計量經(jīng)濟學導論第四版答案中文版

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1、計量經(jīng)濟學導論第四版答案中文版 【篇一：計量經(jīng)濟學導論 (伍德里奇第三版 )課后習題答 案 chapter 1 】 >solutions to problems 1.1 (i) ideally, we could randomly assign students to classes of different sizes. that is, each student is assigned a different class size without regard to any student characteristics such as ability and family bac

2、kground. for reasons we will see in chapter 2, we would like substantial variation in class sizes (subject, of course, to ethical considerations and resource constraints). (ii) a negative correlation means that larger class size is associated with lower performance. we might find a negative correl

3、ation because larger class size actually hurts performance. however, with observational data, there are other reasons we might find a negative relationship. for example, children from more affluent families might be more likely to attend schools with smaller class sizes, and affluent children gene

4、rally score better on standardized tests. another possibility is that, within a school, a principal might assign the better students to smaller classes. or, some parents might insist their children are in the smaller classes, and these same parents tend to be more involved in their children ’s educa

5、tion. (iii) given the potential for confounding factors –some of which are listed in (ii) –finding a negative correlation would not be strong evidence that smaller class sizes actually lead to better performance. some way of controlling for the confounding factors is needed, and this is the subject

6、 of multiple regression analysis. 1.2 (i) here is one way to pose the question: if two firms, say a and b, are identical in all respects except that firm a supplies job training one hour per worker more than firm b, by how much would firm a ’s output differ from firm b ’s? (ii) firms are likel

7、y to choose job training depending on the characteristics of workers. some observed characteristics are years of schooling, years in the workforce, and experience in a particular job. firms might even discriminate based on age, gender, or race. perhaps firms choose to offer training to more or less

8、 able workers, where ―ability ‖ht bmeigdifficult to quantify but where a manager has some idea about the relative abilities of different employees. moreover, different kinds of workers might be attracted to firms that offer more job training on average, and this might not be evident to employers.

9、 (iii) the amount of capital and technology available to workers would also affect output. so, two firms with exactly the same kinds of employees would generally have different outputs if they use different amounts of capital or technology. the quality of managers would also have an effect. (iv) n

10、o, unless the amount of training is randomly assigned. the many factors listed in parts (ii) and (iii) can contribute to finding a positive correlation between output and training even if job training does not improve worker productivity. 1.3 it does not make sense to pose the question in terms of

11、causality. economists would assume that students choose a mix of studying and working (and other activities, such as attending class, leisure, and sleeping) based on rational behavior, such as maximizing utility subject to the constraint that there are only 168 hours in a week. we can then use stat

12、istical methods to measure the association between studying and working, including regression analysis that we cover starting in chapter 2. but we would not be claiming that one variable ―causes ‖ the other. they are both choice variables of the student. chapter 2 solutions to problems 2.1 (i)

13、income, age, and family background (such as number of siblings) are just a few possibilities. it seems that each of these could be correlated with years of education. (income and education are probably positively correlated; age and education may be negatively correlated because women in more rece

14、nt cohorts have, on average, more education; and number of siblings and education are probably negatively correlated.) (ii) not if the factors we listed in part (i) are correlated with educ. because we would like to hold these factors fixed, they are part of the error term. but if u is correlated

15、 with educ then e(u|educ) ? 0, and so slr.4 fails. 2.2 in the equation y = ?0 + ?1x + u, add and subtract ?0 from the right hand side to get y = (?0 + ?0) + ?1x + (u ? ?0). call the new error e = u ? ?0, so that e(e) = 0. the new intercept is ?0 + ?0, but the slope is still ?1. 2.3 (i) let yi = g

16、pai, xi = acti, and n = 8. then = 25.875, = 3.2125, ?(xi – )(yi – ) = i?1n ?= 5.8125, and ?(xi – )2 = 56.875. from equation (2.9), we obtain the slope as ?1 i?1 n ? = – 5.8125/56.875 ? .1022, rounded to four places after the decimal. from (2.17), ?0 ? ? 3.2125 – (.1022)25.875 ? .5681. so we can

17、 write ? 1 ? = .5681 + .1022 act gpan = 8. the intercept does not have a useful interpretation because act is not close to zero for the ? increases by .1022(5) = .511. population of interest. if act is 5 points higher, gpa (ii) the fitted values and residuals — rounded to four decimal places — a

18、re given along with the observation number i and gpa in the following table: you can verify that the residuals, as reported in the table, sum to ?.0002, which is pretty close to zero given the inherent rounding error. ?= .5681 + .1022(20) ? 2.61. (iii) when act = 20, gpa ?i2, is about .4347 (rou

19、nded to four decimal places), (iv) the sum of squared residuals, ?u i?1 n n and the total sum of squares, ?(yi – )2, is about 1.0288. so the r-squared from the i?1 regression is r2 = 1 – ssr/sst ? 1 – (.4347/1.0288) ? .577. therefore, about 57.7% of the variation in gpa is explained by act in

20、this small sample of students. ?2.4 (i) when cigs = 0, predicted birth weight is 119.77 ounces. when cigs = 20, bwght = 109.49. this is about an 8.6% drop. (ii) not necessarily. there are many other factors that can affect birth weight, particularly overall health of the mother and quality of p

21、renatal care. these could be correlated with cigarette smoking during birth. also, something such as caffeine consumption can affect birth weight, and might also be correlated with cigarette smoking. (iii) if we want a predicted bwght of 125, then cigs = (125 – 119.77)/( –.524) ? –10.18, or abou

22、t –10 cigarettes! this is nonsense, of course, and it shows what happens when we are trying to predict something as complicated as birth weight with only a single explanatory variable. the largest predicted birth weight is necessarily 119.77. yet almost 700 of the births in the sample had a birth we

23、ight higher than 119.77. (iv) 1,176 out of 1,388 women did not smoke while pregnant, or about 84.7%. because we are using only cigs to explain birth weight, we have only one predicted birth weight at cigs = 0. the predicted birth weight is necessarily roughly in the middle of the observed birth w

24、eights at cigs = 0, and so we will under predict high birth rates. 2.5 (i) the intercept implies that when inc = 0, cons is predicted to be negative $124.84. this, of course, cannot be true, and reflects that fact that this consumption function might be a poor predictor of consumption at very low-

25、income levels. on the other hand, on an annual basis, $124.84 is not so far from zero. ? = –124.84 + .853(30,000) = 25,465.16 dollars. (ii) just plug 30,000 into the equation: cons (iii) the mpc and the apc are shown in the following graph. even though the intercept is negative, the smallest apc

26、in the sample is positive. the graph starts at an annual income level increases housing prices. (ii) if the city chose to locate the incinerator in an area away from more expensive neighborhoods, then log(dist) is positively correlated with housing quality. this would violate slr.4, and ols esti

27、mation is biased. (iii) size of the house, number of bathrooms, size of the lot, age of the home, and quality of the neighborhood (including school quality), are just a handful of factors. as mentioned in part (ii), these could certainly be correlated with dist [and log(dist)]. 2.7 (i) when we co

28、ndition on inc e(u|inc ?e|inc ) = ?e(e|inc ?0 because e(e|inc) = e(e) = 0. (ii) again, when we condition on inc var(u|inc ?e|inc 2var(e|inc) = ?e2inc because var(e|inc) = ?e2. (iii) families with low incomes do not have much discretion about spending; typically, a low-income family must spend

29、 on food, clothing, housing, and other necessities. higher income people have more discretion, and some might choose more consumption while others more saving. this discretion suggests wider variability in saving among higher income families. 2.8 (i) from equation (2.66), ?n??n2???1 = ??xiyi? / ??

30、xi?. ?i?1??i?1? plugging in yi = ?0 + ?1xi + ui gives ?n??n2???1 = ??xi(?0??1xi?ui)?/ ??xi?. ?i?1??i?1? after standard algebra, the numerator can be written as ?0?xi??1?x??xiui. 2 nnn i?1i?1 i i?1 ? as putting this over the denominator shows we can write ?1 ?n??n2??n??n2???1 = ?0??xi?/ ??xi?

31、 + ?1 + ??xiui?/ ??xi?. ?i?1??i?1??i?1??i?1? conditional on the xi, we have ?n??n2??e(?1) = ?0??xi?/ ??xi? + ?1 ?i?1??i?1? ? is given by the first term in this equation. because e(ui) = 0 for all i. therefore, the bias in ?1this bias is obviously zero when ?0 = 0. it is also zero when ?xi = 0, w

32、hich is the same as i?1n = 0. in the latter case, regression through the origin is identical to regression with an intercept. 【篇二：計量經(jīng)濟學導論第五版第一章上機作業(yè)】 betive statistc* tabstat prate mrate totpart,stat(max min mean p50 sd n) 結(jié)果 stats | prate mratetotpart ---------+-----------------------------

33、- max | 1004.91 58811 min |3 .01 50 mean | 87.36291 .7315124 1354.231 p50 |95.7 .46 276 sd | 16.71654 .7795393 4629.265 n |153415341534 過程 summarize 全部的加總 summarize prate mrate 兩個變量 summarize sole prate,detail 結(jié)果 summarize variable | obs mean std. dev. min max -------------+----------------

34、- --------------------------------------- prate |1534 87.36291 16.71654 3 100 mrate |1534 .7315124 .7795393 .01 4.91totpart |1534 1354.231 4629.2655058811 totelg |1534 1628.535 5370.7195170429 age |1534 13.18123 9.171114 451 -------------+---------------------------------- ----------------------

35、 totemp |1534 3568.495 11217.9458 144387sole |1534 .4876141 .5000096 0 1ltotemp |1534 6.686034 1.4533754.06044311.88025 summarize prate mrate variable | obs mean std. dev. min max -------------+----------------- --------------------------------------- prate |1534 87.36291 16.71654 3 100 mrate

36、 |1534 .7315124 .7795393 .01 4.91 summarize sole prate,detail = 1 if 401k is firms sole plan ------------------------------------------------------------- percentilessmallest 1%0 0 5%0 0 10%0 0 obs 1534 25%0 0 sum of wgt. 1534 50%0 mean .4876141 largest std. dev..5000096 75%1 1 90%1 1 varianc

37、e .2500096 95%1 1 skewness .0495589 99%1 1 kurtosis 1.002456 participation rate, percent ------------------------------------------------------------- percentilessmallest 1%31.4 3 5%53.88.8 10%62.7 14.9 25% 78 17.4 50%95.7 largest 75% 100100 90% 100100 95% 100100 99% 100100 . end of do-file

38、 過程 *cdf* tabulate prate 結(jié)果 累積 participati | on rate, | percent |freq. percent ------------+----------------------------------- 3 | 1 0.07 8.8 | 1 0.07 14.9 | 1 0.07 17.4 | 1 0.07 19.3 | 1 0.07 20.1 | 1 0.07 20.6 | 1 0.07 21 | 1 0.07 21.3 | 1 0.07 22.1 | 1 0.07 25.1 | 1 0.07 26.1 | 1 0.0

39、7 28.6 | 1 0.07 29 | 1 0.07 30.5 | 1 0.07 31.4 | 1 0.07 obssum of wgt. meanstd. dev. variance skewnesskurtosis cum. 0.070.130.200.260.330.390.46 0.520.590.650.720.780.85 0.910.981.04 1534 1534 87.36291 16.71654279.4426-1.5196265.258359 33.5 | 1 0.07 1.17 34.2 | 1 0.07 1.24 35.6 | 1 0.07 1.30 35.

40、8 | 1 0.07 1.37 37 | 1 0.07 1.43 37.5 | 1 0.07 1.50 37.7 | 1 0.07 1.56 38.1 | 1 0.07 1.63 38.4 | 1 0.07 1.69 38.7 | 1 39.4 | 1 39.6 | 1 39.8 | 1 41.5 | 1 42.1 | 1 42.4 | 1 42.5 | 1 43.1 | 1 43.3 | 1 43.6 | 1 43.8 | 1 44.1 | 1 44.3 | 1 44.7 | 2 45.5 | 1 45.8 | 1 46.9 | 3 47.3 | 1 4

41、7.7 | 1 48.2 | 1 48.6 | 2 48.8 | 1 48.9 | 3 49.2 | 1 49.6 | 2 49.7 | 1 50 | 1 50.2 | 1 50.3 | 1 50.7 | 1 50.9 | 2 51 | 1 51.3 | 1 51.5 | 1 51.9 | 1 52.3 | 1 52.4 | 1 52.7 | 1 53.1 | 10.07 1.76 0.07 1.83 0.07 1.89 0.07 1.96 0.07 2.02 0.07 2.09 0.07 2.15 0.07 2.22 0.07 2.28 0.07 2.35 0

42、.07 2.41 0.07 2.48 0.07 2.54 0.07 2.61 0.13 2.74 0.07 2.80 0.07 2.87 0.20 3.06 0.07 3.13 0.07 3.19 0.07 3.26 0.13 3.39 0.07 3.46 0.20 3.65 0.07 3.72 0.13 3.85 0.07 3.91 0.07 3.98 0.07 4.04 0.07 4.11 0.07 4.17 0.13 4.30 0.07 4.37 0.07 4.43 0.07 4.50 0.07 4.56 0.07 4.63 0.07 4.69 0.07 4.76 0.07 4.82

43、 53.7 | 1 0.07 4.95 53.8 | 1 0.07 5.02 54.1 | 1 0.07 5.08 54.2 | 1 0.07 5.15 54.9 | 2 0.13 5.28 55.1 | 1 0.07 5.35 55.5 | 1 0.07 5.41 55.7 | 1 0.07 5.48 55.9 | 1 0.07 5.54 56.2 | 2 56.3 | 2 56.4 | 1 56.7 | 3 56.8 | 1 57 | 2 57.6 | 2 57.7 | 1 57.8 | 2 58 | 1 58.2 | 3 58.3 | 1 58.4 | 2 58.6

44、| 2 58.7 | 1 58.8 | 1 59 | 1 59.1 | 2 59.2 | 2 59.4 | 1 59.6 | 3 59.8 | 1 59.9 | 2 60.1 | 2 60.2 | 1 60.3 | 1 60.4 | 1 60.6 | 2 60.8 | 2 60.9 | 2 61.1 | 1 61.2 | 3 61.3 | 1 61.4 | 1 61.5 | 1 61.6 | 1 61.7 | 2 61.8 | 2 62 | 2 62.2 | 10.13 5.67 0.13 5.80 0.07 5.87 0.20 6.06 0.0

45、7 6.13 0.13 6.26 0.13 6.39 0.07 6.45 0.13 6.58 0.07 6.65 0.20 6.84 0.07 6.91 0.13 7.04 0.13 7.17 0.07 7.24 0.07 7.30 0.07 7.37 0.13 7.50 0.13 7.63 0.07 7.69 0.20 7.89 0.07 7.95 0.13 8.08 0.13 8.21 0.07 8.28 0.07 8.34 0.07 8.41 0.13 8.54 0.13 8.67 0.13 8.80 0.07 8.87 0.20 9.06 0.07 9.13 0.07 9.1

46、9 0.07 9.26 0.07 9.32 0.13 9.45 0.13 9.58 0.13 9.71 0.07 9.78 62.6 | 1 0.07 9.91 62.7 | 2 0.13 10.04 62.9 | 2 0.13 10.17 63 | 3 0.20 10.37 63.3 | 2 0.13 10.50 63.4 | 1 0.07 10.56 63.6 | 1 0.07 10.63 63.7 | 1 0.07 10.69 64 | 1 0.07 10.76 64.3 | 1 64.4 | 3 64.6 | 4 64.7 | 1 64.9 | 2 65 | 2 65.1

47、| 3 65.3 | 1 65.5 | 3 65.6 | 2 65.7 | 1 66.2 | 1 66.3 | 2 66.5 | 1 66.6 | 5 66.9 | 2 67 | 1 67.1 | 1 67.2 | 2 67.3 | 3 67.6 | 2 67.8 | 1 68 | 1 68.3 | 1 68.5 | 1 68.6 | 1 68.7 | 3 68.8 | 2 68.9 | 2 69.2 | 1 69.3 | 1 69.4 | 1 69.6 | 1 69.8 | 1 69.9 | 2 70 | 2 70.2 | 2 70.3 |

48、2 70.5 | 1 70.6 | 20.07 10.82 0.20 11.02 0.26 11.28 0.07 11.34 0.13 11.47 0.13 11.60 0.20 11.80 0.07 11.86 0.20 12.06 0.13 12.19 0.07 12.26 0.07 12.32 0.13 12.45 0.07 12.52 0.33 12.84 0.13 12.97 0.07 13.04 0.07 13.10 0.13 13.23 0.20 13.43 0.13 13.56 0.07 13.62 0.07 13.69 0.07 13.75 0.07 13.82

49、 0.07 13.89 0.20 14.08 0.13 14.21 0.13 14.34 0.07 14.41 0.07 14.47 0.07 14.54 0.07 14.60 0.07 14.67 0.13 14.80 0.13 14.93 0.13 15.06 0.13 15.19 0.07 15.25 0.13 15.38 【篇三：伍德里奇計量經(jīng)濟學導論計算機習題第六章第 13 題 c_6.13 】 >% c6.13 by % 打開文字文件和數(shù)據(jù)文件 importdata(meap00_01.des); data=xlsread(meap00_01); % 檢驗所用數(shù)據(jù)是

50、否為非空 isnan=isnan(data(:,[3,5,8,9])); a=sum(isnan); b=find(a==0); data1=data(b,:); % 變量命名 math4=data1(:,3); lunch=data1(:,5); leoll=data1(:,8); lexppp=data1(:,9); % ols 估計 result1=ols(math4,[ones(length(math4),1),lunch,leoll,lexppp]); vnames=char(math4,constant,lunch,leoll,lexppp); prt(result1,v

51、names) % 回歸結(jié)果 % ordinary least-squares estimates % dependent variable =math4 % r-squared= 0.3729 % rbar-squared= 0.3718 % sigma^2 = 234.1638 % durbin-watson = 1.7006 % nobs, nvars =1692, 4 % *************************************************************** % variablecoefficientt-statistic t-proba

52、bility % constant 91.9324844.6054440.000004 % lunch -0.448743 -30.6476310.000000 % leoll-5.399153 -5.7412650.000000 % lexppp 3.5247421.6801720.093109 % 由回歸結(jié)果中的 p 值發(fā)現(xiàn) lunch,leoll 是在 5%的水平上顯著的 , 而 lexppp 在 5%水平上不顯著 , % 但在 10% 的顯著水平上是顯著的 . % 求出回歸的擬合值及其取值范圍 yhat=result1.yhat; std_yhat=std(yhat);

53、mean_yhat=mean(yhat); yhat_qujian=[mean_yhat-2*std_yhat,mean_yhat+2*std_yhat] % yhat 的取值范圍是 (49.1090 ,96.2665) % 求出 math4 的實際取值范圍 math4_qujian=[min(math4),max(math4)] % math4 的實際取值范圍是 (0 ,100), 可見擬合值的取值范圍要比實 際范圍窄 % 求出回歸殘差 resid=result1.resid; max_resid=max(resid); row=find(resid==max_resid) s

54、chool_code=data1(row,2) % 學校類型是 school_code=1141, 說明該學校的實際數(shù)學考試通過 率要比估計的數(shù)學考試通 % 過率高很多 ,也就這所學校的數(shù)學教學質(zhì) 量較高 % 求取解釋變量數(shù)據(jù)的平方項 leoll2=leoll.^2; lexppp2=lexppp.^2; lunch2=lunch.^2; %在方程中加入所有解釋變量的平方項進行回歸 result2=ols(math4,[ones(length(math4),1),lunch,leoll,lexppp,lu nch2,leoll2,lexppp2]); vnames=char(m

55、ath4,constant,lunch,leoll,lexppp,lunch2,leoll2,l exppp2); prt(result2,vnames) % 檢驗聯(lián)合顯著性 rsqr1=result1.rsqr; rsqr2=result2.rsqr; f=((rsqr2-rsqr1)/3)/((1-rsqr2)/(1692-6-1)) p=1-fcdf(f,3,1685) % f =0.5180 % p =0.6699 % 所以這幾個解釋變量的平方項是聯(lián)合不顯著的 ,所以不應該把他們 放到模型中 % 求所用數(shù)據(jù)值除以其標準差 leoll_new=leoll/std(leo

56、ll); lunch_new=lunch/std(lunch); lexppp_new=lexppp/std(lexppp); math4_new=math4/std(math4); % 重新進行回歸 result3=ols(math4_new,[ones(length(math4_new),1),lunch_ne w,leoll_new,lexppp_new]); vnames=char(math4_new,constant,lunch_new,leoll_new,lexpp p_new); prt(result3,vnames) % 回歸結(jié)果 % ordinary least-

57、squares estimates % dependent variable =math4_new % r-squared= 0.3729 % rbar-squared= 0.3718 % sigma^2 = 0.6282 % durbin-watson = 1.7006 % nobs, nvars =1692, 4 % *************************************************************** % variablecoefficientt-statistic t-probability % constant4.7617594.6054440.000004 % lunch_new -0.612853 -30.6476310.000000 % leoll_new -0.114620 -5.7412650.000000 % lexppp_new 0.0347431.6801720.093109 % 由結(jié)果 lunch,leoll 和 lexppp 分別提高一倍 ,會使數(shù)學通過率分 別變化 0.613,0.115 和 % 0.035 個標準差 ,所以 lunch 對數(shù)學考試通 過率影響最大