斯托克计量经济学第十章第十一章实证练习stata

计量经济学斯托克第四版实证答案十一章1、企业开出的商业汇票为银行承兑汇票，其无力支付票款时，应将应付票据的票面金额转作（）。

[单选题] *A.应付账款B.其他应付款C.预付账款D.短期借款(正确答案)2、某公司为一般纳税人，2019年6月购入商品并取得增值税专用发票，价款100万元，增值税率13％;支付运费取得增值税专用发票，运费不含税价款为30万元，增值税率9％，则该批商品的入账成本为（）。

[单选题] *A.130万元(正确答案)B.7万元C.3万元D.113万元3、企业生产车间发生的固定资产的修理费应计入（）科目。

[单选题] *A.制造费用B.生产成本C.长期待摊费用D.管理费用(正确答案)4、企业因解除与职工的劳动关系给予职工补偿而发生的职工薪酬，应借记的会计科目是（）。

[单选题] *A.管理费用(正确答案)B.计入存货成本或劳务成本C.营业外支出D.计入销售费用5、长期借款利息及外币折算差额，均应记入（）科目。

[单选题] *A.其他业务支出B.长期借款(正确答案)C.投资收益D.其他应付款6、固定资产报废清理后发生的净损失，应计入（）。

[单选题] *A.投资收益B.管理费用C.营业外支出(正确答案)D.其他业务成本7、下列各项税金中不影响企业损益的是（）。

[单选题] *A.消费税B.资源税C.增值税(正确答案)D.企业所得税8、股份有限公司为核算投资者投入的资本应当设置（）科目。

[单选题] *A.“实收资本”B.“股东权益”C.“股本”(正确答案)D.“所有者权益”9、．（年预测）下列属于货币资金转换为生产资金的经济活动的是（）[单选题] * A购买原材料B生产领用原材料C支付工资费用(正确答案)D销售产品10、企业生产经营期间发生的长期借款利息应计入（）科目。

[单选题] *A.在建工程B.财务费用(正确答案)C.开办费D.长期待摊费用11、某企业上年末“利润分配——未分配利润”科目贷方余额为50 000元，本年度实现利润总额为1 000 000元，所得税税率为25％，无纳税调整项目，本年按照10％提取法定盈余公积，应为（）元。

课后习题参考答案第二章教材习题与解析1、 判断下列表达式是否正确：y i =β0+β1x i ,i =1,2,⋯ny ̂i =β̂0+β̂1x i ,i =1,2,⋯nE(y i |x i )=β0+β1x i +u i ,i =1,2,⋯n E(y i |x i )=β0+β1x i ,i =1,2,⋯nE(y i |x i )=β̂0+β̂1x i ,i =1,2,⋯ny i =β0+β1x i +u i ,i =1,2,⋯ny ̂i =β̂0+β̂1x i +u i ,i =1,2,⋯n y i =β̂0+β̂1x i +u i ,i =1,2,⋯n y i =β̂0+β̂1x i +u ̂i ,i =1,2,⋯n y ̂i =β̂0+β̂1x i +u ̂i ,i =1,2,⋯n答案：对于计量经济学模型有两种类型，一是总体回归模型，另一是样本回归模型。

两类回归模型都具有确定形式与随机形式两种表达方式：总体回归模型的确定形式：X X Y E 10)|(ββ+= 总体回归模型的随机形式：μββ++=X Y 10样本回归模型的确定形式：X Y 10ˆˆˆββ+= 样本回归模型的随机形式：e X Y ++=10ˆˆββ 除此之外，其他的表达形式均是错误的2、给定一元线性回归模型：y =β0+β1x +u （1）叙述模型的基本假定；（2）写出参数β0和β1的最小二乘估计公式；（3）说明满足基本假定的最小二乘估计量的统计性质； （4）写出随机扰动项方差的无偏估计公式。

答案：（1）线性回归模型的基本假设有两大类，一类是关于随机误差项的，包括零均值、同方差、不序列相关、满足正态分布等假设；另一类是关于解释变量的，主要是解释变量是非随机的，如果是随机变量，则与随机误差项不相关。

（2）12ˆi iix yxβ=∑∑，01ˆˆY X ββ=- （3）考察总体的估计量，可从如下几个方面考察其优劣性：1）线性性，即它是否是另一个随机变量的线性函数； 2）无偏性，即它的均值或期望是否等于总体的真实值；3）有效值，即它是否在所有线性无偏估计量中具有最小方差；4）渐进无偏性，即样本容量趋于无穷大时，它的均值序列是否趋于总体真值； 5）一致性，即样本容量趋于无穷大时，它是否依概率收敛于总体的真值；6）渐进有效性，即样本容量趋于无穷大时，它在所有的一致估计量中是否具有最小的渐进方差。

计量经济学实证练习作业P106 第四章实证练习1⑴利用Eviews软件得到：Dependent Variable: AHEMethod: Least SquaresDate: 09/21/11 Time: 08:44Sample: 1 7986Included observations: 7986Variable Coefficient Std. Error t-Statistic Prob.AGE 0.451931 0.033526 13.48022 0.0000C 3.324184 1.002230 3.316787 0.0009R-squared 0.022254 Mean dependent var 16.77115Adjusted R-squared 0.022131 S.D. dependent var 8.758696S.E. of regression 8.661234 Akaike info criterion 7.155842Sum squared resid 598935.5 Schwarz criterion 7.157591Log likelihood -28571.28 F-statistic 181.7164Durbin-Watson stat 1.857141 Prob(F-statistic) 0.000000由此得出，平均每小时收入（AHE）对年龄（Age）的回归方程为：= 3.324184 + 0.451931×Age截距的估计值是3.324184，斜率的估计值是0.451931。

当工人年长一岁时，收入增加0.451931美元/小时。

⑵Bob：当Age=26时，AHE = 3.324184 + 0.451931×26 = 15.07439所以利用回归估计预测Bob的收入为15.07439美元/小时。

Alexis：当Age=30时，AHE = 3.324184 + 0.451931×30 = 16.88211所以利用回归估计预测Alexis的收入为16.88211美元/小时。

E4.1E4.21 jse 17 C : \ UsEiaVa sus\Deakt.op\T each! "gRat i"ca.dta172twoway scatter course_eval beauty.「，第一,可画课程评t★和萝三容能的段点匡3reg course_eval beauty, robust 匚1口5七日工山日3口七y)//第二月建立国.三.方程replace .- / -r- [H]4outreg2 asing 2 . docf5mean tiEaut”/算t：—nty的样本均值6logout, save (docJ) word replace : ir.ean beauty ;-zibeautit7sum EEautv“想计算"日口七产的标，佳差「logout, save (doc2) word replace: sum Iieauty9 sum cour s e_eva 1Z / 埴计算corn s e一sal的标准差,结合t―uty的标准差评价效应估计10 logout, save (doc3} word replace : snim. course evalE4.31 'jse :\UBErs\aEij.B\.Des]ctop\CoLLegeE'iBtanceweBt.dta"2leg ed. dist, rotoust. cluster [tlist}3 outreg'2 jsing 3 .doc, replaceE4.41 use n C:\U5er3\a3u3XDe3ktop\GEDwCh.dtfi w2twcway scatter growth tzadeshar一同绘制平启年亳长率对平均贸易额的敖卓民3leg growth cradeshare, robust 匚Luwter 111第三同建立口rmrth对七r^d巳写ti日工己的回！闩d outrea2；asina i. aoc工曰口工30■己，，.出回!3结果r5口工口口IX sun"y_n面ne=n MaJxaT"第四i司剔除马其他的数招6zegress growth tradeahare, robust cluater (trad^shmre) 」/身!除冬至.、二数书.~n^2. gi cwth^j-tr a de aha z e 的回!月~ outregS asxng 5 . doc t上三口J_me;三//导出回结果VARIABLES aheage 0.605(0.0245)Constant 1.082(0.688)Observations 7,711R-squared 0.029Robust standard errors in parentheses *** p<0.01, ** p<0.05, * p<0.11.① 截距估计值estimated intercept: 1,082② 斜率估计值estimated slope: 0,605回归方程：ahe= 1.082+0,605*age③当工人年长1岁，平均每小时工资增加0.605美元。

詹姆斯·斯托克,马克·沃森计量经济学第三章实证练习stata答案⼀、Two-sample t test with equal variancesGroup Obs Mean Std.Err. Std.Dev. 95% Conf. Interval1992 7,612 11.62 0.0644 5.619 11.49 11.742012 7,440 19.80 0.124 10.69 19.56 20.04combined 15,052 15.66 0.0770 9.442 15.51 15.81diff -8,183 0.139 -8.455 -7.911 diff = mean(1992) - mean(2012) t = -58.9871Ho: diff = 0 degrees of freedom = 15050Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000⼆、Two-sample t test with equal variancesGroup Obs Mean Std.Err. Std.Dev. 95% Conf. Interval 1992 7,612 15.64 0.0867 7.564 15.47 15.81 2012 7,440 19.80 0.124 10.69 19.56 20.04 combined 15,052 17.69 0.0772 9.471 17.54 17.85 diff -4.164 0.151 -4.459 -3.869diff = mean(1992) - mean(2012) t = -27.6423Ho: diff = 0 degrees of freedom = 15050Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000三、第⼆题根据通货膨胀率进⾏了调整，反映了购买⼒的变化，所以可⽤利⽤第⼆题的结果进⾏分析。

斯托克,沃森计量经济学第七章实证练习stataE7.2E7.3E7.4-------------------------------------------- (1) (2) ahe ahe -------------------------------------------- age 0.605*** 0.585*** (15.02) (16.02) female -3.664*** (-17.65)bachelor 8.083*** (38.00)_cons 1.082 -0.636 (0.93) (-0.59)（表2）Robust ci in parentheses*** p<0.01, ** p<0.05, * p<0.1-------------------------------------------- N 7711 7711 -------------------------------------------- t statistics in parentheses* p<0.10, ** p<0.05, *** p<0.01 (表1)（1）建⽴ahe 对age 的回归。

截距估计值是1.082，斜率估计值是0.605。

（2）①建⽴ahe 对age ，female 和bachelor 的回归。

Age 对收⼊的效应的估计值是0.585。

② age 回归系数的95%置信区间： (0.514，0.657)（3）设H 0:βa,（2）-βa,（1）=0 H1:βa,（2）-βa （1）≠0由表3，得SE ，SE(βa,（2）-βa,（1）)=√（0.0403）2+(0.0365)2=0.054t=（0.605-0.585）/0.054=0.37<1.96所以不拒绝原假设，即在5%显著⽔平下age 对ahe 的效应估计没有显著差异，所以（1）中的回归没有遭遇遗漏变量偏差。

计量经济学stata操作指南计量经济学stata操作（实验课）第一章stata基本知识1、stata窗口介绍2、基本操作（1）窗口锁定：Edit-preferences-general preferences-windowing-lock splitter （2）数据导入（3）打开文件：use E:\example.dta,clear（4）日期数据导入：gen newvar=date(varname, “ymd”)format newvar %td 年度数据gen newvar=monthly(varname, “ym”)format newvar %tm 月度数据gen newvar=quarterly(varname, “yq”)format newvar %tq 季度数据（5）变量标签Label variable tc ` “total output” ’（6）审视数据describelist x1 x2list x1 x2 in 1/5list x1 x2 if q>=1000drop if q>=1000keep if q>=1000（6）考察变量的统计特征summarize x1su x1 if q>=10000su q,detailsutabulate x1correlate x1 x2 x3 x4 x5 x6（7）画图histogram x1, width(1000) frequency kdensity x1scatter x1 x2twoway (scatter x1 x2) (lfit x1 x2) twoway (scatter x1 x2) (qfit x1 x2) （8）生成新变量gen lnx1=log(x1)gen q2=q^2gen lnx1lnx2=lnx1*lnx2gen larg=(x1>=10000)rename larg largeg large=(q>=6000)replace large=(q>=6000)drop ln*（8）计算功能display log(2)（9）线性回归分析regress y1 x1 x2 x3 x4vce #显示估计系数的协方差矩阵reg y1 x1 x2 x3 x4,noc #不要常数项reg y1 x1 x2 x3 x4 if q>=6000reg y1 x1 x2 x3 x4 if largereg y1 x1 x2 x3 x4 if large==0reg y1 x1 x2 x3 x4 if ～large predict yhatpredict e1,residualdisplay 1/_b[x1]test x1=1 # F检验，变量x1的系数等于1test (x1=1) (x2+x3+x4=1) # F联合假设检验test x1 x2 #系数显著性的联合检验testnl _b[x1]= _b[x2]^2（10）约束回归constraint def 1 x1+x2+x3=1cnsreg y1 x1 x2 x3 x4,c(1)cons def 2 x4=1cnsreg y1 x1 x2 x3 x4,c(1-2)（11）stata的日志File-log-begin-输入文件名log off 暂时关闭log on 恢复使用log close 彻底退出（12）stata命令库更新Update allhelp command第二章有关大样本ols的stata命令及实例（1）ols估计的稳健标准差reg y x1 x2 x3,robust（2）实例use example.dta,clearreg y1 x1 x2 x3 x4test x1=1reg y1 x1 x2 x3 x4,rtestnl _b[x1]=_b[x2]^2第三章最大似然估计法的stata命令及实例（1）最大似然估计help ml（2）LR检验lrtest #对面板数据中的异方差进行检验（3）正态分布检验sysuse auto #调用系统数据集auto.dtahist mpg,normalkdensity mpg,normalqnorm mpg*手工计算JB统计量sum mpg,detaildi (r(N)/6)*((r(skewness)^2)+[(1/4)*(r(kurtosis)-3)^2]) di chi2tail(自由度，上一步计算值)*下载非官方程序ssc install jb6jb6 mpg*正态分布的三个检验sktest mpgswilk mpgsfrancia mpg*取对数后再检验gen lnmpg=log(mpg)kdensity lnmpg, normaljb6 lnmpgsktest lnmpg第四章处理异方差的stata命令及实例（1）画残差图rvfplotrvfplot varname*例题use example.dta,clearreg y x1 x2 x3 x4rvfplot # 与拟合值的散点图rvfplot x1 # 画残差与解释变量的散点图（2）怀特检验estat imtest,white*下载非官方软件ssc install whitetst（3）BP检验estat hettest #默认设置为使用拟合值estat hettest,rhs #使用方程右边的解释变量estat hettest [varlist] #指定使用某些解释变量estat hettest,iidestat hettest,rhs iidestat hettest [varlist],iid（4）WLSreg y x1 x2 x3 x4 [aw=1/var]*例题quietly reg y x1 x2 x3 x4predict e1,resgen e2=e1^2gen lne2=log(e2)reg lne2 x2,nocpredict lne2fgen e2f=exp(lne2f)reg y x1 x2 x3 x4 [aw=1/e2f]（5）stata命令的批处理（写程序）Window-do-file editor-new do-file#WLS for examplelog using E:\wls_example.smcl,replaceset more offuse E:\example.dta,clearreg y x1 x2 x3 x4predict e1,resgen e2=e1^2g lne2=log(e2)reg lne2 x2,nocpredict lne2fg e2f=exp(lne2f)*wls regressionreg y x1 x2 x3 x4 [aw=1/e2f]log closeexit第五章处理自相关的stata命令及实例（1）滞后算子/差分算子tsset yearl.l2.D.D2.LD.（2）画残差图scatter e1 l.e1ac e1pac e1（3）BG检验estat bgodfrey（默认p=1）estat bgodfrey,lags(p)estat bgodfrey,nomiss0（使用不添加0的BG检验）（4）Ljung-Box Q检验reg y x1 x2 x3 x4predict e1,residwntestq e1wntestq e1,lags(p)* wntestq指的是“white noise test Q”，因为白噪声没有自相关（5）DW检验做完OLS回归后，使用estat dwatson（6）HAC稳健标准差newey y x1 x2 x3 x4,lag(p)reg y x1 x2 x3 x4,cluster(varname)（7）处理一阶自相关的FGLSprais y x1 x2 x3 x4 （使用默认的PW估计方法）prais y x1 x2 x3 x4,corc （使用CO估计法）（8）实例use icecream.dta, cleartsset timegraph twoway connect consumption temp100 time, msymbol(circle) msymbol(triangle) reg consumption temp price incomepredict e1, resg e2=l.e1twoway (scatter e1 e2) (lfit e1 e2)ac e1pac e1estat bgodfreywntestq e1estat dwatsonnewey consumption temp price income, lag (3)prais consumption temp price income, corcprais consumption temp price income, nologreg consumption temp l.temp price incomeestat bgodfreyestat dwatson第六章模型设定与数据问题（1）解释变量的选择reg y x1 x2 x3estat ic*例题use icecream.dta, clearreg consumption temp price incomeestat icreg consumption temp l.temp price incomeestat ic（2）对函数形式的检验（reset检验）reg y x1 x2 x3estat ovtest （使用被解释变量的2、3、4次方作为非线性项）estat ovtest, rhs （使用解释变量的幂作为非线性项，ovtest-omitted variable test）*例题use nerlove.dta, clearreg lntc lnq lnpl lnpk lnpfestat ovtestg lnq2=lnq^2reg lntc lnq lnq2 lnpl lnpk lnpfestat ovtest（3）多重共线性estat vif*例题use nerlove.dta, clearreg lntc lnq lnpl lnpk lnpfestat vif（4）极端数据reg y x1 x2 x3predict lev, leverage （列出所有解释变量的lev值）gsort –levsum levlist lev in 1/3*例题use nerlove.dta, clearquietly reg lntc lnq lnpl lnpk lnpfpredict lev, leveragesum levgsort –levlist lev in 1/3（5）虚拟变量gen d=(year>=1978)tabulate province, generate (pr)reg y x1 x2 x3 pr2-pr30（6）经济结构变动的检验方法1：use consumption_china.dta, cleargraph twoway connect c y year, msymbol(circle) msymbol(triangle)reg c yreg c y if year<1992reg c y if year>=1992计算F统计量方法2：gen d=(year>1991)gen yd=y*dreg c y d ydtest d yd第七章工具变量法的stata命令及实例（1）2SLS的stata命令ivregress 2sls depvar [varlist1] (varlist2=instlist)如：ivregress 2sls y x1 (x2=z1 z2)ivregress 2sls y x1 (x2 x3=z1 z2 z3 z4) ,r firstestat firststage,all forcenonrobust （检验弱工具变量的命令）ivregress liml depvar [varlist 1] (varlist2=instlist)estat overid （过度识别检验的命令）*对解释变量内生性的检验（hausman test），缺点：不适合于异方差的情形reg y x1 x2estimates store olsivregress 2sls y x1 (x2=z1 z2)estimates store ivhausman iv ols, constant sigmamore*DWH检验estat endogenous*GMM的过度识别检验ivregress gmm y x1 (x2=z1 z2) （两步GMM）ivregress gmm y x1 (x2=z1 z2),igmm （迭代GMM）estat overid*使用异方差自相关稳健的标准差GMM命令ivregress gmm y x1 (x2=z1 z2), vce (hac nwest[#])（2）实例use grilic.dta,clearsumcorr iq sreg lw s expr tenure rns smsa,rreg lw s iq expr tenure rns smsa,rivregress 2sls lw s expr tenure rns smsa (iq=med kww mrt age),restat overidivregress 2sls lw s expr tenure rns smsa (iq=med kww),r first estat overidestat firststage, all forcenonrobust （检验工具变量与内生变量的相关性）ivregress liml lw s expr tenure rns smsa (iq=med kww),r *内生解释变量检验quietly reg lw s iq expr tenure rns smsaestimates store olsquietly ivregress 2sls lw s expr tenure rns smsa (iq=med kww) estimates store ivhausman iv ols, constant sigmamoreestat endogenous （存在异方差的情形）*存在异方差情形下，GMM比2sls更有效率ivregress gmm lw s expr tenure rns smsa (iq=med kww)estat overidivregress gmm lw s expr tenure rns smsa (iq=med kww),igmm*将各种估计方法的结果存储在一张表中quietly ivregress gmm lw s expr tenure rns smsa (iq=med kww)estimates store gmmquietly ivregress gmm lw s expr tenure rns smsa (iq=med kww),igmmestimates store igmmestimates table gmm igmm第八章短面板的stata命令及实例（1）面板数据的设定xtset panelvar timevarencode country,gen(cntry) （将字符型变量转化为数字型变量）xtdesxtsumxttab varnamextline varname,overlay*实例use traffic.dta,clearxtset state yearxtdesxtsum fatal beertax unrate state yearxtline fatal（2）混合回归reg y x1 x2 x3,vce(cluster id)如：reg fatal beertax unrate perinck,vce(cluster state)estimates store ols对比：reg fatal beertax unrate perinck（3）固定效应xtreg y x1 x2 x3,fe vce(cluster id)xi:reg y x1 x2 x3 i.id,vce(cluster id) （LSDV法）xtserial y x1 x2 x3,output （一阶差分法，同时报告面板一阶自相关）estimates store FD*双向固定效应模型tab year, gen (year)xtreg fatal beertax unrate perinck year2-year7, fe vce (cluster state)estimates store FE_TWtest year2 year3 year4 year5 year6 year7（4）随机效应xtreg y x1 x2 x3,re vce(cluster id) （随机效应FGLS）xtreg y x1 x2 x3,mle （随机效应MLE）xttest0 （在执行命令xtreg, re 后执行，进行LM检验）（5）组间估计量xtreg y x1 x2 x3,be（6）固定效应还是随机效应：hausman testxtreg y x1 x2 x3,feestimates store fextreg y x1 x2 x3,reestimates store rehausman fe re,constant sigmamore （若使用了vce(cluster id)，则无法直接使用该命令，解决办法详见P163）estimates table ols fe_robust fe_tw re be, b se （将主要回归结果列表比较）第九章长面板与动态面板（1）仅解决组内自相关的FGLSxtpcse y x1 x2 x3 ,corr(ar1) （具有共同的自相关系数）xtpcse y x1 x2 x3 ,corr(psar1) （允许每个面板个体有自身的相关系数）例题：use mus08cigar.dta,cleartab state,gen(state)gen t=year-62reg lnc lnp lnpmin lny state2-state10 t,vce(cluster state)estimates store OLSxtpcse lnc lnp lnpmin lny state2-state10 t,corr(ar1) (考虑存在组内自相关，且各组回归系数相同)estimates store AR1xtpcse lnc lnp lnpmin lny state2-state10 t,corr(psar1) (考虑存在组内自相关，且各组回归系数不相同)estimates store PSAR1xtpcse lnc lnp lnpmin lny state2-state10 t, hetonly (仅考虑不同个体扰动性存在异方差，忽略自相关)estimates store HETONL Yestimates table OLS AR1 PSAR1 HETONL Y, b se（2）同时处理组内自相关与组间同期相关的FGLSxtgls y x1 x2 x3,panels (option/iid/het/cor) corr(option/ar1/psar1) igls注：执行上述xtpcse、xtgls命令时，如果没有个体虚拟变量，则为随机效应模型；如果加上个体虚拟变量，则为固定效应模型。

计量经济学分章练习题第一章习题一、判断题1.投入产出模型和数学规划模型都是计量经济模型。

（×）2.弗里希因创立了计量经济学从而获得了诺贝尔经济学奖。

（√）3.丁伯根因创立了建立了第1个计量经济学应用模型从而获得了诺贝尔经济学奖。

（√）4.格兰杰因在协整理论上的贡献而获得了诺贝尔经济学奖。

（√）5.赫克曼因在选择性样本理论上的贡献而获得了诺贝尔经济学奖。

（√）二、名词解释1．计量经济学，经济学的一个分支学科，是对经济问题进行定量实证研究的技术、方法和相关理论。

2．计量经济学模型，是一个或一组方程表示的经济变量关系以及相关条件或假设，是经济问题相关方面之间数量联系和制约关系的基本描述。

3．计量经济检验，由计量经济学理论决定的，目的在于检验模型的计量经济学性质。

通常最主要的检验准则有随机误差项的序列相关检验和异方差性检验，解释变量的多重共线性检验等。

4．截面数据，指在同一个时点上，对不同观测单位观测得到的多个数据构成的数据集。

5．面板数据，是由对许多个体组成的同一个横截面，在不同时点的观测数据构成的数据。

三、单项选择题1.把反映某一单位特征的同一指标的数据，按一定的时间顺序和时间间隔排列起来，这样的数据称为（ B ）A. 横截面数据B. 时间序列数据C. 面板数据D. 原始数据2.同一时间、不同单位按同一统计指标排列的观测数据称为（ C ）A．原始数据 B．时间序列数据C．截面数据 D．面板数据3.不同时间、不同单位按同一统计指标排列的观测数据称为（ D ）A．原始数据 B．时间序列数据C．截面数据 D．面板数据4.对计量经济模型进行的结构分析不包括（ D ）A．乘数分析 B．弹性分析C．比较静态分析 D．随机分析5.一个普通家庭的每月所消费的水费和电费是（ B ）A．因果关系 B．相关关系C．恒等关系 D．不相关关系6.中国的居民消费和GDP是（ C ）A．因果关系 B．相关关系C．相互影响关系 D．不相关关系7. 下列（ B ）是计量经济模型A ．01i Y X ββ=+B ．01i i Y X ββμ=++C ．投入产出模型D ．其他8. 投资是（ A ）经济变量A ．流量B ．存量C ．派生D ．虚拟变量9. 资本是（ B ）经济变量A ．流量B ．存量C ．派生D ．虚拟变量10. 对定性因素进行数量化处理，需要定义和引进（ C ）A ．宏观经济变量B ．微观经济变量C ．虚拟变量D ．派生变量四、计算分析题1.“计量经济模型就是数学”这种说法正确吗，为什么?计量经济学模型不是数学式子，相比数学式子多了一个随机误差项，是随机性的函数关系。

第十章练习题及参考解答10.1表10.10是某国的宏观经济季度数据。

其中, GDP为国内生产总值，PDI为个人可支配收入，PCE为个人消费支出，利润为公司税后利润，红利为公司净红利支出。

表10.10 1980~2001年某国宏观经济季度数据（单位: 亿元）1） 画出利润和红利的散点图，并直观地考察这两个时间序列是否是平稳的。

2) 应用单位根检验分别检验利润和红利两个时间序列是否是平稳的。

3) 分别检验GDP 、PDI 和PCE 等序列是否平稳，并判定其单整阶数是否相同？练习题10.1参考解答:1) 利润和红利的散点图如下,从图中可看出，利润和红利序列均值和方差不稳定，因此可能是非平稳的。

2）利润序列有截距项，在Eviews5.0中选取截距项，同时最大滞后长度取11进行单位根检验，检验结果如下，Null Hypothesis: PFT has a unit root Exogenous: Constant, Linear TrendLag Length: 0 (Automatic based on SIC, MAX LAG=11)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -1.797079 0.6978Test critical values:1% level -4.066981 5% level -3.46229210% level-3.157475t 统计量大于所有显著性水平下的MacKinnon 临界值，故不能拒绝原假设，该序列是不平稳的。

红利序列有截距项和趋势项，在Eviews5.0中选取截距项和趋势项，同时最大滞后长度取11进行单位根检验，检验结果如下，Null Hypothesis: BNU has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic based on SIC, MAX LAG=11)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.893559 0.1698Test critical values: 1% level -4.0682905% level -3.46291210% level -3.157836*MacKinnon (1996) one-sided p-values.t统计量大于所有显著性水平下的MacKinnon临界值，故不能拒绝原假设，该序列是不平稳的。

© 陈强，2015年，《计量经济学及Stata应用》，高等教育出版社。

第11章二值选择模型11.1 二值选择模型如果被解释变量y离散，称为“离散选择模型”(discrete choice model)或“定性反应模型”(qualitative response model)。

最常见的离散选择模型是二值选择行为(binary choices)。

比如：考研或不考研；就业或待业；买房或不买房；买保险或不买保险；贷款申请被批准或拒绝；出国或不出国；回国或不回12国；战争或和平；生或死。

假设个体只有两种选择，比如1y =(考研)或0y =(不考研)。

最简单的建模方法为“线性概率模型”(Linear Probability Model ，LPM)：1122(1,,)i i i K iK i i i y x x x i n βββεε'=+=+= +++x β (11.1)其中，解释变量12()i i i iK x x x '≡ x ，而参数12()K βββ'≡ β。

LPM 的优点是，计算方便，容易得到边际效应(即回归系数)。

3LPM 的缺点是，虽然y 的取值非0即1，但根据线性概率模型所作的预测值却可能出现ˆ1y>或ˆ0y <的不现实情形。

图11.1 线性概率模型4为使y 的预测值介于[0,1]之间，在给定x 的情况下，考虑y 的两点分布概率：P(1|)(,)P(0|)1(,)y F y F ==⎧⎨==-⎩x x x x ββ (11.2)函数(,)F x β称为“连接函数”(link function) ，因为它将x 与y 连接起来。

y 的取值要么为0，要么为1，故y 肯定服从两点分布。

连接函数的选择具有一定灵活性。

通过选择合适的连接函数(,)F x β(比如，某随机变量的累积分布函数)，可保证ˆ01y≤≤，并将ˆy 理解为“1y =”发生的概率，因为5E(|)1P(1|)0P(0|)P(1|)y y y y =⋅=+⋅===x x x x (11.3)如果(,)F x β为标准正态的累积分布函数，则P(1|)(,)()()y F t dt φ'-∞'===Φ≡⎰x x x x βββ (11.4)()φ⋅与()Φ⋅分别为标准正态的密度与累积分布函数；此模型称为“Probit ”。

E10.1(1) (2) (3) (4) (5)lnvio lnvio lnvio lnvio lnvio shall -0.443***-0.368***-0.0461*-0.288***-0.0280(0.0475) (0.0348) (0.0189) (0.0337) (0.0278)incarc_rate 0.00161***-0.0000710 0.00193***0.0000760(0.000181) (0.0000936) (0.000114) (0.0000720)density 0.0267 -0.172*-0.00887 -0.0916(0.0143) (0.0850) (0.0139) (0.0485)avginc 0.00121 -0.00920 0.0129 0.000959(0.00728) (0.00591) (0.00796) (0.00729)pop 0.0427***0.0115 0.0408***-0.00475(0.00315) (0.00872) (0.00252) (0.00781)pb1064 0.0809***0.104***0.1000***0.0292(0.0200) (0.0178) (0.0182) (0.0183)pw1064 0.0312**0.0409***0.0401***0.00925(0.00973) (0.00507) (0.00912) (0.00538)pm1029 0.00887 -0.0503***-0.0444*0.0733***(0.0121) (0.00640) (0.0175) (0.0129)_cons 6.135*** 2.982*** 3.866*** 2.948*** 4.348***(0.0193) (0.609) (0.385) (0.569) (0.435) N 1173 1173 1173 1173 1173R20.087 0.564 0.218 0.580 0.955adj. R2 State Effects Time Effects 0.0859NoNo0.5613NoNo0.1771YesNo0.5690NoYes0.9525YesYesStandard errors in parentheses*p < 0.10, **p < 0.05, ***p < 0.01（1）①回归（2）中shall的系数是-0.368，这意味着隐蔽武器法律，也即“准予”携带法律，约使暴力犯罪减少36.8%。

斯托克计量经济学课后习题实证答案P ART T WO Solutions to EmpiricalExercisesChapter 3Review of StatisticsSolutions to Empirical Exercises1. (a)Average Hourly Earnings, Nominal $’sMean SE(Mean) 95% Confidence Interval AHE199211.63 0.064 11.50 11.75AHE200416.77 0.098 16.58 16.96Difference SE(Difference) 95% Confidence Interval AHE2004 AHE1992 5.14 0.117 4.91 5.37(b)Average Hourly Earnings, Real $2004Mean SE(Mean) 95% Confidence Interval AHE199215.66 0.086 15.49 15.82AHE200416.77 0.098 16.58 16.96Difference SE(Difference) 95% Confidence Interval AHE2004 AHE1992 1.11 0.130 0.85 1.37(c) The results from part (b) adjust for changes in purchasing power. These results should be used.(d)Average Hourly Earnings in 2004Mean SE(Mean) 95% Confidence Interval High School13.81 0.102 13.61 14.01College20.31 0.158 20.00 20.62Difference SE(Difference) 95% Confidence Interval College High School 6.50 0.188 6.13 6.87Solutions to Empirical Exercises in Chapter 3 109(e)Average Hourly Earnings in 1992 (in $2004)Mean SE(Mean) 95% Confidence Interval High School13.48 0.091 13.30 13.65 College19.07 0.148 18.78 19.36Difference SE(Difference) 95% Confidence Interval College High School5.59 0.173 5.25 5.93(f) Average Hourly Earnings in 2004Mean SE(Mean) 95% Confidence Interval AHE HS ,2004AHE HS ,19920.33 0.137 0.06 0.60 AHE Col ,2004AHE Col ,19921.24 0.217 0.82 1.66Col–HS Gap (1992)5.59 0.173 5.25 5.93 Col–HS Gap (2004)6.50 0.188 6.13 6.87Difference SE(Difference) 95% Confidence Interval Gap 2004 Gap 1992 0.91 0.256 0.41 1.41Wages of high school graduates increased by an estimated 0.33 dollars per hour (with a 95%confidence interval of 0.06 0.60); Wages of college graduates increased by an estimated 1.24dollars per hour (with a 95% confidence interval of 0.82 1.66). The College High School gap increased by an estimated 0.91 dollars per hour.(g) Gender Gap in Earnings for High School Graduates Yearm Y s m n m w Y s w n w m Y w Y SE (m Y w Y )95% CI 199214.57 6.55 2770 11.86 5.21 1870 2.71 0.173 2.37 3.05 200414.88 7.16 2772 11.92 5.39 1574 2.96 0.192 2.59 3.34There is a large and statistically significant gender gap in earnings for high school graduates.In 2004 the estimated gap was $2.96 per hour; in 1992 the estimated gap was $2.71 per hour(in $2004). The increase in the gender gap is somewhat smaller for high school graduates thanit is for college graduates.Chapter 4Linear Regression with One RegressorSolutions to Empirical Exercises1. (a) ·AHE 3.32 0.45 u AgeEarnings increase, on average, by 0.45 dollars per hour when workers age by 1 year.(b) Bob’s predicted earnings 3.32 0.45 u 26 $11.70Alexis’s predicted earnings 3.32 0.45 u 30 $13.70(c) The R2 is 0.02.This mean that age explains a small fraction of the variability in earnings acrossindividuals.2. (a)There appears to be a weak positive relationship between course evaluation and the beauty index.Course Eval 4.00 0.133 u Beauty. The variable Beauty has a mean that is equal to 0; the(b) ·_estimated intercept is the mean of the dependent variable (Course_Eval) minus the estimatedslope (0.133) times the mean of the regressor (Beauty). Thus, the estimated intercept is equalto the mean of Course_Eval.(c) The standard deviation of Beauty is 0.789. ThusProfessor Watson’s predicted course evaluations 4.00 0.133 u 0 u 0.789 4.00Professor Stock’s predicted course evaluations 4.00 0.133 u 1 u 0.789 4.105Solutions to Empirical Exercises in Chapter 4 111(d) The standard deviation of course evaluations is 0.55 and the standard deviation of beauty is0.789. A one standard deviation increase in beauty is expected to increase course evaluation by0.133 u 0.789 0.105, or 1/5 of a standard deviation of course evaluations. The effect is small.(e) The regression R2 is 0.036, so that Beauty explains only3.6% of the variance in courseevaluations.3. (a) ?Ed 13.96 0.073 u Dist. The regression predicts that if colleges are built 10 miles closerto where students go to high school, average years of college will increase by 0.073 years.(b) Bob’s predicted years of completed education 13.960.073 u 2 13.81Bob’s predicted years of completed education if he was 10 miles from college 13.96 0.073 u1 13.89(c) The regression R2 is 0.0074, so that distance explains only a very small fraction of years ofcompleted education.(d) SER 1.8074 years.4. (a)Yes, there appears to be a weak positive relationship.(b) Malta is the “outlying” observation with a trade share of 2.(c) ·Growth 0.64 2.31 u TradesharePredicted growth 0.64 2.31 u 1 2.95(d) ·Growth 0.96 1.68 u TradesharePredicted growth 0.96 1.68 u 1 2.74(e) Malta is an island nation in the Mediterranean Sea, south of Sicily. Malta is a freight transportsite, which explains its larg e “trade share”. Many goods coming into Malta (imports into Malta)and immediately transported to other countries (as exports from Malta). Thus, Malta’s importsand exports and unlike the imports and exports of most other countries. Malta should not beincluded in the analysis.Chapter 5Regression with a Single Regressor:Hypothesis Tests and Confidence IntervalsSolutions to Empirical Exercises1. (a) ·AHE 3.32 0.45 u Age(0.97) (0.03)The t -statistic is 0.45/0.03 13.71, which has a p -value of 0.000, so the null hypothesis can berejected at the 1% level (and thus, also at the 10% and 5% levels).(b) 0.45 r 1.96 u 0.03 0.387 to 0.517(c) ·AHE 6.20 0.26 u Age(1.02) (0.03)The t -statistic is 0.26/0.03 7.43, which has a p -value of 0.000, so the null hypothesis can berejected at the 1% level (and thus, also at the 10% and 5% levels).(d) ·AHE 0.23 0.69 u Age(1.54) (0.05)The t -statistic is 0.69/0.05 13.06, which has a p -value of 0.000, so the null hypothesis can berejected at the 1% level (and thus, also at the 10% and 5% levels).(e) The difference in the estimated E 1 coefficients is 1,1,??College HighScool E E 0.69 0.26 0.43. Thestandard error of for the estimated difference is SE 1,1,??()College HighScoolE E (0.032 0.052)1/2 0.06, so that a 95% confidence interval for the difference is 0.43 r 1.96 u 0.06 0.32 to 0.54(dollars per hour).2. ·_ 4.000.13CourseEval Beauty u (0.03) (0.03)The t -statistic is 0.13/0.03 4.12, which has a p -value of 0.000, so the null hypothesis can be rejectedat the 1% level (and thus, also at the 10% and 5% levels).3. (a) ?Ed13.96 0.073 u Dist (0.04) (0.013)The t -statistic is 0.073/0.013 5.46, which has a p -value of 0.000, so the null hypothesis can be rejected at the 1% level (and thus, also at the 10% and 5% levels).(b) The 95% confidence interval is 0.073 r 1.96 u 0.013 or0.100 to 0.047.(c) ?Ed13.94 0.064 u Dist (0.05) (0.018)Solutions to Empirical Exercises in Chapter 5 113(d) ?Ed13.98 0.084 u Dist (0.06) (0.013)(e) The difference in the estimated E 1 coefficients is 1,1,??Female Male E E 0.064 ( 0.084) 0.020.The standard error of for the estimated difference is SE 1,1,??()Female Male E E (0.0182 0.0132)1/20.022, so that a 95% confidence interval for the difference is 0.020 r 1.96 u 0.022 or 0.022 to0.064. The difference is not statistically different.Chapter 6Linear Regression with Multiple RegressorsSolutions to Empirical Exercises1. Regressions used in (a) and (b)Regressor a bBeauty 0.133 0.166Intro 0.011OneCredit 0.634Female 0.173Minority 0.167NNEnglish 0.244Intercept 4.00 4.07SER 0.545 0.513R2 0.036 0.155(a) The estimated slope is 0.133(b) The estimated slope is 0.166. The coefficient does not change by an large amount. Thus, theredoes not appear to be large omitted variable bias.(c) Professor Smith’s predicted course evaluation (0.166 u 0)0.011 u 0) (0.634 u 0) (0.173 u0) (0.167 u 1) (0.244 u 0) 4.068 3.9012. Estimated regressions used in questionModelRegressor a bdist 0.073 0.032bytest 0.093female 0.145black 0.367hispanic 0.398incomehi 0.395ownhome 0.152dadcoll 0.696cue80 0.023stwmfg80 0.051intercept 13.956 8.827SER 1.81 1.84R2 0.007 0.279R0.007 0.277Solutions to Empirical Exercises in Chapter 6 115(a) 0.073(b) 0.032(c) The coefficient has fallen by more than 50%. Thus, it seems that result in (a) did suffer fromomitted variable bias.(d) The regression in (b) fits the data much better as indicated by the R2, 2,R and SER. The R2 and R are similar because the number of observations is large (n 3796).(e) Students with a “dadcoll 1” (so that the student’s father went to college) complete 0.696 moreyears of education, on average, than students with “dadcoll 0” (so that the student’s father didnot go to college).(f) These terms capture the opportunity cost of attending college. As STWMFG increases, forgonewages increase, so that, on average, college attendance declines. The negative sign on thecoefficient is consistent with this. As CUE80 increases, it is more difficult to find a job, whichlowers the opportunity cost of attending college, so that college attendance increases. Thepositive sign on the coefficient is consistent with this.(g) Bob’s predicted years of education 0.0315 u 2 0.093 u58 0.145 u 0 0.367 u 1 0.398 u0 0.395 u 1 0.152 u 1 0.696 u 0 0.023 u 7.5 0.051 u 9.75 8.82714.75(h) Jim’s expected years of education is 2 u 0.0315 0.0630 less than Bob’s. Thus, Jim’s expectedyears of education is 14.75 0.063 14.69.3.Variable Mean StandardDeviation Unitsgrowth 1.86 1.82 Percentage Pointsrgdp60 3131 2523 $1960tradeshare 0.542 0.229 unit freeyearsschool 3.95 2.55 yearsrev_coups 0.170 0.225 coups per yearassasinations 0.281 0.494 assasinations per yearoil 0 0 0–1 indicator variable (b) Estimated Regression (in table format):Regressor Coefficienttradeshare 1.34(0.88)yearsschool 0.56**(0.13)rev_coups 2.15*(0.87)assasinations 0.32(0.38)rgdp60 0.00046**(0.00012)intercept 0.626(0.869)SER 1.59R2 0.29R0.23116 Stock/Watson - Introduction to Econometrics - Second EditionThe coefficient on Rev_Coups is í2.15. An additional coup in a five year period, reduces theaverage year growth rate by (2.15/5) = 0.43% over this 25 year period. This means the GPD in 1995 is expected to be approximately .43×25 = 10.75% lower. This is a larg e effect.(c) The 95% confidence interval is 1.34 r 1.96 u 0.88 or 0.42 to 3.10. The coefficient is notstatistically significant at the 5% level.(d) The F-statistic is 8.18 which is larger than 1% critical value of 3.32.Chapter 7Hypothesis Tests and Confidence Intervals in Multiple RegressionSolutions to Empirical Exercises1. Estimated RegressionsModelRegressor a bAge 0.45(0.03)0.44 (0.03)Female 3.17(0.18)Bachelor 6.87(0.19)Intercept 3.32(0.97)SER 8.66 7.88R20.023 0.1902R0.022 0.190(a) The estimated slope is 0.45(b) The estimated marginal effect of Age on AHE is 0.44 dollars per year. The 95% confidenceinterval is 0.44 r 1.96 u 0.03 or 0.38 to 0.50.(c) The results are quite similar. Evidently the regression in (a) does not suffer from importantomitted variable bias.(d) Bob’s predicted average hourly earnings 0.44 u 26 3.17 u 0 6.87 u 0 3.32 $11.44Alexis’s predicted average hourly earnings 0.44 u 30 3.17 u 1 6.87 u 1 3.32 $20.22 (e) The regression in (b) fits the data much better. Gender and education are important predictors of earnings. The R2 and R are similar because the sample size is large (n 7986).(f) Gender and education are important. The F-statistic is 752, which is (much) larger than the 1%critical value of 4.61.(g) The omitted variables must have non-zero coefficients and must correlated with the includedregressor. From (f) Female and Bachelor have non-zero coefficients; yet there does not seem to be important omittedvariable bias, suggesting that the correlation of Age and Female and Age and Bachelor is small. (The sample correlations are ·Cor(Age, Female) 0.03 and·Cor(Age,Bachelor) 0.00).118 Stock/Watson - Introduction to Econometrics - Second Edition2.ModelRegressor a b cBeauty 0.13**(0.03) 0.17**(0.03)0.17(0.03)Intro 0.01(0.06)OneCredit 0.63**(0.11) 0.64** (0.10)Female 0.17**(0.05) 0.17** (0.05)Minority 0.17**(0.07) 0.16** (0.07)NNEnglish 0.24**(0.09) 0.25** (0.09)Intercept 4.00**(0.03) 4.07**(0.04)4.07**(0.04)SER 0.545 0.513 0.513R2 0.036 0.155 0.1552R0.034 0.144 0.145(a) 0.13 r 0.03 u 1.96 or 0.07 to 0.20(b) See the table above. Intro is not significant in (b), but the other variables are significant.A reasonable 95% confidence interval is 0.17 r 1.96 u 0.03 or0.11 to 0.23.Solutions to Empirical Exercises in Chapter 7 119 3.ModelRegressor (a) (b) (c)dist 0.073**(0.013) 0.031**(0.012)0.033**(0.013)bytest 0.092**(0.003) 0.093** (.003)female 0.143**(0.050) 0.144** (0.050)black 0.354**(0.067) 0.338** (0.069)hispanic 0.402**(0.074) 0.349** (0.077)incomehi 0.367**(0.062) 0.374** (0.062)ownhome 0.146*(0.065) 0.143* (0.065)dadcoll 0.570**(0.076) 0.574** (0.076)momcoll 0.379**(0.084) 0.379** (0.084)cue80 0.024**(0.009) 0.028** (0.010)stwmfg80 0.050*(0.020) 0.043* (0.020)urban 0.0652(0.063) tuition 0.184(0.099)intercept 13.956**(0.038) 8.861**(0.241)8.893**(0.243)F-statitisticfor urban and tuitionSER 1.81 1.54 1.54R2 0.007 0.282 0.284R0.007 0.281 0.281(a) The group’s claim is that the coefficien t on Dist is 0.075 ( 0.15/2). The 95% confidence forE Dist from column (a) is 0.073 r 1.96 u 0.013 or 0.099 to 0.046. The group’s claim is includedin the 95% confidence interval so that it is consistent with the estimated regression.120 Stock/Watson - Introduction to Econometrics - Second Edition(b) Column (b) shows the base specification controlling for other important factors. Here thecoefficient on Dist is 0.031, much different than the resultsfrom the simple regression in (a);when additional variables are added (column (c)), the coefficient on Dist changes little from the result in (b). From the base specification (b), the 95% confidence interval for E Dist is0.031 r1.96 u 0.012 or 0.055 to 0.008. Similar results are obtained from the regression in (c).(c) Yes, the estimated coefficients E Black and E Hispanic are positive, large, and statistically significant.Chapter 8Nonlinear Regression FunctionsSolutions to Empirical Exercises1. This table contains the results from seven regressions that are referenced in these answers.Data from 2004(1) (2) (3) (4) (5) (6) (7) (8)Dependent VariableAHE ln(AHE) ln(AHE) ln(AHE) ln(AHE) ln(AHE) ln(AHE) ln(AHE) Age 0.439**(0.030) 0.024**(0.002)0.147**(0.042)0.146**(0.042)0.190**(0.056)0.117*(0.056)0.160Age2 0.0021**(0.0007) 0.0021** (0.0007)0.0027**(0.0009)0.0017(0.0009)0.0023(0.0011)ln(Age) 0.725**(0.052)Female u Age 0.097 (0.084) 0.123 (0.084) Female u Age2 0.0015 (0.0014)0.0019 (0.0014) Bachelor u Age 0.064 (0.083)0.091 (0.084) Bachelor u Age2 0.0009 (0.0014) 0.0013 (0.0014) Female 3.158**(0.176) 0.180**(0.010)0.180**(0.010)0.180**(0.010)(0.014)1.358*(1.230)0.210**(0.014)1.764(1.239)Bachelor 6.865**(0.185) 0.405**(0.010)0.405**(0.010)0.405**(0.010)0.378**(0.014)0.378**(0.014)0.769(1.228)1.186(1.239)Female u Bachelor 0.064** (0.021) 0.063**(0.021)0.066**(0.021)0.066**(0.021)Intercept 1.884(0.897) 1.856**(0.053)0.128(0.177)0.059(0.613)0.078(0.612)0.633(0.819)0.604(0.819)0.095(0.945)F-statistic and p-values on joint hypotheses(a) F-statistic on terms involving Age 98.54(0.00)100.30(0.00)51.42(0.00)53.04(0.00)36.72(0.00)(b) Interaction termswithAge24.12(0.02)7.15(0.00)6.43(0.00)SER 7.884 0.457 0.457 0.457 0.457 0.456 0.456 0.456 R0.1897 0.1921 0.1924 0.1929 0.1937 0.1943 0.1950 0.1959 Significant at the *5% and **1% significance level.122 Stock/Watson - Introduction to Econometrics - Second Edition(a) The regression results for this question are shown in column (1) of the table. If Age increasesfrom 25 to 26, earnings are predicted to increase by $0.439 per hour. If Age increases from33 to 34, earnings are predicted to increase by $0.439 per hour. These values are the samebecause the regression is a linear function relating AHE and Age .(b) The regression results for this question are shown in column (2) of the table. If Age increasesfrom 25 to 26, ln(AHE ) is predicted to increase by 0.024. This means that earnings are predicted to increase by 2.4%. If Age increases from 34 to 35, ln(AHE ) is predicted to increase by 0.024.This means that earnings are predicted to increase by 2.4%. These values, in percentage terms,are the same because the regression is a linear function relating ln(AHE ) and Age .(c) The regression results for this question are shown in column (3) of the table. If Age increasesfrom 25 to 26, then ln(Age ) has increased by ln(26) ln(25) 0.0392 (or 3.92%). The predictedincrease in ln(AHE ) is 0.725 u (.0392) 0.0284. This means that earnings are predicted toincrease by 2.8%. If Age increases from 34 to 35, then ln(Age ) has increased by ln(35) ln(34) .0290 (or 2.90%). The predicted increase in ln(AHE ) is 0.725 u (0.0290) 0.0210. This means that earnings are predicted to increase by 2.10%.(d) When Age increases from 25 to 26, the predicted change in ln(AHE ) is(0.147 u 26 0.0021 u 262) (0.147 u 25 0.0021 u 252) 0.0399.This means that earnings are predicted to increase by 3.99%.When Age increases from 34 to 35, the predicted change in ln(AHE ) is(0. 147 u 35 0.0021 u 352) (0. 147 u 34 0.0021 u 342) 0.0063.This means that earnings are predicted to increase by 0.63%.(e) The regressions differ in their choice of one of the regressors. They can be compared on the basis of the .R The regression in (3) has a (marginally) higher 2,R so it is preferred.(f) The regression in (4) adds the variable Age 2 to regression(2). The coefficient on Age 2 isstatistically significant ( t 2.91), and this suggests that the addition of Age 2 is important. Thus,(4) is preferred to (2).(g) The regressions differ in their choice of one of the regressors. They can be compared on the basis of the .R The regression in (4) has a (marginally) higher 2,R so it is preferred.(h)Solutions to Empirical Exercises in Chapter 8 123 The regression functions using Age (2) and ln(Age) (3) are similar. The quadratic regression (4) is different. It shows a decreasing effect of Age on ln(AHE) as workers age.The regression functions for a female with a high school diploma will look just like these, but they will be shifted by the amount of the coefficient on the binary regressor Female. The regression functions for workers with a bachelor’s degree will also look just like these, but they would be shifted by the amount of the coefficient on the binary variable Bachelor.(i) This regression is shown in column (5). The coefficient on the interaction term Female uBachelor shows the “extra effect” of Bachelor on ln(AHE) for women relative the effect for men.Predicted values of ln(AHE):Alexis: 0.146 u 30 0.0021 u 302 0.180 u 1 0.405 u 1 0.064 u 1 0.078 4.504Jane: 0.146 u 30 0.0021 u 302 0.180 u 1 0.405 u 0 0.064 u 0 0.078 4.063Bob: 0.146 u 30 0.0021 u 302 0.180 u 0 0.405 u 1 0.064 u 0 0.078 4.651Jim: 0.146 u 30 0.0021 u 302 0.180 u 0 0.405 u 0 0.064 u 0 0.078 4.273Difference in ln(AHE): Alexis Jane 4.504 4.063 0.441Difference in ln(AHE): Bob Jim 4.651 4.273 0.378Notice that the difference in the difference predicted effects is 0.441 0.378 0.063, which is the value of the coefficient on the interaction term.(j) This regression is shown in (6), which includes two additional regressors: the interactions of Female and the age variables, Age and Age2. The F-statistic testing the restriction that the coefficients on these interaction terms is equal to zero is F 4.12 with a p-value of 0.02. This implies that there is statistically significant evidence (at the 5% level) that there is a different effect of Age on ln(AHE) for men and women.(k) This regression is shown in (7), which includes two additional regressors that are interactions of Bachelor and the age variables, Age and Age2. The F-statistic testing the restriction that the coefficients on these interaction terms is zero is 7.15 with a p-value of 0.00. This implies that there is statistically significant evidence (at the 1% level) that there is a different effect of Age on ln(AHE) for high school and college graduates.(l) Regression (8) includes Age and Age2 and interactions terms involving Female and Bachelor.The figure below shows the regressions predicted value of ln(AHE) for male and females with high school and college degrees.124 Stock/Watson - Introduction to Econometrics - Second EditionThe estimated regressions suggest that earnings increase as workers age from 25–35, the rangeof age studied in this sample. There is evidence that the quadratic term Age2 belongs in theregression. Curvature in the regression functions in particularly important for men.Gender and education are significant predictors of earnings, and there are statistically significant interaction effects between age and gender and age and education. The table below summarizes the regressions predictions for increases in earnings as a person ages from 25 to 32 and 32 to 35Gender, Education Predicted ln(AHE) at Age(Percent per year)25 32 35 25 to 32 32 to 35Males, High School 2.46 2.65 2.67 2.8% 0.5%Females, BA 2.68 2.89 2.93 3.0% 1.3%Males, BA 2.74 3.06 3.09 4.6% 1.0%Earnings for those with a college education are higher than those with a high school degree, andearnings of the college educated increase more rapidly early in their careers (age 25–32). Earnings for men are higher than those of women, and earnings of men increase more rapidly early in theircareers (age 25–32). For all categories of workers (men/women, high school/college) earningsincrease more rapidly from age 25–32 than from 32–35.。

P313 11.1解：（1）G u Y Y u Y G I C Y t t t t t t t t t t +++++++=++=-2122212011110ββββββββββββββββ21112121112111122211120101111--++--+--+--+=-u uGY ttttυπππt t t G Y 11211110+++=-ββ1110+=C t(βββββββββββ21112121112111122211120101111--++--+--+--+-uu GY tttt ) +ut1u u u G Y t t t t t 1211111211112211112010101)()(+--++++++=-ββββββββββ υπππt t t G Y 22212120++=+-(2120ββ+=Itβββββββββββ21112121112111122211120101111--++--+--+--+-uu GY tttt ）+u Yt t 2122+-β)1)((1)1()1)((22111212121112112111222122211120102120u u u G Y t tttt +--++--+--++--++=-βββββββββββββββββυπππtt t G Y33213130+++=-即有简化模型：=Y t υπππtt t G Y11211110+++- C tυπππttt G Y 22212120++=+-I tυπππttt G Y33213130+++=-其中内生变量有Yt、Ct、It即M=3，前定变量有Yt 1-、G t即K=2 。

对模型：uY C ttt 11110++=ββ——①u YY It t tt21222120+++=-βββ——②G I C Yt t t t++= ——③用阶条件进行方程识别：对方程①：m1=2，k=0，因为K-k=2-0 > 2-1=1，所以方程①过度识别 对方程②：m2=2，k=1，因为K-k=2-1-1=m2-1=1，所以方程②恰好识别 对方程③为定义式，故可以不判断其识别性。

第十章误差项自相关与异方差习题参考答案1. 由于自相关和异方差都是对随机误差项球形假定cov（μ）=σμ2I的违背，因此对OLS回归的影响是基本相同。

具体而言，虽然随机误差项自相关不会影响OLSE的线性、无偏性和一致性，但是会导致OLSE不再具备有效性，并造成统计推断不可靠以及因变量的预测精度降低。

2.（1）错误。

自相关系数为1的准（广义）差分变换相当于一阶差分变换。

（2）错误。

当模型的误差项自相关时，OLS估计量是无偏的和无效的。

（3）正确。

当d L<DW<d U时，处于检验盲区，无法确定是否存在自相关。

（4）正确。

被排除的应该包含在模型中自变量带来的系统性误差被并入随机误差项中，此时系统误差的自相关性表现为误差项的自相关。

显著的LM统计量也意味着随机误差项在不同时期取值之间存在相关性。

因此，从回归模型中排除一个（或多个）应含的自变量可能导致LM统计量显著。

（5）错误。

两个模型有不同的因变量，不可以直接比较两个模型的拟合优度。

3.（1）样本容量T=21，自变量个数k=3，按照显著性水平0.05，查DW临界值表可知，d L=1.026，d U=1.669。

由于d=0.81< d L，所以存在一阶正自相关问题。

（2）样本容量T=15，自变量个数k=2，按照显著性水平0.05，查DW临界值表可知，d L=0.946，d U=1.543。

由于4-d L<d=3.48<4，所以存在一阶负自相关问题。

（3）样本容量T=30，自变量个数k=5，按照显著性水平0.05，查DW临界值表可知，d L=1.071，d U=1.883。

由于d L<d=1.56<d U，所以无法确定是否存在自相关问题。

（4）样本容量T=35，自变量个数k=4，按照显著性水平0.05，查DW临界值表可知，d L=1.222，d U =1.726。

由于4- d U<d=2.64<4-d L，所以无法确定是否存在自相关问题。

一、Two-sample t test with equal variancesGroup Obs Mean Std.Err. Std.Dev. 95% Conf. Interval1992 7,612 11.62 0.0644 5.619 11.49 11.742012 7,440 19.80 0.124 10.69 19.56 20.04combined 15,052 15.66 0.0770 9.442 15.51 15.81diff -8,183 0.139 -8.455 -7.911 diff = mean(1992) - mean(2012) t = -58.9871Ho: diff = 0 degrees of freedom = 15050Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000二、Two-sample t test with equal variancesGroup Obs Mean Std.Err. Std.Dev. 95% Conf. Interval 1992 7,612 15.64 0.0867 7.564 15.47 15.81 2012 7,440 19.80 0.124 10.69 19.56 20.04 combined 15,052 17.69 0.0772 9.471 17.54 17.85 diff -4.164 0.151 -4.459 -3.869diff = mean(1992) - mean(2012) t = -27.6423Ho: diff = 0 degrees of freedom = 15050Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000三、第二题根据通货膨胀率进行了调整，反映了购买力的变化，所以可用利用第二题的结果进行分析。

计量经济学第⼗章习题（龚志民）fixed第10章模型设定与实践问题10.1 模型设定误差有哪些类型？如何诊断？答：模型设定误差主要有以下四种类型：1.漏掉⼀个相关变量；2.包含⼀个⽆关的变量；3.错误的函数形式；4.对误差项的错误假定。

诊断的⽅法有：1.侦察是否含有⽆关变量；2.残差分析，拉姆齐（Ramsey）的RESET检验法，DM（Davidsion-MacKinnon：戴维森麦-克⾦龙）检验；3.拟合优度、校正拟合优度、系数显著性、系数符合的合理性。

10.2 模型遗漏相关变量的后果是什么？答：模型遗漏相关变量的后果是：所有回归系数的估计量是有偏的，除⾮这个被去除的变量与每⼀个放⼊的变量都不相关。

常数估计量通常也是有偏的，从⽽预测值是有偏的。

由于放⼊变量的回归系数估计量是有偏的，所以假设检验是⽆效的。

系数估计量的⽅差估计量是有偏的。

10.3 模型包含不相关变量的后果是什么？答：模型包含不相关变量的后果是：系数估计量的⽅差变⼤，从⽽估计量的精度下降。

10.4 什么是嵌套模型？什么是⾮嵌套模型？答：如果两个模型不能被互相包容，即任何⼀个都不是另⼀个的特殊情形，便称这两个模型是⾮嵌套的。

如果两个模型能互相包容，即其中⼀个是另⼀个的特殊情形，便称这两个模型是嵌套的。

10.5 ⾮嵌套模型之间的⽐较有哪些⽅法？答：⾮嵌套模型之间的⽐较⽅法有：拟合优度或校正拟合优度、AIC（Akaike’s information criterion）准则、SIC（Schwarz’s information criterion）准则和HQ（Hannnan-Qinn criterion）准则。

拉姆齐（Ramsey）的RESET检验法，DM（Davidsion-MacKinnon：戴维森麦-克⾦龙）检验。

习题10.6 对数线性模型在⼈⼒资源⽂献中有⽐较⼴泛的应⽤，其理论建议把⼯资或收⼊的对数作为因变量。

如果教育投资收益率为r ，则接受⼀年教育的⼯资为10(1)w r w =+，0w 是基准⼯资（未接受教育）。

E8.1(1) (2) (3) (4)ahe lnahe lnahe lnahe age 0.585***0.0273***0.0814(0.0365) (0.00186) (0.0434) female -3.664***-0.186***-0.186***-0.186***(0.208) (0.0108) (0.0108) (0.0108) bachelor 8.083***0.428***0.428***0.428***(0.213) (0.0108) (0.0108) (0.0108) lnage 0.804***(0.0545)age2 -0.000915(0.000735) _cons -0.636 1.876***-0.0345 1.085(1.083) (0.0559) (0.185) (0.635) N7711 7711 7711 7711R20.200 0.201 0.201 0.201 adj. R20.1995 0.2003 0.2005 0.2004 Standard errors in parentheses （表1）*p < 0.05, **p < 0.01, ***p < 0.001（1）该问的回归结果如表第（1）列所示。

如果age从25增加到26岁，则预期收入每小时增加0.585美元。

如果age从33增加到34岁，则预期收入也是每小时增加0.585美元。

（2）该问的回归结果如表第（2）列所示。

如果age从25增加到26，则lnahe预计增加0.0273，即预期收入每小时增加2.73%。

如果age从33增加到34岁，则预期收入也是每小时增加2.73%。

（3）该问的回归结果如表第（3）列所示。

如果age从25增加到26岁，则lnage增加ln26-ln25≈0.04，预计lnahe增加0.04×0.804=0.03216，所以预期收入每小时增加3.216%。

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R20.0859 0.5613 0.1771 0.5690 0.9525State Effects No No Yes No YesTime Effects No No No Yes YesStandard errors in parentheses* p < 0.10, ** p < 0.05, *** p < 0.01（1）①回归（2）中shall 的系数是-0.368 ，这意味着隐蔽武器法律，也即“准予”携带法律，约使暴力犯罪减少36.8% 。