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.Error----------------------------------------------------x80.2318750.01460.0051y100.20970.00970.0031HypothesisTestNullhypothesis:Mean1-Mean2=0Alternative:Mean1-Mean2^=0IfVariancesAretstatisticDfPrt----------------------------------------------------Equal3.878160.0013NotEqual3.70411.670.0032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/Variance2^=1-DegreesofFreedom-FNumer.Denom.PrF----------------------------------------------2.27790.2501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:ySumofSourceDFSquaresMeanSquareFValuePrFModel41480.823000370.20575040.88.0001Error15135.8225009.054833CorrectedTotal191616.645500R-SquareCoeffVarRootMSEyMean0.91598513.120233.00912522.93500SourceDFAnovaSSMeanSquareFValuePrFc41480.823000370.20575040.88.0001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:RSumofSourceDFSquaresMeanSquareFValuePrFModel1182.83333337.53030301.390.2895Error1265.00000005.4166667CorrectedTotal23147.8333333R-SquareCoeffVarRootMSERMean0.56031622.342782.32737310.41667SourceDFTypeISSMeanSquareFValuePrFm244.3333333322.166666674.090.0442n311.500000003.833333330.710.5657m*n627.000000004.500000000.830.5684SourceDFTypeIIISSMeanSquareFValuePrFm244.3333333322.166666674.090.0442n311.500000003.833333330.710.5657m*n627.000000004.500000000.830.5684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。(2)引入协变量作方差分析由分析结果可见，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ABCD1234567891（0.1份）112（0.3份）223（0.2份）331（0.3份）2（0.4份）3（0.5份）1231231（0.2份）2（0.1份）3（0.1份）2313121（0.5份）2（0.3份）3（0.1份）312231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091.05.02.03.01.011.0A272.05.02.03.01.013.0B182.05.02.03.01.012.0C455.05.02.03.01.015.0D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：由分析结果得回归方程为：32115000.155000.057500.090000.9xxxy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）由分析结果可知，在a=0.05下，仅有x3和x1应当引入方程。故所求方程为：3115000.157500.090000.9xxy6-9解：分析结果如下：DependentVariable:yStepwiseSelection:Step1Variablet9Entered:R-Square=0.3473andC(p)=175.7517AnalysisofVarianceSumofMeanSourceDFSquaresSquareFValuePrFModel176.2438976.243897.450.0163Error14143.2837110.23455CorrectedTotal15219.52760ParameterStandardVariableEstimateErrorTypeIISSFValuePrFIntercept8.229801.38618360.7494935.25.0001t90.010560.0038776.243897.450.0163Boundsonconditionnumber:1,1------------------------------------------------------------------------------------------------------StepwiseSelection:Step2Variablet13Entered:R-Square=0.6717andC(p)=84.4265AnalysisofVarianceSumofMeanSourceDFSquaresSquareFValuePrFModel2147.4655173.7327613.300.0007Error1372.062095.54324CorrectedTotal15219.52760ParameterStandardVariableEstimateErrorTypeIISSFValuePrFIntercept18.334832.99803207.3226437.40.0001t90.011730.0028792.8173316.740.0013t13-1.899380.5298971.2216212.850.0033------------------------------------------------------------------------------------------------------StepwiseSelection:Step3Variablet5Entered:R-Square=0.7627andC(p)=60.2727AnalysisofVarianceSumofMeanSourceDFSquaresSquareFValuePrFModel3167.4249255.8083112.850.0005Error1252.102684.34189CorrectedTotal15219.52760ParameterStandardVariableEstimateErrorTypeIISSFValuePrFIntercept19.499412.70837225.0645251.84.0001t50.001630.0007617619.959414.600.0532t90.007280.0032821.446264.940.0462t13-2.193050.4885687.4851520.150.0007Boundsonconditionnumber:1.8185,13.825------------------------------------------------------------------------------------------------------Allvariablesleftinthemodelaresignificantatthe0.1500level.Noothervariablemetthe0.1500significancelevelforentryintothemodel.SummaryofStepwiseSelectionVariableVariableNumberPartialModelStepEnteredRemovedVarsInR-SquareR-SquareC(p)FValuePrF1t910.34730.3473175.7527.450.01632t1320.32440.671784.426512.850.00333t530.09090.762760.27274.600.0532由结果可知，y=19.49941+0.0016321xx+0.0072842xx-2.193053x6-10解：（1）散点图如下：可以采用Logistic拟合此数据。（2）用Logistic模拟结果为：DependentVariableyMethod:Gauss-NewtonSumofIterbcaSquares03.71802.000021.00001124.113.64081.849314.8393570.923.54751.668414.9977534.733.47041.520815.2362499.443.40461.396315.4814464.353.34691.288715.7348429.263.29551.194315.9985394.173.24911.110716.2735359.183.20691.036016.5601324.493.16840.969116.8579290.5103.13310.909117.1660257.6113.10080.855117.4829226.3123.07120.806717.8067196.8133.04420.763218.1349169.7143.01950.724218.4653145.1152.99710.689218.7951123.1162.97680.658019.1220104.0172.95840.630019.443887.4241182.94200.605019.758673.4129192.92720.582720.064861.7148202.91400.562820.361052.0895212.90230.545020.646444.2826222.89200.529120.920338.0422232.88280.514921.182233.1304242.87480.502121.431929.3307252.86790.490721.669326.4509262.86180.480421.894624.3243272.85660.471122.1078