Solution Manual for Introductory Econometrics: A Modern Approach , 7th Edition

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1CHAPTER 1The Nature of Econometrics and Economic Data

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2SOLUTIONS TOPROBLEMS1.1(i)Ideally, we could randomly assign students to classes of different sizes. That is,eachstudent is assigned a different class size without regard to any student characteristics suchasability and family background. For reasons we will see in Chapter 2, we would likesubstantialvariation in class sizes (subject, of course, to ethical considerations and resourceconstraints).(ii)A negative correlation means thatalarger class size is associated with lowerperformance.We might find a negative correlation becausealarger class size actually hurtsperformance.However, with observational data, there are other reasons we might find anegativerelationship. For example, children from more affluent families might be more likely toattend schoolswith smaller class sizes, and affluent children generallymightscore better onstandardized tests.Another possibility is that, within a school, a principal might assign thebetter students to smallerclasses. Or, some parents mightinsist their childrento be placed insmaller classes, and these sameparents tend to be more involved in their children’seducation.(iii)Given the potential for confounding factors–some of which are listed in (ii)–findinganegative correlation would not be strong evidence that smaller class sizes actually lead tobetterperformance. Some way of controlling for the confounding factors is needed, and this isthesubject of multiple regressionanalysis.1.3Itdoes notmake sense to pose the question in terms of causality. Economists wouldassumethat students choose a mix of studying and working (and other activities, such as attendingclass,leisure, and sleeping) based on rational behavior, such as maximizing utility subject totheconstraint that there are only 168 hours in a week. We can then use statistical methodstomeasure the association between studying and working, including regression analysis, whichwecover starting in Chapter 2. But we would not be claiming that one variable “causes” theother.They are both choice variables of thestudent.SOLUTIONS TO COMPUTEREXERCISESC1.1(i) The average ofeducis about 12.6 years. There are two people reporting zero yearsofeducation and 19 peoplereporting 18 years ofeducation.(ii)The average ofwagein the sampleis about $5.90, which seems low.(iii)Using Table B-60 in the 2004Economic Report of the President, the CPI was 56.9in1976 and233in2013.(iv)To convert 1976 dollars into 2013dollars, we use the ratio of the CPIs, whichis233/ 56.94.09.Therefore, the average hourly wage in2013dollars isroughly4.09($5.90)$24.13, which is a reasonablefigure.

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3(v)The sample contains 252 women (the number of observations withfemale= 1) and274men.C1.3(i) The largest is 100, the smallest is0.(ii)289out of 1,823, or about15.85percent of thesample.(iii)17(iv)The average ofmath4is about 71.9 and the average ofread4is about 60.1. So, atleastin 2001, the reading test was harder topass.(v)The sample correlation betweenmath4andread4is about .843, which is a veryhighdegree of (linear) association. Notsurprisingly, schools that have high pass rates on onetesthave a strong tendency to have high pass rates on the othertest.(vi)The average ofexpppis about $5,194.87. The standard deviation is $1,091.89,whichshows rather wide variation inspending per pupil. [The minimum is $1,206.88 andthemaximum is$11,957.64.](vii) The percentage by which school A outspends school B is(vii)The percentage by which school A outspends school Bis(6,000 −5,500)100∙5,500≈9.09%.

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4When we use the approximation based on the differenceinthe natural logs we get asomewhatsmallernumber:100 ∙ [log(6,000)−log(5,500)] ≈8.71%.C1.5(i) The smallest and largest values ofchildrenare 0 and 13, respectively. The averageisabout2.27.(ii)Out of 4,358 women, only 611 have electricity in the home, or about 14.02percent.(iii)The average ofchildrenfor women without electricity is about 2.33, and for thosewithelectricity it is about 1.90. So, on average, women with electricity have .43 fewer childrenthanthose who donot.(iv)We cannot infer causality here. There are many confounding factors that may berelatedto the number ofchildren and the presence of electricity in the home; householdincomeand level of education are two possibilities. For example, it could be that women withmoreeducation have fewer children and are more likely to have electricity in the home (thelatterdue to an incomeeffect).C1.7(i) The percentage of men in the sample report abusing alcohol is 9.9. Theemploymentrate is 24.3.(ii) Theemploymentrateof menwho abuse alcohol is 22.6.(iii) Theemploymentrate who do not abuse alcohol is24.5.(iv)The employment ratesof men who abuse alcohol and who do not are 22.6 and 24.5,respectively. The differenceintheseemployment rates is very less, which means that alcoholabuse does not cause unemployment.

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5otherwise ona password-protected website or school-approved learning management system for classroom use.

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8CHAPTER 2The Simple Regression ModelSOLUTIONS TO PROBLEMS2.1(i) Income, age, and family background (such as number of siblings) are just a fewpossibilities. It seems that each of these could be correlated withyears of education. (Incomeand education are probably positively correlated; age and education may be negatively correlatedbecause women in more recent cohorts have, on average, more education; and number of siblingsand education are probably negatively correlated.)(ii) Not if the factors we listed in part (i) are correlated witheduc. Because we would like tohold these factors fixed, they are part of the error term. But ifuis correlated witheduc,thenE(u|educ)0, and so SLR.4 fails.

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92.3(i) Letyi=GPAi,xi=ACTi, andn= 8. Thenx= 25.875,y= 3.2125,1ni(xi–x)(yi–y)=5.8125, and1ni(xi–x)2= 56.875. From equation (2.19), we obtain the slope as1ˆ=5.8125/56.875.1022, rounded to four places after the decimal. From (2.17),0ˆ=y–1ˆx3.2125–(.1022)25.875.5681. So we can writeGPA= .5681 + .1022ACTn= 8.The intercept does not have a useful interpretation becauseACTis not close to zero for thepopulation of interest. IfACTis 5 points higher,GPAincreases by .1022(5)= .511.(ii) The fitted values and residuals—rounded to four decimal places—are given along withthe observation numberiandGPAin the following table:iGPAGPAˆu12.82.7143.085723.43.0209.379133.03.2253–.225343.53.3275.172553.63.5319.068163.03.1231–.123172.73.1231–.423183.73.6341.0659You can verify that the residuals, as reported in the table, sum to.0002, which is pretty close tozero given the inherent rounding error.(iii) WhenACT= 20,GPA= .5681 + .1022(20)2.61.(iv) The sum of squared residuals,21ˆniiu, is about .4347 (rounded to four decimal places),and the total sum of squares,1ni(yi–y)2, is about 1.0288. So theR-squared from the regressionis

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10R2= 1–SSR/SST1–(.4347/1.0288).577.Therefore, about 57.7% of the variation inGPAis explained byACTin this small sample ofstudents.2.5(i) The intercept implies that wheninc= 0,consis predicted to be negative $124.84. This, ofcourse, cannot be true, and reflectsthefact that this consumption function might be a poorpredictor of consumption at very low-income levels. On the other hand, on an annual basis,$124.84 is not so far from zero.(ii) Just plug 30,000 into the equation:cons=–124.84 + .853(30,000)= 25,465.16 dollars.(iii) The MPC and the APC are shown in the following graph. Even though the intercept isnegative, the smallest APC in the sample is positive. The graph starts at an annual income levelof $1,000 (in 1970 dollars).2.7(i) When we condition onincin computing an expectation,incbecomes a constant. SoE(u|inc)= E(ince|inc) =incE(e|inc)=inc0 because E(e|inc)= E(e)= 0.inc1000100002000030000.7.728.853APCMPC.9APCMPC

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11(ii) Again, when we condition onincin computing a variance,incbecomes a constant. SoVar(u|inc)= Var(ince|inc)= (inc)2Var(e|inc)=2eincbecause Var(e|inc)=2e.(iii) Families with low incomes do not have much discretion about spending; typically, alow-income family must spend on food, clothing, housing, and other necessities. Higher-incomepeople have more discretion, and some might choose more consumption while others moresaving. This discretion suggests wider variability in saving among higher income families.2.9(i) We follow the hint, noting that1c y=1c y(the sample average of1ic yisc1times thesample average ofyi) and2c x=2c x. When we regressc1yionc2xi(including an intercept),weuse equation (2.19) to obtain the slope:From (2.17), we obtain the intercept as0= (c1y)–1(c2x)= (c1y)–[(c1/c2)1ˆ](c2x)=c1(y–1ˆx)=c10ˆ) because the intercept from regressingyionxiis (y–1ˆx).(ii) We use the same approach from part (i) along with the fact that1()cy=c1+yand2()cx=c2+x. Therefore,11()()icycy= (c1+yi)–(c1+y)=yi–yand (c2+xi)–2()cx=xi–x. Soc1andc2entirely drop out of the slope formula for the regression of (c1+yi) on (c2+xi), and1=1ˆ. The intercept is0=1()cy–12()cx= (c1+y)–1ˆ(c2+x)= (1ˆyx)+c1–c21ˆ=0ˆ+c1–c21ˆ, which is what we wanted to show.(iii) We can simply apply part (ii) because11log()log()log()iic ycy. In other words,replacec1with log(c1),replaceyiwith log(yi), and setc2= 0.(iv) Again, we can apply part (ii) withc1= 0 and replacingc2with log(c2) andxiwith log(xi).If01ˆˆandare the original intercept and slope, then11ˆand0021ˆˆlog()c.

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122.11(i) We would want to randomly assign the number of hours in the preparation course so thathoursis independent of other factors that affect performance on the SAT. Then, we wouldcollect information on SAT score for each student in the experiment, yielding a data set{(,) :1,..., }iisathoursin, wherenis the number of students we can afford to have in the study.From equation (2.7), we should try to get as much variation inihoursas is feasible.(ii) Here are three factors: innate ability, family income, and general health on the day of theexam. If we think students with higher native intelligence think they do not need to prepare forthe SAT, then ability andhourswill be negatively correlated. Family income would probably bepositively correlated withhours, because higher income families can more easily affordpreparation courses. Ruling out chronic health problems, health on the day of the exam shouldbe roughly uncorrelated with hours spent in a preparation course.(iii) If preparation courses are effective,1should be positive;other factors equal, anincrease inhoursshould increasesat.(iv) The intercept,0, has a useful interpretation in this example: because E(u) = 0,0is theaverage SAT score for students in the population withhours= 0.2.13(i) Since𝑥𝑖is a binary variable, it is equal to either 0 or 1. Thus, the number of observationswith𝑥𝑖=0will be𝑛0=∑(1−𝑥𝑖)since the value in the summation is equal to 1 whenever𝑥𝑖=0and equal to 0 whenever𝑥𝑖=1. Similarly,𝑛1=∑𝑥𝑖will give us the number ofobservations with𝑥𝑖=1since we are only going to be counting instances in which𝑥𝑖is notequal to 0.We know that𝑥̅=1𝑛∑𝑥𝑖. We also have shown that∑𝑥𝑖=𝑛1. Thus,𝑥̅=𝑛1𝑛.We can write1−𝑥̅=1−𝑛1𝑛=𝑛−𝑛1𝑛=𝑛0𝑛since𝑛0+𝑛1=𝑛.𝑥̅=𝑛1𝑛tells us the fraction of observations for which𝑥𝑖=1.(ii) We know that𝑦ത0=𝑛0−1∑𝑦𝑖𝑛0𝑖=1, where we are just looking at values of𝑦𝑖for which𝑥𝑖=0.One way we can formalize this is by recognizing that(1−𝑥𝑖)𝑦𝑖=𝑦𝑖if𝑥𝑖=0and 0 otherwise.Thus,𝑦ത0=𝑛0−1∑𝑦𝑖𝑛0𝑖=1=𝑛0−1∑(1−𝑥𝑖)𝑦𝑖𝑛𝑖=1.Similarly, we can show that𝑦ത1=𝑛1−1∑𝑦𝑖𝑛1𝑖=1=𝑛1−1∑𝑥𝑖𝑦𝑖𝑛𝑖=1, since𝑥𝑖𝑦𝑖=𝑦𝑖if𝑥𝑖=1and 0otherwise.(iii) We can express𝑦ത=1𝑛∑𝑦𝑖𝑛𝑖=1=1𝑛൫∑𝑦0,𝑖𝑛0𝑖=1+∑𝑦1,𝑖𝑛1𝑖=1൯=𝑛0𝑛𝑦ത0+𝑛1𝑛𝑦ത1We have shown that1−𝑥̅=𝑛0𝑛and𝑥̅=𝑛1𝑛. Thus,𝑦ത=(1−𝑥̅)𝑦ത0+𝑥̅𝑦ത1

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13(iv) If𝑥𝑖is binary, then𝑥𝑖2=𝑥𝑖. Thus,𝑛−1∑𝑥𝑖2−𝑥̅2=𝑛−1∑𝑥𝑖−𝑥̅2=𝑥̅−𝑥̅2=𝑥̅(1−𝑥̅)(v) First note that𝑛−1∑𝑥𝑖𝑦𝑖𝑛𝑖=1=𝑛−1∑𝑦𝑖𝑛1𝑖=1=𝑛1𝑛𝑦ത1since we will only be counting instancesin which𝑥𝑖=1. Using the result from part i, we show that𝑛−1∑𝑥𝑖𝑦𝑖𝑛𝑖=1=𝑥̅𝑦ത1Therefore𝑛−1∑𝑥𝑖𝑦𝑖𝑛𝑖=1−𝑥̅𝑦ത=𝑥̅𝑦ത1−𝑥̅𝑦ത=𝑥̅𝑦ത1−𝑥̅[(1−𝑥̅)𝑦ത0+𝑥̅𝑦ത1], using the result frompart iii.With some manipulation, we can thus show that𝑛−1∑𝑥𝑖𝑦𝑖𝑛𝑖=1−𝑥̅𝑦ത=𝑥̅(1−𝑥̅)(𝑦ത1−𝑦ത0)(vi) Equation 2.74 states that𝛽መ1=𝑦ത1−𝑦ത0. This is derived in the same way as the usual OLSestimator: minimizing the sum of the squared residuals to get𝛽መ1=∑(𝑥𝑖−𝑥̅)(𝑦𝑖−𝑦ത)∑(𝑥𝑖−𝑥̅)2Expanding the numerator and denominator:𝛽መ1=∑𝑥𝑖𝑦𝑖−𝑛𝑥̅𝑦ത∑𝑥𝑖2−𝑛𝑥̅2Using the results from parts v and vi:𝛽መ1=𝑥̅(1−𝑥̅)(𝑦ത1−𝑦ത0)𝑥̅(1−𝑥̅)=𝑦ത1−𝑦ത0(vii) We know that𝛽መ0=𝑦ത−𝛽መ1𝑥̅. Using the results from parts iii and vi, we have:𝛽መ0=(1−𝑥̅)𝑦ത0+𝑥̅𝑦ത1−(𝑦ത1−𝑦ത0)𝑥̅=𝑦ത0−𝑥̅𝑦ത0+𝑥̅𝑦ത1−𝑥̅𝑦ത1+𝑥̅𝑦ത0=𝑦ത02.15(i) The population average treatment effect𝜏𝑎௧𝑒=𝐸[𝑦(1)−𝑦(0)]. If we are able toobserve𝑦𝑖(0)and𝑦𝑖(1)for each observation𝑖(e.g. we can observe the same observation in bothstates of the world), then𝐸ቂ1𝑛∑𝑦𝑖(1)ቃ=1𝑛∑𝐸[𝑦𝑖(1)]=1𝑛∗𝑛𝐸[𝑦(1)]=𝐸[𝑦(1)].Similarly,1𝑛∑𝑦𝑖(0)is an unbiased estimator for𝐸[𝑦(0)].Therefore𝐸ቄ1𝑛∑[𝑦𝑖(1)−𝑦𝑖(0)]ቅ=𝐸[𝑦(1)−𝑦(0)](ii) We can write𝑦𝑖=(1−𝑥𝑖)𝑦𝑖(0)+𝑥𝑖𝑦𝑖(1). Then,𝑦ത0=1𝑛∑[(1−𝑥𝑖)𝑦𝑖(0)+𝑥𝑖𝑦𝑖(1)]𝑛𝑖=1=1𝑛0∑𝑦𝑖(0)𝑛0𝑖=1since𝑥𝑖=0for all observations used to calculate𝑦ത0. By a similar logic,𝑦ത1=1𝑛1∑𝑦𝑖(1)𝑛1𝑖=1since𝑥1=1for all observations used to calculate𝑦ത1. Thus, both𝑦ത0and𝑦ത1arecomputed using a subset of the sample. By contrast𝑦ത(0)and𝑦ത(1)would be computed using theentire sample, using the outcomes of each observation in both states of the world.2.17(i)𝑉𝑎𝑟(𝑢𝑖|𝑥𝑖)=𝑉𝑎𝑟[(1−𝑥𝑖)𝑢𝑖(0)+𝑥𝑖𝑢𝑖(1)]=(1−𝑥𝑖)2𝜎02+𝑥𝑖2𝜎12. Recall that we areconditioning on𝑥𝑖, so we can treat it as a deterministic variable. Finally, note that since𝑥𝑖isbinary, we can write𝑉𝑎𝑟(𝑢𝑖|𝑥𝑖)=(1−𝑥𝑖)𝜎02+𝑥𝑖𝜎12. The conditional variance is a weightedaverage of the variances from the two different states of the world.(ii) There are three ways in which𝑉𝑎𝑟(𝑢𝑖|𝑥𝑖)could be constant:1.𝑥𝑖=0for all𝑖2.𝑥𝑖=1for all𝑖3.𝜎02=𝜎12

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14The first two scenarios imply that we have no observations in either the treatment or controlgroups and thus the experiment is not of any use. The last case suggests that outcomes for non-treated state have exactly the same variability as outcomes for thetreated state. While this iscertainly a possibility, it would imply that the only effect of the treatment is to change the meanoutcome, not any other aspects of its distribution.SOLUTIONS TO COMPUTER EXERCISESC2.1(i) The averageprateis about 87.36,and the averagemrateis about .732.(ii) The estimated equation isprate= 83.08 + 5.86mraten= 1,534,R2= .075.(iii) The intercept implies that, even ifmrate= 0, the predicted participation rate is 83.08percent. The coefficient onmrateimplies that a one-dollar increase in the match rate–a fairlylarge increase–is estimated to increaseprateby 5.86 percentage points. This assumes, ofcourse, that this changeprateis possible (if, say,prateis already at 98, this interpretation makesno sense).(iv) If we plugmrate= 3.5 into the equation,we getˆprate= 83.08+ 5.86(3.5)= 103.59.This is impossible, as we can have at most a 100 percent participation rate. This illustrates that,especially when dependent variables are bounded, a simple regression model can give strangepredictions for extreme values of the independent variable. (In the sample of 1,534 firms, only34 havemrate3.5.)(v)mrateexplains about 7.5% of the variation inprate. This is not much and suggests thatmany other factors influence 401(k) plan participation rates.C2.3(i) The estimated equation issleep= 3,586.4–.151totwrkn= 706,R2= .103.The intercept implies that the estimated amount of sleep per week for someone who does notwork is 3,586.4 minutes, or about 59.77 hours. This comes to about 8.5hours per night.

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15(ii) If someone works two more hours per week,thentotwrk= 120 (becausetotwrkismeasured in minutes), and sosleep=–.151(120)=–18.12 minutes. This is only a few minutesa night. If someone were to workone more hour on each of five working days,sleep=–.151(300)=–45.3 minutes, or about five minutes a night.C2.5(i) The constant elasticity model is a log-log model:log(rd) =0+1log(sales) +u,where1is the elasticity ofrdwith respect tosales.(ii) The estimated equation islog()rd=–4.105 + 1.076 log(sales)n= 32,R2= .910.The estimated elasticity ofrdwith respect tosalesis 1.076, which is just above one. A onepercent increase insalesis estimated to increaserdby about 1.08%.C2.7(i) The average gift is about 7.44 Dutch guilders. Out of 4,268 respondents, 2,561 did notgive a gift, or about 60 percent.(ii) The average mailings per year is about 2.05. The minimum value is .25 (whichpresumably means that someone has been on the mailing list for at least four years),and themaximum value is 3.5.(iii) The estimated equation is22.012.654,268,.0138.giftmailsyearnR(iv) The slope coefficient from part (iii) means that each mailing per year is associated with–perhaps even “causes”–an estimated 2.65 additional guilders, on average. Therefore, if eachmailing costs one guilder, the expected profit from each mailingis estimated to be 1.65 guilders.This is only the average, however. Some mailings generate no contributions, or a contributionless than the mailing cost; other mailings generated much more than the mailing cost.(v) Because the smallestmailsyearin the sample is .25, the smallest predicted value ofgiftsis 2.01 + 2.65(.25)2.67. Even if we look at the overall population, where some people have

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16received no mailings, the smallest predicted value is about two. So, with this estimated equation,we never predict zero charitable gifts.C2.9(i) In 1996, 1,051 counties had zero murders. Out of 2,197 counties, 31 counties had atleast one execution and the largest number of executions is 3.(ii)The estimated equationis𝑚𝑢𝑟𝑑𝑒𝑟𝑠=5.46+58.56𝑒𝑥𝑒𝑐𝑠𝑛=2197,𝑅2=0.0439.(iii)The slope coefficient onexecsimplies that if the number of executions increases byone, theestimated number of murders increases largely by about 59. No, theestimated equation does not suggest a deterrent effect of capital punishment.(iv)The smallest number of murders can be predicted by the equation is 5.46,that isabout 5 murders. The residual for a county with zero executions and zero murders is-5.46.(v)This simple linear regression equation predicts that if the number of executionsincreases by one, the estimated number of murders increases largely by about 59,which means capital punishment does not have a deterrent effect on murders—capital punishment is not discouraging people from doing murders. The sign andmagnitude of the estimate +58.56 makes us suspect that the error termuand theindependent variableexecsare correlated. Therefore, the regression model is notwell suited for prediction.C2.11(i) There are 141 students in the sample. The average college GPA is a 3.057, while themaximum GPA is a 4.0 (1 student with this GPA).(ii) 56 students (39.7% of the sample) owned their own PC.(iii)𝑐𝑜𝑙𝐺෠𝑃𝐴=2.989(0.040)+0.170(0.063)∗𝑃𝐶. We estimates that students who do not own a PC have anaverage college GPA of 2.989. Those who own a PC have college GPA’s that are on average0.17 points higher than those who do not. Both coefficient estimates are statistically significant atthe 1% level. In terms of magnitude, a 0.17 point increase in GPA represents about a 5.7%increase from the mean college GPA for those without a computer.(iv) The𝑅2=0.05. We are only explaining about 5% of the variation in college GPAs withcomputer ownership. Thus, there are a lot of other factors that influence college GPA.

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17(v) If we truly believed that PC ownership was randomly assigned, then we could infer a causaleffect on college GPA. However, it is highly unlikely that we have random assignment in thisexperiment. There are a wide range of factors that could both influence college GPA and PCownership that we are overlooking. For example, consider parent’s income. Students fromwealthier families are more likely to own PCs. These same students are also less likely to have towork during college, freeing up more time fortheir studies and thus earning higher GPAs. Assuch, we cannot disentangle the effects of PC ownership from parental income. There are anynumber of other omitted variables that could cause us to violate the random assignmentassumption.

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