計(jì)量經(jīng)濟(jì)學(xué)試驗(yàn)報(bào)告-LOGIT模型

XXX學(xué)院實(shí)驗(yàn)報(bào)告課程名稱計(jì)量經(jīng)濟(jì)學(xué)成績實(shí)驗(yàn)名稱虛擬變量模型專業(yè)投資學(xué)年級(jí)/班級(jí)B18投資學(xué)一班指導(dǎo)教師劉某某實(shí)驗(yàn)地點(diǎn)文博樓實(shí)驗(yàn)室名稱經(jīng)濟(jì)學(xué)實(shí)驗(yàn)中心實(shí)驗(yàn)時(shí)間實(shí)驗(yàn)儀器計(jì)算機(jī)、eviews計(jì)量分析軟件姓名學(xué)號(hào)實(shí)驗(yàn)?zāi)康?運(yùn)用習(xí)題7-1數(shù)據(jù)，建立人^丫人模型，并解釋相關(guān)變量。.運(yùn)用習(xí)題7-1數(shù)據(jù)，建立人尤。丫人模型，并解釋相關(guān)變量。.運(yùn)用習(xí)題7-1數(shù)據(jù)，建立LPM模型，并解釋相關(guān)變量。.運(yùn)用7-1數(shù)據(jù)，建立LOGIT模型，并解釋相關(guān)變量。.運(yùn)用7-1數(shù)據(jù)，建立PROBIT模型，并解釋相關(guān)變量。.運(yùn)用7-2數(shù)據(jù)，建立審察回歸模型TOBIT模型，并解釋相關(guān)變量。實(shí)驗(yàn)內(nèi)容于驟.建立ANOVA模型：打開Eviews，選擇數(shù)據(jù)GPA、PSI打開，點(diǎn)擊Quick-EstimateEquation進(jìn)行回歸分析。.建立ANCOVA模型：選擇GPAGRADEPSITUCE作為一組數(shù)據(jù)打開，點(diǎn)擊Quick-EstimateEquation進(jìn)行回歸分析。.建立LPM模型：在EstimateEquation命令框中輸入GRADECGPAPSITUCE進(jìn)行回歸，得到LPM模型。.建立LOGIT模型：在Method下拉列表中選擇BINARY選項(xiàng)，并在BinaryEstimate選項(xiàng)組中選擇LOGIT，同時(shí)在命令框中輸入GRADEGPAPSITUCEC得到模型。.建立PROBIT模型：在Method下拉列表選擇BINARY選項(xiàng)，并在BinaryEstimate選項(xiàng)組中選擇PROBIT，同時(shí)在命令框中輸入GRADEGPAPSITUCEC得到模型。.建立TOBIT模型：在Method下拉列表中選擇CENSORED選項(xiàng)，在EstimateEquation命令框中輸入YX1X2X3X4C，在Distribution中選擇Normal。第1頁

.建立ANOVA模型如下：回歸分析如下：DependentVariable:GPAMethod:LeastSquaresDate:11/21/21Time:11:24Sample:132Includedobservations:32實(shí)驗(yàn)結(jié)果分析及討論Variable實(shí)驗(yàn)結(jié)果分析及討論VariableCoefficientStd.Errort-StatisticProb.PSI0.0367460.168928 0.2175240.8293C3.1011110.111736 27.754020.0000R-squared0.001575Meandependentvar3.117188AdjustedR-squared-0.031706S.D.dependentvar0.466713S.E.ofregression0.474054Akaikeinfocriterion1.405470Sumsquaredresid6.741813Schwarzcriterion1.497079Loglikelihood-20.48752Hannan-Quinncriter.1.435836F-statistic0.047317Durbin-Watsonstat2.225416Prob(F-statistic)0.829272根據(jù)回歸數(shù)據(jù)建立模型，結(jié)果為：GPA=3.101111+0.036746PSIt=（27.75402）（0.217524）R2=0.001575,F=0.047317由回歸結(jié)果可知，PSI所對(duì)應(yīng)的P值為0.8293，P>0.05,P值不顯著，這說明PSI對(duì)GPA沒有顯著影響，即是否接受新的教學(xué)方法對(duì)學(xué)生的平均分?jǐn)?shù)沒有很大的關(guān)系。

.建立ANCOVA模型如下:回歸分析結(jié)果如下：DependentVariable:GPAMethod:LeastSquaresDate:11/21/21Time:11:48Sample:132Includedobservations:32實(shí)驗(yàn)結(jié)果分析及討論(續(xù))Variable實(shí)驗(yàn)結(jié)果分析及討論(續(xù))VariableCoefficientStd.Errort-StatisticProb.GRADE0.4884730.1705532.8640540.0078PSI-0.1877880.156612-1.1990610.2406TUCE0.0307190.0192401.5966380.1216C2.3575330.4182445.6367380.0000R-squared0.342418Meandependentvar3.117188AdjustedR-squared0.271963S.D.dependentvar0.466713S.E.ofregression0.398223Akaikeinfocriterion1.112861Sumsquaredresid4.440289Schwarzcriterion1.296078Loglikelihood-13.80577Hannan-Quinncriter.1.173592F-statistic4.860075Durbin-Watsonstat1.831719Prob(F-statistic)0.007604根據(jù)回歸數(shù)據(jù)建立模型，結(jié)果為：GPA=2.357533+0.488473GRADE-0.187788PSI+0.030719TUCEt~5.636738）（2.864054）（—1.199061）（1.596638）R2=0.342418,F=4.860075由模型結(jié)果可知，當(dāng)GRADE為常量時(shí)，接受新方法比不接受新方法的平均分?jǐn)?shù)低0.187788分；當(dāng)GRADE和PSI為常量時(shí)，測驗(yàn)得分每增加一個(gè)單位，學(xué)生的平均分?jǐn)?shù)就增加2.388252分。另外GRADE的P值較小，小于0.05,說明是否接受新方法對(duì)學(xué)生的學(xué)習(xí)成績影響較大。

.建立LPM模型如下：DependentVariable:GRADEMethod:LeastSquaresDate:11/21/21Time:11:27Sample:132Includedobservations:32實(shí)驗(yàn)結(jié)果分析及討論(續(xù))VariableCoefficientStd.Errort-StatisticProb.GPA0.4638520.1619562.8640540.0078PSI0.3785550.1391732.7200350.0111TUCE0.010495實(shí)驗(yàn)結(jié)果分析及討論(續(xù))VariableCoefficientStd.Errort-StatisticProb.GPA0.4638520.1619562.8640540.0078PSI0.3785550.1391732.7200350.0111TUCE0.0104950.0194830.5386850.5944C-1.4980170.523889-2.8594190.0079R-squared0.415900Meandependentvar0.343750AdjustedR-squared0.353318S.D.dependentvar0.482559S.E.ofregression0.388057Akaikeinfocriterion1.061140Sumsquaredresid4.216474Schwarzcriterion1.244357Loglikelihood-12.97825Hannan-Quinncriter.1.121872F-statistic6.645658Durbin-Watsonstat2.346447Prob(F-statistic)0.001571tk-2.859419）（2.864054）（2.720035）（0.538685）R2=0.4159,F=6.645658模型回歸結(jié)果表明，模型總體的統(tǒng)計(jì)是顯著的?？芍狦PA的斜率估計(jì)量為0.463852,這說明平均分?jǐn)?shù)每增加1時(shí)，接受新教學(xué)方法后成績改善的概率上升46.39%；TUCE的斜率估計(jì)量為0.010495,這說明測驗(yàn)得分每增加1個(gè)單位時(shí)，接受新教學(xué)方法后成績改善的概率上升1.05%；PSI的斜率估計(jì)量為0.378555,這接受新方法的意愿每增加1個(gè)單位時(shí)，接受新教學(xué)方法后成績改善的概率上升37.86%。.建立LOGIT模型如下：DependentVariable:GRADEMethod:ML-BinaryLogit(Quadratichillclimbing)Date:11/21/21Time:11:32Sample:132Includedobservations:32Convergenceachievedafter4iterationsCovariancematrixcomputedusingsecondderivatives實(shí)驗(yàn)結(jié)果分析及討論(續(xù))Variable實(shí)驗(yàn)結(jié)果分析及討論(續(xù))VariableCoefficientStd.Errorz-StatisticProb.GPA0.2993360.681542 0.4392040.6605PSI1.6363570.812620 2.0136810.0440TUCE-0.1014720.099563 -1.0191830.3081Meandependentvar0.343750S.D.dependentvar0.482559S.E.ofregression0.467281Akaikeinfocriterion1.360661Sumsquaredresid6.332187Schwarzcriterion1.498074Loglikelihood-18.77057Hannan-Quinncriter.1.406209Deviance37.54114Restr.deviance41.18346Avg.loglikelihood-0.586580ObswithDep=021Totalobs32ObswithDep=111根據(jù)上述結(jié)果，可以得出二元Logit模型結(jié)果為：1P-ln(—i-1-pi=0.29933GPA+1.63635PSI-0.10147TUCEzk0.439204)(2.013681)(-1.019183)由二元Logit模型可得：i1+e(0.299336GPA+1.636357PSI-0.101472TUCE)i i i可知解釋變量PSI的Z統(tǒng)計(jì)量較大且其相應(yīng)的概率值比較小，這說明其在統(tǒng)計(jì)上是顯著的，從而說明PSI對(duì)提高學(xué)生成績有顯著的影響。然而解釋變量GPA，TUCE參數(shù)估計(jì)值相應(yīng)的概率值較大，統(tǒng)計(jì)上不顯著，說明這兩個(gè)變量對(duì)提高學(xué)生成績沒有顯著影響。 第5頁

.建立PROBIT模型如下：DependentVariable:GRADEMethod:ML-BinaryProbit(Quadratichillclimbing)Date:11/21/21Time:11:31Sample:132Includedobservations:32Convergenceachievedafter3iterationsCovariancematrixcomputedusingsecondderivatives實(shí)驗(yàn)結(jié)果分析及討論(續(xù))Variable實(shí)驗(yàn)結(jié)果分析及討論(續(xù))VariableCoefficientStd.Errorz-StatisticProb.GPA0.1763780.408920 0.4313260.6662PSI0.9786390.474634 2.0618820.0392TUCE-0.0602020.059089 -1.0188220.3083Meandependentvar0.343750S.D.dependentvar0.482559S.E.ofregression0.467831Akaikeinfocriterion1.363158Sumsquaredresid6.347100Schwarzcriterion1.500570Loglikelihood-18.81052Hannan-Quinncriter.1.408706Deviance37.62104Restr.deviance41.18346Avg.loglikelihood-0.587829ObswithDep=021Totalobs32ObswithDep=111根據(jù)上述結(jié)果，可以得出二元Probit模型結(jié)果為：Q(P=0.176378GPA+0.978639PSI-0.060202TUCEz~0.431326)(2.061882)(-1.018822)由回歸模型結(jié)果可得：P=①(0.176378GPA+0.978639PSI-0.060202TUCE)可知PSI對(duì)應(yīng)的P值為0.0392，小于0.05，從而表明PSI對(duì)提高學(xué)生成績有明顯的影響。而解釋變量GPA和TUCE相應(yīng)的概率較大，統(tǒng)計(jì)上不顯著，說明其對(duì)提高學(xué)生成績沒有明顯的影響。

.建立TOBIT模型如下:DependentVariable:YMethod:ML-CensoredNormal(TOBIT)(Quadratichillclimbing)Date:11/21/21Time:11:34Sample:150Includedobservations:50Leftcensoring(value)atzeroConvergenceachievedafter5iterationsCovariancematrixcomputedusingsecondderivatives實(shí)驗(yàn)結(jié)果分析及討論(續(xù))X1-6.2048323.179638實(shí)驗(yàn)結(jié)果分析及討論(續(xù))X1-6.2048323.179638-1.9514270.0510X2-0.9415340.326982-2.8794710.0040X33.0494971.3462202.2652300.0235X40.0001560.0001700.9153960.3600C18.6330320.644960.9025460.3668ErrorDistributionSCALE:C(6)22.790783.1028637.3450830.0000Meandependentvar19.16000S.D.dependentvar19.02454S.E.ofregression17.27325Akaikeinfocriterion6.628115Sumsquaredresid13128.06Schwarzcriterion6.857558Loglikelihood-159.7029Hannan-Quinncriter.6.715488Avg.loglikelihood-3.194058VariableCoefficientStd.Errorz-StatisticProb.Leftcensoredobs18Rightcensoredobs0Uncensoredobs32Totalobs50根據(jù)結(jié)果可以得出回歸方程：Y=18.63303”—6.204832)X1”—0.941534)X2+3.049497X3+0.000156X4可知XJX2的回歸系數(shù)值為-6.204832和-0.941534,且P值都比較顯著，