应用统计学试题2005.6(A)(参考答案)

1华东理工大学2004–2005学年第二学期《应用统计学》课程期末考试试卷A2005.6开课学院：商学院，考试形式：开卷，所需时间：120分钟考生姓名：学号：专业：班级题序一二三总分得分评卷人一、某橡胶配方试验分析（30分）对某种橡胶配方进行研究，目的是提高其弯曲次数（越多越好），现考察三个二水平因子，选取因素水平表如下。试验除考察三个因素的作用外，还要考察B×C的交互作用。因素水平A促进剂总量B炭墨品种C硫磺分量11.5天津耐高磨2.522.5天津耐高磨与长春硬炭黑并用2.0选用L8(27)表安排试验，试验计划和试验结果见表1。表1试验计划和试验结果列号试验号ABCB×C弯曲（万次）1234567yi111111111.5211122222.0312211222.0412222111.5521212122.0621221213.0722112212.5822121122.02经SPSS软件的计算，列表如下，请根据计算结果表格进行分析。表2ANOVNSourceTypeIIISumofSquaresdfMeanSquareFSig.A0.78110.015B0.03110.0311.0000.391C10.0311.0000.391B*C0.78110.78125.0000.015Error0.09430.031Total1.7187表3EstimatedMarginalMeansAMeanStd.Error1.001.7500.0882.002.3750.088BMeanStd.Error1.002.1250.0882.002.0000.088CMeanStd.Error1.002.0000.0882.002.1250.088BCMeanStd.Error1.001.001.7500.1252.002.5000.1252.001.002.2500.1252.001.7500.125（1）表2中有些数据没给出，请根据方差分析表的原理将其计算出来，给出计算过程。源于因子C的误差平方和＝1.718-0.781-0.031-0.781-0.094=0.031均方和VA＝0.781/1=0.781FA=0.781/0.031=253（2）由表2分析哪些因子是显著的，给出理由。因子A是显著的，因为其P=0.0150.1因子B*C是显著的，因为其P=0.0150.1（3）由表2和表3分析出该橡胶最优配方条件，给出理由。该橡胶最优配方条件：A2B1C2因为121.7502.375AA在因子B和C的组合中，122.5BC最大（4）根据表1，得出在该橡胶最优配方条件下生产出的橡胶弯曲次数，给出理由。在该橡胶最优配方条件下生产出的橡胶弯曲次数为3万次，因为因子A、B和C的水平分别取2、1和2时所对应的试验结果为3万次。4二、医院工作满意程度研究（30分）某医院管理工作者希望了解病人对医院工作的满意程度Y和病人的年龄X1、病情的严重程度X2和忧虑程度X3之间的关系。他们随机选取了23位病人，得到下表所列数据：nX1X2X3YRX1RX2RX3RY150.0051.002.3048.0020.00014.00013.0005.000236.0046.002.3057.009.5003.50013.00010.500340.0048.002.2080.0012.0006.0008.50019.500441.0044.001.8090.0013.0002.0001.50023.000528.0043.001.8060.001.0001.0001.50012.500649.0054.002.9036.0019.00018.50022.0002.000742.0050.002.2046.0014.00011.0008.5003.000845.0048.002.4054.0018.0006.00017.5009.000952.0062.002.9026.0021.00023.00022.0001.0001029.0050.002.1077.003.50011.0005.50016.5001129.0048.002.4089.003.5006.00017.50022.0001243.0053.002.4067.0015.50017.00017.50015.0001338.0055.002.2047.0011.00020.0008.5004.0001434.0051.002.3051.008.00014.00013.0007.0001553.0054.002.2057.0022.00018.5008.50010.5001636.0049.002.0066.009.5008.5004.00014.0001733.0056.002.5079.006.50021.00020.00018.0001829.0046.001.9088.003.5003.5003.00021.0001933.0049.002.1080.006.5008.5005.50019.5002055.0051.002.4049.0023.00014.00017.5006.0002129.0052.002.3077.003.50016.00013.00016.5002244.0058.002.9052.0017.00022.00022.0008.0002343.0050.002.3060.0015.50011.00013.00012.500经SPSS软件的计算，得到下列计算结果：表4VariablesEntered/Removed(a)ModelVariablesEnteredVariablesRemovedMethod1X1Stepwise(Criteria:Probability-of-F-to-enter=0.090,Probability-of-F-to-remove=0.100).2X3.Stepwise(Criteria:Probability-of-F-to-enter=0.090,Probability-of-F-to-remove=0.100).a:DependentVariable:Y5表5ANOVA(c)ModelSumofSquaresdfMeanSquareFSig.1Regression2988.07212988.07216.5760.001(a)Residual3785.58021180.266Total6773.652222Regression3550.28521775.14311.0140.001(b)Residual3223.36720161.168Total6773.65222a:Predictors:(Constant),X1;b:Predictors:(Constant),X1,X3;c:DependentVariable:Y表6Coefficients(a)ModelUnstandardizedCoefficientstSig.BStd.Error1(Constant)116.94813.6798.5490.000X1-1.3760.338-4.0710.0012(Constant)147.43120.8257.0800.000X1-1.0340.369-2.8050.011X3-19.18910.274-1.8680.077a:DependentVariable:Y表7VariablesEntered/Removed(a)ModelVariablesEnteredVariablesRemovedMethod1RX1*RX2Stepwise(Criteria:Probability-of-F-to-enter=0.090,Probability-of-F-to-remove=0.100).a:DependentVariable:RY表8ANOVA(b)ModelSumofSquaresdfMeanSquareFSig.1Regression494.2851494.28520.1270.000(a)Residual515.7152124.558Total1010.00022a:Predictors:(Constant),RX1*RX2;b:DependentVariable:RY表9Coefficients(a)ModelUnstandardizedCoefficientstSig.BStd.Error1(Constant)17.3901.58510.9740.000RX1*RX2-0.0330.007-4.4860.000a:DependentVariable:RY6（1）利用SPSS计算结果，建立Y关于X1，X2，X3的逐步回归方程13ˆ147.4311.034X19.189XY（2）利用SPSS计算结果，建立RY关于RX1、RX2、RX3及它们的平方项RX11,RX22,RX33，相互乘积项RX12,RX13,RX23的R逐步回归方程。1217.3900.033RYRXRX＝（3）试用上面所求得的两个回归方程，计算第3点、第22点的残差。①由13ˆ147.4311.034X19.189XY得322ˆˆ63.86575,46.29844YY则第3点残差33ˆ80.0063.8657516.13425YY第22点残差2222ˆ52.0046.298445.70156YY②根据1217.3900.033RYRXRX＝计算残差第3点对应的12()12.000,()6.000RXRX，则3()17.3900.03312615.014RY，其对应的3(15)(16)(15)3ˆ()(()15)67(7767)(15.01415)67.14YYYYRY则第3点残差33ˆ80.0067.1412.86YY第22点对应的12()17.000,()22.000RXRX，则2217.3900.03317225.048RY＝，其对应的22(5)(6)(5)22ˆ()(()5)48(4948)(5.0485)48.048YYYYRY则第22点残差2222ˆ5248.0483.952YY（4）试利用R回归方程，求出X1=54，X2=52，X3=2.2时，Y的预测值。X1=54，X2=52，X3=2.2对应的1(22)11(23)1(22)15453()222222.55553XXRXXX，2()16RX则17.3900.03322.5165.51RY＝，其预测值为(5)(6)(5)ˆ()(()5)48(4948)(5.515)48.51YYYYRY7三、主成分回归分析（40分）某科学基金会的管理人员希望估价从事数学研究工作的中等或较高水平的数学家的年工资额Y与他们的研究成果（论文、著作）的质量指标X1，从事研究工作的时间X2以及能成功获得资助的指标X3之间的关系，为此按一定的试验设计方法调查了24位此类型的数学家，得到下列数据：nX1X2X3YnX1X2X3Y13.509.004.0033.20138.0023.008.3043.3025.3020.006.0040.30146.5035.007.0044.1035.1018.005.9038.70156.6039.007.4042.8045.8033.006.4046.80163.7021.004.3033.6054.2031.005.0041.40176.207.007.0034.2066.0013.006.7037.50187.0040.007.6048.0076.8025.007.5039.00194.0035.004.9038.0085.5030.006.0040.70204.5023.005.0035.9093.105.003.5030.10215.9033.006.4040.