实验设计作业2

SAS课后作业3-7.分别使用金球和铂球测定引力常数（单位：）1.用金球测定观察值为6.683，6.681,6.676,6.678,6.679,6.6722.用铂球测定观察值为6.661,6.661，6.667,6.667,6.664设测定值总体为N（u，）试就1,2两种情况求u的置信度为0.9的置信区间，并求的置信度为0.9的置信区间。解：用sas分析结果如下：第一组：第二组:3-10．某批矿砂的5个样品的镍含量，经测定为（％）3.25，3.27,3.24，3.26,3.24设测定值总体服从正态分布，问在a=0.01下能否接受假设：这批矿砂的镍含量的均值为3.25。解：用sas分析结果如下：HypothesisTestNullhypothesis:Meanofx=3.25Alternative:Meanofx^=3.25tStatisticDfProbt---------------------------------0.34340.7489因为p值大于0.01（显著性水平），故可认为接受原假设，这批砂的镍含量为3.25。3-13.下表分别给出两个文学家马克吐温的8篇小品文以及斯诺特格拉斯的10篇小品文中由3个字母组成的词的比例：马克吐温：0.2250.2620.2170.2400.2300.2290.2350.217斯诺特格拉斯：0.2090.2050.1960.2100.2020.2070.2240.2230.2200.201设两组数据分别来自正态总体，且两个总体方差相等，两个样本相互独立，问两个作家所写的小品文中包含由3个字母组成的词的比例是否有显著差异（a=0.05）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.在13题中分别记两个总体的方差为和。试检验假设：（取a=0.05）H0：，H1：以说明在第13题中我们假设2方差相等是合理的。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。4-1.将抗生素注入人体会产生抗生素与血浆蛋白质结合的现象，以致减少了药效。下表列出5种常用的抗生素注入到牛的体内时，抗生素与血浆蛋白质结合的百分比。试在水平a=0.05下检验这些百分比的均值有无显著差异。设个总体服从正态分布，且方差相等。解：青霉素四环素链霉素红霉素氯霉素29.627.35.821.629.224.332.66.217.432.828.530.811.018.325.032.034.88.319.024.2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下表给出某种化工生产过程在三种浓度、四种温度水平下得率的数据：浓度（%）温度（℃）1024385221411131010119124910767811106513121411141310假设在诸水平搭配下得率的总体服从正态分布，且方差相等。试在a=0.05下检验：在不同浓度下得率有无显著差异；在不同温度下的率是否有显著差异；交互作用的效应是否显著。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。4-4.①不用协变量做方差分析解：TheGLMProcedureDependentVariable:ySumofSourceDFSquaresMeanSquareFValuePrFModel31041.791667347.2638891.380.2766Error205018.166667250.908333CorrectedTotal236059.958333R-SquareCoeffVarRootMSEyMean0.17191422.8875415.8400969.20833SourceDFTypeISSMeanSquareFValuePrFv1330.0416667330.04166671.320.2650m1693.3750000693.37500002.760.1120v*m118.375000018.37500000.070.7895SourceDFTypeIIISSMeanSquareFValuePrFv1330.0416667330.04166671.320.2650m1693.3750000693.37500002.760.1120v*m118.375000018.37500000.070.7895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TheGLMProcedureDependentVariable:ySumofSourceDFSquaresMeanSquareFValuePrFModel45832.7190861458.179772121.92.0001Error19227.23924711.959960CorrectedTotal236059.958333R-SquareCoeffVarRootMSEyMean0.9625024.9969673.45831869.20833SourceDFTypeISSMeanSquareFValuePrFv1330.041667330.04166727.60.0001m1693.375000693.37500057.97.0001v*m118.37500018.3750001.540.2303x14790.9274194790.927419400.58.0001SourceDFTypeIIISSMeanSquareFValuePrFv123.94428123.9442812.000.1733m1479.288865479.28886540.07.0001v*m175.07199875.0719986.280.0215x14790.9274194790.927419400.58.0001由此可见，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5-3.配比试验。四因素ABCD的水平表如下（因素C用了一个拟水平）：因素ABCD水平10.10.30.20.5水平20.30.40.10.3水平30.20.5（0.1）0.1试用L9（）排出配比方案（要求各行四个比值之和为1）因素试验号ABCD10.12500.37500.12500.375020.23080.23080.15380.384630.28560.42860.14290.142940.09090.36360.09090.454650.33330.44450.11110.111160.18180.36370.18180.272770.11110.55560.22220.111180.25000.41670.08330.250090.15380.38460.07700.3846第六章6-5.一种合金在某种添加剂的不同浓度下，各做三次试验，得数据如下：浓度x10.015.020.025.030.0抗压强度y25.229.831.231.729.427.331.132.630.130.828.727.829.732.332.8(1)做散点图（2）以模型,拟合数据，其中b0,b1,b2,与x无关，求回归方程.作出散点图如下：Sas分析结果如下：DependentVariable:yAnalysisofVarianceSumofMeanSourceDFSquaresSquareFValuePrFModel238.9371419.468579.540.0033Error1224.476192.03968CorrectedTotal1463.41333RootMSE1.42817R-Square0.6140DependentMean30.03333AdjR-Sq0.5497CoeffVar4.75530ParameterEstimatesParameterStandardVariableDFEstimateErrortValuePr|t|Intercept119.033333.277555.81.0001t111.008570.356432.830.0152t21-0.020380.00881-2.310.0393故所求的方程为：。6-6.某化工产品的得率y与反应温度x1、反应时间x2及某反应物浓度x3有关，设对于给定的x1、x2、x3得率y服从正态分布且方差与x1、x2、x3无关，今得实验结果如下表所示，其中x1、x2、x3均为2水平且均已编码形式表达，X1-1-1-1-11111X2-1-111-1-111X3-11-11-11-11得率7.610.39.210.28.411.19.812.6（1）设,求y的多元线性回归方程，并