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CHAPTER 2 Exercise Solutions 1 Chapter 2, Exercise Solutions, Principles of Econometrics,23e EXERCISE 2.1 (a) x y x − x () − x 2 y − y () −xy() 3 5 2 4 3 6 2 2 1 1 0 0 1 3 0 0 1 0 − 1 2 −2 4 0 0 0 − 2 − 1 1 − 4 4 x = y = x − x = xx− 2= y − y = x −xyy ∑ i ∑ i ∑ ( i ) ∑ ()i ∑ ( ) ∑ () () 5 10 0 10 0 10 x =1=, 2 ∑ (x−y−)( ) 10 (b)= b2= = 2 1. b2is the estimated slope of the fitted line. ∑ ()− 10 b1 2 =−×= 211.1 b1is the estimated value of y when x = 0, it is the estimated intercept of the fitted line. 5 2 2 2 2 2 2 (c) ∑ xi=+ 3+1− +1=051 () i= 5 ∑ xii=× 3+ 2 +1+3− ×1+×−0 ()2 2 ( ) i=1 5 2 2 2 2 ∑ ∑i−==−i− 15 5 1 10 () x i= i= 5 5 ∑ ∑iix××=−− −20 5 1 2x x 10y i) i() i= =1 (d) xi yi yi ei ei xii 3 5 4 1 1 3 2 2 3 − 1 1 − 2 1 3 2 1 1 1 − 1 2 0 2 4 − 2 0 − 2 1 − 3 9 0 2 ∑ xi = ∑ yi= ∑ yi= ∑ ei= ∑ ei= ∑ xii = 5 101106 0 (e) Refer to Figure xr2.1 below. Chapter 2, Exercise Solutions, Principles of Econometrics, 3e3 Exercise 2.1 (continued) (f) 6 5 4 3 2 y 1 0 -1 -2 -3 -1.2 -0.8 -0.4 0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 x Figure xr2.1 Fitted line, mean and observations (g) yb=+ y x = b = b2, =1, 1, 1 1 2 1 2 Therefore: 2=11+× = =+ (+) + y+ = ∑ yi (43205 /2 ) y 2 ∑ ei 16 (i) σ= = = 5.3333 N − 23 2 n σ 5.3333 (j) var()2 === 2 10 .53333 ∑ ()i− Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 4 EXERCISE 2.2 (a) Using equation (B.30), ⎛110 ⎞−μ y|x =$100=0 =μ yx| $1000 140−μ yx |$1000 P ( )0 << 140 = < P < ⎟ ⎜ σ σ σ2 2 ⎝ ⎠ yx =1=00 = yx $1000 yx| $1000 ⎛11⎞−1− 25 140 125 =< < − < = = <P ⎜ ⎟ P Z ( )2.1429 2.1429 0.9679 ⎝ ⎠ 49 49 .06 .05 .04 FY.03 .02 .01 .00 100 110 120 130 140 150 Y Figure xr2.2 Sketch of PDF (b) Using the same formula as above: ⎛ 11⎞−1− 25 140 125 =P ( )0 << − 140= = <P⎜Z ⎟ P Z ( ).6667 1.6667 0.9044 ⎝ ⎠ 81 81 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 5 EXERCISE 2.3 (a) Theservations y and x and the estimated least-squares line are graphed in part (b). The line drawn for part (a) will depend on each student’s subjective choice about the position of the line. For this reason, it has been omitted. (b) Preliminacylculatiniseld: 2 ∑ ∑ i=i21−−i iy i4 ∑ xxy () () xx 22 () 17.5 y =x7.3333 3.5 The least squares estimates are ∑ ()x−y−y() 22 = b2= = 2 1.257 ∑ ()x− 17.5 by=− = 7.3333×1.=2571 3.5 2.9333 1 2 12 11 10 9 8 y 7 6 5 4 3 0 1 2 3 4 5 6 7 x Figure xr2.3 Observations and fitted line (c) N= ∑ i 44 6 7.3333 xN= = 21 6 3.5 ∑ i The predicted value fory at x = x is yb=+x1 2 +.9333× 1=.2571 3.5 7.3333 We observe that y =b1 2= . That is, the predicted value at the sample meax is the sample mean of the dependent variable y . This implies that the least-squares estimated line passes through the point (,x)y Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 6 Exercise 2.3 (continued) (d) The values of the least squares residuals, computed fromei ixb −−1i 2 , are: e1=− 0.19048 e2 =0.55238 e3= 0.29524 e4=− 0.96190 e5 = −0.21905 e6= 0.52381 Their sum is e = 0. ∑ i (e) xe =×1− 0.190 2 0.552 3 0.295 4 + − 0.962 5 = 0.291 6 0.524 0 ∑ ii Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 7 EXERCISE 2.4 (a) If β1 0, the simple linear regression model becomes y =β xe + i ii 2 (b) Graphicallse,tting β=10 implies the mean of the simple linear regression model E()y x=β passes through the origin (0, 0). i i 2 (c) To save on subscript notation we set β 2β. The sum of squares function becomes N N 2 2 2 22 22 S() = (∑ ∑ βi i= i(i i i ixy =i i i− β y2 + β xy x i= i= 2 2 =β−52 2 176 91 352 352 91 40 35 30 25 SUM_SQ 20 15 10 1.6 1.8 2.0 2.2 2.4 BETA Figure xr2.4(a) Sum of squares for β2 The minimum of this function is approximately 12 and occurs at approximately β= 1.95. 2 The significance of this value is that it is the least-squares estimate. (d) To find the value of β that minimizes S(β) we obtain dS =− 2 2 x y+ x β 2 dβ ∑∑ ii i Setting this derivative equal to zero, we have 2 ∑ xii b∑∑i i i or b= 2 ∑ xi Thus, the least-squares estimate is 176 b == 1.9341 2 91 which agrees with the approximate value of 1.95 that we obtained geometrically. Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 8 Exercise 2.4 (Continued) (e) 12 10 8 * (3.5, 7.333) Y16 4 2 00 1 2 3 4 5 6 X1 Figure xr2.4(b) Fitted regression line and mean The fitted regression line is plotted in Figure xr2.4 (b). Note that the point (,x)y does not lie on the fitted line in this instance. (f) The least squares residuals, obtained from ei i i− 2 are: e = 2.0659 e = 2.1319 eˆ = 1.1978 1 2 3 e 4− 0.7363 e5= − 0.6703 e6= − 0.6044 Their sum is e = 3.3846. Note this value is not equal to zero as it was for β≠ 0. ∑ i 1 (g) ∑ xii+=××.0659 1 2.1319 2 1.1978 3 =−0×.7363 4 0.6703 5 0.6044 6 0 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 9 EXERCISE 2.5 (a) The consultant’s report implies that the least squares estimates satisfy the following two equations b1 24=50 7500 b +6 =00 8500 1 2 Solving these two equations yields 1000 b = = 6.6667 b = 4500 2 150 1 9000 8000 * weekly averages 7000 SAL6000 5000 4000 0 100 200 300 400 500 600 ADVERT Figure xr2.5 Graph of sales-advertising regression line Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 10 EXERCISE 2.6 (a) The intercept estimate b 1− 240 is an estimate of the number of sodas sold when the temperature is 0 degrees Fahrenheit. A common problem when interpreting the estimated intercept is that we often do not have any data points near X = 0. If we have no observations in the region where temperature is 0, then the estimated relationship may not be a good approximation to reality in that region. Clearly, it is impossible to sell −240 sodas and so this estimate should not be accepted as a sensible one. The slope estimate b = 6 is an estimate of the increase in sodas sold when temperature 2 increases by 1 Fahrenheit degree. This estima te does make sense. One would expect the number of sodas sold to increase as temperature increases. (b) If temperature is 80 degrees, the predicted number of sodas sold is y =− 240+ 6 × 8= 0 240 (c) If no sodas are sold, y = 0, and 0 =− 240+ 6 × x or x = 40 Thus, she predicts no sodas will be sold below 40°F. (d) A graph of the estimated regression line: 300 200 100 0 SODAS -100 -200 -300 20 40 60 80 TEMP Figure xr2.6 Graph of regression line for soda sales and temperature Chapter 2, Exercise Solutions, Principles of Econometrics,11e EXERCISE 2.7 (a) Since 2 ∑ ei σ= = 2.04672 N − 2 foliat s 2 = ∑ =×i= −.04672( 2) 2.04672 49 100.29 (b) The standard error fo2 bs n se(b2b2 =var( ) 0.00098 0.031305 Also, n σ2 var( 2= 2 ∑ ( i− x Thus, 2 σ2 2.04672 ∑ ()i−= n = 2088.5 var()2 0.00098 (c) Thelue b2= 0.18 suggests that a 1% increase in th e percentage of males 18 years or older who are high school graduates will lead to an increase of $180 in the mean income of males who are 18 years or older. (d) b1 2 = 15.187 0.18= 69.139 2.742 2 2 2 (e) Since ∑ ∑)i−iNx − , we have 2 2 22 ∑∑ Ni i−+() ×+ 2088.5 51 69.139 = 245,879 (f) Aorrkansas ei i i i− i= 1 2 − 1× −=4 2.742 0.18 58.3 0.962 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 12 EXERCISE 2.8 (a) The EZ estimator can be written as y k y b y = =2 1 − ⎛ ⎞ 1 = i EZ xx xx− ⎜ ⎟−xx 2 1 ∑ 2 1 2 1⎝ ⎠ ⎠ 2 1 where k = −1 , k = 1 , and k = k = ... = k = 0 1 x − x 2 x − x 3 4 N 2 1 2 1 Thus, bEZ is a linear estimator. (b) Takingpectationelds ⎡y2 1 1 1 E ()EZ E = ⎢ ⎥ − Ey () 2 1 Ey () ⎣ ⎦ 1x−2 1 − x2 1 1 1 = xx− − ( 122 xxx 1 12 ( )β 2 1 2 1 β22 x 2 1x ⎛2⎞ 1 β−x x xx = xx x x 2⎜ ⎟2 2 1 2 1 2 1 2 1 ⎝ ⎠ Thus, bEZ is an unbiased estimator. (c) The variance is given by 22 2 ek k yvai ii i iZ =ar∑∑= ) ∑var( ) ⎛ ⎞ 2 =σ 2⎜ ⎟ 1 1 2+ = σ ⎜ ⎟ 2 2 2 ⎝ ⎠2 1 −xx2 1() 2 1 () ⎡ 2σ 2 ⎤ (d) If ei,~0 ( σ2) , then EZ ~ , ⎢β2 2⎥ ⎣ ()x2 1 ⎦ Exercise 2.8 (continued) (e) To convince E.Z. Stuff that var(b ) < var(b ), we need to show that 2 2 2σ σ 2 ( 2 1x−x Consider x x x x x x x x − x 2 ( 2 1 2 2 Thus, we need to show that N 2∑ ()i− >x i=1 thator 2 ()1−2x−x+ −xxx−x thator ⎣ ⎦1 2x⎤x This last inequality clearly holds. Thus, Rather than prove the result directly, as we have done above, we could also refer Professor E.Z. Stuff to the Gauss Markov theorem. Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 14 EXERCISE 2.9 (a) Plootfs UNITCOST against CUMPROD and ln UNITCOST ( ) against ln CU( )OD t t t t appear in Figure xr2.9(a) & (b). The two plots are quite similar in nature. 26 24 22 20 UNITCOST 18 16 1000 2000 3000 4000 CUMPROD Figure xr2.9(a) The learning curve data 3.3 3.2 3.1 3.0 ln(UNITCOST) 2.8 2.7 7.0 7.2 7.4 7.6 7.8 8.0 8.2 ln(CUMPROD) Figure xr2.9(b) Learning curve data with logs Chapter 2, Exercise Solutions, Principles of Econometrics153e Exercise 2.9 (continued) (b) The least squares estimates are b 1 6.0191 b2= −0.3857 li(ce UNITCOST ) 1 β ,1an estimate of 1 is n UNITCOST =1 1=xp() = ex( 6.019) 411.208 This result suggests that 411.2 was the cost of producing the first unit. Theb2stimate −0.3857 suggests that a 1% increase in cumulative production will decrease costs by 0.386%. The numbers seem sensible. 3.4 3.3 3.2 3.1 3.0 ln2.9ITCOST) 2.8 2.7 7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 8.0 8.1 8.2 ln(CUMPROD) Figure xr2.9(c) Observations and fitted line (c) The coefficient covariance matrix has the elements b var−(21=b 20755531= var() 0.001297 cov (), 0.009888 (d) The error variance estimate is σ= 0.049930= 0.002493. (e) When CUMPROD = 2000, the predicted unit cost is 0 UNITCOST =0xp 6.(1909−0.= 385696ln ()00 ) 21.921 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 16 EXERCISE 2.10 (a) The model is a simple regression model because it can be written as y =β +β x +e 1 2 where y =− rj f , x =−rm f , β 1α j and β =2 . j (b) General General Exxon- Firm Microsoft IBM Disney Electric Motors Mobil ˆ b2=β j 1.430 0.983 1.074 1.268 0.9590.403 The stocks Microsoft, General Motors and IBM are aggressive with Microsoft being the most aggressive with a beta value of β=21.430 . General Electric, Disney and Exxon- Mobil are defensive with Exxon-Mobil being the most defensive since it has a beta value of β= 0.403. 2 (c) General General Exxon- Firm Microsoft IBM Disney Electric Motors Mobil b = α ˆ 1 j 0.010 0.006 -0.002 0.007-0.001 0.007 All the estimates of α j are close to zero and are therefore consistent with finance theory. In the case of Microsoft, Figure xr2.10 illustra tes how close the fitted line is to passing through the origin. .5 .4 .3 .2 .1 .0 MSFT-RKFREE -.2 -.3 -.4 -.20 -.15 -.10 -.05 .00 .05 .10 MKT-RKFREE Figure xr2.10 Observations and fitted line for microsoft Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 17 Exercise 2.10 (continued) (d) The estimates for β gijen α= 0 arj as follows. General General Exxon- Firm Microsoft IBM Disney Electric Motors Mobil ˆ βj 1.464 1.003 1.067 1.291 0.956 0.427 resthiction α j 0 has led to only small changes in the β j Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 18 EXERCISE 2.11 (a) 1600000 1200000 800000 PRICE 400000 0 0 2000 4000 6000 8000 SQFT Figure xr2.11(a) Price against square feet – all houses 1200000 1000000 800000 600000 PRICE 400000 200000 0 0 2000 4000 6000 8000 SQFT Figure xr2.11(b) Price against square feet for houses of traditional style Chapter 2, Exercise Solutions, Principles of Econometrics, 319 Exercise 2.11 (continued) (b) The estimated equation for all houses is PRICE =− 60,861 + 92.747SQFT The coefficient 92.747 suggests house price increases by approximately $92.75 for each additional square foot of house size. The intercept, if taken literally, suggests a house with zero square feet would cost − $60,861, a meaningless value. The model should not be accepted as a serious one in the region of zero square feet. 1,600,000 1,400,000 1,200,000 1,000,000 800,000 PRICE 600,000 400,000 200,000 0 0 2,000 4,000 6,000 8,000 SQFT Figure xr2.11(c) Fitted line for Exercise 2.11(b) (c) The estimated equation for traditional style houses is PRICE =− 28,408+ 73.772SQFT The slope of 73.772 suggests that house price increases by approximately $73.77 for each additional square foot of house size. The intercept term is− 28,408 which would be interpreted as the dollar price of a traditional house of zero square feet. Once again, this estimate should not be accepted as a serious one. A negative value is meaningless and there is no data in the region of zero square feet. Comparing the estimates to those in part (b), we see that extra square feet are not worth as much in traditional style houses as they are for houses in general ($77.77 < $92.75). A comparison of intercepts is not meaningful, but we can compare equations to see which type of house is more expensive. The prices are equal when −28,408 73.−77+2SQFT = 60,861 92.747SQFT Sofving SQFT yields 60861−28408 SQFT = = 1710 92.747 −73.772 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 20 Exercise 2.11(c) (continued) (c) Thus, we predict that the price of traditional style houses is greater than the price of houses in general when SQFT <1710. Traditional style houses are cheaper when SQFT >1710. (d) Residu pallots: 500000 400000 300000 200000 100000 RESID 0 -100000 -200000 -300000 -400000 0 2000 4000 6000 8000 SQFT Figure xr2.11(d) Residuals against square feet – all houses 600000 500000 400000 300000 200000 RESID000 0 -100000 -200000 -300000 0 2000 4000 6000 8000 SQFT Figure xr2.11(e) Residuals against square feet for houses of traditional style The magnitude of the residuals tends to increase as housing size increases suggesting that SR3 var () | =σ 2 [homoskedasticity] could be violated. The larger residuals for larger i houses imply the spread or variance of the errors is larger as SQFT increases. Or, in other words, there is not a constant variance of the error term for all house sizes. Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 21 EXERCISE 2.12 (a) We can see a positive relationship between price and house size. 600000 500000 400000 300000 PRICE 200000 100000 0 0 1000 2000 3000 4000 5000 SQFT Figure xr2.12(a) Price against square feet (b) The estimated equation for all houses in the sample is PRICE =− 18,386+ 81.389SQFT The coefficient 81.389 suggests house price increases by approximately $81 for each additional square foot in size. The intercept, if taken literally, suggests a house with zero square feet would cost − $18,386, a meaningless value. The model should not be accepted as a serious one in the region of zero square feet. 600,000 500,000 400,000 300,000 PRICE 200,000 100,000 0 0 1,000 2,000 3,000 4,000 5,000 SQFT Figure xr2.12(b) Fitted regression line Chapter 2, Exercise Solutions, Principles of Econometrics, 22 Exercise 2.12 (continued) (c) The estimated equation when a house is vacant at the time of sale is PRICE =− 4792.70+ 69.908SQFT For houses that are occupied it is n PRICE =− 27,169+ 89.259SQFT These results suggest that price increases by $69.91 for each additional square foot in size for vacant houses and by $89.26 for each additional square foot of house size for houses that are occupied. Also, the two estimated lines will cross such that vacant houses will have a lower price than occupied houses when the house size is large, and occupied houses will be cheaper for small houses. To obtain the break-even size where prices are equal we write −4+792.+0 =69.908SQFT 27,169 89.259SQFT Sofving SQFT yields 27169− 4792.7 SQFT = = 1156 89.259−69.908 Thus, we estimate that occupied houses have a lower price per square foot when SQFT <1156 and a higher price per square foot when SQFT >1156. (d) Residpallots 200,000 160,000 120,000 80,000 40,000 RESID 0 -40,000 -80,000 -120,000 0 1,000 2,000 3,000 4,000 5,000 SQFT Figure xr2.12(c) Residuals against square feet for occupied houses Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 23 Exercise 2.12(d) (continued) (d) 250,000 200,000 150,000 100,000 RES50,000 0 -50,000 -100,000 0 1,000 2,000 3,000 4,000 5,000 SQFT Figure xr2.12(d) Residuals against square feet for vacant houses The magnitude of the residuals tends to be la rger for larger-sized houses suggesting that SR3 var ex 2 [the homoskedasticity assumption of the model] could be violated. () | i =σ As the size of the house increases, the spread of distribution of residuals increases, indicating that there is not a constant variance of the error term with respect to house size. (e) Using the model estimated in part (b), the predicted price when SQFT = 2000 is PRICE =− 18,386+ 81.389 × 200= 0 $144,392 Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 24 EXERCISE 2.13 (a) 11 10 9 8 7 6 5 1990 1995 2000 2005 FIXED_RATE Figure xr2.13(a) Fixed rate against time 140 120 100 80 60 40 20 90 92 94 96 98 00 02 04 SOLD Figure xr2.13(b) Houses sold (1000’s ) against time Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 25 Exercise 2.13(a) (continued) (a) 2400 2000 1600 1200 800 400 90 92 94 96 98 00 02 04 STARTS Figure xr2.13(c) New privately owned houses started against time (b) Refer to Figure xr2.13(d). (c) The estimated model is STARTS = 2 −992.739 194.233FIXED_RATE cTehfficient −194.233 suggests that the number of new privately owned housing starts decreases by 194,233 for a 1% increase in the 30 year fixed interest rate for home loans. The intercept suggests that when the 30 year fixed interest rate is 0%, 2,992,739 will be started. Caution should be exercise with this interpretation, however, because it is beyond the range of the data. Figure xr2.13(d) shows us where the fitted lin e lies among the data points. The fitted line appears to go evenly through the centre of data and the residuals appear be of relatively equal magnitude as we move along the fitted line. 2400 2000 1600 1200 STARTS 800 400 5 6 7 8 9 10 11 FIXED_RATE Figure xr2.13 (d) Fitted line and observations for housing starts against the fixed rate Chapter 2, Exercise Solutions, Principles of Econometrics, 326 Exercise 2.13 (continued) (d) Refer to Figure xr2.13(e). (e) The estimated model is SOLD =1 −67.548 13.034FIXED_RATE cTehficient −13.034 suggests that a 1% increase in the 30 year fixed interest rate for home loans is associated with a decrease of around 13,034 houses sold. The intercept suggests that when the 30 year fixed interest rate is 0%, 167,548 houses will be sold over a period of 1 month. Caution should be exercise with this interpretation, however, because it is beyond the range of the data. 140 120 100 80 SOLD 60 40 20 5 6 7 8 9 10 11 FIXED_RATE Figure xr2.13(e) Fitted line and observations for houses sold against fixed rate Figure xr2.13(e) shows us where the fitted line lies amongst the da ta points. From this figure we can see that the data appear slightly convex relative to the fitted line suggesting that a different functional form might be suitable. A plot of the residuals against the fixed rate might shed more light oin this question. We can see also that the residuals appear to have a constant distribution over the majority of fixed rates. (f) Using the model estimated in part (c), the predicted number of monthly housing starts for FIXED_ RATE = 6 is n (STARTS ×)= 000 ( )92.××3=9 1×4.233 6 1000 1827.34 1000 1,827,340 There will be 1,827,340 new privately owned houses started at a 30 year fixed interest rate of 6%. This is a seasonally adjusted annual rate. On a monthly basis we estimate 155,278 starts. Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 27 EXERCISE 2.14 (a) 65 60 55 50 VOTE 45 40 35 -16 -12 -8 -4 0 4 8 12 GROWTH Figure xr2.14(a) Incumbent share against growth rate of real GDP per capita There appears to be a positive association between VOTE and GROWTH. (b) The estimated equation is n VOTE = 5 +1.939 0.660GROWTH The coefficient 0.660 suggests that for an increase in 1% of the annual growth rate of GDP per capita, there is an associated increase in the share of votes of the incumbent party of 0.660. The coefficient 51.939 indicates that the incumbent party receives 51.9% of the votes on average, when the growth rate in real GDP is zero. This suggests that when there is no real GDP growth, the incumbent party will still maintain the majority vote. A graph of the fitted line and data is shown in Figure xr2.14(b). 65 60 55 50 VOTE 45 40 35 -16 -12 -8 -4 0 4 8 12 GROWTH Figure xr2.14(b) Graph of vote-growth regression Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 28 Exercise 2.14 (continued) (c) Figure xr2.14(c) shows a plot of VOTE against INFLATION. It shows a negative correlation between the two variables. The estimated equation is: VOTE= 53.496−0.445INFLATION cTehficient −0.445 indicates that a 1% increase in inflation, the GDP deflator, during the incumbent party’s first 15 quarters, is asso ciated with a 0.445 drop in the share of votes. The coefficient 53.496 suggest that on average, when inflation is at 0% for that party’s first 15 quarters, the associated share of votes won by the incumbent party is 53.496%; the incumbent party maintains the majority vote when inflation, during their first 15 quarters, is at 0%. 65 60 55 50 VOTE 45 40 35 0 1 2 3 4 5 6 7 8 INFLATION Figure xr2.14(c) Graph of vote-inflation regression line and observations Chapter 2, Exercise Solutions, Principles of Econometrics, 3e 29 EXERCISE 2.15 (a) 400 Series: EDUC 350 Sample 1 1000 Observations 1000 300 Mean 13.28500 250 Median 13.00000 Maximum18.00000 200 Minimum.000000 150 Std. Dev. 2.468171 Skewness -0.211646 100 Kurtosis 4.525053 Jarque-Bera 104.3734 50 Probability 0.000000 0 2 4 6 8 10 12 14 16 18 Figure xr2.15(a) Histogram and statistics for EDUC From Figure xr2.15 we can see that the observations of EDUC are skewed to the left indicating that there are few observations with less than 12 years of education. Half of the sample has more than 13 years of educati on, with the average being 13.29 years of education. The maximum year of education received is 18 years, and the lowest level of education achieved is 1 year. 240 Series: WAGE_HISTOGRAM_STATS 200 Sample 1 1000 Observations 1000 160 Mean 10.21302 Median 8.790000 120 Maximu6m0.19000 Minimu2.030000 Std. Dev. 6.246641 80 Skewness 1.953258 Kurtosis 10.01028 40 Jarque-Bera 2683.539 Probability 0.000000 0 0 10 20 30 40 50 60 Figure xr2.15(b) Histogram and statistics for WAGE Figure xr2.15(b) shows us that the observations for WAGE are skewed to the right indicating that most of the observations lie between the hourly wages of 5 to 20, and that there are few observations with an hourly wage greater than 20. Half of the sample earns an hourly wage of more than 8.79 dollars an hour, with the average being 10.21 dollars an hour. The maximum earned in this sample is 60.19 dollars an hour and the least earned in this sample is 2.03 dollars an hour. Chapter 2, Exercise Solutions, Principles of Economet30cs, 3e Exercise 2.15 (continued) (b) The estimated equation is WAGE =− 4.912+ 1.139EDUC The coefficient 1.139 represents the associated increase in the hourly wage rate for an extra year of education. The coeff−4.912 represents the estimated wage rate of a worker with no years of education. It should not be considered meaningful as it is not possible to have a negative hourly wage rate. Also, as shown in the histogram, there are no data points at or close to the region EDUC = 0. (c) The residuals are plotted against education in Figure xr2.15(c). There is a pattern evident; as EDUC increases, the magnitude of the residuals also increases. If the assumptions SR1-SR5 hold, there should not be any patterns evident in the least squares residuals. 50 40 30 20 RESID 0 -10 -20 0 4 8 12 16 20 EDUC Figure xr2.15(c) Residuals against education (d) The estimated regressions are n femIafle: WAGE =− 5.963+ 1.121EDUC malef: WAGE =− 3.562+ 1.131EDUC n blacf: WAGE = 0+.653 0.590EDUC whitf: WAGE =− 5.151+ 1.167EDUC From these regression results we can see that the hourly wage of a white worker increases significantly more, per additional year of edu cation, compared to that of a black worker. Similarly, the hourly wage of a male worker increases more per additional year of education than that of a female worker; although this difference is relatively small.

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