斯托克,沃森计量经济学第四章实证练习stata操作及答案

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(2) 3.060 1.657ln() 1.057ln()
(0.337) (0.092) (0.215)0.992 0.991 F 1275.093
GDP CPI R =-+-===进口居民消费价格指数的回归系数的符号不能进行合理的经济意义解释可能数据中有多重共线性。

计算相关系数:
22ln Y 4.09071.2186ln () t= (-10.6458) (34.6222)
0.9828 0.9820 1198.698
GDP R R F =-+===ln Y 5.4424 2.6637ln (PI)C =-+
从修正的可决系数和F统计量可以看出，全部变量对数线性多元回归整体对样本拟合很好，著。

可是其中的lnX3、lnX4、lnX6对lnY影响不显著,而且lnX2、lnX5
可以看出lnx1与lnx2、lnx3、lnx4、lnx5、lnx6之间高度相关，许多相关系数高于作为解释变量，很可能会出现严重多重共线性问题。

在本章开始的“引子”提出的“农业的发展反而会减少财政收入吗？
表4.13 1978-2007
财政收入(亿元)CS农业增加值(亿元)NZ工业增加值(亿元)GZ建筑业增加值
1132.31027.51607
1146.41270.21769.7
1159.91371.61996.5
1175.81559.52048.4
(1)根据样本数据得到各解释变量的样本相关系数矩阵如下：样本相关系数矩阵
解释变量之间相关系数较高，特别是农业增加值、工业增加值、建筑业增加值、最终消费之间，相关系数都在这显然与第三章对模型的无多重共线性假定不符合。

《计量经济学》习题（第四章）

《计量经济学》习题（第四章）第四章习题⼀、单选题1、如果回归模型违背了同⽅差假定，最⼩⼆乘估计量＿＿＿＿A ．⽆偏的，⾮有效的 B.有偏的，⾮有效的C ．⽆偏的，有效的 D.有偏的，有效的2、Goldfeld-Quandt ⽅法⽤于检验＿＿＿＿A ．异⽅差性 B.⾃相关性C ．随机解释变量 D.多重共线性3、DW 检验⽅法⽤于检验＿＿＿＿A ．异⽅差性 B.⾃相关性C ．随机解释变量 D.多重共线性4、在异⽅差性情况下，常⽤的估计⽅法是＿＿＿＿A ．⼀阶差分法 B.⼴义差分法C ．⼯具变量法 D.加权最⼩⼆乘法5、在以下选项中，正确表达了序列⾃相关的是＿＿＿＿j i u x Cov D j i x x Cov C ji u u Cov B ji u u Cov A j i j i j i j i ≠≠≠≠≠=≠≠,0),(.,0),(.,0),(.,0),(.6、如果回归模型违背了⽆⾃相关假定，最⼩⼆乘估计量＿＿＿＿A ．⽆偏的，⾮有效的 B.有偏的，⾮有效的C ．⽆偏的，有效的 D.有偏的，有效的7、在⾃相关情况下，常⽤的估计⽅法＿＿＿＿A ．普通最⼩⼆乘法 B.⼴义差分法C ．⼯具变量法 D.加权最⼩⼆乘法8、White 检验⽅法主要⽤于检验＿＿＿＿A ．异⽅差性 B.⾃相关性C ．随机解释变量 D.多重共线性9、Glejser 检验⽅法主要⽤于检验＿＿＿＿A ．异⽅差性 B.⾃相关性C ．随机解释变量 D.多重共线性10、简单相关系数矩阵⽅法主要⽤于检验＿＿＿＿A ．异⽅差性 B.⾃相关性C ．随机解释变量 D.多重共线性2222)(.)(.)(.)(.σσσσ==≠≠i i i i x Var D u Var C x Var B u Var A12、所谓不完全多重共线性是指存在不全为零的数k λλλ,,,21 ，有＿＿＿＿1112211221221122.0.0..k k k k k x x x k k k k A x x x v B x x x C x x x v e D x x x v e v λλλλλλλλλλλλ++++=+++=∑?++++=++++=式中是随机误差项13、设21,x x 为解释变量，则完全多重共线性是＿＿＿＿0.(021.0.021.22121121=+=++==+x x e x D v v x x C e x B x x A 为随机误差项）14、⼴义差分法是对＿＿＿＿⽤最⼩⼆乘法估计其参数 11211211121121)()1(....-------+-+-=-++=++=++=t t t t t t t t t t t t t t t u u x x y y D u x y C u x y B u x y A ρρβρβρρρβρβρββββ15、在DW 检验中要求有假定条件，在下列条件中不正确的是＿＿＿＿A ．解释变量为⾮随机的 B.随机误差项为⼀阶⾃回归形式C ．线性回归模型中不应含有滞后内⽣变量为解释变量D.线性回归模型为⼀元回归形式16、在下例引起序列⾃相关的原因中，不正确的是＿＿＿＿A.经济变量具有惯性作⽤B.经济⾏为的滞后性C.设定偏误D.解释变量之间的共线性17、在DW 检验中，当d 统计量为2时，表明＿＿＿＿A.存在完全的正⾃相关B.存在完全的负⾃相关C.不存在⾃相关D.不能判定18、在DW 检验中，当d 统计量为4时，表明＿＿＿＿A.存在完全的正⾃相关B.存在完全的负⾃相关C.不存在⾃相关D.不能判定19、在DW 检验中，当d 统计量为0时，表明＿＿＿＿A.存在完全的正⾃相关C.不存在⾃相关D.不能判定20、在DW 检验中，存在不能判定的区域是＿＿＿＿A. 0﹤d ﹤l d ,4-l d ﹤d ﹤4B. u d ﹤d ﹤4-u dC. l d ﹤d ﹤u d ,4-u d ﹤d ﹤4-l dD. 上述都不对21、在修正序列⾃相关的⽅法中，能修正⾼阶⾃相关的⽅法是＿＿＿＿A. 利⽤DW 统计量值求出ρB. Cochrane-Orcutt 法C. Durbin 两步法D. 移动平均法22、在下列多重共线性产⽣的原因中，不正确的是＿＿＿＿A.经济本变量⼤多存在共同变化趋势B.模型中⼤量采⽤滞后变量C.由于认识上的局限使得选择变量不当D.解释变量与随机误差项相关23、在DW 检验中，存在正⾃相关的区域是＿＿＿＿A. 4-l d ﹤d ﹤4B. 0﹤d ﹤l dC. u d ﹤d ﹤4-u dD. l d ﹤d ﹤u d ,4-u d ﹤d ﹤4-l d24、逐步回归法既检验⼜修正了＿＿＿＿A ．异⽅差性 B.⾃相关性 C ．随机解释变量 D.多重共线性25、设)()(,2221i i i i i ix f u Var u x y σσββ==++=，则对原模型变换的正确形式为＿＿＿＿ )()()()(.)()()()(.)()()()(..212222122121i i i i i i i i i i i i i i i i i i i i i i i i x f u x f x x f x f y D x f u x f x x f x f y C x f u x f x x f x f y B u x y A ++=++=++=++=ββββββββ 26、在修正序列⾃相关的⽅法中，不正确的是＿＿＿＿A.⼴义差分法B.普通最⼩⼆乘法C.⼀阶差分法D. Durbin 两步法27、在检验异⽅差的⽅法中，不正确的是＿＿＿＿A. Goldfeld-Quandt ⽅法B. spearman 检验法C. White 检验法28、在DW 检验中，存在零⾃相关的区域是＿＿＿＿A. 4-l d ﹤d ﹤4B. 0﹤d ﹤l dC. u d ﹤d ﹤4-u dD. l d ﹤d ﹤u d ,4-u d ﹤d ﹤4-l d29．如果模型中的解释变量存在完全的多重共线性，参数的最⼩⼆乘估计量是（）A ．⽆偏的 B. 有偏的 C. 不确定 D. 确定的30. 已知模型的形式为u x y 21+β+β=,在⽤实际数据对模型的参数进⾏估计的时候,测得DW 统计量为0.6453,则⼴义差分变量是( )A. 1t t ,1t t x 6453.0x y 6453.0y ----B. 1t t 1t t x 6774.0x ,y 6774.0y ----C. 1t t 1t t x x ,y y ----D. 1t t 1t t x 05.0x ,y 05.0y ----31. 在具体运⽤加权最⼩⼆乘法时,如果变换的结果是x u x x x 1xy 21+β+β=,则Var(u)是下列形式中的哪⼀种?( )A. 2σxB. 2σ2x B. 2σx D. 2σLog(x)32. 在线性回归模型中,若解释变量1x 和2x 的观测值成⽐例,即有i 2i 1kx x =,其中k 为⾮零常数,则表明模型中存在( )A. 异⽅差B. 多重共线性C. 序列⾃相关D. 设定误差33. 已知DW 统计量的值接近于2，则样本回归模型残差的⼀阶⾃相关系数ρ近似等于( ) A. 0 B. –1 C. 1 D. 4⼆、多项选择1、能够检验多重共线性的⽅法有＿＿＿＿A.简单相关系数法B. DW检验法C. 判定系数检验法D. ⽅差膨胀因⼦检验E.逐步回归法2、能够修正多重共线性的⽅法有＿＿＿＿A.增加样本容量B.岭回归法C.剔除多余变量E.差分模型3、如果模型中存在异⽅差现象，则会引起如下后果＿＿＿＿A. 参数估计值有偏B. 参数估计值的⽅差不能正确确定C. 变量的显著性检验失效D. 预测精度降低E. 参数估计值仍是⽆偏的4、能够检验异⽅差的⽅法是＿＿＿＿A. gleiser检验法B. White检验法C. 图形法D. spearman检验法E. DW检验法F. Goldfeld-Quandt检验法5、如果模型中存在序列⾃相关现象，则会引起如下后果＿＿＿＿A. 参数估计值有偏B. 参数估计值的⽅差不能正确确定C. 变量的显著性检验失效D. 预测精度降低E. 参数估计值仍是⽆偏的6、检验序列⾃相关的⽅法是＿＿＿＿A. gleiser检验法B. White检验法C. 图形法D. DW检验法E. Goldfeld-Quandt检验法7、能够修正序列⾃相关的⽅法有＿＿＿＿A. 加权最⼩⼆乘法B. Durbin两步法C. ⼴义最⼩⼆乘法D. ⼀阶差分法E. ⼴义差分法8、Goldfeld-Quandt检验法的应⽤条件是＿＿＿＿A. 将观测值按解释变量的⼤⼩顺序排列B. 样本容量尽可能⼤C. 随机误差项服从正态分布D. 将排列在中间的约1/4的观测值删除掉9、在DW检验中，存在不能判定的区域是＿＿＿＿A. 0﹤d﹤l dB. u d﹤d﹤4-u dC. l d﹤d﹤u dD. 4-u d﹤d﹤4-l dE. 4-l d﹤d﹤4。

计量经济学第四章习题详解word精品

第四章习题4.1没有进行t 检验，并且调整的可决系数也没有写出来，也就是没有考虑自由度的影响，会使结果存在一研究的目的和要求我们知道，商品进口额与很多因素有关，了解其变化对进出口产品有很大帮助。

为了探究和预测商品进口额的变化，需要定量地分析影响商品进口额变化的主要因素。

二、模型的设定及其估计经分析，商品进口额可能与国内生产总值、居民消费价格指数有关。

为此，考虑国内生产总值居民消费价格指数 CPI 为主要因素。

各影响变量与商品进口额呈正相关。

为此，设定如下形式的计量经济模型：4.3199511048.160793.7302.8+ In+ InCP1996 11557.4 71176.6 327.9 1997 11806.5 78973.0 337.1 1998 11626.1 84402.3 334.4 1999 13736.4 89677.1 329.7 2000 18638.8 99214.6 331.0 2001 20159.2 109655.2 333.3 2002 24430.3 120332.7 330.6 2003 34195.6 135822.8 334.6 2004 46435.8 159878.3 I 347.7 2005 54273.7 183084.8 353.9 2006 63376.9 211923.5 359.2 2007 73284.6 249529.9 376.5 2008 79526.5 314045.4 398.7 2009 68618.4 340902.8 395.9 201094699.3 401512.8 408.9 2011113161.4472881.6431.0GDP 、式中，为第年中国商品进口额（亿元）；In GDP 为第年国内生产总值（亿元）；In CPI 为居民消费价格指数（以1985年为100）。

各解释变量前的回归系数预期都大于零。

第四章计量经济学答案范文

第四章一元线性回归第一部分学习目的和要求本章主要介绍一元线性回归模型、回归系数的确定和回归方程的有效性检验方法。

回归方程的有效性检验方法包括方差分析法、t检验方法和相关性系数检验方法。

本章还介绍了如何应用线性模型来建立预测和控制。

需要掌握和理解以下问题：1 一元线性回归模型2 最小二乘方法3 一元线性回归的假设条件4 方差分析方法5 t检验方法6 相关系数检验方法7 参数的区间估计8 应用线性回归方程控制与预测9 线性回归方程的经济解释第二部分练习题一、术语解释1 解释变量2 被解释变量3 线性回归模型4 最小二乘法5 方差分析6 参数估计7 控制8 预测二、填空ξ，目的在于使模型更1 在经济计量模型中引入反映（）因素影响的随机扰动项t符合（）活动。

2 在经济计量模型中引入随机扰动项的理由可以归纳为如下几条：（1）因为人的行为的（）、社会环境与自然环境的（）决定了经济变量本身的（）；（２）建立模型时其他被省略的经济因素的影响都归入了（）中；（３）在模型估计时，（）与归并误差也归入随机扰动项中；（4）由于我们认识的不足，错误的设定了（）与（）之间的数学形式，例如将非线性的函数形式设定为线性的函数形式，由此产生的误差也包含在随机扰动项中了。

3 （）是因变量离差平方和，它度量因变量的总变动。

就因变量总变动的变异来源看，它由两部分因素所组成。

一个是自变量，另一个是除自变量以外的其他因素。

（）是拟合值的离散程度的度量。

它是由自变量的变化引起的因变量的变化，或称自变量对因变量变化的贡献。

（）是度量实际值与拟合值之间的差异，它是由自变量以外的其他因素所致，它又叫残差或剩余。

4 回归方程中的回归系数是自变量对因变量的（）。

某自变量回归系数β的意义，指的是该自变量变化一个单位引起因变量平均变化（）个单位。

5 模型线性的含义，就变量而言，指的是回归模型中变量的（）；就参数而言，指的是回归模型中的参数的（）；通常线性回归模型的线性含义是就（）而言的。

计量经济学实证练习作业

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。

(完整word版)计量经济学第四章习题详解

第四章习题4.1 没有进行t检验,并且调整的可决系数也没有写出来，也就是没有考虑自由度的影响，会使结果存在误差.4.3200224430.3120332。

7 330.6200334195。

6135822.8 334。

6200446435.8159878.3 l347.7200554273.7183084.8 353.9200663376.9211923。

5 359。

2200773284。

6249529。

9 376.5200879526.5314045.4 398.7200968618。

4340902。

8 395。

9201094699.3401512.8 408。

92011113161.4472881.6 431.0一研究的目的和要求我们知道，商品进口额与很多因素有关，了解其变化对进出口产品有很大帮助。

为了探究和预测商品进口额的变化，需要定量地分析影响商品进口额变化的主要因素。

二、模型的设定及其估计经分析，商品进口额可能与国内生产总值、居民消费价格指数有关。

为此,考虑国内生产总值GDP、居民消费价格指数CPI为主要因素。

各影响变量与商品进口额呈正相关。

为此，设定如下形式的计量经济模型：=+ln+lnCP式中,亿元）；lnGDP为国内生产总值(亿元）；lnCPI为居民消费价格指数（以1985年为100）。

各解释变量前的回归系数预期都大于零。

为估计模型，根据上表的数据，利用EViews软件，生成Y、lnGDP、lnCPI等数据，采用OLS方法估计模型参数,得到的回归结果如下图所示:模型方程为：lnY=-3。

111486+1。

338533lnGDP-0.421791lnCPI（0。

463010）(0。

088610）(0。

233295）t= (—6。

720126) (15。

10582）（—1。

807975)=0.988051 =0.987055 F=992。

2582该模型=0.988051,=0。

987055,可决系数很高，F检验值为992.2582，明显显著。

斯托克,沃森计量经济学实证练习stata操作及答案

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

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。

计量经济学第四章练习题及参考解答

第四章练习题及参考解答4.1 假设在模型i i i iu X X Y +++=33221βββ中，32X X 与之间的相关系数为零，于是有人建议你进行如下回归：ii i i i i u X Y u X Y 23311221++=++=γγαα(1)是否存在3322ˆˆˆˆβγβα==且？为什么？ (2)111ˆˆˆβαγ会等于或或两者的某个线性组合吗？ (3)是否有()()()()3322ˆvar ˆvar ˆvar ˆvarγβαβ==且？练习题4.1参考解答：(1) 存在3322ˆˆˆˆβγβα==且。

因为()()()()()()()23223223232322ˆ∑∑∑∑∑∑∑--=iiiii iii iii x x x x x xx y x x y β当32X X 与之间的相关系数为零时，离差形式的032=∑i i x x有()()()()222223222322ˆˆαβ===∑∑∑∑∑∑iiiiiiii xx y x x x x y 同理有：33ˆˆβγ= (2) 111ˆˆˆβαγ会等于或的某个线性组合因为12233ˆˆˆY X X βββ=--，且122ˆˆY X αα=-，133ˆˆY X γγ=- 由于3322ˆˆˆˆβγβα==且，则 11222222ˆˆˆˆˆY Y X Y X X αααββ-=-=-=11333333ˆˆˆˆˆY Y X Y X X γγγββ-=-=-=则 1112233231123ˆˆˆˆˆˆˆY Y Y X X Y X X Y X X αγβββαγ--=--=--=+- (3) 存在()()()()3322ˆvar ˆvar ˆvar ˆvarγβαβ==且。

因为()()∑-=22322221ˆvarr x iσβ当023=r 时，()()()22222232222ˆvar 1ˆvar ασσβ==-=∑∑iixr x 同理，有()()33ˆvar ˆvar γβ=4.2在决定一个回归模型的“最优”解释变量集时人们常用逐步回归的方法。

计量经济学课后答案第四、五章（内容参考）

计量经济学课后答案第四、五章（内容参考）第四章随机解释变量问题1. 随机解释变量的来源有哪些?答：随机解释变量的来源有：经济变量的不可控，使得解释变量观测值具有随机性；由于随机干扰项中包括了模型略去的解释变量，而略去的解释变量与模型中的解释变量往往是相关的；模型中含有被解释变量的滞后项，而被解释变量本身就是随机的。

2．随机解释变量有几种情形? 分情形说明随机解释变量对最小二乘估计的影响与后果?答：随机解释变量有三种情形，不同情形下最小二乘估计的影响和后果也不同。

(1)解释变量是随机的，但与随机干扰项不相关；这时采用OLS估计得到的参数估计量仍为无偏估计量；(2)解释变量与随机干扰项同期无关、不同期相关；这时OLS估计得到的参数估计量是有偏但一致的估计量；(3)解释变量与随机干扰项同期相关；这时OLS估计得到的参数估计量是有偏且非一致的估计量。

3. 选择作为工具变量的变量必须满足那些条件?答：选择作为工具变量的变量需满足以下三个条件：(1)与所替代的随机解释变量高度相关；(2)与随机干扰项不相关；(3)与模型中其他解释变量不相关，以避免出现多重共线性。

4．对模型Y t =β+β1X1t+β2X2t+β3Yt-1+μt假设Yt-1与μt相关。

为了消除该相关性，采用工具变量法：先求Y t关于X1t与 X2t回归，得到Yt，再做如下回归：Y t =β+β1X1t+β2X2t+β3Y t?1-+μt试问：这一方法能否消除原模型中Yt的相关性? 为什么?解答：能消除。

在基本假设下，X1t，X2t与μt应是不相关的，由此知，由X1t 与X2t估计出的Yt应与μt不相关。

5．对于一元回归模型Y t =β+β1Xt*+μt假设解释变量Xt *的实测值Xt与之有偏误：Xt= Xt*+et,其中et是具有零均值、无序列相关，且与Xt不相关的随机变量。

试问：(1) 能否将X t= X t*+e t代入原模型，使之变换成Y t=β0+β1X t+νt后进行估计? 其中，νt为变换后模型的随机干扰项。

斯托克,沃森计量经济学第四章实证练习stata操作及答案

斯托克,沃森计量经济学第四章实证练习stata操作及答案E4.1E4.2E4.3E4.4VARIABLES 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。

2.Bob: 0.605*26+1.082=16.812（美元）Alexis: 0.605*30+1.082=19.232（美元）答：预测Bob的收入为每小时16.812美元，Alexis为19.232美元。

3.年龄不能解释不同个体收入变化的大部分。

因为R-squared反映了因变量的全部变化能通过回归关系被自变量充分解释的比例，而分析得R-squared的值为0.029，解释度低，说明年龄不能解释不同个体收入变化的大部分。

1.答：两者看上去有微弱的正相关关系2.VARIABLES course_evalbeauty 0.133(0.0550)Constant 3.998(0.0449)Observations 463R-squared 0.036Robust standard errors in parentheses*** p<0.01, ** p<0.05, * p<0.1①截距估计值：3.998斜率估计值：0.133回归方程：Course_Eval=3.998+0.133*beauty//mean beautyMean estimation Number of obs = 463Mean Std.Err. 95% Conf. Interval beauty 4.75e-08 0.0367 -0.0720 0.0720②截距的估计值=Course_Eval的样本均值-斜率估计值*Beauty 的样本均值计算得Beauty的样本均值趋近于零，所以截距的估计值等于Course_Eval的样本均值。

计量经济学第4章课后答案

17CHAPTER 4SOLUTIONS TO PROBLEMS4.2 (i) and (iii) generally cause the t statistics not to have a t distribution under H 0.Homoskedasticity is one of the CLM assumptions. An important omitted variable violates Assumption MLR.3. The CLM assumptions contain no mention of the sample correlations among independent variables, except to rule out the case where the correlation is one.4.3 (i) While the standard error on hrsemp has not changed, the magnitude of the coefficient has increased by half. The t statistic on hrsemp has gone from about –1.47 to –2.21, so now the coefficient is statistically less than zero at the 5% level. (From Table G.2 the 5% critical value with 40 df is –1.684. The 1% critical value is –2.423, so the p -value is between .01 and .05.)(ii) If we add and subtract 2βlog(employ ) from the right-hand-side and collect terms, we havelog(scrap ) = 0β + 1βhrsemp + [2βlog(sales) – 2βlog(employ )] + [2βlog(employ ) + 3βlog(employ )] + u = 0β + 1βhrsemp + 2βlog(sales /employ ) + (2β + 3β)log(employ ) + u ,where the second equality follows from the fact that log(sales /employ ) = log(sales ) – log(employ ). Defining 3θ ≡ 2β + 3β gives the result.(iii) No. We are interested in the coefficient on log(employ ), which has a t statistic of .2, which is very small. Therefore, we conclude that the size of the firm, as measured by employees, does not matter, once we control for training and sales per employee (in a logarithmic functional form).(iv) The null hypothesis in the model from part (ii) is H 0:2β = –1. The t statistic is [–.951 – (–1)]/.37 = (1 – .951)/.37 ≈ .132; this is very small, and we fail to reject whether we specify a one- or two-sided alternative.4.4 (i) In columns (2) and (3), the coefficient on profmarg is actually negative, although its t statistic is only about –1. It appears that, once firm sales and market value have been controlled for, profit margin has no effect on CEO salary.(ii) We use column (3), which controls for the most factors affecting salary. The t statistic on log(mktval ) is about 2.05, which is just significant at the 5% level against a two-sided alternative.18(We can use the standard normal critical value, 1.96.) So log(mktval ) is statistically significant. Because the coefficient is an elasticity, a ceteris paribus 10% increase in market value is predicted to increase salary by 1%. This is not a huge effect, but it is not negligible, either.(iii) These variables are individually significant at low significance levels, with t ceoten ≈ 3.11 and t comten ≈ –2.79. Other factors fixed, another year as CEO with the company increases salary by about 1.71%. On the other hand, another year with the company, but not as CEO, lowers salary by about .92%. This second finding at first seems surprising, but could be related to the “superstar” effect: firms that hire CEOs from outside the company often go after a small pool of highly regarded candidates, and salaries of these people are bid up. More non-CEO years with a company makes it less likely the person was hired as an outside superstar.4.7 (i) .412 ± 1.96(.094), or about .228 to .596.(ii) No, because the value .4 is well inside the 95% CI.(iii) Yes, because 1 is well outside the 95% CI.4.8 (i) With df = 706 – 4 = 702, we use the standard normal critical value (df = ∞ in Table G.2), which is 1.96 for a two-tailed test at the 5% level. Now t educ = −11.13/5.88 ≈ −1.89, so |t educ | = 1.89 < 1.96, and we fail to reject H 0: educ β = 0 at the 5% level. Also, t age ≈ 1.52, so age is also statistically insignificant at the 5% level.(ii) We need to compute the R -squared form of the F statistic for joint significance. But F = [(.113 − .103)/(1 − .113)](702/2) ≈ 3.96. The 5% critical value in the F 2,702 distribution can be obtained from Table G.3b with denominator df = ∞: cv = 3.00. Therefore, educ and age are jointly significant at the 5% level (3.96 > 3.00). In fact, the p -value is about .019, and so educ and age are jointly significant at the 2% level.(iii) Not really. These variables are jointly significant, but including them only changes the coefficient on totwrk from –.151 to –.148.(iv) The standard t and F statistics that we used assume homoskedasticity, in addition to the other CLM assumptions. If there is heteroskedasticity in the equation, the tests are no longer valid.4.11 (i) Holding profmarg fixed, n rdintensΔ = .321 Δlog(sales ) = (.321/100)[100log()sales ⋅Δ] ≈ .00321(%Δsales ). Therefore, if %Δsales = 10, n rdintens Δ ≈ .032, or only about 3/100 of a percentage point. For such a large percentage increase in sales,this seems like a practically small effect.(ii) H 0:1β = 0 versus H 1:1β > 0, where 1β is the population slope on log(sales ). The t statistic is .321/.216 ≈ 1.486. The 5% critical value for a one-tailed test, with df = 32 – 3 = 29, is obtained from Table G.2 as 1.699; so we cannot reject H 0 at the 5% level. But the 10% criticalvalue is 1.311; since the t statistic is above this value, we reject H0 in favor of H1 at the 10% level.(iii) Not really. Its t statistic is only 1.087, which is well below even the 10% critical value for a one-tailed test.1920SOLUTIONS TO COMPUTER EXERCISESC4.1 (i) Holding other factors fixed,111log()(/100)[100log()](/100)(%),voteA expendA expendA expendA βββΔ=Δ=⋅Δ≈Δwhere we use the fact that 100log()expendA ⋅Δ ≈ %expendA Δ. So 1β/100 is the (ceteris paribus) percentage point change in voteA when expendA increases by one percent.(ii) The null hypothesis is H 0: 2β = –1β, which means a z% increase in expenditure by A and a z% increase in expenditure by B leaves voteA unchanged. We can equivalently write H 0: 1β + 2β = 0.(iii) The estimated equation (with standard errors in parentheses below estimates) isn voteA = 45.08 + 6.083 log(expendA ) – 6.615 log(expendB ) + .152 prtystrA(3.93) (0.382) (0.379) (.062) n = 173, R 2 = .793.The coefficient on log(expendA ) is very significant (t statistic ≈ 15.92), as is the coefficient on log(expendB ) (t statistic ≈ –17.45). The estimates imply that a 10% ceteris paribus increase in spending by candidate A increases the predicted share of the vote going to A by about .61percentage points. [Recall that, holding other factors fixed, n voteAΔ≈(6.083/100)%ΔexpendA ).] Similarly, a 10% ceteris paribus increase in spending by B reduces n voteAby about .66 percentage points. These effects certainly cannot be ignored.While the coefficients on log(expendA ) and log(expendB ) are of similar magnitudes (andopposite in sign, as we expect), we do not have the standard error of 1ˆβ + 2ˆβ, which is what we would need to test the hypothesis from part (ii).(iv) Write 1θ = 1β +2β, or 1β = 1θ– 2β. Plugging this into the original equation, and rearranging, givesn voteA = 0β + 1θlog(expendA ) + 2β[log(expendB ) – log(expendA )] +3βprtystrA + u ,When we estimate this equation we obtain 1θ≈ –.532 and se( 1θ)≈ .533. The t statistic for the hypothesis in part (ii) is –.532/.533 ≈ –1. Therefore, we fail to reject H 0: 2β = –1β.21C4.3 (i) The estimated model isn log()price = 11.67 + .000379 sqrft + .0289 bdrms (0.10) (.000043) (.0296)n = 88, R 2 = .588.Therefore, 1ˆθ= 150(.000379) + .0289 = .0858, which means that an additional 150 square foot bedroom increases the predicted price by about 8.6%.(ii) 2β= 1θ – 1501β, and solog(price ) = 0β+ 1βsqrft + (1θ – 1501β)bdrms + u= 0β+ 1β(sqrft – 150 bdrms ) + 1θbdrms + u .(iii) From part (ii), we run the regressionlog(price ) on (sqrft – 150 bdrms ), bdrms ,and obtain the standard error on bdrms . We already know that 1ˆθ= .0858; now we also getse(1ˆθ) = .0268. The 95% confidence interval reported by my software package is .0326 to .1390(or about 3.3% to 13.9%).C4.5 (i) If we drop rbisyr the estimated equation becomesn log()salary = 11.02 + .0677 years + .0158 gamesyr (0.27) (.0121) (.0016)+ .0014 bavg + .0359 hrunsyr (.0011) (.0072)n = 353, R 2= .625.Now hrunsyr is very statistically significant (t statistic ≈ 4.99), and its coefficient has increased by about two and one-half times.(ii) The equation with runsyr , fldperc , and sbasesyr added is22n log()salary = 10.41 + .0700 years + .0079 gamesyr(2.00) (.0120) (.0027)+ .00053 bavg + .0232 hrunsyr (.00110) (.0086)+ .0174 runsyr + .0010 fldperc – .0064 sbasesyr (.0051) (.0020) (.0052) n = 353, R 2 = .639.Of the three additional independent variables, only runsyr is statistically significant (t statistic = .0174/.0051 ≈ 3.41). The estimate implies that one more run per year, other factors fixed,increases predicted salary by about 1.74%, a substantial increase. The stolen bases variable even has the “wrong” sign with a t statistic of about –1.23, while fldperc has a t statistic of only .5. Most major league baseball players are pretty good fielders; in fact, the smallest fldperc is 800 (which means .800). With relatively little variation in fldperc , it is perhaps not surprising that its effect is hard to estimate.(iii) From their t statistics, bavg , fldperc , and sbasesyr are individually insignificant. The F statistic for their joint significance (with 3 and 345 df ) is about .69 with p -value ≈ .56. Therefore, these variables are jointly very insignificant.C4.7 (i) The minimum value is 0, the maximum is 99, and the average is about 56.16. (ii) When phsrank is added to (4.26), we get the following:n log() wage = 1.459 − .0093 jc + .0755 totcoll + .0049 exper + .00030 phsrank (0.024) (.0070) (.0026) (.0002) (.00024)n = 6,763, R 2 = .223So phsrank has a t statistic equal to only 1.25; it is not statistically significant. If we increase phsrank by 10, log(wage ) is predicted to increase by (.0003)10 = .003. This implies a .3% increase in wage , which seems a modest increase given a 10 percentage point increase in phsrank . (However, the sample standard deviation of phsrank is about 24.)(iii) Adding phsrank makes the t statistic on jc even smaller in absolute value, about 1.33, but the coefficient magnitude is similar to (4.26). Therefore, the base point remains unchanged: the return to a junior college is estimated to be somewhat smaller, but the difference is not significant and standard significant levels.(iv) The variable id is just a worker identification number, which should be randomly assigned (at least roughly). Therefore, id should not be correlated with any variable in the regression equation. It should be insignificant when added to (4.17) or (4.26). In fact, its t statistic is about .54.23C4.9 (i) The results from the OLS regression, with standard errors in parentheses, aren log() psoda =−1.46 + .073 prpblck + .137 log(income ) + .380 prppov (0.29) (.031) (.027) (.133)n = 401, R 2 = .087The p -value for testing H 0: 10β= against the two-sided alternative is about .018, so that we reject H 0 at the 5% level but not at the 1% level.(ii) The correlation is about −.84, indicating a strong degree of multicollinearity. Yet eachcoefficient is very statistically significant: the t statistic for log()ˆincome β is about 5.1 and that forˆprppovβ is about 2.86 (two-sided p -value = .004).(iii) The OLS regression results when log(hseval ) is added aren log() psoda =−.84 + .098 prpblck − .053 log(income ) (.29) (.029) (.038) + .052 prppov + .121 log(hseval ) (.134) (.018)n = 401, R 2 = .184The coefficient on log(hseval ) is an elasticity: a one percent increase in housing value, holding the other variables fixed, increases the predicted price by about .12 percent. The two-sided p -value is zero to three decimal places.(iv) Adding log(hseval ) makes log(income ) and prppov individually insignificant (at even the 15% significance level against a two-sided alternative for log(income ), and prppov is does not have a t statistic even close to one in absolute value). Nevertheless, they are jointly significant at the 5% level because the outcome of the F 2,396 statistic is about 3.52 with p -value = .030. All of the control variables – log(income ), prppov , and log(hseval ) – are highly correlated, so it is not surprising that some are individually insignificant.(v) Because the regression in (iii) contains the most controls, log(hseval ) is individually significant, and log(income ) and prppov are jointly significant, (iii) seems the most reliable. It holds fixed three measure of income and affluence. Therefore, a reasonable estimate is that if the proportion of blacks increases by .10, psoda is estimated to increase by 1%, other factors held fixed.。

计量经济学第四章练习题及参考解答

第四章练习题及参考解答4.1 假设在模型i i i i u X X Y +++=33221βββ中，32X X 与之间的相关系数为零，于是有人建议你进行如下回归：ii i i i i u X Y u X Y 23311221++=++=γγαα(1)是否存在3322ˆˆˆˆβγβα==且？为什么？ (2)111ˆˆˆβαγ会等于或或两者的某个线性组合吗？ (3)是否有()()()()3322ˆvar ˆvar ˆvar ˆvar γβαβ==且？练习题4.1参考解答：(1) 存在3322ˆˆˆˆβγβα==且。

因为()()()()()()()23223223232322ˆ∑∑∑∑∑∑∑--=iiiii iii iii x x x x x x x y x x y β当32X X 与之间的相关系数为零时，离差形式的032=∑i ix x有()()()()222223222322ˆˆαβ===∑∑∑∑∑∑iiiiiiii xx y x x x x y 同理有：33ˆˆβγ= (2) 111ˆˆˆβαγ会等于或的某个线性组合因为 12233ˆˆˆY X X βββ=--，且122ˆˆY X αα=-，133ˆˆY X γγ=- 由于3322ˆˆˆˆβγβα==且，则 11222222ˆˆˆˆˆY Y X Y X X αααββ-=-=-= 11333333ˆˆˆˆˆY Y X Y X X γγγββ-=-=-= 则 1112233231123ˆˆˆˆˆˆˆY Y Y X X Y X X Y X X αγβββαγ--=--=--=+- (3) 存在()()()()3322ˆvar ˆvar ˆvar ˆvar γβαβ==且。

因为()()∑-=22322221ˆvar r x iσβ当023=r 时，()()()22222232222ˆvar 1ˆvar ασσβ==-=∑∑iixr x 同理，有()()33ˆvar ˆvar γβ=4.2在决定一个回归模型的“最优”解释变量集时人们常用逐步回归的方法。

计量经济学第四章第6和10题答案

第4章练习6解：（1）答：不能，因为将代入原模型中使其变换后的模型为，显然,由于与同期相关，则说明变换后的模型中的随机干扰项与同期相关。

解：（2）对于多数经济变量的时间序列，除非它们是以一阶差分的形式或变化率的形式出现，往往具有较强的相关性，因此，当和直接表示经济规模或水平的经济变量时，它们之间很可能相关；如果变量是一阶差分的形式或以变化率的形态出现，则它们间的相关性就会降低，但仍有一定程度的相关性。

解：（3）由（2）的结论知，，即与变换后的模型的随机干扰项不相关，而且与有较强的相关性，因此可用作为的工具变量对变换后的模型进行估计。

第4章练习10编编号号170080081006115018001876026501000100907120020002052039001200127308140022002201049501400142509155024002435051100160016930101500260026860解：根据eview软件操作得：Dependent Variable: YMethod: Least SquaresDate: 04/17/11 Time: 22:28Sample: 1 10Included observations: 10Variable CoefficientStd.Error t-Statistic Prob.C245.515869.523483.5314080.0096 X10.5684250.7160980.7937810.4534 X2-0.0058330.070294-0.0829750.9362R-squared0.962099 Mean dependentvar1110.000Adjusted R-squared0.951270 S.D. dependentvar314.2893S.E. ofregression69.37901 Akaike infocriterion11.56037Sum squaredresid33694.13 Schwarzcriterion11.65115Log likelihood-54.80185 Hannan-Quinncriter.11.46079F-statistic88.84545 Durbin-Watsonstat 2.708154Prob(F-statistic)0.000011根据以上表格可得估计的回归模型为：（3.53）（0.79）（-0.083）分析：1.从回归估计的结果看，模型拟合较好。

斯托克,沃森计量经济学第五章第六章实证练习stata操作及答案

E5.2E5.3E6.2E6.3(1) VARIABLES ahe age 0.605*** (1.40e-09) Constant 1.082 (0.150) Observations 7,711 R-squared 0.029Robust pval in parentheses *** p<0.01, ** p<0.05, * p<0.1Robust t-statistics in parentheses *** p<0.01, ** p<0.05, * p<0.1a. 在对双边备择检验中，系数的t 统计量为24.70>2.58，与系数对应的p 值是0.0000000014趋近于0<0.01，所以可以在1%显著水平下拒绝原假设，自然可以在5%、10%水平下拒绝原假设。

(1) VARIABLES ahe age 0.605*** (0.550 - 0.660) Constant 1.082 (-0.473 - 2.638) Observations 7,711 R-squared 0.029Robust ci in parentheses *** p<0.01, ** p<0.05, * p<0.1b. 斜率系数95%的置信区间是(0.550，0.660)m1 VARIABLES ahe age 0.605*** (24.70) Constant 1.082 (1.574) Observations 7,711 R-squared 0.029 Ajusted R2 0.0289(1) m2 VARIABLES aheage 0.298***(7.513)Constant 6.522***(5.585)Observations 4,002R-squared 0.012Ajusted R2 0.0117Robust t-statistics in parentheses Robust pval in parentheses *** p<0.01, ** p<0.05, * p<0.1 *** p<0.01, ** p<0.05, * p<0.1只利用高中毕业生的数据，系数的t 统计量为7.513>2.58，与系数对应的p 值是0.0000364趋近于0<0.01，所以可以在1%显著水平下拒绝原假设，自然可以在5%、10%水平下拒绝原假设。

《计量经济学》第三版例题stata解答

第二章例2.1．1(p24)（1）表2.1.2中E(Y|X=800)即条件均值的求法，将数据直接复制到stata 中。

程序： sum y if x==800程序：程序：（2）图2.1.1的做法：程序：twoway(scatter y x )(lfit y x ),title("不同可支配收入水平组家庭消费支出的条件分布图")xtitle("每月可支配收入（元）")ytitle("每月消费支出（元）")xtick(500(500)4000)ytick(0(500)3500)例2.3．1（p37）将数据直接复制到stata 中程序：（1）total xiyireturn listscalars:r(skip) = 0r(first) = 1r(k_term) = 0r(k_operator) = 0r(k) = 0r(k_level) = 0r(output) = 1r(b) = 4974750r(se) = 1507820.761894463g a=r(b) in 1 total xi2 xiyi 4974750 1507821 1563822 8385678Total Std. Err. [95% Conf. Interval]Scatter 表示散点图选项，lfit 表示回归线，title 表示题目，xtick 表示刻度，（500（500）4000）分别表示起始刻度，中间数表示以单位刻度，4000表示最后的刻度。

要注意的是命令中的符号都要用英文字符，否则命令无效。

return listg b=r(b) in 1di a/b.67(2)mean Yigen m=r(b) in 1mean Xig n=r(b) in 1di m-n*0.67142.4由此得到回归方程：Y=142.4+0.67Xi例2.6．2(p53)程序：（1）回归reg y x（2）求X的样本均值和样本方差:mean xMean estimation Number of obs = 31 Mean Std. Err. [95% Conf. Interval] x 11363.69 591.7041 10155.27 12572.11sum x ,d（d表示detail的省略，这个命令会产生更多的信息）xPercentiles Smallest1% 8871.27 8871.275% 8920.59 8920.5910% 9000.35 8941.08 Obs 3125% 9267.7 9000.35 Sum of Wgt. 3150% 9898.75 Mean 11363.69Largest Std. Dev. 3294.46975% 12192.24 16015.5890% 16015.58 18265.1 Variance 1.09e+0795% 19977.52 19977.52 Skewness 1.69197399% 20667.91 20667.91 Kurtosis 4.739267di r(Var)（特别注意Var的大小写）10853528例2.6.2（P56）（1）reg Y XSource SS df MS Number of obs = 29F( 1, 27) = 2214.60Model 2.4819e+09 1 2.4819e+09 Prob > F = 0.0000Residual 30259023.9 27 1120704.59 R-squared = 0.9880Adj R-squared = 0.9875 Total2.5122e*************.8RootMSE=1058.6Y Coef. Std. Err. t P>|t| [95% Conf. Interval]X .4375268 .0092973 47.06 0.000 .4184503 .4566033_cons 2091.295 334.987 6.24 0.000 1403.959 2778.632（2）图2.6.1的绘制：twoway (line Y X year),title("中国居民可支配总收入X与消费总支出Y 的变动图")第三章例3.2.2（p72）reg Y X1 X2Source SS df MS Number of obs = 31F( 2, 28) = 560.57Model 166971988 2 83485994.2 Prob > F = 0.0000Residual 4170092.27 28 148931.867 R-squared = 0.9756Adj R-squared = 0.9739Total 171142081 30 5704736.02 Root MSE = 385.92Y Coef. Std. Err. t P>|t| [95% Conf. Interval]X1 .5556438 .0753076 7.38 0.000 .4013831 .7099046X2 .2500854 .1136343 2.20 0.036 .0173161 .4828547_cons 143.3266 260.4032 0.55 0.586 -390.0851 676.7383例3.5．1（p85）g lnP1=ln(P1)g lnP0=ln(P0)g lnQ=ln(Q)g lnX=ln(X)Source SS df MS Number of obs = 22 F( 3, 18) = 258.84 Model .765670868 3 .255223623 Prob > F = 0.0000 Residual .017748183 18 .00098601 R-squared = 0.9773 Adj R-squared = 0.9736 Total .783419051 21 .037305669 Root MSE = .0314 lnQ Coef. Std. Err. t P>|t| [95% Conf. Interval]lnX .5399167 .0365299 14.78 0.000 .4631703 .6166631 lnP1 -.2580119 .1781856 -1.45 0.165 -.632366 .1163422 lnP0 -.2885609 .2051844 -1.41 0.177 -.7196373 .1425155 _cons 5.53195 .0931071 59.41 0.000 5.336339 5.727561 drop lnX lnP1 lnP0g lnXP0=ln(X/P0)g lnP1P0=ln(P1/P0)reg lnQ lnXP0 lnP1P0Source SS df MS Number of obs = 22F( 2, 19) = 408.93Model .765632331 2 .382816165 Prob > F = 0.0000Residual .01778672 19 .000936143 R-squared = 0.9773Adj R-squared = 0.9749Total .783419051 21 .037305669 Root MSE = .0306lnQ Coef. Std. Err. t P>|t| [95% Conf. Interval]lnXP0 .5344394 .0231984 23.04 0.000 .4858846 .5829942lnP1P0 -.2753473 .1511432 -1.82 0.084 -.5916936 .040999_cons 5.524569 .0831077 66.47 0.000 5.350622 5.698515练习题13（p105）g lnY=ln(Y)g lnK=ln(K)g lnL=ln(L)reg lnY lnK lnLSource SS df MS Number of obs = 31 F( 2, 28) = 59.66 Model 21.6049266 2 10.8024633 Prob > F = 0.0000 Residual 5.07030244 28 .18108223 R-squared = 0.8099 Adj R-squared = 0.7963 Total 26.6752291 30 .889174303 Root MSE = .42554 lnY Coef. Std. Err. t P>|t| [95% Conf. Interval]lnK .6092356 .1763779 3.45 0.002 .2479419 .9705293 lnL .3607965 .2015915 1.79 0.084 -.0521449 .7737378 _cons 1.153994 .7276114 1.59 0.124 -.33645 2.644439第四章例4.1．4 (P116)（1）回归g lnY=ln(Y)g lnX1=ln(X1)g lnX2=ln(X2)reg lnY lnX1 lnX2Source SS df MS Number of obs = 31 F( 2, 28) = 49.60 Model 2.9609923 2 1.48049615 Prob > F = 0.0000 Residual .835744123 28 .029848004 R-squared = 0.7799 Adj R-squared = 0.7642 Total 3.79673642 30 .126557881 Root MSE = .17277 lnY Coef. Std. Err. t P>|t| [95% Conf. Interval]lnX1 .1502137 .1085379 1.38 0.177 -.072116 .3725435 lnX2 .4774534 .0515951 9.25 0.000 .3717657 .5831412 _cons 3.266068 1.041591 3.14 0.004 1.132465 5.39967于是得到方程：lnY=3.266+0.1502lnX1+0.4775lnX2（2）绘制参差图：predict e, residg ei2=e^2scatter ei2 lnX2,title("图4.1.3 异方差性检验图")xtick(6(0.4)9.2)ytick(0(0.04)0.24)predict在回归结束后，需要对拟合值以及残差进行分析，需要使用此命令。

斯托克计量经济学课后习题实证答案

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.。

第四章习题参考答案

4第四章习题参考答案(总20页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--第四章习题参考答案 P 1357. 1）用OLS法建立居民人均消费支出与可支配收入的线性模型。

create u 20； data consump income；ls consump c incomeDependent Variable: CONSUMPMethod: Least SquaresSample: 1 20Included observations: 20INCOMER-squared Mean dependent varAdjusted R-squared . dependent var. of regression Akaike info criterionSum squared resid Schwarz criterionLog likelihood F-statisticDurbin-Watson stat Prob(F-statistic)线性模型如下：CONSUMP = 5389 + *INCOME2）检验模型是否存在异方差性图：是否有明显的散点扩大/缩小/复杂型趋势i) X Yscat income consumpii)解释变量—残差图：是否形成一条斜率为0的直线scat income resid^2 或者genr ei2=resid^2； scat income ei2由两个图形，均可判定存在递增型异方差。

还可以用帕克检验，戈里瑟检验，戈德菲尔德-匡特检验，怀特检验等方法。

iii) 戈德菲尔德-匡特检验：共有20个样本，去掉中间1/4个样本（4个），剩余大样本、小样本各8个。

Sort income； smpl 1 8； ls consump C incomeSmpl 13 20； ls consump C income210.050.05615472.0126528.34.86(,)(81,81) 4.28118118111111RSS RSS F F F n k n k n k n k ===--=>=--------------，存在异方差。

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