(Method2toMethod1),weobtain=6.31.464;thusMethod2producesanintervalthatis1.464timesaslargeasMethod1.UnderMethod2,supposewetakenmeasurementsateachofthesixdoselevels.ItisnotdifficulttoshowthatnowSxx=6.3n.So,inorderfortheintervalstobeequivalent,wemusthavethat6.3n=13.5,andson=2.14.So,rou
6、ghlytwiceasmanyobservationsarerequired.12.7AlthoughitwasassumedthattheresponsevariableYistrulylinearovertherangeofx,theexperimenterhasnowaytoverifythisusingMethod2.Byassigningafewpointsatx=3.5,theexperimentercouldcheckforcurvatureintheresponsefunction.257258Chapter12:ConsiderationsinDesigni
7、ngExperimentsInstructor’sSolutionsManual12.8Checkingfortruelinearityandconstanterrorvariancecannotbeperformedifthedatapointsarespreadoutasfaraspossible.12.9a.Eachhalfoftheironoresampleshouldbereasonablysimilar,andassumingthetwomethodsaresimilar,thedatapa
