计量经济学导论第五版第一章上机作业
计量经济学上机作业答案(2)
非平稳时间序列建模一.确定消费与收入是否为平稳序列，如果是非平稳的，请确定单整的阶数；对消费进行单位根检验H0：r=1 H1：r<1（常数）检验P值为0.2593，p值很大，接受原假设，消费是非平稳序列。
Null Hypothesis: CONS has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.665442 0.2593Test critical values: 1% level -4.5325985% level -3.67361610% level -3.277364另外，在同时有常数和变量或两者都没有（none）的检验中，p值都很大，所以也接受原假设，消费是非平稳序列。
Null Hypothesis: CONS has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.665442 0.2593Test critical values: 1% level -4.5325985% level -3.67361610% level -3.277364*MacKinnon (1996) one-sided p-values.Warning: Probabilities and critical values calculated for 20 observationsand may not be accurate for a sample size of 19Null Hypothesis: CONS has a unit rootExogenous: NoneLag Length: 2 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic 2.082985 0.9874Test critical values: 1% level -2.6997695% level -1.96140910% level -1.606610*MacKinnon (1996) one-sided p-values.Warning: Probabilities and critical values calculated for 20 observationsand may not be accurate for a sample size of 18对消费数据一阶差分后回归的p值仍旧很大，所以仍不平稳形式一：Null Hypothesis: D(CONS) has a unit rootExogenous: ConstantLag Length: 1 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.423252 0.1496 Test critical values: 1% level -3.8573865% level -3.04039110% level -2.660551*MacKinnon (1996) one-sided p-values.Warning: Probabilities and critical values calculated for 20 observationsAnd may not be accurate for a sample size of 18形式二：Null Hypothesis: D(CONS) has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.757380 0.2283 Test critical values: 1% level -4.5715595% level -3.69081410% level -3.286909*MacKinnon (1996) one-sided p-values.Warning: Probabilities and critical values calculated for 20 observationsAnd may not be accurate for a sample size of 18形式三：Null Hypothesis: D(CONS) has a unit rootExogenous: Constant, Linear TrendLag Length: 1 (Automatic - based on SIC, maxlag=4)t-Statistic Prob.*Augmented Dickey-Fuller test statistic -2.757380 0.2283 Test critical values: 1% level -4.5715595% level -3.69081410% level -3.286909*MacKinnon (1996) one-sided p-values.Warning: Probabilities and critical values calculated for 20 observationsand may not be accurate for a sample size of 18再进行二阶差分检验（none），P值为0.0013，p值很小，消费是二阶非平稳序列。
计量经济学导论 第五版 答案
APPENDIX ASOLUTIONS TO PROBLEMSA.1 (i) $566.(ii) The two middle numbers are 480 and 530; when these are averaged, we obtain 505, or $505.(iii) 5.66 and 5.05, respectively.(iv) The average increases to $586 while the median is unchanged ($505).A.3 If price = 15 and income = 200, quantity = 120 – 9.8(15) + .03(200) = –21, which is nonsense. This shows that linear demand functions generally cannot describe demand over a wide range of prices and income.A.5 The majority shareholder is referring to the percentage point increase in the stock return, while the CEO is referring to the change relative to the initial return of 15%. To be precise, the shareholder should specifically refer to a 3 percentage point increase.$45,935.80.≈ $40,134.84. When exper = 5, salary = exp[10.6 + .027(5)] ≈A.7 (i) When exper = 0, log(salary) = 10.6; therefore, salary = exp(10.6) (ii) The approximate proportionate increase is .027(5) = .135, so the approximate percentage change is 13.5%.14.5%, so the exact percentage increase is about one percentage point higher.≈(iii) 100[(45,935.80 – 40,134.84)/40,134.84)A.9 (i) The relationship between yield and fertilizer is graphed below. (ii) Compared with a linear function, the functionyieldhas a diminishing effect, and the slope approaches zero as fertilizer gets large. The initial pound of fertilizer has the largest effect, and each additional pound has an effect smaller than the previous pound.APPENDIX BSOLUTIONS TO PROBLEMSB.1 Before the student takes the SAT exam, we do not know – nor can we predict with certainty – what the score will be. The actual score depends on numerous factors, many of which we cannot even list, let alone know ahead of time. (The student’s innate ability, how the student feels on exam day, and which particular questions were asked, are just a few.) The eventual SAT score clearly satisfies the requirements of a random variable.B.3 (i) Let Yit be the binary variable equal to one if fund i outperforms the market in year t. By assumption, P(Yit = 1) = .5 (a 50-50 chance of outperforming the market for each fund in each year). Now, for any fund, we are also assuming that performance relative to the market isP(Yi2 = 1) P(Yi,10 = 1) = (.5)10 = 1/1024 (which is slightly less than .001). In fact, if we define a binary random variable Yi such that Yi = 1 if and only if fund i outperformed the market in all 10 years, then P(Yi = 1) =1/1024.⋅independent across years. But then the probability that fund i outperforms the market in all 10 years, P(Yi1 = 1,Yi2 = 1, , Yi,10 = 1), is just the product of the probabilities: P(Yi1 = 1).983. This means, if performance relative to the market is random and independent across funds, it is almost certain that at least one fund will outperform the market in all 10 years.≈P(Y4,170 = 0) = 1 –(1023/1024)4170 ⋅⋅⋅ P(Y2 = 0)⋅= 1/1024. We want to compute P(X ≥ 1) =1 – P(X = 0) = 1 –P(Y1 = 0, Y2 = 0, …, Y4,170 = 0) = 1 – P(Y1 = 0)θ)distribution with n = 4,170 and θ(ii) Let X denote the number of funds out of 4,170 that outperform the market in all 10 years. Then X = Y1 + Y2 + + Y4,170. If we assume that performance relative to the market is independent across funds, then X has the Binomial (n,(iii) Using the Stata command Binomial(4170,5,1/1024), the answer is about .385. So there is a nontrivial chance that at least five funds will outperform the market in all 10 years..931.≈ 1) = 1 – P(X = 0) = 1 – (.8)12 ≥B.5 (i) As stated in the hint, if X is the number of jurors convinced of Simpson’s innocence, then X ~ Binomial(12,.20). We want P(X(ii) Above, we computed P(X = 0) as about .069. We need P(X = 1), which we obtain from1 – (.069 + .206) = .725, so there is almost a three in four chance that the jury had at least two members convinced of Simpson’s innocence prior to the trial.≈ 2) ≥ .206. Therefore, P(X ≈ (.2)(.8)11 ⋅ = .2, and x = 1:P(X = 1) = 12θ(B.14) with n = 12,B.7 In eight attempts the expected number of free throws is 8(.74) = 5.92, or about six free throws.X, and so the expected value of Y is 1,000 times the expected value of X, and the standard deviation of Y is 1,000 times the standard deviation of X. Therefore, the expected value and standard deviation of salary, measured in dollars, are $52,300 and $14,600, respectively.⋅B.9 If Y issalary in dollars then Y = 1000。
计量经济学导论第一次作业（第四组）伍德里奇
计量经济学导论第一次作业（第四组）伍德里奇
计量经济学导论第一次作业
（第4组）
第一题：设计一假想的理想化随机对照试验来研究系上安全带对高速公路上交通死亡事故产生的影响。
试提出实施这个实验可能遇到的障碍。
两组试验：随机抽取两组汽车租赁公司，一组司机每天驾驶必须系安全带，一组司机未系安全带。
进行一个月的有效数据跟踪监测。
统计两组事故发生率。
障碍：1、道德约束，不可能让所有司机违反交通规则。
2、司机会有性别、驾驶员驾驶的驾驶年龄不同等其他因素的影响。
第二题：
（1）use "D:\计量经济学\高级计量第1次作业+数据PT\401K.DTA", clear
Prate 均值为87.36291mrate均值为0.7315124
（2）
prate=83.07546 + 5.861079 mrate样本容量为1534, R2为0.0747。
（3）截距表示，当mrate取值为零时，prate的值为83.07546。
而mrate的系数表示它每增加一个单位，prate将变化5.861079个单位。
（4）prate= 83.07546 + 5.861079 *3.5=103.5892
预测值为103.5892不合适，最高为100%，还有就是一般取值应该在均值附近，3.5这个值太靠右。
（5）方程拟合程度0.0747调整后0.0741具体prate的变异可能并不都是由mrate造成，还需要进一步进行检验及考虑其他变量。
对于样本数量，养老金是一个大群体，人口占国家数量很大。
该数量太少。
计量经济学第一次上机作业(1)
计量经济学实验报告班级：数理金融1一元线性回归模型研究内容：研究全国各地区城市居民的消费是否有显著差异以及城市居民消费与人均收入之间的关系。
研究方法：最小二乘法（实验原理略）研究数据：详细数据见附录1.研究工具：Matlab2008研究过程：我们以城市居民的人均消费支出为被解释变量，以城市居民每人每年可支配收入为解释变量的一元回归分析模型。
21、散点图2、求解过程matlab代码：[n,k]=size(x);X=[ones(n,1),x];%构建结构阵X，A=X'*X; %求算信息阵A，C=inv(A); %求算信息阵的逆阵，b=X\y, % 求算回归统计数向量，其中第一行为回归截距a,RSS=y'*y-b'*X'*y, %求算离回归平方和，MSe=RSS/(n-k-1),%求算离回归方差，Up=b.*b./diag(C);%求算偏回归平方和，其中第一行是a与0差异的偏平方和，F=Up/MSe,%F测验，其中第一行为a与0差异的F值，34sb=sqrt(MSe*diag(C)); %求算回归统计数标准误，t=b./sb, % 回归统计数的 t 测验，其中第一行为a 与0差异的t 测验值。
[t, t.^2, F],%验证t^2=F SSy=var(y)*(n-1) R2=(SSy-RSS)/SSy3、回归方程Y =−554.5943+0.2484x4、回归曲线图5、指标分析估计值 t 值 F 值 b0 −554.5943 -16.8745 284.7 b 10.2484 35.46841258.0 R 20.9836分析：我们可以很清楚的看到，t值和F值都表明回归的截距项和系数都比较大，显然这一定落在拒绝域中，那么我们可以知道回归方程的截距项和斜率系数都是显著的。
上述的R^2=0.9836,已经很接近于1了，也就是说被解释变量有98.36%的变异可以被解释变量解释，那么这个比例已经很高了，这也从一定程度上说明回归的拟合效果比较好。
计量经济学第1章 计量经济学导论
1.2 经验经济分析的步骤
修订模型 不符合
设定计量模型
参数估计
模型检验
是否符合标准 符合
模型应用
结构分析
经济预测
验证理论
经济理论 实际经济活动 搜集统计数据
政策评价
1.3 经济数据的结构
㈠横截面数据(cross-sectional data) 定义：在给定时点对个人、家庭、企业、地 区等单位采集的数据。 特点：可假定为从样本背后的总体中通过随 机抽样而得到的。 例：2012我国31个省/直辖市城镇居民人均可支配
1.1 什么是计量经济学
创立
建立第1个应用模型 经
典
计
建立概率论基础
量
经
发展数据基础
济
学
发展应用模型
建立投入产出模型
Frisch Tinbergen Haavelmo
Stone Klein Leontief
1.1 什么是计量经济学
微观计量：选择性样本模型
Heckman
非
微观计量：离散选择模型
McFadden
收入和人均消费支出。
1.3 经济数据的结构
㈡时间序列数据(time series data) 定义：对同一个体（个人、家庭、企业、地 区等）在多个时期内收集到的数据。 特点： 按时间顺序排列 时间跨度一致 例：1978-2012年我国城镇居民人均可支配收入 和人均消费支出(以1978年价格核算)。
配套教材
习题解答： 案例数据：
第1章 计量经济学的性质与经济数据
1.1什么是计量经济学 1.2经验经济分析的步骤 1.3经济数据的结构 1.4计量经济分析中的因果关系和其他条件不
变的概念
展
计量经济学导论 第五版 答案
APPENDIX ASOLUTIONS TO PROBLEMSA.1 (i) $566.(ii) The two middle numbers are 480 and 530; when these are averaged, we obtain 505, or $505.(iii) 5.66 and 5.05, respectively.(iv) The average increases to $586 while the median is unchanged ($505).A.3 If price = 15 and income = 200, quantity = 120 – 9.8(15) + .03(200) = –21, which is nonsense. This shows that linear demand functions generally cannot describe demand over a wide range of prices and income.A.5 The majority shareholder is referring to the percentage point increase in the stock return, while the CEO is referring to the change relative to the initial return of 15%. To be precise, the shareholder should specifically refer to a 3 percentage point increase.$45,935.80.≈ $40,134.84. When exper = 5, salary = exp[10.6 + .027(5)] ≈A.7 (i) When exper = 0, log(salary) = 10.6; therefore, salary = exp(10.6) (ii) The approximate proportionate increase is .027(5) = .135, so the approximate percentage change is 13.5%.14.5%, so the exact percentage increase is about one percentage point higher.≈(iii) 100[(45,935.80 – 40,134.84)/40,134.84)A.9 (i) The relationship between yield and fertilizer is graphed below. (ii) Compared with a linear function, the functionyieldhas a diminishing effect, and the slope approaches zero as fertilizer gets large. The initial pound of fertilizer has the largest effect, and each additional pound has an effect smaller than the previous pound.APPENDIX BSOLUTIONS TO PROBLEMSB.1 Before the student takes the SAT exam, we do not know – nor can we predict with certainty – what the score will be. The actual score depends on numerous factors, many of which we cannot even list, let alone know ahead of time. (The student’s innate ability, how the student feels on exam day, and which particular questions were asked, are just a few.) The eventual SAT score clearly satisfies the requirements of a random variable.B.3 (i) Let Yit be the binary variable equal to one if fund i outperforms the market in year t. By assumption, P(Yit = 1) = .5 (a 50-50 chance of outperforming the market for each fund in each year). Now, for any fund, we are also assuming that performance relative to the market isP(Yi2 = 1) P(Yi,10 = 1) = (.5)10 = 1/1024 (which is slightly less than .001). In fact, if we define a binary random variable Yi such that Yi = 1 if and only if fund i outperformed the market in all 10 years, then P(Yi = 1) =1/1024.⋅independent across years. But then the probability that fund i outperforms the market in all 10 years, P(Yi1 = 1,Yi2 = 1, , Yi,10 = 1), is just the product of the probabilities: P(Yi1 = 1).983. This means, if performance relative to the market is random and independent across funds, it is almost certain that at least one fund will outperform the market in all 10 years.≈P(Y4,170 = 0) = 1 –(1023/1024)4170 ⋅⋅⋅ P(Y2 = 0)⋅= 1/1024. We want to compute P(X ≥ 1) =1 – P(X = 0) = 1 –P(Y1 = 0, Y2 = 0, …, Y4,170 = 0) = 1 – P(Y1 = 0)θ)distribution with n = 4,170 and θ(ii) Let X denote the number of funds out of 4,170 that outperform the market in all 10 years. Then X = Y1 + Y2 + + Y4,170. If we assume that performance relative to the market is independent across funds, then X has the Binomial (n,(iii) Using the Stata command Binomial(4170,5,1/1024), the answer is about .385. So there is a nontrivial chance that at least five funds will outperform the market in all 10 years..931.≈ 1) = 1 – P(X = 0) = 1 – (.8)12 ≥B.5 (i) As stated in the hint, if X is the number of jurors convinced of Simpson’s innocence, then X ~ Binomial(12,.20). We want P(X(ii) Above, we computed P(X = 0) as about .069. We need P(X = 1), which we obtain from1 – (.069 + .206) = .725, so there is almost a three in four chance that the jury had at least two members convinced of Simpson’s innocence prior to the trial.≈ 2) ≥ .206. Therefore, P(X ≈ (.2)(.8)11 ⋅ = .2, and x = 1:P(X = 1) = 12θ(B.14) with n = 12,B.7 In eight attempts the expected number of free throws is 8(.74) = 5.92, or about six free throws.X, and so the expected value of Y is 1,000 times the expected value of X, and the standard deviation of Y is 1,000 times the standard deviation of X. Therefore, the expected value and standard deviation of salary, measured in dollars, are $52,300 and $14,600, respectively.⋅B.9 If Y issalary in dollars then Y = 1000。
计量经济学上机作业试题以及答案
题目：第二题：下表中，Y代表新客车出售量，X1代表新车价格指数，X2代表消费者价格指数，X3代表个人可支配收入，X4代表利率，X5代表就业人数。
试建模并估计结果。
年度 1X2X3 X451971 10227 112 121.3 776.8 4.89 793671972 10872 111 125.3 839.6 4.55 821531973 11350 111.1 133.1 949.8 7.38 850641974 8775 117.5 147.7 1038.4 8.61 867941975 8539 127.6 161.2 1142.8 6.16 858461976 9994 135.7 170.5 1252.6 5.22 887521977 11046 142.9 181.5 1379.3 5.5 920171978 11164 153.8 195.3 1551.2 7.78 960481979 10559 166 217.7 1729.3 10.25 988241980 8979 179.3 247 1918 11.28 993031981 8535 190.2 272.3 2127.6 13.73 1003971982 7980 197.6 286.6 2261.4 11.2 995261983 9179 202.6 297.4 2428.1 8.69 1008341984 10394 208.5 307.6 2670.6 9.65 1050051985 11039 215.2 318.5 2841.1 7.75 1071501986 11450 224.4 323.4 3022.1 6.31 109597第三题为了了解影响电信业务的发展情况，特收集了如下数据，请建模并估计合理的结果。
年电信业务总量邮政业务总量中国人口数市镇人口比重人均GDP人均消费水平1991 1.5163 0.5275 11.5823 0.2637 1.879 0.896 1992 2.2657 0.6367 11.7171 0.2763 2.287 1.070 1993 3.8245 0.8026 11.8517 0.2814 2.939 1.331 1994 5.9230 0.9589 11.9850 0.2862 3.923 1.746 1995 8.7551 1.1334 12.1121 0.2904 4.854 2.236 1996 12.0875 1.3329 12.2389 0.2937 5.576 2.641 1997 12.6895 1.4434 12.3626 0.2992 6.053 2.834 1998 22.6494 1.6628 12.4810 0.3040 6.307 2.972 1999 31.3238 1.9844 12.5909 0.3089 6.534 3.143第四题：X代表职工的工龄，Y代表薪水。
古扎拉蒂《计量经济学基础》(第5版)笔记和课后习题详解
资料来源：EconomicReport ofthe President，2007，Table13-110，P.356．
答：a．把汇率的对数作为纵轴并把时间作为横轴进行描点，如图1-4所示，汇率的波动性很大。比如，在1985年，1美元只能兑换0.257比索，但到了2004年，它能兑换约11.29比索。
三、回归与因果关系
从逻辑上说，回归得到的统计关系式本身不可能意味着任何因果关系。肯德尔和斯图亚特认为，一个统计关系式永远不能确立因果方面的联系：对因果关系的理念，必须来自统计学以外的某种理论。
四、回归与相关
1．相关分析与相关系数
相关分析是以测度两个变量之间的线性关联程度为其主要目的。
相关系数是用来测度线性关联强度的。
6.01
1986
1.86
4.13
0.67
2.53
-0.10
6.11
3.42
1987
3.65
4.32
0.00
3.24
0.19
4.59
4.18
1988
4.14
4.05
0.67
2.73
1.33
4.99
4.93
1989
4.82
4.95
2.27
3.46
2.73
6.59
7.72
1990
5.40
4.80
3.15
3.34
-0.08
1.78
1.83
5.37
3.36
1996
2.95
1.59
0.08
2.02
1.50
3.87
2.46
1997
2.29
计量经济学(伍德里奇第五版中文版)答案
第1 章解决问题的办法1.1（一）理想的情况下，我们可以随机分配学生到不同尺寸的类。
也就是说，每个学生被分配一个不同的类的大小，而不考虑任何学生的特点，能力和家庭背景。
对于原因，我们将看到在第 2 章中，我们想的巨大变化，班级规模（主题，当然，伦理方面的考虑和资源约束）。
（二）呈负相关关系意味着，较大的一类大小是与较低的性能。
因为班级规模较大的性能实际上伤害，我们可能会发现呈负相关。
然而，随着观测数据，还有其他的原因，我们可能会发现负相关关系。
例如，来自较富裕家庭的儿童可能更有可能参加班级规模较小的学校，和富裕的孩子一般在标准化考试中成绩更好。
另一种可能性是，在学校，校长可能分配更好的学生，以小班授课。
或者，有些家长可能会坚持他们的孩子都在较小的类，这些家长往往是更多地参与子女的教育。
（三）鉴于潜在的混杂因素- 其中一些是第（ii）上市- 寻找负相关关系不会是有力的证据，缩小班级规模，实际上带来更好的性能。
在某种方式的混杂因素的控制是必要的，这是多元回归分析的主题。
1.2（一）这里是构成问题的一种方法：如果两家公司，说 A 和B，相同的在各方面比 B 公司à用品工作培训之一小时每名工人，坚定除外，多少会坚定的输出从 B 公司的不同？（二）公司很可能取决于工人的特点选择在职培训。
一些观察到的特点是多年的教育，多年的劳动力，在一个特定的工作经验。
企业甚至可能歧视根据年龄，性别或种族。
也许企业选择提供培训，工人或多或少能力，其中，“能力”可能是难以量化，但其中一个经理的相对能力不同的员工有一些想法。
此外，不同种类的工人可能被吸引到企业，提供更多的就业培训，平均，这可能不是很明显，向雇主。
（iii ）该金额的资金和技术工人也将影响输出。
所以，两家公司具有完全相同的各类员工一般都会有不同的输出，如果他们使用不同数额的资金或技术。
管理者的素质也有效果。
（iv）无，除非训练量是随机分配。
许多因素上市部分（二）及（iii ）可有助于寻找输出和培训的正相关关系，即使不在职培训提高工人的生产力。
伍德里奇计量经济学导论课后题计算机操作
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伍德里奇计量经济学导论第5版笔记和课后习题详解
伍德里奇《计量经济学导论》（第5版）笔记和课后习题详解目录第1章计量经济学的性质与经济数据1.1复习笔记1.2课后习题详解第一篇横截面数据的回归分析第2章简单回归模型2.1复习笔记2.2课后习题详解第3章多元回归分析：估计3.1复习笔记3.2课后习题详解第4章多元回归分析：推断4.1复习笔记4.2课后习题详解第5章多元回归分析：OLS的渐近性5.1复习笔记5.2课后习题详解第6章多元回归分析：深入专题6.1复习笔记6.2课后习题详解第7章含有定性信息的多元回归分析：二值（或虚拟）变量7.1复习笔记7.2课后习题详解第8章异方差性8.1复习笔记8.2课后习题详解第9章模型设定和数据问题的深入探讨9.1复习笔记9.2课后习题详解第二篇时间序列数据的回归分析第10章时间序列数据的基本回归分析10.1复习笔记10.2课后习题详解第11章OLS用于时间序列数据的其他问题11.1复习笔记11.2课后习题详解第12章时间序列回归中的序列相关和异方差性12.1复习笔记12.2课后习题详解第三篇高级专题讨论第13章跨时横截面的混合：简单面板数据方法13.1复习笔记13.2课后习题详解第14章高级的面板数据方法14.2课后习题详解第15章工具变量估计与两阶段最小二乘法15.1复习笔记15.2课后习题详解第16章联立方程模型16.1复习笔记16.2课后习题详解第17章限值因变量模型和样本选择纠正17.1复习笔记17.2课后习题详解第18章时间序列高级专题18.1复习笔记18.2课后习题详解第19章一个经验项目的实施19.2课后习题详解本书是伍德里奇《计量经济学导论》（第5版）教材的学习辅导书，主要包括以下内容：（1）整理名校笔记，浓缩内容精华。
每章的复习笔记以伍德里奇所著的《计量经济学导论》（第5版）为主，并结合国内外其他计量经济学经典教材对各章的重难点进行了整理，因此，本书的内容几乎浓缩了经典教材的知识精华。
（2）解析课后习题，提供详尽答案。
计量经济学导论习题及答案
计量经济学导论一、单项选择题1、计量经济学是__________的一个分支学科。
CA统计学B数学C经济学D数理统计学2、计量经济学成为一门独立学科的标志是__________。
BA 1930年世界计量经济学会成立B 1933年《计量经济学》会刊出版C 1969年诺贝尔经济学奖设立D 1926年计量经济学（E conomi cs）一词构造出来3、外生变量和滞后变量统称为__________。
DA控制变量B解释变量C被解释变量D前定变量4、横截面数据是指__________。
AA同一时点上不同统计单位相同统计指标组成的数据B同一时点上相同统计单位相同统计指标组成的数据C同一时点上相同统计单位不同统计指标组成的数据D同一时点上不同统计单位不同统计指标组成的数据5、同一统计指标，同一统计单位按时间顺序记录形成的数据列是__________。
CA时期数据B混合数据C时间序列数据D横截面数据6、在计量经济模型中，由模型系统内部因素决定，表现为具有一定的概率分布的随机变量，其数值受模型中其他变量影响的变量是__________。
BA内生变量B 外生变量C 滞后变量D 前定变量7、描述微观主体经济活动中的变量关系的计量经济模型是__________。
AA微观计量经济模型B 宏观计量经济模型C 理论计量经济模型D 应用计量经济模型8、经济计量模型的被解释变量一定是__________。
CA控制变量B 政策变量C 内生变量D 外生变量9、下面属于横截面数据的是__________。
DA1991－2003年各年某地区20个乡镇企业的平均工业产值B1991－2003年各年某地区20个乡镇企业各镇的工业产值C某年某地区20个乡镇工业产值的合计数D某年某地区20个乡镇各镇的工业产值10、经济计量分析工作的基本步骤是__________。
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过程*describetive statistc*tabstat prate mrate totpart,stat(max min mean p50 sd n)结果stats | prate mrate totpart---------+------------------------------max | 100 4.91 58811min | 3 .01 50mean | 87.36291 .7315124 1354.231p50 | 95.7 .46 276sd | 16.71654 .7795393 4629.265N | 1534 1534 1534过程summarize 全部的加总summarize prate mrate 两个变量summarize sole prate,detail结果summarizeVariable | Obs Mean Std. Dev. Min Max -------------+--------------------------------------------------------prate | 1534 87.36291 16.71654 3 100mrate | 1534 .7315124 .7795393 .01 4.91 totpart | 1534 1354.231 4629.265 50 58811totelg | 1534 1628.535 5370.719 51 70429 age | 1534 13.18123 9.171114 4 51 -------------+--------------------------------------------------------totemp | 1534 3568.495 11217.94 58 144387sole | 1534 .4876141 .5000096 0 1 ltotemp | 1534 6.686034 1.453375 4.060443 11.88025 summarize prate mrateVariable | Obs Mean Std. Dev. Min Max -------------+--------------------------------------------------------prate | 1534 87.36291 16.71654 3 100mrate | 1534 .7315124 .7795393 .01 4.91 summarize sole prate,detail= 1 if 401k is firm's sole plan-------------------------------------------------------------Percentiles Smallest1% 0 05% 0 010% 0 0 Obs 153425% 0 0 Sum of Wgt. 153450% 0 Mean .4876141Largest Std. Dev. .5000096 75% 1 190% 1 1 Variance .2500096 95% 1 1 Skewness .0495589 99% 1 1 Kurtosis 1.002456participation rate, percent-------------------------------------------------------------Percentiles Smallest1% 31.4 35% 53.8 8.810% 62.7 14.9 Obs 1534 25% 78 17.4 Sum of Wgt. 1534 50% 95.7 Mean 87.36291Largest Std. Dev. 16.71654 75% 100 10090% 100 100 Variance 279.4426 95% 100 100 Skewness -1.519626 99% 100 100 Kurtosis 5.258359 .end of do-file过程*cdf*tabulate prate结果累积participati |on rate, |percent | Freq. Percent Cum.------------+-----------------------------------3 | 1 0.07 0.078.8 | 1 0.07 0.1314.9 | 1 0.07 0.2017.4 | 1 0.07 0.2619.3 | 1 0.07 0.3320.1 | 1 0.07 0.3920.6 | 1 0.07 0.4621 | 1 0.07 0.5221.3 | 1 0.07 0.5922.1 | 1 0.07 0.6525.1 | 1 0.07 0.7226.1 | 1 0.07 0.7828.6 | 1 0.07 0.8529 | 1 0.07 0.9130.5 | 1 0.07 0.9831.4 | 1 0.07 1.0433.5 | 1 0.07 1.1734.2 | 1 0.07 1.2435.6 | 1 0.07 1.30 35.8 | 1 0.07 1.37 37 | 1 0.07 1.43 37.5 | 1 0.07 1.5037.7 | 1 0.07 1.5638.1 | 1 0.07 1.63 38.4 | 1 0.07 1.6938.7 | 1 0.07 1.7639.4 | 1 0.07 1.83 39.6 | 1 0.07 1.89 39.8 | 1 0.07 1.9641.5 | 1 0.07 2.0242.1 | 1 0.07 2.09 42.4 | 1 0.07 2.1542.5 | 1 0.07 2.2243.1 | 1 0.07 2.28 43.3 | 1 0.07 2.35 43.6 | 1 0.07 2.4143.8 | 1 0.07 2.4844.1 | 1 0.07 2.54 44.3 | 1 0.07 2.6144.7 | 2 0.13 2.7445.5 | 1 0.07 2.8045.8 | 1 0.07 2.8746.9 | 3 0.20 3.0647.3 | 1 0.07 3.1347.7 | 1 0.07 3.1948.2 | 1 0.07 3.26 48.6 | 2 0.13 3.39 48.8 | 1 0.07 3.4648.9 | 3 0.20 3.6549.2 | 1 0.07 3.72 49.6 | 2 0.13 3.8549.7 | 1 0.07 3.9150 | 1 0.07 3.98 50.2 | 1 0.07 4.04 50.3 | 1 0.07 4.11 50.7 | 1 0.07 4.1750.9 | 2 0.13 4.3051 | 1 0.07 4.37 51.3 | 1 0.07 4.43 51.5 | 1 0.07 4.5051.9 | 1 0.07 4.5652.3 | 1 0.07 4.63 52.4 | 1 0.07 4.6952.7 | 1 0.07 4.7653.1 | 1 0.07 4.8253.7 | 1 0.07 4.9553.8 | 1 0.07 5.0254.1 | 1 0.07 5.08 54.2 | 1 0.07 5.1554.9 | 2 0.13 5.2855.1 | 1 0.07 5.35 55.5 | 1 0.07 5.41 55.7 | 1 0.07 5.4855.9 | 1 0.07 5.5456.2 | 2 0.13 5.67 56.3 | 2 0.13 5.80 56.4 | 1 0.07 5.87 56.7 | 3 0.20 6.0656.8 | 1 0.07 6.1357 | 2 0.13 6.26 57.6 | 2 0.13 6.39 57.7 | 1 0.07 6.4557.8 | 2 0.13 6.5858 | 1 0.07 6.65 58.2 | 3 0.20 6.84 58.3 | 1 0.07 6.91 58.4 | 2 0.13 7.04 58.6 | 2 0.13 7.17 58.7 | 1 0.07 7.2458.8 | 1 0.07 7.3059 | 1 0.07 7.37 59.1 | 2 0.13 7.50 59.2 | 2 0.13 7.63 59.4 | 1 0.07 7.69 59.6 | 3 0.20 7.89 59.8 | 1 0.07 7.9559.9 | 2 0.13 8.0860.1 | 2 0.13 8.21 60.2 | 1 0.07 8.28 60.3 | 1 0.07 8.34 60.4 | 1 0.07 8.41 60.6 | 2 0.13 8.54 60.8 | 2 0.13 8.6760.9 | 2 0.13 8.8061.1 | 1 0.07 8.87 61.2 | 3 0.20 9.06 61.3 | 1 0.07 9.13 61.4 | 1 0.07 9.19 61.5 | 1 0.07 9.26 61.6 | 1 0.07 9.32 61.7 | 2 0.13 9.4561.8 | 2 0.13 9.5862 | 2 0.13 9.71 62.2 | 1 0.07 9.7862.6 | 1 0.07 9.91 62.7 | 2 0.13 10.0462.9 | 2 0.13 10.1763 | 3 0.20 10.37 63.3 | 2 0.13 10.50 63.4 | 1 0.07 10.56 63.6 | 1 0.07 10.6363.7 | 1 0.07 10.6964 | 1 0.07 10.76 64.3 | 1 0.07 10.82 64.4 | 3 0.20 11.02 64.6 | 4 0.26 11.28 64.7 | 1 0.07 11.3464.9 | 2 0.13 11.4765 | 2 0.13 11.60 65.1 | 3 0.20 11.80 65.3 | 1 0.07 11.86 65.5 | 3 0.20 12.06 65.6 | 2 0.13 12.1965.7 | 1 0.07 12.2666.2 | 1 0.07 12.32 66.3 | 2 0.13 12.45 66.5 | 1 0.07 12.52 66.6 | 5 0.33 12.8466.9 | 2 0.13 12.9767 | 1 0.07 13.04 67.1 | 1 0.07 13.10 67.2 | 2 0.13 13.23 67.3 | 3 0.20 13.43 67.6 | 2 0.13 13.5667.8 | 1 0.07 13.6268 | 1 0.07 13.69 68.3 | 1 0.07 13.75 68.5 | 1 0.07 13.82 68.6 | 1 0.07 13.89 68.7 | 3 0.20 14.08 68.8 | 2 0.13 14.2168.9 | 2 0.13 14.3469.2 | 1 0.07 14.41 69.3 | 1 0.07 14.47 69.4 | 1 0.07 14.54 69.6 | 1 0.07 14.60 69.8 | 1 0.07 14.6769.9 | 2 0.13 14.8070 | 2 0.13 14.93 70.2 | 2 0.13 15.06 70.3 | 2 0.13 15.19 70.5 | 1 0.07 15.25 70.6 | 2 0.13 15.3870.9 | 3 0.20 15.6571.1 | 1 0.07 15.71 71.4 | 1 0.07 15.78 71.5 | 1 0.07 15.84 71.6 | 2 0.13 15.97 71.7 | 4 0.26 16.2371.9 | 2 0.13 16.3672 | 2 0.13 16.49 72.1 | 1 0.07 16.56 72.3 | 1 0.07 16.62 72.5 | 2 0.13 16.75 72.6 | 2 0.13 16.88 72.7 | 1 0.07 16.95 72.8 | 2 0.13 17.0872.9 | 3 0.20 17.2873 | 4 0.26 17.54 73.2 | 1 0.07 17.60 73.4 | 3 0.20 17.80 73.5 | 5 0.33 18.12 73.6 | 1 0.07 18.19 73.7 | 3 0.20 18.38 73.8 | 4 0.26 18.6473.9 | 4 0.26 18.9074 | 4 0.26 19.17 74.1 | 2 0.13 19.30 74.3 | 2 0.13 19.43 74.4 | 1 0.07 19.49 74.5 | 1 0.07 19.56 74.6 | 2 0.13 19.69 74.7 | 4 0.26 19.9574.8 | 1 0.07 20.0175 | 1 0.07 20.08 75.1 | 3 0.20 20.27 75.2 | 3 0.20 20.47 75.3 | 7 0.46 20.93 75.4 | 2 0.13 21.06 75.5 | 3 0.20 21.25 75.6 | 1 0.07 21.32 75.7 | 4 0.26 21.58 75.8 | 1 0.07 21.6475.9 | 3 0.20 21.8476.1 | 1 0.07 21.90 76.2 | 1 0.07 21.97 76.3 | 3 0.20 22.16 76.4 | 5 0.33 22.49 76.5 | 3 0.20 22.6976.9 | 2 0.13 22.8277 | 4 0.26 23.08 77.2 | 4 0.26 23.3477.4 | 3 0.20 23.92 77.5 | 2 0.13 24.05 77.6 | 2 0.13 24.19 77.7 | 3 0.20 24.38 77.8 | 1 0.07 24.4577.9 | 3 0.20 24.6478 | 6 0.39 25.03 78.1 | 2 0.13 25.16 78.2 | 2 0.13 25.29 78.3 | 1 0.07 25.36 78.4 | 1 0.07 25.42 78.5 | 4 0.26 25.68 78.6 | 2 0.13 25.81 78.7 | 3 0.20 26.0178.8 | 4 0.26 26.2779 | 4 0.26 26.53 79.1 | 1 0.07 26.60 79.2 | 2 0.13 26.73 79.3 | 5 0.33 27.05 79.4 | 1 0.07 27.12 79.5 | 2 0.13 27.25 79.6 | 1 0.07 27.31 79.7 | 4 0.26 27.57 79.8 | 3 0.20 27.7779.9 | 1 0.07 27.8480 | 2 0.13 27.97 80.1 | 1 0.07 28.03 80.2 | 1 0.07 28.10 80.4 | 2 0.13 28.23 80.5 | 2 0.13 28.36 80.6 | 4 0.26 28.62 80.7 | 2 0.13 28.75 80.8 | 1 0.07 28.8180.9 | 5 0.33 29.1481 | 4 0.26 29.40 81.1 | 2 0.13 29.53 81.2 | 2 0.13 29.66 81.3 | 3 0.20 29.86 81.5 | 2 0.13 29.99 81.6 | 1 0.07 30.05 81.7 | 2 0.13 30.1881.8 | 3 0.20 30.3882 | 2 0.13 30.51 82.1 | 2 0.13 30.64 82.2 | 2 0.13 30.77 82.3 | 1 0.07 30.83 82.5 | 6 0.39 31.23 82.6 | 5 0.33 31.55 82.7 | 2 0.13 31.6882.9 | 2 0.13 31.9483 | 1 0.07 32.01 83.2 | 1 0.07 32.07 83.3 | 2 0.13 32.20 83.4 | 1 0.07 32.27 83.5 | 3 0.20 32.46 83.6 | 2 0.13 32.59 83.7 | 3 0.20 32.79 83.8 | 2 0.13 32.9283.9 | 2 0.13 33.0584 | 1 0.07 33.12 84.1 | 2 0.13 33.25 84.2 | 2 0.13 33.38 84.3 | 3 0.20 33.57 84.5 | 2 0.13 33.70 84.6 | 4 0.26 33.96 84.7 | 2 0.13 34.0984.9 | 4 0.26 34.3585 | 2 0.13 34.49 85.1 | 5 0.33 34.81 85.2 | 2 0.13 34.94 85.3 | 4 0.26 35.20 85.4 | 1 0.07 35.27 85.5 | 3 0.20 35.46 85.6 | 2 0.13 35.59 85.7 | 5 0.33 35.9285.8 | 5 0.33 36.2586 | 2 0.13 36.38 86.1 | 2 0.13 36.51 86.2 | 1 0.07 36.57 86.3 | 4 0.26 36.83 86.4 | 2 0.13 36.96 86.5 | 3 0.20 37.16 86.6 | 1 0.07 37.22 86.7 | 1 0.07 37.29 86.8 | 1 0.07 37.3586.9 | 2 0.13 37.4887 | 2 0.13 37.61 87.1 | 2 0.13 37.74 87.2 | 2 0.13 37.87 87.3 | 2 0.13 38.01 87.4 | 4 0.26 38.27 87.5 | 2 0.13 38.40 87.6 | 4 0.26 38.66 87.7 | 2 0.13 38.79 87.8 | 1 0.07 38.8587.9 | 1 0.07 38.9288 | 3 0.20 39.11 88.1 | 3 0.20 39.3188.3 | 2 0.13 39.63 88.4 | 1 0.07 39.70 88.5 | 2 0.13 39.83 88.6 | 1 0.07 39.9088.8 | 3 0.20 40.0989 | 6 0.39 40.48 89.1 | 2 0.13 40.61 89.2 | 1 0.07 40.68 89.3 | 1 0.07 40.74 89.4 | 3 0.20 40.94 89.5 | 1 0.07 41.00 89.6 | 3 0.20 41.20 89.7 | 2 0.13 41.33 89.8 | 7 0.46 41.7989.9 | 6 0.39 42.1890 | 1 0.07 42.24 90.1 | 2 0.13 42.37 90.2 | 1 0.07 42.44 90.3 | 3 0.20 42.63 90.4 | 2 0.13 42.76 90.5 | 4 0.26 43.02 90.7 | 2 0.13 43.1690.8 | 5 0.33 43.4891 | 1 0.07 43.55 91.1 | 1 0.07 43.61 91.2 | 2 0.13 43.74 91.3 | 1 0.07 43.81 91.4 | 2 0.13 43.94 91.5 | 2 0.13 44.07 91.6 | 5 0.33 44.39 91.7 | 5 0.33 44.72 91.8 | 3 0.20 44.9291.9 | 3 0.20 45.1192 | 2 0.13 45.24 92.1 | 1 0.07 45.31 92.2 | 2 0.13 45.44 92.3 | 3 0.20 45.63 92.4 | 1 0.07 45.70 92.5 | 3 0.20 45.89 92.6 | 3 0.20 46.09 92.7 | 2 0.13 46.22 92.8 | 1 0.07 46.2892.9 | 2 0.13 46.4193 | 2 0.13 46.54 93.1 | 3 0.20 46.74 93.2 | 3 0.20 46.94 93.3 | 3 0.20 47.13 93.4 | 2 0.13 47.26 93.5 | 2 0.13 47.3993.7 | 3 0.20 47.65 93.8 | 2 0.13 47.7893.9 | 4 0.26 48.0494 | 2 0.13 48.17 94.2 | 4 0.26 48.44 94.4 | 4 0.26 48.70 94.5 | 1 0.07 48.76 94.6 | 2 0.13 48.89 94.8 | 2 0.13 49.0294.9 | 1 0.07 49.0995 | 5 0.33 49.41 95.1 | 1 0.07 49.48 95.2 | 1 0.07 49.54 95.3 | 2 0.13 49.67 95.4 | 2 0.13 49.80 95.6 | 2 0.13 49.93 95.7 | 6 0.39 50.33 95.8 | 3 0.20 50.5295.9 | 2 0.13 50.6596 | 3 0.20 50.85 96.1 | 3 0.20 51.04 96.2 | 1 0.07 51.11 96.3 | 3 0.20 51.30 96.4 | 1 0.07 51.37 96.5 | 3 0.20 51.56 96.6 | 1 0.07 51.6396.8 | 3 0.20 51.8397 | 2 0.13 51.96 97.1 | 1 0.07 52.02 97.2 | 2 0.13 52.15 97.4 | 1 0.07 52.22 97.6 | 2 0.13 52.35 97.7 | 3 0.20 52.54 97.8 | 3 0.20 52.7497.9 | 1 0.07 52.8098.1 | 2 0.13 52.93 98.3 | 1 0.07 53.00 98.4 | 1 0.07 53.06 98.5 | 1 0.07 53.13 98.6 | 2 0.13 53.26 98.7 | 1 0.07 53.32 98.8 | 2 0.13 53.4698.9 | 4 0.26 53.7299 | 3 0.20 53.91 99.1 | 4 0.26 54.17 99.2 | 4 0.26 54.43 99.3 | 2 0.13 54.56 99.4 | 1 0.07 54.63 99.5 | 2 0.13 54.7699.6 | 4 0.26 55.0299.7 | 1 0.07 55.0899.8 | 3 0.20 55.2899.9 | 4 0.26 55.54100 | 682 44.46 100.00 ------------+-----------------------------------Total | 1,534 100.00过程*pairwise correlation*pwcorr prate mrate totpart totelgpwcorr prate mrate totpart totelg,sig star(0.05)结果两两相比较| prate mrate totpart totelg -------------+------------------------------------prate | 1.0000mrate | 0.2733 1.0000totpart | 0.0042 0.0186 1.0000totelg | -0.0764 -0.0007 0.9761 1.0000 显著性pwcorr prate mrate totpart totelg,sig star(0.05)| prate mrate totpart totelg -------------+------------------------------------prate | 1.0000||mrate | 0.2733* 1.0000| 0.0000|totpart | 0.0042 0.0186 1.0000| 0.8703 0.4658|totelg | -0.0764* -0.0007 0.9761* 1.0000| 0.0028 0.9770 0.0000|.end of do-file过程*graph*histogram prate,width(5)frequencykdensity pratescatter prate mratetwoway(scatter prate mrate)(qfit prate mrate)graph save graphpractice1结果*graph*. histogram prate,width(5)frequency (bin=20, start=3, width=5). kdensity prate. scatter prate mrate. twoway(scatter prate mrate)(qfit prate mrate). graph save graphpractice1( saved).end of do-file过程twoway(scatter prate mrate)(lfit prate mrate) graph save graphpractice2graph combine graph1.gph graph2.gph结果 050100012345401k plan match rate participation rate, percentFitted values。