教程说明分析markscheme units46686 june

June6686StatisticsS4MarkSchemeQuestion1E(1)=E(X3)E(X4)E(X53=3= ias allowE(2)=E(X10)-E(X13=1/3(= Bias=- allowE(3)=3E(X1)2E(X2)E(X106=326= ias allowVar(1)=1{(VarX2)+Var(X3)+9=1{9 13Var(2)=29Var(3)=1{9 71isthebetterestimator.Ithasalowervar.andno2istheworstestimator.Itisb2H1: S2= s2= s2=s2F(8,6)=1.04<4.15donotrejectH0.ThevariancesaretheAssumethesamplesareselectedatrandom,s2p=8(22.5)6(21.62)= awrtH0: t=40.66722.121 = 0.42–Criticalvalue=t14(2.5%)=0.462<2.145NoevidencetorejectH0.ThemeansaretheM1M1bcM1dMusichasnoeffecton3Differences2.1-0.72.6-1.73.31.61.71.21.6d=H0:d=0H1:d40.6510s =9t = awrt2.941.519... 10 t9(1%)=2.935..>2.821EvidencetorejectH0.Therehasbeenanincreaseinthemeanweightofthemice.M12sampletestcanscoreM0M0B1forH0:A=BH1:A<M1924.59M0B1B1ftiex=668.125s=T7(5%)=Confidencelimits=668.1251.8958=611.6andConfidenceinterval=(612,725)Normaldistribution£650iswithintheconfidenceinterval.NoneedtoM1bcB1√B15Confidenceinterval=150.003,150.003 =(0.00164,0.0720.0049iswithinthe95%confidenceThereisnoevidencetorejecttheideathatthestandarddeviationofthevolumesisnot0.07orThemachineisworkingwell.A1b6H0p HpB1bLetX=NumbercuredthenXB(20,Pypeerror)=P(x3)+P(x11)givenp==0.0444+=c=P(TypeIIerror)=P(4x dPower=1– eNotagoodBetterfurtherawayfrom0.35Thisisnotaverypowerfultest(power=1-7H0:=230H1:<Fromtablecriticalvaluex=228.3S=txsn=228.3230=0.301>1.833.NoevidencetorejectH0.Meanis230N/mm2SincethetensilestrengthisthesameandthepriceischeapermendusenewB1b20086EdexcelGCE数学2008年6月6686统计S41.aE(X)+E(X)+E(X)E(θ1)=34533μ=μΒias=0允许无偏E(X)-E(X)101E(θ2)=3=1/3(μ.μ)0μμ3E(X2E(X+E(X)1210E(θ3)63μ2μμ+6μΒias0B1bVar(θ1)=1{(VarX2)+Var(X3)+Var(X4)}9=1{σ2+σ2+σ2}9σ23σ2Var(θ2)=91Var(θ3)={9σ2+4σ2+σ2}367=σ218ci)θ1B1θ2是估计量。偏向B1a222H1:σA=σBHT0:σA2≠σB2SA2=22.5sB=21.6awrtM12s1=1.042s2M1A1F(8,6)=4.15B11.04<4.15不H0。方差是相同的。b，（独立）B1c28(22.5)+6(21.62)sp==22.12awrt22.114M1A1H0:μA=μBHT1:μA≠μBB140.667。39.57t=1122.120.4620.42–0.47临界值=t14(2.5%)2.1450.462<2.145H0B1dB1差异2.1-0.72.6-1.73.31.61.71.21.62.4.d=1.41H0:μd=0H1:μd240.6510.×1.41s==1.5191…91.41t==2.935…awrt2.94/2.93. ..10.t9(1%)=2.8212.935..>2.821H0的。小鼠的平均体重有所增加。M1M1B1M1M1A1B1B1英尺(8)2M0M0B1H0:μA=μBH1:μA<M0A0B12.552B1英尺即4a.x=668.125s=84.428M1M1T7(5%)=1.895B11.895×84.428置信限=668.125±8611.6和724.7置信区间=(612,bB1c650英镑在置信区间内。不用担心。B1√B16aH0p=0.35H1:p≠0.35B1b令X=治愈数则X～B(20,0.35)αP(ΤiiP(x3)+P(x11)p=0.350.04440.0532M1cβ=P(II类误差)=P(4≤x≤10)p0.20.30.40.5β0.58800.87580.85650.5868d1ΒM10.41200.1435A1e这不是一个好的程序。0.35B1dep（1β）(2)5ab.15×0.00315×0.003.置信区间=,...27.4886.262。=(0.00164,0.072=0.00490.0049在95%置信区间内。没有表明体积的标准偏差不是0.07或机器运行良好。M1B1B1A1A1(5)M1A