ch2

Solutions, Chapter 2/HL

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ANSWERS TO CHAPTER 2

The Simple Regression Model

Econometrics Economics of Innovation and Growth

A = Problems B = Examples (from chapter 2)

C = Cumputer Exercises


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A: Problems 2.1 Let kids denote the number of children born to a woman, and let educ denote years of education for the woman. A simple model relating fertility to years of education is

ueduckids ++= 10 ββ where u is the unobserved error. (i) What kind of factors are contained in u? Are these likely to be correlated with

level of education? (ii) Will a simple regression analysis uncover ceteris paribus effects of education on

fertility? Explain. (i) 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 recent 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 with educ then E(u|educ) ≠ 0, and so SLR.3 fails. --------------------------------------------------------------------------------------------------------------- 2.2 In the simple linear regression model y=β0+β1x + u, suppose that E(u) ≠ 0. Letting α0=e(u), show that the model can always be rewritten with the same slope, but new intercept and error, where the new error has a zero expected value. Answers In the equation y = β0 + β1x + u, add and subtract α0 from the right hand side to get y = (α0 + β0) + β1x + (u − α0). Call the new error e = u − α0, so that E(e) = 0. The new intercept is α0 + β0, but the slope is still β1.

≠


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2.3 The following table contains the ATC scores and the GPA (grade point average) for 8 college students. Grade point average is based on a four-point scale and has been rounded to the one digit after the decimal.

Student GPA ACT 1 2.8 21 2 3.4 24 3 3.0 26 4 3.5 27 5 3.6 39 6 3.0 25 7 2.7 25 8 3.7 30

(i) Estimate the relationship between GPA and ACT using ols; that is, obtain the intercept and slope in the equation

ACTAPG 10ˆˆˆ ββ +=

Comment on the direction of the relationship. Does the intercept have a useful interpretation here? Explain. How much higher is GPA predicted to be if the ACT score is increased by 5 points? (ii) Compute the fitted valued and the residuals for each observation, and verify that the residuals (approximately) sum to zero. (iii) What is the predicted value of GPA when ACT =20? (iv) How much of the variation in GPA for the 8 students is explained by ACT. Explain.

2.3 (i) Let yi = GPAi, xi = ACTi, and n = 8. Then x = 25.875, y = 3.2125, 1

n

i=∑ (xi – x )(yi –

y ) = 5.8125, and 1

n

i=∑ (xi – x )2 = 56.875. From equation (2.9), we obtain the slope as 1β =

5.8125/56.875 ≈ .1022, rounded to four places after the decimal. From (2.17), 0β = y –

1β x ≈ 3.2125 – (.1022)25.875 ≈ .5681. So we can write

ˆGPA = .5681 + .1022 ACT

n = 8.

The intercept does not have a useful interpretation because ACT is not close to zero for the population of interest. If ACT is 5 points higher, ˆGPA increases by .1022(5) = .511.


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(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:

i GPA ˆGPA u

1 2.8 2.7143 .0857 2 3.4 3.0209 .3791 3 3.0 3.2253 –.2253 4 3.5 3.3275 .1725 5 3.6 3.5319 .0681 6 3.0 3.1231 –.1231 7 2.7 3.1231 –.4231 8 3.7 3.6341 .0659

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. (iii) When ACT = 20, ˆGPA= .5681 + .1022(20) ≈ 2.61.

(iv) The sum of squared residuals, 2

1

ˆn

ii

u=∑ , is about .4347 (rounded to four decimal places),

and the total sum of squares, 1

n

i=∑ (yi – y )2, is about 1.0288. So the R-squared from the

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 this small sample of students. --------------------------------------------------------------------------------------------------------------- 2.4 The data set BWGHT.DTA contains data on births to women in the United States. Two variables of interest are the dependent variable infant birth weight on ounces (bwght), and an explanatory variable, average number of cigarettes the mother smoked per day during pregnancy (cigs). The following simple regression was estimated using data on n=1,388 births

cigshtgbw 514.077.119ˆ −= (i) What is the predicted birth weight when cig s= 0? What about when cigs = 20? Comment on the difference. (ii) Does the simple regression necessarily capture a causal relationship between the chid’s birth weight and the mother’s smoking habits? Explain.


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(iii) The predict a birth weight of 125 ounces, what would cigs have to be? Comment. (iv) What fraction of the women in the sample do not smoke while pregnant? Does this help reconcile your finding from part (iii)? (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 prenatal 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 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 predicted birth weight is necessarily 119.77. Yet almost 700 of the births in the sample had a birth weight higher than 119.77. (iii) 1,176 out of 1,388 women did not smoke while pregnant, or about 84.7%. --------------------------------------------------------------------------------------------------------------- 2.5 In the linear consumption function

incnsoc 10ˆˆˆ ββ +=

the (estimated) marginal propensity to consume (MPC) out of the income is simply the slope

,ˆ1β while the average propensity to consume (APC) is 10

ˆ/ˆ/ˆ ββ += incincnsoc Using observations for 100 families, the annual income and consumption (both measured in dollars), the following equation is obtained: Cons(est)=-124.84 + 0.853 inc N = 100, R2=0.692 (i) Interpret the intercept in this equation, and comment onb its sign and magnitude. (ii) What is the predicted consumption when family income is $30,000? (iii) Whit inc on the x-axis, draw the graph of the estimated MPC and APC (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-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.


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(iii) The MPC and the APC 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 level of $1,000 (in 1970 dollars).

inc1000 10000 20000 30000

.7

.728

.853

APCMPC .9

APC

MPC

--------------------------------------------------------------------------------------------------------------- 2.6 Using data from 1988 for houses sold in Andover, Massachusetts, from Kiel and McClain (1995), the following equation relates housing price (price) to the distance from the recently built garbage incinerator (dist): Log (price) = 0.9.40 + 0.213 log(dist) N = 135, R2=0.162 (i) Interpret the coefficient on log (dist). Is the sign of this estimate what you expected it to be? (ii) Do you think simple regression provides an unbiased estimator of the ceteris paribus elasticity of price with respect to dist? (Think about the city’s decision on where to put the incinerator.)


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(iii) What other factors about a house affect its price? Might these be correlated with distance from the incinerator? (i) Yes. If living closer to an incinerator depresses housing prices, then being farther away 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.3, 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), these could certainly be correlated with dist [and log(dist)].


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B: Examples

Example 2.3: CEO Salary and Return on Equity Data: CEOSAL1 sum salary roe Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------- salary | 209 1281.12 1372.345 223 14822 roe | 209 17.18421 8.518509 .5 56.3 reg salary roe Source | SS df MS Number of obs = 209 ---------+------------------------------ F( 1, 207) = 2.77 Model | 5166419.04 1 5166419.04 Prob > F = 0.0978 Residual | 386566563 207 1867471.32 R-squared = 0.0132 ---------+------------------------------ Adj R-squared = 0.0084 Total | 391732982 208 1883331.64 Root MSE = 1366.6 ------------------------------------------------------------------------------ salary | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- roe | 18.50119 11.12325 1.663 0.098 -3.428195 40.43057 _cons | 963.1913 213.2403 4.517 0.000 542.7902 1383.592 ------------------------------------------------------------------------------ Salary for ROE = 0 display _b[roe]*0+_b[_cons] 963.19134 Salary for ROE = 30 display _b[roe]*30+_b[_cons] 1518.2269

Example 2.4: Wage and Education Data WAGE1 summ wage Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------- wage | 526 5.896103 3.693086 .53 24.98 reg wage educ Source | SS df MS Number of obs = 526 ---------+------------------------------ F( 1, 524) = 103.36 Model | 1179.73204 1 1179.73204 Prob > F = 0.0000 Residual | 5980.68225 524 11.4135158 R-squared = 0.1648 ---------+------------------------------ Adj R-squared = 0.1632 Total | 7160.41429 525 13.6388844 Root MSE = 3.3784 ------------------------------------------------------------------------------ wage | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- educ | .5413593 .053248 10.167 0.000 .4367534 .6459651 _cons | -.9048516 .6849678 -1.321 0.187 -2.250472 .4407687 ------------------------------------------------------------------------------ Wage for educ = 0 display _b[educ]*0+_b[_cons] -.90485161 Wage for educ = 8 display _b[educ]*8+_b[_cons] 3.4260224 Return to 4 years education display _b[educ]*4 2.165437


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Example 2.5: Voting Outcomes and Campaign Expenditures Data VOTE1 reg voteA shareA Source | SS df MS Number of obs = 173 ---------+------------------------------ F( 1, 171) = 1017.70 Model | 41486.4749 1 41486.4749 Prob > F = 0.0000 Residual | 6970.77363 171 40.7647581 R-squared = 0.8561 ---------+------------------------------ Adj R-squared = 0.8553 Total | 48457.2486 172 281.728189 Root MSE = 6.3847 ------------------------------------------------------------------------------ voteA | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- shareA | .4638239 .0145393 31.901 0.000 .4351243 .4925234 _cons | 26.81254 .8871887 30.222 0.000 25.06129 28.56379 ------------------------------------------------------------------------------

Example 2.6: CEO Salary and Return on Equity

Data: CEOSAL1 summ salary roe Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------- salary | 209 1281.12 1372.345 223 14822 roe | 209 17.18421 8.518509 .5 56.3 reg salary roe Source | SS df MS Number of obs = 209 ---------+------------------------------ F( 1, 207) = 2.77 Model | 5166419.04 1 5166419.04 Prob > F = 0.0978 Residual | 386566563 207 1867471.32 R-squared = 0.0132 ---------+------------------------------ Adj R-squared = 0.0084 Total | 391732982 208 1883331.64 Root MSE = 1366.6 ------------------------------------------------------------------------------ salary | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- roe | 18.50119 11.12325 1.663 0.098 -3.428195 40.43057 _cons | 963.1913 213.2403 4.517 0.000 542.7902 1383.592 ------------------------------------------------------------------------------

Fitted Values and Residuals for the First 15 CEOs predict salhat, xb gen uhat=salary-salhat list roe salary salhat uhat in 1/15 roe salary salhat uhat 1. 14.1 1095 1224.058 -129.0581 2. 10.9 1001 1164.854 -163.8542 3. 23.5 1122 1397.969 -275.9692 4. 5.9 578 1072.348 -494.3484 5. 13.8 1368 1218.508 149.4923 6. 20 1145 1333.215 -188.2151 7. 16.4 1078 1266.611 -188.6108 8. 16.3 1094 1264.761 -170.7606 9. 10.5 1237 1157.454 79.54626 10. 26.3 833 1449.773 -616.7726 11. 25.9 567 1442.372 -875.3721 12. 26.8 933 1459.023 -526.0231 13. 14.8 1339 1237.009 101.9911 14. 22.3 937 1375.768 -438.7678 15. 56.3 2011 2004.808 6.191895


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Example 2.7: Wage and Education Data: WAGE1 summ wage educ Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------- wage | 526 5.896103 3.693086 .53 24.98 educ | 526 12.56274 2.769022 0 18 reg wage educ Source | SS df MS Number of obs = 526 ---------+------------------------------ F( 1, 524) = 103.36 Model | 1179.73204 1 1179.73204 Prob > F = 0.0000 Residual | 5980.68225 524 11.4135158 R-squared = 0.1648 ---------+------------------------------ Adj R-squared = 0.1632 Total | 7160.41429 525 13.6388844 Root MSE = 3.3784 ------------------------------------------------------------------------------ wage | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- educ | .5413593 .053248 10.167 0.000 .4367534 .6459651 _cons | -.9048516 .6849678 -1.321 0.187 -2.250472 .4407687 ------------------------------------------------------------------------------

Wage for educ = 12.56

display _b[educ]*12.56+_b[_cons] 5.8824

Example 2.8: CEO Salary and Return on Equity Data: CEOSAL1 reg salary roe Source | SS df MS Number of obs = 209 ---------+------------------------------ F( 1, 207) = 2.77 Model | 5166419.04 1 5166419.04 Prob > F = 0.0978 Residual | 386566563 207 1867471.32 R-squared = 0.0132 ---------+------------------------------ Adj R-squared = 0.0084 Total | 391732982 208 1883331.64 Root MSE = 1366.6 ------------------------------------------------------------------------------ salary | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- roe | 18.50119 11.12325 1.663 0.098 -3.428195 40.43057 _cons | 963.1913 213.2403 4.517 0.000 542.7902 1383.592 ------------------------------------------------------------------------------

Example 2.9: Voting Outcomes and Campaign Expenditures Data: VOTE1 reg voteA shareA Source | SS df MS Number of obs = 173 ---------+------------------------------ F( 1, 171) = 1017.70 Model | 41486.4749 1 41486.4749 Prob > F = 0.0000 Residual | 6970.77363 171 40.7647581 R-squared = 0.8561 ---------+------------------------------ Adj R-squared = 0.8553 Total | 48457.2486 172 281.728189 Root MSE = 6.3847 ------------------------------------------------------------------------------ voteA | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- shareA | .4638239 .0145393 31.901 0.000 .4351243 .4925234 _cons | 26.81254 .8871887 30.222 0.000 25.06129 28.56379 ------------------------------------------------------------------------------


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Example 2.10: A Log Wage Equation Data:WAGE1 reg lwage educ Source | SS df MS Number of obs = 526 ---------+------------------------------ F( 1, 524) = 119.58 Model | 27.5606296 1 27.5606296 Prob > F = 0.0000 Residual | 120.769132 524 .230475443 R-squared = 0.1858 ---------+------------------------------ Adj R-squared = 0.1843 Total | 148.329762 525 .28253288 Root MSE = .48008 ------------------------------------------------------------------------------ lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- educ | .0827444 .0075667 10.935 0.000 .0678796 .0976092 _cons | .5837726 .0973358 5.998 0.000 .3925562 .774989 ------------------------------------------------------------------------------

Example 2.11: CEO Salary and Firm Sales Data CEOSAL1 reg lsalary lsales Source | SS df MS Number of obs = 209 ---------+------------------------------ F( 1, 207) = 55.30 Model | 14.0661711 1 14.0661711 Prob > F = 0.0000 Residual | 52.6559988 207 .254376806 R-squared = 0.2108 ---------+------------------------------ Adj R-squared = 0.2070 Total | 66.7221699 208 .320779663 Root MSE = .50436 ------------------------------------------------------------------------------ lsalary | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- lsales | .2566717 .0345167 7.436 0.000 .1886225 .324721 _cons | 4.821996 .2883397 16.723 0.000 4.253537 5.390455 ------------------------------------------------------------------------------

Example 2.12: Student Math Performance and the School Lunch Program Data: MEAP93 reg math10 lnchprg Source | SS df MS Number of obs = 408 ---------+------------------------------ F( 1, 406) = 83.77 Model | 7665.26597 1 7665.26597 Prob > F = 0.0000 Residual | 37151.9145 406 91.5071786 R-squared = 0.1710 ---------+------------------------------ Adj R-squared = 0.1690 Total | 44817.1805 407 110.115923 Root MSE = 9.5659 ------------------------------------------------------------------------------ math10 | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- lnchprg | -.3188643 .0348393 -9.152 0.000 -.3873523 -.2503763 _cons | 32.14271 .9975824 32.221 0.000 30.18164 34.10378 ------------------------------------------------------------------------------


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C: Computer exercises Question 2.10* Data: 401K.DTA

The data in 401K.DTA are a subset of data analysed by Papke (1995) to study the relationship between participation in a 401and the generosity of the plan. The variable prate is the percentage of eligible workers with an active account: This is the variable we would like to explain. The measure of generosity is the plan match rate, mrate. This variable gives the average amount the firm contributes to each worker’s plan for each $1 dollar contribution by thw workers. For example, if mrate=0.50, then $1 contribution by the worker is matched by a 50 cents contribution by the firm (i) Find the average participation rate and the average match rate in the sample of plan.

Answer *i sum prate mrate Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- prate | 1534 87.36291 16.71654 3 100 mrate | 1534 .7315124 .7795393 .01 4.91

The average prate is about 87.36 and the average mrate is about .732. (ii) Estimate the simple regression equation mrateteapr 10

ˆˆˆ ββ += and report the results along the sample size and R-squared. *ii reg prate mrate Source | SS df MS Number of obs = 1534 -------------+------------------------------ F( 1, 1532) = 123.68 Model | 32001.7271 1 32001.7271 Prob > F = 0.0000 Residual | 396383.812 1532 258.73617 R-squared = 0.0747 -------------+------------------------------ Adj R-squared = 0.0741 Total | 428385.539 1533 279.442622 Root MSE = 16.085 ------------------------------------------------------------------------------ prate | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mrate | 5.861079 .5270107 11.12 0.000 4.82734 6.894818 _cons | 83.07546 .5632844 147.48 0.000 81.97057 84.18035 ------------------------------------------------------------------------------

The estimated equation is teapr ˆ = 83.05 + 5.86 mrate

n = 1,534, R2 = .075

(iii) Interpret the intercept in your equation. Interpret the coefficient on mrate. The intercept implies that, even if mrate = 0, the predicted participation rate is 83.05 percent. The coefficient on mrate implies that a one-dollar increase in the match rate – a fairly large increase – is estimated to increase prate by 5.86 percentage points. This assumes, of course, that this change prate is possible (if, say, prate is already at 98, this interpretation makes no sense).


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(iv) Find the predicted prate when mrate =3.5. Is this a reasonable prediction? Explain what is happening here.

If we plug mrate = 3.5 into the equation we get ˆprate = 83.05 + 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 strange predictions for extreme values of the independent variable. (In the sample of 1,534 firms, only 34 have mrate ≥ 3.5.)

(v) How much of the variation in prate is explained by mrate. Is this a lot in your opinion?

(v) mrate explains about 7.5% of the variation in prate. This is not much, and suggests that many other factors influence 401(k) plan participation rates.

Question 2.11

The data set in CEOSAL2.DTA contains information of chief executive officers for U.S corporations. The variable salary is annual compensation, in thousand of dollars, and ceoten is prior number of years as company CEO. (i) Find the average salary and the average tenure in the sample sum salary ceoten Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- salary | 177 865.8644 587.5893 100 5299 ceoten | 177 7.954802 7.150826 0 37

Average salary is about 865.864, which means $865,864 because salary is in thousands of dollars. Average ceoten is about 7.95.

(ii) How many CEOs are in their first year as CEO (that is, ceoten=0)?, what is the longest tenure as CEO

There are five CEOs with ceoten = 0. The longest tenure is 37 years. *ii tab ceoten years as | ceo with | company | Freq. Percent Cum. ------------+----------------------------------- 0 | 5 2.82 2.82 1 | 19 10.73 13.56 2 | 10 5.65 19.21 3 | 21 11.86 31.07 4 | 21 11.86 42.94 5 | 10 5.65 48.59 6 | 11 6.21 54.80 7 | 6 3.39 58.19 8 | 11 6.21 64.41 9 | 8 4.52 68.93 10 | 8 4.52 73.45 11 | 4 2.26 75.71


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12 | 7 3.95 79.66 13 | 7 3.95 83.62 14 | 5 2.82 86.44 15 | 2 1.13 87.57 16 | 2 1.13 88.70 17 | 2 1.13 89.83 18 | 1 0.56 90.40 19 | 2 1.13 91.53 20 | 4 2.26 93.79 21 | 1 0.56 94.35 22 | 1 0.56 94.92 24 | 3 1.69 96.61 26 | 2 1.13 97.74 28 | 1 0.56 98.31 34 | 1 0.56 98.87 37 | 2 1.13 100.00 ------------+----------------------------------- Total | 177 100.00

(iii) Estimate the simple regression model uceotensalary ++= 10)log( ββ and report your results in the usual form. What is the (approcimate) predicted percentage increase in salary given one more year as CEO

*iii reg lsalary ceoten Source | SS df MS Number of obs = 177 -------------+------------------------------ F( 1, 175) = 2.33 Model | .850907024 1 .850907024 Prob > F = 0.1284 Residual | 63.795306 175 .364544606 R-squared = 0.0132 -------------+------------------------------ Adj R-squared = 0.0075 Total | 64.6462131 176 .367308029 Root MSE = .60378 ------------------------------------------------------------------------------ lsalary | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- ceoten | .0097236 .0063645 1.53 0.128 -.0028374 .0222846 _cons | 6.505498 .0679911 95.68 0.000 6.37131 6.639686 ------------------------------------------------------------------------------

(iii) The estimated equation is log ( )salary = 6.51 + .0097 ceoten

n = 177, R2 = .013.

We obtain the approximate percentage change in salary given ∆ceoten = 1 by multiplying the coefficient on ceoten by 100, 100(.0097) = 0.97%. Therefore, one more year as CEO is predicted to increase salary by almost 1%.

2.12 USE the data SLEEP 75.DTA to study whether there is a trade off between time spent sleeping per week and the time spent in paid work. We could use either variable is the dependent variable. For concreteness, estimate the model: utotwrksleep ++= 10 ββ where sleep is minutes sleeping per week and totwrk is total minutes working during the week. (i) Report your results in equation form along with the number of observations and R2. What does the intercept in this equation mean?


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*i reg sleep totwrk Source | SS df MS Number of obs = 706 -------------+------------------------------ F( 1, 704) = 81.09 Model | 14381717.2 1 14381717.2 Prob > F = 0.0000 Residual | 124858119 704 177355.282 R-squared = 0.1033 -------------+------------------------------ Adj R-squared = 0.1020 Total | 139239836 705 197503.313 Root MSE = 421.14 ------------------------------------------------------------------------------ sleep | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- totwrk | -.1507458 .0167403 -9.00 0.000 -.1836126 -.117879 _cons | 3586.377 38.91243 92.17 0.000 3509.979 3662.775 ------------------------------------------------------------------------------

The estimated equation is Sleep = 3,586.4 – .151 totwrk

n = 706, R2 = .103.

The intercept implies that the estimated amount of sleep per week for someone who does not work is 3,586.4 minutes, or about 59.77 hours. This comes to about 8.5 hours per night. (ii) If totwrk increases by 2 hours, by how much is sleep estimated to fall? Do you find this to be a large effect? If someone works two more hours per week then ∆totwrk = 120 (because totwrk is measured in minutes), and so sleep∆ = –.151(120) = –18.12 minutes. This is only a few minutes a

night. If someone were to work one more hour on each of five working days, sleep∆ = –.151(300) = –45.3 minutes, or about five minutes a night.

2.13 Use the data in WAGE.DTA to estimate a simple regression explaining monthly salary (wage) in terms of IQ score (IQ) (i) Find the average salary and average IQ in the sample. What is the standard deviation of IQ? (IQ scores are standardized so that the average in the population is 100 with the standard deviation equal to 15.) sum wage IQ Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- wage | 935 957.9455 404.3608 115 3078 IQ | 935 101.2824 15.05264 50 145

Average salary is about $957.95 and average IQ is about 101.28. The sample standard deviation of IQ is about 15.05, which is pretty close to the population value of 15.


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(ii) Estimate the simple regression model where a one-point increase in IQ change wage by a constant dollar amount. Use this model to find the predicted increase in wage for an increase in IQ of 15 points. Does IQ explain most of the variation in wage? *ii reg wage IQ Source | SS df MS Number of obs = 935 -------------+------------------------------ F( 1, 933) = 98.55 Model | 14589782.6 1 14589782.6 Prob > F = 0.0000 Residual | 138126386 933 148045.429 R-squared = 0.0955 -------------+------------------------------ Adj R-squared = 0.0946 Total | 152716168 934 163507.675 Root MSE = 384.77 ------------------------------------------------------------------------------ wage | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- IQ | 8.303064 .8363951 9.93 0.000 6.661631 9.944498 _cons | 116.9916 85.64153 1.37 0.172 -51.08078 285.0639 ------------------------------------------------------------------------------

This calls for a level-level model: wage = 116.99 + 8.30 IQ

n = 935, R2 = .096.

An increase in IQ of 15 increases predicted monthly salary by 8.30(15) = $124.50 (in 1980 dollars). IQ score does not even explain 10% of the variation in wage.

(iii) Now, estimate a model where each one-point increase in IQ has the same percentage effect on wage. If IQ increases by 15 points, what is the approximate percentage increase in the predicted wage? reg lwage IQ Source | SS df MS Number of obs = 935 -------------+------------------------------ F( 1, 933) = 102.62 Model | 16.4150939 1 16.4150939 Prob > F = 0.0000 Residual | 149.241189 933 .159958402 R-squared = 0.0991 -------------+------------------------------ Adj R-squared = 0.0981 Total | 165.656283 934 .177362188 Root MSE = .39995 ------------------------------------------------------------------------------ lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- IQ | .0088072 .0008694 10.13 0.000 .007101 .0105134 _cons | 5.886994 .0890206 66.13 0.000 5.712291 6.061698 ------------------------------------------------------------------------------

This calls for a log-level model:

log ( )wage = 5.89 + .0088 IQ

n = 935, R2 = .099.


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If ∆IQ = 15 then log ( )wage∆ = .0088(15) = .132, which is the (approximate) proportionate change in predicted wage. The percentage increase is therefore approximately 13.2.

2.14 For the population of firm in the chemical industry, let rd denote annual expenditures on research and development, and let sales denote annual sales (both are in million of dollars). (i) Write down a model (not an estimated equation) that implies a constant elasticity between rd and sales. Which parameter is the elasticity? The constant elasticity model is a log-log model:

log(rd) = 0β + 1β log(sales) + u,

where 1β is the elasticity of rd with respect to sales. (ii) Now, estimate the model using RDCHEM.DTA. Write out the the estimated equation in the usual form. What is the elasticity of rd with respect to sales? Explain what this elasticity means. i reg lrd lsales *ii reg lrd lsales Source | SS df MS Number of obs = 32 -------------+------------------------------ F( 1, 30) = 302.72 Model | 84.8395785 1 84.8395785 Prob > F = 0.0000 Residual | 8.40768588 30 .280256196 R-squared = 0.9098 -------------+------------------------------ Adj R-squared = 0.9068 Total | 93.2472644 31 3.00797627 Root MSE = .52939 ------------------------------------------------------------------------------ lrd | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- lsales | 1.075731 .0618275 17.40 0.000 .9494619 1.201999 _cons | -4.104722 .4527678 -9.07 0.000 -5.029398 -3.180047 ------------------------------------------------------------------------------

The estimated elasticity of rd with respect to sales is 1.076, which is just above one. A one percent increase in sales is estimated to increase rd by about 1.08

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