计量经济学导论(伍德里奇第三版)课后习题答案 CHAPTER 1.doc

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1、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 background. For reasons we will see in Chapter 2, we wou

2、ld 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 correlation because larger class size actually hurts performan

3、ce.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 generally score better on standardized tests. Another possibili

4、ty 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 childrens education. (iii) Given the potential for confounding factors some

5、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 of multiple regression analysis. 1.2 (i) Here is one way to pos

6、e the question: If two firms, say A and B, are identical in allrespects except that firm A supplies job training one hour per worker more than firm B, by how much would firm As output differ from firm Bs? (ii) Firms are likely to choose job training depending on the characteristics of workers. Some

7、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 able workers, where ability might be difficult toquantify but where a manag

8、er 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. (iii) The amount of capital and technology available to workers would also affec

9、t 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) No, unless the amount of training is randomly assigned. The many factors listed in p

10、arts (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 causality. Economists would assume that students choose a mix of studying and workin

11、g (and other activities, such as attending class,1leisure, 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 statistical methods to measure the association between studying and working, including re

12、gression 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 2SOLUTIONS TO PROBLEMS 2.1 (i) Income, age, and family background (such as number of siblings) are just a fewpossibilities. I

13、t 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 recent cohorts have, on average, more education; and number of siblings and education are probably ne

14、gatively 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 with educ then E(u|educ) 0, and so SLR.4 fails. 2.2 In the equation y = b0 + b1x + u, add and subtr

15、act a0 from the right hand side to get y = (a0 + b0) + b1x + (u - a0). Call the new error e = u - a0, so that E(e) = 0. The new intercept is a0 + b0, but the slope is still b1. 2.3 (i) Let yi = GPAi, xi = ACTi, and n = 8. Then = 25.875, = 3.2125, (xi )(yi ) = i=1n= 5.8125, and (xi )2 = 56.875. From

16、equation (2.9), we obtain the slope as b1i=1n = 5.8125/56.875 .1022, rounded to four places after the decimal. From (2.17), b0 3.2125 (.1022)25.875 .5681. So we can write b1 = .5681 + .1022 ACT GPAn = 8. The intercept does not have a useful interpretation because ACT is not close to zero for the inc

17、reases 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 are given along with the observation number i and GPA in the following table: 2 You can verify that the residuals, as reported in the table, sum to -.

18、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 (rounded to four decimal places), (iv) The sum of squared residuals, ui=1nnand the total sum of squares, (yi )2, is about 1.0288. So the R-squared from thei=1r

19、egression is R2 = 1 SSR/SST 1 (.4347/1.0288) .577. Therefore, about 57.7% of the variation in GPA is explained by ACT in 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. The

20、re are many other factors that can affect birth weight, particularly overall health of the mother and quality of prenatal care. These could be correlated withcigarette smoking during birth. Also, something such as caffeine consumption can affect birth weight, and might also be correlated with cigare

21、tte smoking. (iii) If we want a predicted bwght of 125, then cigs = (125 119.77)/( .524) 10.18, or about 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 p

22、redicted birth weight is necessarily 119.77. Yet almost 700 of the births in the sample had a birth weight higher than 119.77. 3 (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

23、 at cigs = 0. The predicted birth weight is necessarily roughly in the middle of the observed birth weights 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 refl

24、ects that fact that this consumption function might be a poor predictor of consumption at very low-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 A

25、PC are shown in the following graph. Even though the intercept is negative, the smallest APC in the sample is positive. The graph starts at an annual income levelincreases housing prices. (ii) If the city chose to locate the incinerator in an area away from more expensiveneighborhoods, then log(dist

26、) is positively correlated with housing quality. This would violate SLR.4, and OLS estimation 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

27、), these could certainly be correlated with dist and log(dist).4 2.7 (i) When we condition on incE(u|ince|inc) = E(e|inc0 because E(e|inc) = E(e) = 0. (ii) Again, when we condition on incVar(u|ince|inc2Var(e|inc) = se2inc because Var(e|inc) = se2. (iii) Families with low incomes do not have much dis

28、cretion about spending; typically, a low-income family must spend 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 fa

29、milies. 2.8 (i) From equation (2.66), nn2%b1 = xiyi / xi. i=1i=1 Plugging in yi = b0 + b1xi + ui gives nn2%b1 = xi(b0+b1xi+ui)/ xi. i=1i=1 After standard algebra, the numerator can be written as b0xi+b1x+xiui. 2nnni=1i=1ii=1% as Putting this over the denominator shows we can write b1 nn2nn2%b1 = b0x

30、i/ xi + b1 + xiui/ xi. i=1i=1i=1i=1 Conditional on the xi, we have nn2%E(b1) = b0xi/ xi + b1 i=1i=1% is given by the first term in this equation. because E(ui) = 0 for all i. Therefore, the bias in b1This bias is obviously zero when b0 = 0. It is also zero when xi = 0, which is the same asi=1n= 0. I

31、n the latter case, regression through the origin is identical to regression with an intercept.5 %in part (i) we have, conditional on the xi, (ii) From the last expression for b1%) Var(b1n2nn2n2= xiVarxiui = xixiVar(ui) i=1i=1i=1i=1-2-2-2 nn22n22= xisxi = s/ xi2. i=1i=1i=1nnn222(iii) From (2.57), Var

32、(b1) = s/(xi-). From the hint, xi (xi-)2, and soi=1i=1i=1nn%) Var(b). A more direct way to see this is to write (x-)2 = x2-n()2, which Var(bii11i=1i=1is less than xi2 unless = 0.i=1n % increases as increases (holding the sum of the (iv) For a given sample size, the bias in b1increases relative to Va

33、r(b%). The bias in b% x2 fixed). But as increases, the variance of bi111% or b on a mean squared error is also small when b0 is small. Therefore, whether we prefer b11basis depends on the sizes of b0, , and n (in addition to the size of xi2).i=1n 2.9 (i) We follow the hint, noting that c1y = c1 (the

34、 sample average of c1yi is c1 times the sample average of yi) and c2x = c2. When we regress c1yi on c2xi (including an intercept) we use equation (2.19) to obtain the slope:%=b1(cx-cx)(c-c)cc(x-)(y-)2in21i112niii=1nn=i=1(cx-c)2i2i=1nn2c(x-)22ii=12 c1=i=1c2(xi-)(yi-)=2i(x-)i=1c1b1.c2% = (c1) b%(c2) =

35、 (c1) (c1/c2)b(c2) = From (2.17), we obtain the intercept as b011) because the intercept from regressing yi on xi is ( b) = c1b). c1( b1 1 (ii) We use the same approach from part (i) along with the fact that (c1+y) = c1 + and(c2+x) = c2 + . Therefore, (c1+yi)-(c1+y) = (c1 + yi) (c1 + ) = yi and (c2

36、+ xi) 6(c2+x) = xi . So c1 and c2 entirely drop out of the slope formula for the regression of (c1 + yi)% = (c+y) b% = b. The intercept is b%(c+x) = (c1 + ) b(c2 + ) = on (c2 + xi), and b1101121 + c1 c2b) + c1 c2b = b, which is what we wanted to show. (-b0111 (iii) We can simply apply part (ii) beca

37、use log(c1yi)=log(c1)+log(yi). In other words, replace c1 with log(c1), yi with log(yi), and set c2 = 0. (iv) Again, we can apply part (ii) with c1 = 0 and replacing c2 with log(c2) and xi with log(xi). and b are the original intercept and slope, then b%=b and b%=b-log(c)b. If b01110021 2.10 (i) Thi

38、s derivation is essentially done in equation (2.52), once (1/SSTx) is brought inside the summation (which is valid because SSTx does not depend on i). Then, just definewi=di/SSTx. ,)=E(b-b) , we show that the latter is zero. But, from part (i), (ii) Because Cov(b111n=nwE(u). Because the u are pairwi

39、se uncorrelated -b) =EE(bwui11ii=1iii=1i(they are independent), E(ui)=E(ui2/n)=s2/n (because E(uiuh)=0, ih). Therefore, ()i=1wiE(ui)=i=1wi(s2/n)=(s2/n)i=1wi=0. nnn=-b and, plugging in =b+b+ (iii) The formula for the OLS intercept isb010=(b+b+)-b=b+-(b-b) gives b0011011 and are uncorrelated, (iv) Bec

40、ause b1)=Var()+Var(b)2=s2/n+(s2/SST)2=s2/n+s22/SST, Var(b01xxwhich is what we wanted to show. )=s2(SST/n)+2/SST (v) Using the hint and substitution gives Var(b0xx-12222-12=s2nx-+/SST=snx/SSTx. xi=1ii=1i 2.11 (i) We would want to randomly assign the number of hours in the preparation course so that h

41、ours is independent of other factors that affect performance on the SAT. Then, we would collect information on SAT score for each student in the experiment, yielding a data set (sati,hoursi):i=1,.,n, where n is the number of students we can afford to have in the study. From equation (2.7), we should

42、 try to get as much variation in hoursi as is feasible.nn()()7(ii) Here are three factors: innate ability, family income, and general health on the day of the exam. If we think students with higher native intelligence think they do not need to prepare for the SAT, then ability and hours will be nega

43、tively correlated. Family income would probably be positively correlated with hours, because higher income families can more easily affordpreparation courses. Ruling out chronic health problems, health on the day of the exam should be roughly uncorrelated with hours spent in a preparation course. (i

44、ii) If preparation courses are effective,b1 should be positive: other factors equal, anincrease in hours should increase sat. (iv) The intercept, b0, has a useful interpretation in this example: because E(u) = 0, b0 is the average SAT score for students in the population with hours = 0. CHAPTER 3SOL

45、UTIONS TO PROBLEMS 3.1 (i) hsperc is defined so that the smaller it is, the lower the students standing in high school. Everything else equal, the worse the students standing in high school, the lower is his/her expected college GPA. (ii) Just plug these values into the equation: = 1.392 - .0135(20)

46、 + .00148(1050) = 2.676. colgpa (iii) The difference between A and B is simply 140 times the coefficient on sat, because hsperc is the same for both students. So A is predicted to have a score .00148(140) .207 higher. (iv) With hsperc fixed, Dcolgpa = .00148Dsat. Now, we want to find Dsat such that = .5, so .5 = .00148(Dsat) or Dsat = .5/(.00148) 338. Perhaps not surprisingly, a Dcolgpalarge ceteris paribus difference in SAT score almost two and one-half standard deviations i