试验设计与数据处理课后答案

《试验设计与数据处理》第三章：统计推断3-13解：取假设H0：u1-u2^0和假设H1：u1-u2>0用sas分析结果如下:SampleStatisticsGroupNMeanStd.Dev.Std.Errorx80.2318750.01460.0051y100.20970.00970.0031HypothesisTestNullhypothesis:Mean1-Mean2=0Alternative:Mean1-Mean2a=0IfVariancesAretstatisticDfPr>tEqual3.878160.0013NotEqual3.70411.670.0032由此可见p值远小于0.05,可认为拒绝原假设，即认为2个作家所写的小品文中由3个字母组成的词的比例均值差异显著。3-14解：用sas分析如下：HypothesisTestNullhypothesis: Variance1/Variance2=1Alternative: Variance1/Variance2A=1-DegreesofFreedom-F Numer.Denom. Pr>F2.27 7 9 0.2501由p值为0.2501>0.05（显著性水平），所以接受原假设，两方差无显著差异第四章：方差分析和协方差分析4-1解：Sas分析结果如下：DependentVariable:ySourceModelDF4SumofSourceModelDF4SumofSquaresMeanSquare1480.823000 370.205750FValuePr>F40.88 <.0001ErrorCorrectedTotal1519135.8225001616.6455009.054833R-Square CoeffVar RootMSE yMean0.91598513.12023 3.009125 22.93500SourceDFAnovaSSMeanSquareFValuePr>Fc41480.823000370.205750 40.88 <,0001由结果可知，p值小于0.001，故可认为在水平a=0.05下，这些百分比的均值有显著差异。4-2解：TheGLMProcedureDependentVariable:RSumofSourceDFSquaresMeanSquareFValuePr>FModel1182.83333337.5303030 1.39 0.2895Error1265.00000005.4166667CorrectedTotal23147.8333333R-Square CoeffVar RootMSE RMean0.56031622.34278 2.327373 10.41667SourceDFTypeISSMeanSquareFValuePr>Fm244.3333333322.16666667 4.09 0.0442n311.500000003.83333333 0.71 0.5657m*n627.000000004.50000000 0.83 0.5684SourceDFTypeIIISSMeanSquareFValuePr>Fm244.3333333322.16666667 4.09 0.0442n311.500000003.83333333 0.71 0.5657m*n627.000000004.50000000 0.83 0.5684由结果可知，在不同浓度下得率有显著差异，在不同温度下得率差异不明显，交互作用的效应不显著。4-4解：（1）不用协变量做方差分析

TheGLMProcedureDependentVanable:ySimiofSourceDFSquaresMeanSquareFValuePr>FModelErrorCorrectedTotal3 1041791667 347.263S89 1.3S20 J01SH56667 250.^0833323 <5059.938333R-Square 知^蹒 RootMSE yMeaM0171914 22.SS754 15.S4009 包2083302766SourceDFTypelSSMeanSquareFValuePr>Fy13300416&671.32026501的3.3方00002.7601120118J7500000070.7895SourceDFTypedSSMeanSquareFValuePr>FV1□□004166671J202650由分析结果可知，花的品种、温度和两者的交互作用对鲜花产量的影响都是不显著的。1ir37500000070JS9J(2)引入协变量作万差分析TheGLMProceduieDependentVariable:ySumofSourceDFSquaresMeanSquareFValuePr>FModel4□832.7190861458.179772121.92<0001Error19227.2392471L9J9M0CorrectsdTotal23W59.9jS333R-Square依液场 RootMSE yMean096250249MW7 3.45831S 69.20833SourceDFTypelSSMeanSquareFValuePr>Fy133004166727.150<0001勇1693.3烦。。.57.97<0001118.3750001.5402303514790.92741940058<0001SourceDFTypeHISSMeanSquareFValuePr>Fy1239442812000.17331479.288S6D4007<0001由分析结果可见，引入协变量后，-v、m、和x对鲜花产量的影响都是显著地。第五章:正交试验设计

5-3解:用L9（34）确定配比试验方案：试验方案因素试验号ABCD11（0.1份）1（0.3份）1（0.2份）1（0.5份）212（0.4份）2（0.1份）2（0.3份）313（0.5份）3（0.1份）3（0.1份）42（0.3份）123522316231273（0.2份）1328321393321以1号条件为例，表中四个数值的组成比为：A：B：C：D=0.1：0.3：0.2：0.5配比方案中，要求各行四个比值之和为1。在1号条件中，四种数值分别是1 1A=0.1x =0.091B=0.3x =0.2720.1+0.3+0.2+0.5 0.1+0.3+0.2+0.511

C=0.2x =0.182D=0.5x =0.4550.1+0.3+0.2+0.5 0.1+0.3+0.2+0.5其余实验条件可按照相同方法得出。第六章：回归分析6-6解：（1）作线性回归分析结果如下：Depaident'Variable:yAnalysisofVaianoeSumofMeanSource DFSquaresSquareFValuePr>FModel 315.645005.21» 1J.17 00119Error 41375000.34375CorrectedTotal '：7.02000RwtMSE0JS630R-Square 09192DependaitMean9900000.8586g堀 5.92224ParameterEstimatesParameterStandardVariableDFEstimateErrortValueIntercept19900000207294?.76<000110J7J000.207292.7700J011OL550000.207292.6500J6S11.15000020729000J2

由分析结果得回归方程为：y=9.90000+0.57500气+0.55000x2+1.15000x3由p值都小于0.1可知，每项都是显著的，方程也是显著的。(2)StepwiseSelection:Step3VariableEntered:R-Square=0.9192and0(p)=4.0000AnalysisofVarianceSumofMeanPSourceDFSquaresSquareFValuePr>FModel315.645003215001117 00119Error41375000.34375CorrectsdTotal7n.02000ParameterStandardVariableEstimateErrorTypeUSSFValuePr>FIntercept9.900000.20729784-080002280.96<0001057JOO0207292.64:007.69O.OjOlOjjOOO0.207292.420007.0400j6Sx3.IL50000.2072910.j800030.780.0052BoundsonconditiDnminiber:1.9由分析结果可知，故所求方程为:6-9x3+0.57500x+由分析结果可知，故所求方程为:6-9x3+0.57500x+1.15000x己二二二二［.・：1二…：m己三二3二：】NumberPartialDependentVar理1e:yRanged加In R-Square晾敏螂Op)FValuePr11 06216 城徵14.73459.S6002012xl- Stepwise.Selection:Step-1904003.48012093qVariablet9Entered:R-Sqyare=0.3473^nd(p(p)=175.75177.040.0568ModelAnalysisofVarianceSumofMeanSourceDF SquaresSquareFValuePr>FModel1 76.2438976.24389 7.45 0.0163Error14 143.2837110.23455CorrectedTotal15 219.52760Parameter StandardVariableEstimate ErrorTypeIISSFValuePr>FIntercept8.22980 1.38618360.74949 35.25<.0001t90.01056 0.0038776.24389 7.450.0163

Boundsonconditionnumber:1,1StepwiseSelection:Step2Variablet13Entered:R-Square=0.6717andC(p)=84.4265AnalysisofVarianceSumofMeanSourceDF SquaresSquareFValuePr>FModel2 147.4655173.73276 13.30 0.0007Error13 72.062095.54324CorrectedTotal15 219.52760Parameter StandardVariableEstimate ErrorTypeIISSFValuePr>FIntercept18.33483 2.99803207.32264 37.40<.0001t90.01173 0.0028792.81733 16.740.0013t13-1.89938 0.5298971.22162 12.850.0033StepwiseSelection:Step3Variablet5Entered:R-Square=0.7627andC(p)=60.2727AnalysisofVarianceSumofMeanSourceDF SquaresSquareFValuePr>FModel3 167.4249255.80831 12.85 0.0005Error12 52.102684.34189CorrectedTotal15 219.52760Parameter StandardVariableEstimate ErrorTypeIISSFValuePr>FIntercept19.49941 2.70837225.06452 51.84<.0001t50.00163 0.0007617619.95941 4.600.0532t90.00728 0.0032821.44626 4.940.0462t13-2.19305 0.4885687.48515 20.150.0007Boundsonconditionnumber:1.8185,13.825Allvariablesleftinthemodelaresignificantatthe0.1500level.Noothervariablemetthe0.1500significancelevelforentryintothemodel.SummaryofStepwiseSelectionVariableVariableNumberPartialModelStepEnteredRemovedVarsInR-SquareR-SquareC(p)FValuePr>F

Step1t910.34730.3473175.7527.450.01632t1320.32440.671784.426512.850.00333t530.09090.762760.27274.600.0532由结果可知，y=19.49941+0.00163xx+0.00728XXx-2.19305X-12 24 310解：（1）散点图如下:2015-v105-・.・・・246810 12 14可以采用Logistic拟合此数据。(2)用Logistic模拟结果为：DependentVariableyMethod:Gauss-NewtonSumofIterbcaSquares03.71802.000021.00001124.113.64081.849314.8393570.923.54751.668414.9977534.733.47041.520815.2362499.443.40461.396315.4814464.353.34691.288715.7348429.263.29551.194315.9985394.173.24911.110716.2735359.183.20691.036016.5601324.493.16840.969116.8579290.5103.13310.909117.1660257.6113.10080.855117.4829226.3123.07120.806717.8067196.8133.04420.763218.1349169.7143.01950.724218.4653145.1152.99710.689218.7951123.1162.97680.658019.1220104.0

172.95840.630019.443887.4241182.94200.605019.758673.4129192.92720.582720.064861.7148202.91400.562820.361052.0895212.90230.545020.646444.2826222.89200.529120.920338.0422232.88280.514921.182233.1304242.87480.502121.431929.3307252.86790.490721.669326.4509262.86180.480421.894624.3243272.85660.471122.107822.8087282.85220.462822.309421.7843292.84840.455322.499521.1514302.84530.448622.678720.8276312.84280.442522.847220.7456WARNING:Stepsizeshowsnoimprovement.WARNING:PROCNLINfailedtoconverge.EstimationSummary(NotConverged)MethodGauss-NewtonIterations31Subiterations29AverageSubiterations0.935484R0.653053PPC(c)0.012486TheNLINProcedureEstimationSummary(NotConverged)RPC.Object0.00394Objective20.74557ObservationsRead15ObservationsUsed15ObservationsMissing0NOTE:Aninterceptwasnotspecifiedforthismodel.SumofMeanApproxSourceDFSquaresSquareFValuePr>FRegression33245.51081.8625.77 <,0001Residual1220.74561.7288UncorrectedTotal153266.3CorrectedTotal14949.9ApproxParameterEstimateStdErrorApproximate95%ConfidenceLimitsb2.84280.004882.83212.8534c0.44250.003210.43550.4494a22.84720.574921.594724.0998

ApproximateCorrelationMatrixbcab1.00000000.7483922-0.2191675c0.74839221.0000000-0.4085662a-0.2191675-0.40856621.0000000故得二=二壬72p=2.8428y=0.4425第七章：回归正交设计7-1解：n=9作变换X=+1=z—38.5=2z-77，则x1=1，x2=2，x3=3，…，x9=9。0.5 0.5并可设y=b0+b101(x)+b202(x)+b202(x)+b4中4(x)对于n=9，查附表6,利用SAS软件进行回归多项式分析，结果如下：根据结果分析b0，b1，b2的P值小于0.05，是显著的，TheSASSystem21:16Wednesday,November14,2010TheORTHOREGProcedureDependentVariable：y12.072222222222220.030668717512.072222222222220.030668717567.57<.000110.11950.011877943210.060.000510.015277777777770.00174751248.740.00091-0.006292929292920.0029241475-2.150.097810.000174825174820.00205629220.090.9363b3,b4的P值大于0.05是不显著的，所以只需要配要二次项就行了。当n=9时，有：中(x)=入。(x)=x—511中(x)=m(X)=3X(x2—10x+758)=3x2—30x+37922 6SourceDFSumofSquaresMeanSquareFValuePr>FModelErrorCorrectedTotal1.54309502720.03386052841.57695555560.38577375680.008465132145.570.0014RootMSE 0.0920061525R-Square 0.9785278121VariableDFParameterEstimateStandardError\:ValuePr>|t|y=b+b甲(x)+b中(x)=2.07222 +0.1195x(x—5)+0.01528x(3x2—30x+379)将X=2乙-77代入上式得所求多项式回归方程：y=304.5489-14.79452z+0.18336z23解：本题属于一次回归的正交设计采用二水平，做变换：X1=土辿，x2=土史，、3=土奏1，x4=土史10 5 25 10试验方案试验号x1x2x3x4y'111115.4211-1-13.431-11-10.84-1-1-110.35-111-1-2.56-11-11-17-1-111-3.38-1-1-1-1-3.6程序如下:dataE1;inputx1-x4y;cards;11115.411-1-13.41-11-10.81-1-110.3-111-1-2.5-11-11-1-1-111-3.3-1-1-1-1-3.6;procregdata=E1;modely=x1-x4;run;TheSASSystem22:08Wednesday,November14,2010 1TheREGProcedureModel:MODEL1DependentVariable：yAnalysisofVarianceSourceDFSumofSquaresMeanSquareFValuePr>FModel468.4850017.1212512.730.0317Error34.033751.34458CorrectedTotal 772.51875RootMSE1.15956R-Square0.9444DependentMean-0.06250AdjR-Sq0.8702CoeffVar-1855.29872ParameterEstimatesVariableDFParameterEstimateStandardErrortValuePr>ItlIntercept1-0.062500.40997-0.150.8885xl12.537500.409976.190.0085x211.387500.409973.380.0430x310.162500.409970.400.7183x410.412500.409971.010.3885y"=-0.0625+2.5370x1+1.38750x2+0.16250x3+0.41250x4即y—87=-0.0625+2.5370"—310+1.38750七2—25+0.16250七3—225+0.41250'4—9010 5 25 10y=—3.822+0.2537z1+0.2775z2+0.0065z3+0.041250z4第八章：均匀设计1解：安排方案如下:^素试验号^^\_Z1Z2Z312.06002423.48002234.85002046.270018第九章：单纯形优化设计9-1解：（1）反射点E的坐标（5，10，0）3（2）（i）扩大，a>1（ii）内收缩，a<0（iii）收缩，0<a<1（3） 新单纯形各点坐标B（2,4,3），A’（1.5,3,3.5），C’（2.5,2.5,2.5），D”（3,3.5，2）。

第十章：析因试验设计10-4解：分析结果如下:ANOVAforY1SourceMasterModelPredictiveModelDF88MSFPr>FDF88MSFPr>FXIX2X3X1*X1X1*X2X1*X3X2*X2X2*X3X3*X310.4050.781255.6112534.441632.4926.0131.68006109.202538.50160.4050.781255.6112534.441632.4926.0131.68006109.202538.50160.3959590.763815.48598733.6727431.7647125.4293630.97285106.764737.642110.5568110.4221230.0661970.0021430.0024380.0039580.0025780.0001460.0016710.4050.781255.6112534.441632.4926.0131.68006109.202538.50160.4050.781255.6112534.441632.4926.0131.68006109.202538.50160.3959590.763815.48598733.6727431.7647125.4293630.97285106.764737.642110.5568110.4221230.0661970.0021430.0024380.0039580.0025780.0001460.00167ModelErrorTotal265.20985.114167270.32429.467761.02283328.809930.000874265.20985.114167270.32429.467761.02283328.809930.000874FitStatisticsforY1MasterMode1PredictiveMode1MeanR-squareAdj.R-squareRMSE 1.CV 3.32.5298.11需94.70^01135210993932.5298.11^94.70^1.0113523.109939AliasStructureforY1MasterModelPredictiveModelNoeffect;jaliased.Noeffectsaliased.PredictiveModelforY1Pnrl^rlIajuIsf-11、■EffectEstimatesforY1TermMasterModelPredictiveModelEstimateStdErrtPr>|t|EstimateStdErrtPr>|t|XI-0.2250.357567-0.629250.556811-0.2250.357567-0.629250.556811X2-0.31250.357567-0.873960.422123-0.31250.357567-0.873960.422123X3-0.83750.357567-2.342220.066197-0.83750.357567-2.342220.066197-3.0541670.526324-5.802820.002143-3.0541670.526324-5.802820.002143-2.050.505676-5.636020.002438-2.850.505676-5.636020.002438-2.550.505676-5.042750.003958-2.550.505676-5.042750.003958心观-2.9291670.526324-5.565330.002578-2.9291670.526324-5.565330.002578海楸3-5.2250.505676-10.33270.000146-5.2250.505676-10.33270.000146X3楸3-3.2291670.526324-6.135320.00167-3.2291670.526324-6.135320.00167经分析可知，在a=0.05的情况下，一次项均不显著，二次项、交互项均是显著的。第十二章：多指标综合评价概论12-4解：记P(U<Ua)=«，其中优等品10件，一等品30件，二等品40件，20件三等品，所有标准分依次是—U；-U;U;U，查标准正态表得到标准分依次是：-1.64，-0.67，0.25；1.28；第十三章：主成分分析法和因子分析法ppt第一题解：利用SAS软件进行主成分分析主要输出结果如下：EigenvalueDifferenceProportionCumulative3.104912520.207470900.38810.38812.897441621.967226080.36220.75030.930215550.288093290.11630.06660.642122260.338038130.08030.94680.304084130.217486370.03800.98480.006597760.054413330.01080.99570.032184380.028742610.00400.99870.002441780.00031.000014:39Monday,December5,2010EigenvaluesoftheCorrelationMatrix12345678ThePRINCOMPProcedureEigenvectorsPrinlPrin2Prin3Prin4Prin5Prin6Prin7Prin8xl0.4766500.2959910.1041900.045303-.184219-.0658540.7576190.245000x20.4728080.2778940.162883-.1744310.305448-.048451-.5184130.527105x30.4230450.3779510.1562550.0506700.0174750.099048-.174045-.780540x4-.2128930.451408-.0085440.516086-.5394070.287855-.2494270.220126x5-.3884600.3308450.321133-.1894160.4488980.5822880.2328690.030623x6-.3524270.4027370.1451440.2792570.316035-.7135710.056436-.042355x70.214835-.3774150.1404590.7581690.4182010.1935870.0528420.041160x80.0550340.272736-.8911620.0718550.3222010.1221680.067111-.003300要求累积方差贡献率N85%，所以取前三个作为主成分即可，Prin1=0.476650x1+0.472808x2+0.423845x3...+0.055034x8Prin2=0.295991x1+0.277894x2+0.377951x3...+0.272736x8Prin3=0.104190x1+0.162983x2+0.156255x3...-0.891162x8第一主成分，各项数值彼此相差不大，表示各单项指标对综合的经济效益起着相当的作用，因此第一主成分可以理解为全面能力综合指标；第二主成分，指标x4的系数明显比其它的指示系数大，而且系数为正，表明第二主成分主要由x4决定；第三主成分，指标x8的系数明显比其它的指示系数大，而且系数为负的，表明第三主成分主要是反应与x8相反的指标。

November23,2010 8排序结果为如下:November23,2010 8ObsnamePrinlPrin2Prin3prin123rankprin12314(?m22.789091.413571.12043i.99053'l31口避纫)2.118340.20704-0.010271.03463241.106820.658430.356070.81879351.12000-0.277290.303220.42604460.88156-0.528740.072080.18359571(若耄)-1.052181.276860.659050.1506668-3.372333.706520.108860.053117913位教艺术用品)0.413730.44054-2.92320-0.022220100.37240-0.77274-0.43574-0.214569116(&)-0.93583-1.38400-0.20544-1.0252910122Si)-1.33533-2.164340.45585-1.4410411133(麒)-2.10809-2.575850.49910-1.9542512TheSASSystem20:40Friday,第十四章：模糊综合评价14-1解：单指标评价结果归一化处理得：0414一999151一—0—399142R=——0399210——033180——099权向量a=(0.300.250.150.200.10)a・R=(0.1333,0.4833,0.2223,0.1611)根据最大隶属度原则0.48333所对应的评语为一般。根据秩加权平均原则，用1、2、3、4分别代表差、一般、良好、优，A=1X0.1611+2X0.2223+3X0.4833+4X0.1333=2.588得A=2.2588，即位于一般与良好之间，且该食品一般偏良。100X0.1611+80X0.2223+60X0.4833+40X0.1333=68.2245评分值为：P=68.2245分。第十五章：聚类分析与判别分析

15-3解：用动态聚类的方法来分类，结果如下：TheFASTCLUSProcedureInitialSeedsClusterxlqx3x4x5x6114-0000000013.000000002810000000014-0000000022.0000000039.00000000220_(MW(W016.000000002.0000000038.OOOOOOOO的.3000000037.000000009-0000000020.0000000023.00000000ClusterListing如ClusterSeed:103463212.0539312.1323430163676619.9145120S110L6D8S9113173710197364第一类:学生编号｛1，2，3,5，6,8，9,10｝；第二类:学生编号｛7｝；第三类:学生编号｛4｝。15-6解：PosteriorProbabi1ityofMembershipinspeciesClaaaifiedFromintoyearspeciesspecies1231951110.98040.00280.0168185223卅0.00020.82210.6777185821卅0.67880.18270.12751954220.00400.59610.39991955220.10370.80650.08981956330.00060.12410.87531857110.88890.00060.01061858880.07140.16580.76271858880.00120.08720.8015196032卅0.00010.62700.37301961330.11350.41560.47091962110.88510.09130.02361868220.00010.81510.08481864110.84470.01640.08881965110.81630.07700.10661966110.99370.00060.00571967110.99070.00220.00711968330.06900.02530.9056186882卅0.18240.55780.26081870880.04120.81640.64241971330.03610.05130.91261972330.06200.02340.91461973330.01150.25870.72971874220.01520.50580.47801875220.00010.68850.81641976330.00060.09800.90141977110.57740.03440.3882197823卅0.00390.17860.81751979330.00010.40200.5979

NumberofObservationsandPercentClassifiedintospeciesFromspecies123Total1100.000.000.00100.00212.5062.5025.00100.0030.0015.3884.6213100.00Total31.0324.141344.8329100.00Priors0.275860.275860.44828123TotalRate0.00000.37500.15380.1724Priors0.27590.27590.4483ErrorCountEstimatesforspecies第十六章：典型相关分析ppt第一题解：对44名学生期末成绩开卷和并卷进行典型相关分析，结果如下:5,2010 1ionsinthearezeroDFPr>F780.00225,2010 1ionsinthearezeroDFPr>F780.0022400.8312TheCANCORRProcedureCanonicaICorrelationAnalysisCanonica1CorrelationAdjustedCanonica1CorrelationApproximateStandardEr