中级计量中计

1、中级计量作业国际216020206004习题8.61. logit模型首先进行logit回归，得以下结果：Dependent Variable: GRADEMethod: ML - Binary Logit (Quadratic hill climbing)Date: 05/18/17Time: 19:07Sle: 1 32Included observations: 32Convergence achieved after 5 iterationsCovariance matrix computed using second derivativesVariableCoefficientStd

2、. Errorz-SisticProb.C GPA PSITUCE-14.345443.2535893.1197210.0615375.6578441.4318911.2586740.146202-2.5354962.2722322.4785770.4209080.01120.02310.01320.6738McFadden R-squaredS.D. dependent var Akaike info criterion Schwarz criterion Hannan-Quinn criter. Restr. devianceLR sisticProb(LR sistic)0.441494

3、0.4709290.9437611.1269781.00449239.7495317.549180.000545Mean dependent varS.E. of regres Sum squared resid Log likelihood DevianceRestr. log likelihoodAvg. log likelihood0.3125000.3554413.537473-11.1001822.20035-19.87476-0.346880Obs with Dep=0Obs with Dep=12210Total obs32因此，模型为：1) = 1 (1 + 其中 = 14.3

4、454 + 3.2536 + 3.1197 + 0.0615 经济学意义检验：保持其他变量不变的情况下，平均绩点（GPA）每上升 0.5 个绩点，学生学习成绩提高的概率平均提高 0.5*0.15*3.253589=0.2440保持其他变量不变的情况下，如果采取新的教学方式（PSI），学生学习成绩提高的概率平均提高 1*0.15*3.119721=0.4679保持其他变量不变的情况下，非期末成绩（TUCE）每上升 0.5 个分，学生学习成绩提高的概率平均提高 0.5*0.15*0.061537=0.0046统计检验：（1） 拟合优度：McFadden R2=0.441494，说明模型对数据的拟合

5、程度较好（2）LR sistic=17.54918，由于LR 17.54918 2(3) = 7.81473，故在0.05给定显著性水平 = 0.05原假设 0: 2 = 3 = 4 = 0 ，表明离散选择模型中所有解释变量的系数不全为 0。（3）判断变量 GPAi（平均绩点）系数的显著性：z sistic = 2.272232 0.025 = 1.96在给定显著性水平 = 0.05时，原假设0: 2 = 0，即在其他解释变量不变的情况下，解释变量“平均绩点”（GPA）对被解释变量 GRADE 有显著影响。判断变量 PSIi（是否采取新的教学方法）系数的显著性：z sistic = 2.478

6、577 0.025 = 1.96在给定显著性水平 = 0.05时，原假设0: 3 = 0，即在其他解释变量不变的情况下，解释变量“采用新的学习方法”（PSI）对被解释变量 GRADE 有显著影响。判断变量 TUCEi（非期末z s成绩）系数的显著性：istic = 0.420908 0.025 = 1.96在给定显著性水平 = 0.05时，接受原假设0: 4 = 0，即在其他解释变量不变的情况下，解释变量“非期末响。（4）计算变量 GPAi（非期末成绩”（TUCE）对被解释变量 GRADE 没有显著影成绩分数）对 GRADEi 的边际效应：f()= 0.15 3.253589 = 0.4880

7、计算变量 PSIi（采用新的教学方法）对 GRADEi 的边际效应：f() = 0.15 3.119721 = 0.4679计算变量 TUCEi（非期末成绩）对 GRADEi 的边际效应：f() = 0.15 0.061537 = 0.0092检验：（1）“朴素方法”：以 0.5 为临界值。Expec ion-Prediction Evaluation for Binary SpecificationEquation: UNTITLEDDate: 05/18/17Time: 19:55Sucs cutoff: C = 0.5Estimated EquationConstant Probabil

8、ityDep=0Dep=1TotalDep=0Dep=1TotalP(Dep=1)CTotal Correct% Correct% Incorrect Total Gain* Percent Gain*3221986.3613.64-13.64NA710770.0030.0070.0070.0010322681.2518.7512.5040.0002222100.000.0001000.00100.000322268.7531.25Estimated EquationConstant ProbabilityDep=0Dep=1TotalDep=0Dep=1TotalE(# of Dep=0)

9、E(# of Dep=1) TotalCorrect% Correct% Incorrect Total Gain* Percent Gain*18.483.5222.0018.4884.0215.9815.2748.873.526.4810.006.4864.8535.1533.6048.8722.0010.0032.0024.9778.0321.9721.0048.8715.136.8822.0015.1368.7531.256.883.1310.003.1331.2568.7522.0010.0032.0018.2557.0342.97*Change in % Correct from

10、default (constant probability) specification*Percent of incorrect (default) prediction corrected by equation（2）“先验方法”：以全部样本中选择 1 的样本所占比例为临界值。10 32 = 0.3125。Expec ion-Prediction Evaluation for Binary SpecificationEquation: UNTITLEDDate: 05/18/17Time: 20:06Sucs cutoff: C = 0.3125Estimated EquationCons

11、tant ProbabilityDep=0Dep=1TotalDep=0Dep=1TotalP(Dep=1)CTotal Correct% Correct% Incorrect3221986.3613.64910990.0010.0012322887.5012.5002222100.000.0001000.00100.000322268.7531.25Total Gain*Percent Gain*-13.64NA90.0090.0018.7560.00Estimated EquationConstant ProbabilityDep=0Dep=1TotalDep=0Dep=1TotalE(#

12、 of Dep=0) E(# of Dep=1) TotalCorrect% Correct% Incorrect Total Gain* Percent Gain*18.483.5222.0018.4884.0215.9815.2748.873.526.4810.006.4864.8535.1533.6048.8722.0010.0032.0024.9778.0321.9721.0048.8715.136.8822.0015.1368.7531.256.883.1310.003.1331.2568.7522.0010.0032.0018.2557.0342.97*Change in % Co

13、rrect from default (constant probability) specification*Percent of incorrect (default) prediction corrected by equation2. probit 模型再进行 probit 回归，得以下结果：Dependent Variable: GRADEMethod: ML - Binary Probit (Quadratic hill climbing)Date: 05/18/17Time: 20:14Sle: 1 32Included observations: 32Convergence a

14、chieved after 4 iterationsCovariance matrix computed using second derivativesVariableCoefficientStd. Errorz-SisticProb.C GPA PSITUCE-7.9263261.8200841.8602630.0266812.8092350.7511490.6978790.086340-2.8215252.4230682.6655970.3090210.00480.01540.00770.7573McFadden R-squaredS.D. dependent var0.4458420.

15、470929Mean dependent varS.E. of regres0.3125000.359021Akaike info criterion Schwarz criterion Hannan-Quinn criter. Restr. devianceLR sisticProb(LR sistic)0.9383601.1215770.99909139.7495317.722000.000502Sum squared resid Log likelihood DevianceRestr. log likelihoodAvg. log likelihood3.609091-11.01376

16、22.02752-19.87476-0.344180Obs with Dep=0Obs with Dep=12210Total obs32因此，模型为：21 = 1 2 2其中 = 7.9263 + 1.8201 + 1.8603 + 0.0267 经济学意义检验：保持其他变量不变的情况下，平均绩点（GPA）每上升 0.5 个绩点，学生学习成绩提高的概率平均提高 0.5*0.2771*1.820084=0.252保持其他变量不变的情况下，如果采取新的教学方式（PSI），学生学习成绩提高的概率平均提高 1*0.2771*1.860263=0.515保持其他变量不变的情况下，非期末成绩（TUCE）

17、每上升 0.5 个分，学生学习成绩提高的概率平均提高 0.5*0.2771*0.026681=0.0037统计检验：（1）拟合优度：McFadden R2=0.445842，说明模型对数据的拟合程度较好（2）LR sistic=17.72200，由于LR 17.72200 2(3) = 7.81473，故在0.05给定显著性水平 = 0.05原假设 0: 2 = 3 = 4 = 0 ，表明离散选择模型中所有解释变量的系数不全为 0。（3）判断变量 GPAi（平均绩点）系数的显著性：z sistic = 2.423068 0.025 = 1.96在给定显著性水平 = 0.05时，原假设0: 2

18、= 0，即在其他解释变量不变的情况下，解释变量“平均绩点”（GPA）对被解释变量 GRADE 有显著影响。判断变量 PSIi（是否采取新的教学方法）系数的显著性：z sistic = 2.665597 0.025 = 1.96在给定显著性水平 = 0.05时，原假设0: 3 = 0，即在其他解释变量不变的情况下，解释变量“采用新的学习方法”（PSI）对被解释变量 GRADE 有显著影响。判断变量 TUCEi（非期末z s成绩）系数的显著性：istic = 0.309021 0.025 = 1.96在给定显著性水平 = 0.05时，接受原假设0: 4 = 0，即在其他解释变量不变的情况下，解释变

19、量“非期末响。（4）计算变量 GPAi（非期末成绩”（TUCE）对被解释变量 GRADE 没有显著影成绩分数）对 GRADEi 的边际效应：f() = 0.2771 1.820084 = 0.504计算变量 PSIi（采用新的教学方法）对 GRADEi 的边际效应：f() = 0.2771 1.860263 = 0.515计算变量 TUCEi（非期末成绩）对 GRADEi 的边际效应：f() = 0.2771 0.026681 = 0.0074检验：（1）“朴素方法”：以 0.5 为临界值。Expec ion-Prediction Evaluation for Binary Specifica

20、tionEquation: UNTITLEDDate: 05/18/17Time: 21:05Sucs cutoff: C = 0.5Estimated EquationConstant ProbabilityDep=0Dep=1TotalDep=0Dep=1TotalP(Dep=1)CTotal Correct% Correct% Incorrect Total Gain* Percent Gain*3221986.3613.64-13.64NA610660.0040.0060.0060.009322578.1321.889.3830.0002222100.000.0001000.00100

21、.000322268.7531.25Estimated EquationConstant ProbabilityDep=0Dep=1TotalDep=0Dep=1TotalE(# of Dep=0) E(# of Dep=1) TotalCorrect% Correct% Incorrect18.503.5022.0018.5084.0815.923.576.4310.006.4364.2635.7422.079.9332.0024.9277.8922.1115.136.8822.0015.1368.7531.256.883.1310.003.1331.2568.7522.0010.0032.

22、0018.2557.0342.97Total Gain*Percent Gain*15.3349.0733.0148.0120.8648.54*Change in % Correct from default (constant probability) specification*Percent of incorrect (default) prediction corrected by equation（2）“先验方法”：以全部样本中选择 1 的样本所占比例为临界值。10 32 = 0.3125。Expec ion-Prediction Evaluation for Binary SpecificationEquation: UNTITLEDDate: 05/18/17Time: 21:07Sucs cutoff: C = 0.3125E