CHAPTER 8. Exercise Solutions

CHAPTER 8 Exercse Solutons 77

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 78 EXERCISE 8. When = N N N ( x x) ( x x) ( x x) = = = N = = = N N N ( x ) ( ) ( ) ( x x ) x x x x x = = = =

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 79 EXERCISE 8. (a) Multplyng the frst normal equaton by ( x ) and the second one by ( ) ( )( ) ( ) ( ) ( )( ) β + ( )( ) β = ( ) x β + x β = x y x x x y * Subtractng the frst of these two equatons from the second yelds Thus, ( )( ) ( ) ( ) ( ) x x β = x y x y * * ( ) x y ( x )( y ) ( )( x ) ( x ) β = = yx y x x x yelds In ths last expresson, the second lne s obtaned from the frst by makng the substtutons y = y and x = x, and by dvdng numerator and denomnator by ( ). Solvng the frst normal equaton ( ) ( x ) and makng the substtutons y = y and x y x β = β β + β = y for = x, yelds β (b),, x, When = for all, yx = yx y = y x = and = N. Makng these substtutons nto the expresson for β yelds yx y x yx y x N N N β N = = x x x x N N N and that for β becomes y x β = β = β N N y x These formulas are equal to those for the least squares estmators b and b. See pages and 4-44 of POE.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 80 Exercse 8. (contnued) (c) The least squares estmators b and b are functons of the followng averages x = x N y = y N xy N N x For the generalzed least squares estmator for β and β, these unweghted averages are replaced by the weghted averages x y y x x In these weghted averages each observaton s weghted by the nverse of the error varance. Relable observatons wth small error varances are weghted more heavly than those wth hgher error varances that make them more unrelable.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 8 EXERCISE 8.3 For the model y =β +β x + e where constant error varance s y =β x +β + e * * * var( e ) = x, the transformed model that gves a where y* = y x, x* = x, and e* = e x. Ths model can be estmated by least squares wth the usual smple regresson formulas, but wth β and β reversed. Thus, the generalzed least squares estmators for β and β are N x y x y β = β = β * * * * * * and y x * N ( x ) * x ( ) Usng observatons on the transformed varables, we fnd * y = 7, * x = 37, * * xy = 47 8, Wth N = 5, the generalzed least squares estmates are and 5(47 8) (37 )(7) β = =.984 5(349 44) (37 ) (37 ) * * β (7 5).984 0.44 = y β x = = 5 * ( x ) = 349 44

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 8 EXERCISE 8.4 (a) (b) In the plot of the resduals aganst ncome the absolute value of the resduals ncreases as ncome ncreases, but the same effect s not apparent n the plot of the resduals aganst age. In ths latter case there s no apparent relatonshp between the magntude of the resduals and age. Thus, the graphs suggest that the error varance depends on ncome, but not age. Snce the resdual plot shows that the error varance may ncrease when ncome ncreases, and ths s a reasonable outcome snce greater ncome mples greater flexblty n travel, we set up the null and alternatve hypotheses as the one tal test H0 : = versus H: >, where and are artfcal varance parameters for hgh and low ncome households. The value of the test statstc s F (.947 0 ) (00 4) = = =.84 (.0479 0 ) (00 4) 7 7 The 5% crtcal value for (96, 96) degrees of freedom s F (0.95,96,96) =.40. Thus, we reject H 0 and conclude that the error varance depends on ncome. Remark: An nspecton of the fle vacaton.dat after the observatons have been ordered accordng to INCOME reveals 7 mddle observatons wth the same value for INCOME, namely 6. Thus, when the data are ordered only on the bass of INCOME, there s not one unque orderng, and the values for SSE and SSE wll depend on the orderng chosen. Those specfed n the queston were obtaned by orderng frst by INCOME and then by AGE. (c) () All three sets of estmates suggest that vacaton mles travelled are drectly related to household ncome and average age of all adults members but nversely related to the number of kds n the household. () The Whte standard errors are slghtly larger but very smlar n magntude to the conventonal ones from least squares. Thus, usng Whte s standard errors leads one to conclude estmaton s less precse, but t does not have a bg mpact on assessment of the precson of estmaton. () The generalzed least squares standard errors are less than the Whte standard errors for least squares, suggestng that generalzed least squares s a better estmaton technque.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 83 EXERCISE 8.5 (a) (b) (c) The table below dsplays the 95% confdence ntervals obtaned usng the crtcal t-value t (0.975,497) =.965 and both the least squares standard errors and the Whte s standard errors. After recognzng heteroskedastcty and usng Whte s standard errors, the confdence ntervals for CRIME, AGE and TAX are narrower whle the confdence nterval for ROOMS s wder. However, n terms of the magntudes of the ntervals, there s very lttle dfference, and the nferences that would be drawn from each case are smlar. In partcular, none of the ntervals contan zero and so all of the varables have coeffcents that would be judged to be sgnfcant no matter what procedure s used. 95% confdence ntervals Least squares standard errors Whte s standard errors Lower Upper Lower Upper CRIME 0.55 0. 0.5 0.4 ROOMS 5.600 7.43 5.065 7.679 AGE 0.076 0.00 0.070 0.06 TAX 0.00 0.005 0.09 0.007 Most of the standard errors dd not change dramatcally when Whte s procedure was used. Those whch changed the most were for the varables ROOMS, TAX, and PTRATIO. Thus, heteroskedastcty does not appear to present major problems, but t could lead to slghtly msleadng nformaton on the relablty of the estmates for ROOMS, TAX and PTRATIO. As mentoned n parts (a) and (b), the nferences drawn from use of the two sets of standard errors are lkely to be smlar. However, keepng n mnd that the dfferences are not great, we can say that, after recognzng heteroskedastcty and usng Whte s standard errors, the standard errors for CRIME, AGE, DIST, TAX and PTRATIO decrease whle the others ncrease. Therefore, usng ncorrect standard errors (least squares) understates the relablty of the estmates for CRIME, AGE, DIST, TAX and PTRATIO and overstates the relablty of the estmates for the other varables. Remark: Because the estmates and standard errors are reported to 4 decmal places n Exercse 5.5 (Table 5.7), but only 3 n ths exercse (Table 8.), there wll be some roundng error dfferences n the nterval estmates n the above table. These dfferences, when they occur, are no greater than 0.00.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 84 EXERCISE 8.6 (a) (b) ROOMS sgnfcantly effects the varance of house prces through a relatonshp that s quadratc n nature. The coeffcents for ROOMS and ROOMS are both sgnfcantly dfferent from zero at a % level of sgnfcance. Because the coeffcent of ROOMS s postve, the quadratc functon has a mnmum whch occurs at the number of rooms for whch e =α + α 3ROOMS = 0 ROOMS Usng the estmated equaton, ths number of rooms s ROOMS α 305.3 mn = = = α 3 3.8 6.4 Thus, for houses of 6 rooms or less the varance of house prces decreases as the number of rooms ncreases and for houses of 7 rooms or more the varance of house prces ncreases as the number of rooms ncreases. The varance of house prces s also a quadratc functon of CRIME, but ths tme the quadratc functon has a maxmum. The crme rate for whch t s a maxmum s CRIME α.85 4 max = = = α 5 0.039 9.3 Thus, the varance of house prces ncreases wth the crme rate up to crme rates of around 30 and then declnes. There are very few observatons for whch CRIME 30, and so we can say that, generally, the varance ncreases as the crme rate ncreases, but at a decreasng rate. The varance of house prces s negatvely related to DIST, suggestng that the further the house s from the employment centre, the smaller the varaton n house prces. We can test for heteroskedastcty usng the Whte test. The null and alternatve hypotheses are H : α =α = =α = 0 0 3 6 H : not all α n H are zero s 0 The test statstc s test value s H f χ = N R. We reject 0 χ = N R = = 506 0.08467 4.84 χ >χ (0.95,5) where Snce 4.84 >.07, we reject H 0 and conclude that heteroskedastcty exsts. χ (0.95,5) =.07. The

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 85 EXERCISE 8.7 (a) Hand calculatons yeld x y x y x = 0 = 3. = 89.35 = 5.34 x = 0 y = 3.8875 The least squares estmates are gven by Nxy xy 8 89.35 0 3. b = = =.707 N x x 8 5.34 0 and ( ) ( ) b = y bx = 3.8875.707 0 = 3.8875 (b) The least squares resduals e = y y and other nformaton useful for part (c) follow observaton ê ln( e ) z ln( e ).933946.395 4.3533 0.7338 0.68977 0.85693 3 9.549756 4.5303 3.599 4.74707.078484 5.068875 5 3.9665.38787 4.5795 6 3.887376.75469 8.46587 7 3.484558.49668 5.74369 8 3.746079.6449 6.90508 (c) To estmate α, we begn by takng logs of both sdes of = exp( α z), that yelds ln( ) =α z. Then, we replace the unknown wth e to gve the estmatng equaton e =α z + v ln( ) Usng least squares to estmate α from ths model s equvalent to a smple lnear regresson wthout a constant term. See, for example, Exercse.4. The least squares estmate for α s 8 ( z ln( e )) 86.4674 = α= = = 0.4853 8 z 78.7 =

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 86 Exercse 8.7 (contnued) (d) Varance estmates are gven by the predctons exp( = α z) = exp(0.4853 z). These values and those for the transformed varables y * x, * y = x = are gven n the followng table. observaton 4.960560 0.493887 0.4494.5675 0.464895.78937 3 9.87947 3.45764 0.58548 4 9.78598 0.87700 0.57540 5.5453 4.036003.446 6 7.535 0.345673 0.674 7 3.05360.57536.37350 8.330994 0.0433 0.0433 (e) From Exercse 8., the generalzed least squares estmate for β s x x β = yx y x y * 5.33594.938 ( 0.38385) =.00863 5.4437 ( 0.38385).00863 8.47748 = 7.540580 =.4 The generalzed least squares estmate for β s y x β.938 ( 0.38385).4.653 = β = = x *

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 87 EXERCISE 8.8 (a) The regresson results wth standard errors n parenthess are PRICE = 593.5 + 68.3907SQFT 7.8433AGE (se) 3586.64.687 35.0976 ( ) ( ) ( ) These results tell us that an ncrease n the house sze by one square foot leads to an ncrease n house prce of $63.39. Also, relatve to new houses of the same sze, each year of age of a house reduces ts prce by $7.84. (b) For SQFT = 400 and AGE = 0 PRICE = 593.5 + 68.3907 400 7.8433 0 = 96,583 The estmated prce for a 400 square foot house, whch s 0 years old, s $96,583. For SQFT = 800 and AGE = 0 PRICE = 593.5 + 68.3907 800 7.8433 0 = 3,940 The estmated prce for a 800 square foot house, whch s 0 years old, s $3,940. (c) For the Whte test we estmate the equaton e =α +α SQFT +α AGE +α SQFT +α AGE +α SQFT AGE + v 3 4 5 6 and test the null hypothess H0 : α =α 3 = =α 6 = 0. The value of the test statstc s Snce N R 940 0.0375 35.5 χ = = = χ (0.95,5) =.07, the calculated value s larger than the crtcal value. That s, χ >χ. (0.95,5) Thus, we reject the null hypothess and conclude that heteroskedastcty exsts. (d) Estmatng the regresson log( e ) = α +α SQFT + v gves the results α = 6.3786, α = 0.0044 Wth these results we can estmate as exp(6.3786 0.0044 SQFT ) = +

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 88 Exercse 8.8 (contnued) (e) Generalzed least squares requres us to estmate the equaton PRICE SQFT AGE e =β +β +β + When estmatng ths model, we replace the unknown wth the estmated standard devatons. The regresson results, wth standard errors n parenthess, are PRICE = 849.4 + 65.369SQFT 87.6587AGE (se) 309.43.085 9.844 ( ) ( ) ( ) These results tell us that an ncrease n the house sze by one square foot leads to an ncrease n house prce of $65.33. Also, relatve to new houses of the same sze, each year of age of a house reduces ts prce by $87.66. (f) For SQFT = 400 and AGE = 0 PRICE = 849.4 + 65.369 400 87.6587 0 = 96,96 The estmated prce for a 400 square foot house, whch s 0 years old, s $96,96. For SQFT = 800 and AGE = 0 PRICE = 849.4 + 65.369 800 87.6587 0 =,36 The estmated prce for a 800 square foot house, whch s 0 years old, s $,36.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 89 EXERCISE 8.9 (a) () Under the assumptons of Exercse 8.8 part (a), the mean and varance of house prces for houses of sze SQFT = 400 and AGE = 0 are EPRICE ( ) =β + 400β + 0β 3 Replacng the parameters wth ther estmates gves EPRICE ( ) = 96583 Assumng the errors are normally dstrbuted, var( PRICE ) = var( PRICE ) = 539.63 5000 96583 P( PRICE > 5000) = P Z > 539.6 = P Z > ( 0.87) = 0.07 where Z s the standard normal random varable Z N(0,). The probablty s depcted as an area under the standard normal densty n the followng dagram. The probablty that your 400 square feet house sells for more than $5,000 s 0.07. () For houses of sze SQFT = 800 and AGE = 0, the mean and varance of house prces from Exercse 8.8(a) are EPRICE ( ) = 3940 The requred probablty s 0000 3940 P( PRICE< 0000) = P Z < 539.6 = P Z < ( 0.685) = 0.68 var( PRICE ) = 539.63 The probablty that your 800 square feet house sells for less than $0,000 s 0.68.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 90 Exercse 8.9 (contnued) (b) () Usng the generalzed least squares estmates as the values for β, β and β 3, the mean of house prces for houses of sze SQFT = 400 and AGE = 0 s, from Exercse 8.8(f), EPRICE ( ) = 9696. Usng estmates of α and α from Exercse 8.8(d), the varance of these house types s Thus, var( PRICE ) = exp( α +.704 +α 400) = exp(6.378549 +.704 + 0.0044769 400) = 3.3477 0 = (895.3) 5000 9696 P( PRICE > 5000) = P Z > 895.3 = P Z > 8 (.078) = 0.5 The probablty that your 400 square feet house sells for more than $5,000 s 0.5. () For your larger house where SQFT = 800, we fnd that EPRICE ( ) = 36 and Thus, var( PRICE ) = exp( α +.704 +α 800) = exp(6.378549 +.704 + 0.0044769 800) = 5.8937 0 = (475.8) 0000 36 P( PRICE < 0000) = P Z < 475.8 = P Z < 8 ( 0.5077) = 0.306 The probablty that your 800 square feet house sells for less than $0,000 s 0.306. (c) In part (a) where the heteroskedastc nature of the error term was not recognzed, the same standard devaton of prces was used to compute the probabltes for both house types. In part (b) recognton of the heteroskedastcty has led to a standard devaton of prces that s smaller than that n part (a) for the case of the smaller house, and larger than that n part (a) for the case of the larger house. These dfferences have n turn led to a smaller probablty for part () where the dstrbuton s less spread out and a larger probablty for part () where the dstrbuton has more spread.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 9 EXERCISE 8.0 (a) The transformed model correspondng to the varance assumpton = x s y e =β +β x + e where e = x x x We obtan the resduals from ths model, square them, and regress the squares on x to obtan e = 3.79 + 3.35x R = 0.3977 To test for heteroskedastcty, we compute a value of the χ = N R = = 40 0.3977 5.59 χ test statstc as A null hypothess of no heteroskedastcty s rejected because 5.59 s greater than the 5% crtcal value χ (0.95,) = 3.84. Thus, the varance assumpton = x was not adequate to elmnate heteroskedastcty. (b) The transformed model used to obtan the estmates n (8.7) s and y x e =β +β + e where e = = exp(0.93779596 +.39387 ln( x ) We obtan the resduals from ths model, square them, and regress the squares on x to obtan e =.7 + 0.05896x R = 0.074 To test for heteroskedastcty, we compute a value of the χ = N R = = 40 0.074.09 χ test statstc as A null hypothess of no heteroskedastcty s not rejected because.09 s less than the 5% crtcal value χ (0.95,) = 3.84. Thus, the varance assumpton = x γ s adequate to elmnate heteroskedastcty.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 9 EXERCISE 8. The results are summarzed n the followng table and dscussed below. part (a) part (b) part (c) β 8.000 76.70 8.009 se( β ) 3.8.004 33.806 β 0.38 0.6 0.33 se( β ).706.04.733 χ = N R 6.64.665 6.955 The transformed models used to obtan the generalzed estmates are as follows. y x (a) 0.5 =β 0.5 +β e 0.5 + x x x e where e = 0.5 x y x (b) = β +β + e x x x where e = e x y x (c) =β +β + e ln( x) ln( x) ln( x) where e = e ln( x ) In each case the resduals from the transformed model were squared and regressed on ncome and ncome squared to obtan the R values used to compute the χ values. These equatons were of the form ê = α +α x+α x + v 3 For the Whte test we are testng the hypothess H0: α =α 3 = 0 aganst the alternatve hypothess H: α 0 and/or α3 0. The crtcal ch-squared value for the Whte test at a 5% level of sgnfcance s χ (0.95,) = 5.99. After comparng the crtcal value wth our test statstc values, we reject the null hypothess for parts (a) and (c) because, n these cases, χ >χ (0.95,). The assumptons var( e ) = x and var( e) = ln( x) do not elmnate heteroskedastcty n the food expendture model. On the other hand, we do not reject the null hypothess n part (b) because χ <χ (0.95,). Heteroskedastcty has been elmnated wth the assumpton that var( e ) = x. In the two cases where heteroskedastcty has not been elmnated (parts (a) and (c)), the coeffcent estmates and ther standard errors are almost dentcal. The two transformatons have smlar effects. The results are substantally dfferent for part (b), however, partcularly the standard errors. Thus, the results can be senstve to the assumpton made about the heteroskedastcty, and, mportantly, whether that assumpton s adequate to elmnate heteroskedastcty.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 93 EXERCISE 8. (a) (b) Ths suspcon mght be reasonable because rcher countres, countres wth a hgher GDP per capta, have more money to dstrbute, and thus they have greater flexblty n terms of how much they can spend on educaton. In comparson, a country wth a smaller GDP wll have fewer budget optons, and therefore the amount they spend on educaton s lkely to vary less. The regresson results, wth the standard errors n parentheses are EE GDP = 0.46 + 0.073 P P (se) 0.0485 0.005 ( ) ( ) The ftted regresson lne and data ponts appear n the followng fgure. There s evdence of heteroskedastcty. The plotted values are more dspersed about the ftted regresson lne for larger values of GDP per capta. Ths suggests that heteroskedastcty exsts and that the varance of the error terms s ncreasng wth GDP per capta..4..0 0.8 0.6 0.4 0. 0.0-0. 0 4 6 8 0 4 6 8 GDP per capta (c) For the Whte test we estmate the equaton GDP GDP e =α +α +α 3 + v P P Ths regresson returns an R value of 0.998. For the Whte test we are testng the hypothess H0: α =α 3 = 0 aganst the alternatve hypothess H: α 0 and/or α3 0. The Whte test statstc s N R χ = = = 34 0.998 9.96 The crtcal ch-squared value for the Whte test at a 5% level of sgnfcance s χ (0.95,) = 5.99. Snce 9.96 s greater than 5.99, we reject the null hypothess and conclude that heteroskedastcty exsts.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 94 Exercse 8. (contnued) (d) Usng Whte s formula: ( b ) ( b ) se = 0.04044, se = 0.006 The 95% confdence nterval for β usng the conventonal least squares standard errors s b ± t se( b ) = 0.07373 ±.0369 0.0057947 = (0.066, 0.0837) (0.975,3) The 95% confdence nterval for β usng Whte s standard errors s b ± t se( b ) = 0.07373 ±.0369 0.0066 = (0.0605, 0.0858) (0.975,3) In ths case, gnorng heteroskedastcty tends to overstate the precson of least squares estmaton. The confdence nterval from Whte s standard errors s wder. (e) Re-estmatng the equaton under the assumpton that var( e ) = x, we obtan EE GDP = 0.099 + 0.0693 P P (se) 0.089 0.0044 ( ) ( ) Usng these estmates, the 95% confdence nterval for β s b ± t se( b ) = 0.0693±.0369 0.00447 = (0.0603, 0.0783) (0.975,3) The wdth of ths confdence nterval s less than both confdence ntervals calculated n part (d). Gven the assumpton var( e ) = x s true, we expect the generalzed least squares confdence nterval to be narrower than that obtaned from Whte s standard errors, reflectng that generalzed least squares s more precse than least squares when heteroskedastcty s present. A drect comparson of the generalzed least squares nterval wth that obtaned usng the conventonal least squares standard errors s not meanngful, however, because the least squares standard errors are based n the presence of heteroskedastcty.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 95 EXERCISE 8.3 (a) For the model C Q Q Q e =β +β +β +β +, where ( e ) 3 t t 3 t 4 t t least squares estmates of β, β, β 3 and β 4 are: var t = Q t, the generalzed estmated coeffcent standard error β 93.595 3.4 β 68.59 7.484 β 3 0.744 3.774 β 4.0086 0.45 (b) (c) (d) The calculated F value for testng the hypothess that β = β 4 = 0 s 08.4. The 5% crtcal value from the F (,4) dstrbuton s 3.40. Snce the calculated F s greater than the crtcal F, we reject the null hypothess that β = β 4 = 0. The F value can be calculated from ( SSER SSEU ) ( 637.65 6.34) F = = = 08.4 SSE 4 6.34 4 ( ) The average cost functon s gven by U ( ) C e =β +β +β Q +β Q + Q Q Q t t 3 t 4 t t t t Thus, f β =β 4 = 0, average cost s a lnear functon of output. The average cost functon s an approprate transformed model for estmaton when heteroskedastcty s of the form var ( e t) = Q t.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 96 EXERCISE 8.4 (a) (b) The least squares estmated equatons are 3 C = 7.774 + 83.659Q 3.796Q +.9Q = 34.85 (se) 3.655 4.597 0.7 SSE = 7796.49 ( ) ( ) ( ) 3 C = 5.85 + 08.9Q 0.05Q +.63Q = 847.66 (se) 8.933 6.56 0.380 SSE = 0343.83 ( ) ( ) ( ) To see whether the estmated coeffcents have the expected sgns consder the margnal cost functon dc MC = =β + β 3Q+ 3β 4Q dq We expect MC > 0 when Q = 0; thus, we expect β > 0. Also, we expect the quadratc MC functon to have a mnmum, for whch we requre β 4 > 0. The slope of the MC functon s dmc ( ) dq= β 3 + 6β 4Q. For ths slope to be negatve for small Q (decreasng MC), and postve for large Q (ncreasng MC), we requre β 3 < 0. Both our least-squares estmated equatons have these expected sgns. Furthermore, the standard errors of all the coeffcents except the constants are qute small ndcatng relable estmates. Comparng the two estmated equatons, we see that the estmated coeffcents and ther standard errors are of smlar magntudes, but the estmated error varances are qute dfferent. Testng H0: = aganst H: s a two-tal test. The crtcal values for performng a two-tal test at the 0% sgnfcance level are F (0.05,4,4) = 0.0504 and F (0.95,4,4) =.984. The value of the F statstc s 847.66 F = = =.6 34.85 Snce F F(0.95,4,4) >, we reject H 0 and conclude that the data do not support the proposton that =. (c) Snce the test outcome n (b) suggests, but we are assumng both frms have the same coeffcents, we apply generalzed least squares to the combned set of data, wth the observatons transformed usng and. The estmated equaton s C = 67.70 + 89.90Q 5.408Q +.306Q (se) 6.973 3.45 0.065 3 ( ) ( ) ( ) Remark: Some automatc software commands wll produce slghtly dfferent results f the transformed error varance s restrcted to be unty or f the varables are transformed usng varance estmates from a pooled regresson nstead of those from part (a).

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 97 Exercse 8.4 (contnued) (d) Although we have establshed that H0 : β =δ, β =δ, β 3 =δ3, β 4 =δ 4 under the assumpton that, t s nstructve to frst carry out the test for =, and then under the assumpton that. Assumng that =, the test s equvalent to the Chow test dscussed on pages 79-8 of POE. The test statstc s ( SSER SSEU ) J F = SSEU ( N K ) where SSE U s the sum of squared errors from the full dummy varable model. The dummy varable model does not have to be estmated, however. We can also calculate SSE U as the sum of the SSE from separate least squares estmaton of each equaton. In ths case SSE = SSE + SSE = 7796.49 + 0343.83 = 840.3 U The restrcted model has not yet been estmated under the assumpton that =. Dong so by combnng all 56 observatons yelds SSE R = 8874.34. The F-value s gven by ( SSER SSEU ) J (8874.34 840.3) 4 F = = = 0.33 SSE N K 840.3 (56 8) The correspondng U ( ) χ -value s χ = 4 F =.5. These values are both much less than ther respectve 5% crtcal values F (0.95,4,48) =.565 and χ (0.95,4) = 9.488. There s no evdence to suggest that the frms have dfferent coeffcents. In the formula for F, note that the number of observatons N s the total number from both frms, and K s the number of coeffcents from both frms. The above test s not vald n the presence of heteroskedastcty. It could gve msleadng results. To perform the test under the assumpton that, we follow the same steps, but we use values for SSE computed from transformed resduals. For restrcted estmaton from part (c) the result s SSE R = 49.4. For unrestrcted estmaton, we have the nterestng result Thus, SSE SSE ( N K ) ( N K ) SSE = + = + = N K + N K = 48 * U (49.4 48) 4 F = = 0.303 and 48 48 χ =.4 The same concluson s reached. There s no evdence to suggest that the frms have dfferent coeffcents.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 98 EXERCISE 8.5 (a) To estmate the two varances usng the varance model specfed, we frst estmate the equaton WAGE =β +β EDUC +β EXPER +β METRO + e 3 4 From ths equaton we use the squared resduals to estmate the equaton e =α +α METRO + v ln( ) The estmated parameters from ths regresson are α =.508448 and α = 0.33804. Usng these estmates, we have METRO = 0 R = exp(.508448 + 0.33804 0) = 4.597 METRO =, M = exp(.508448 + 0.33804 ) = 6.33759 These error varance estmates are much smaller than those obtaned from separate subsamples ( M = 3.84 and R = 5.43 ). One reason s the bas factor from the exponental functon see page 06 of POE. Multplyng M = 6.3375 and R = 4.597 by the bas factor exp(.704) yelds M =.576 and R = 6.00. These values are closer, but stll dfferent from those obtaned usng separate sub-samples. The dfferences occur because the resduals from the combned model are dfferent from those from the separate sub-samples. (b) (c) To use generalzed least squares, we use the estmated varances above to transform the model n the same way as n (8.3). After dong so the regresson results are, wth standard errors n parentheses WAGE = 9.705 +.85EDUC + 0.38EDUC +.530METRO (se).0485 0.0694 0.050 0.3858 ( ) ( ) ( ) ( ) The magntudes of these estmates and ther standard errors are almost dentcal to those n equaton (8.33). Thus, although the varance estmates can be senstve to the estmaton technque, the resultng generalzed least squares estmates of the mean functon are much less senstve. The regresson output usng Whte standard errors s WAGE = 9.940 +.340EDUC + 0.33EDUC +.54METRO (se).4 0.0835 0.058 0.3445 ( ) ( ) ( ) ( ) Wth the excepton of that for METRO, these standard errors are larger than those n part (b), reflectng the lower precson of least squares estmaton.

Chapter 8, Exercse Solutons, Prncples of Econometrcs, 3e 99 EXERCISE 8.6 (a) Separate least squares estmaton gves the error varance estmates -4 and = 5.363 0. A =.8995 0 4 G (b) (c) The crtcal values for testng the hypothess H : 0 G = A aganst the alternatve H : G A at a 5% level of sgnfcance are F (0.05,5,5) = 0.349 and F (0.975,5,5) =.86. The value of the F-statstc s -4 A 5.363 0 F = = = 5.98-4 G.8995 0 Snce 5.98 >.86, we reject the null hypothess and conclude that the error varances of the two countres, Austra and Germany, are not the same. The estmates of the coeffcents usng generalzed least squares are estmated coeffcent standard error β [const].068 0.4005 β [ln(inc)] 0.4466 0.838 β [ln(price)] 0.954 0.6 3 β [ln(cars)] 0.039 0.38 4 (d) Testng the null hypothess that demand s prce nelastc,.e., H0 : β3 aganst the alternatve H: β 3 <, s a one-tal t test. The value of our test statstc s 0.954 ( ) t = = 5.58 0.6 The crtcal t value for a one-tal test and 34 degrees of freedom s t (0.05,34) =.69. Snce 5.58 >.69, we do not reject the null hypothess and conclude that there s not enough evdence to suggest that demand s elastc.