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4.5* Download from the website the OECD data set on employment growth rates and GDP growth rates tabulated in Exercise 1.1, plot a scatter diagram and investigate whether a nonlinear specification might be superior to a linear one. 1 EXERCISE 4.5

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2 The figure shows the employment growth rates and GDP growth rates for the sample of 25 OECD countries. EXERCISE 4.5

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. reg e g Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 1, 23) = 33.22 Model | 14.2762167 1 14.2762167 Prob > F = 0.0000 Residual | 9.88359869 23.429721682 R-squared = 0.5909 ---------+------------------------------ Adj R-squared = 0.5731 Total | 24.1598154 24 1.00665898 Root MSE =.65553 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g |.4846863.0840907 5.764 0.000.3107315.6586411 _cons | -.5208643.2707298 -1.924 0.067 -1.080912.039183 ------------------------------------------------------------------------------ 3 A linear regression produces quite plausible results. The slope coefficient indicates that e increases by 0.48 percent for each percent increase in g. This makes sense because part of the variation in g is due to variations in the growth of efficiency. EXERCISE 4.5

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. reg e g Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 1, 23) = 33.22 Model | 14.2762167 1 14.2762167 Prob > F = 0.0000 Residual | 9.88359869 23.429721682 R-squared = 0.5909 ---------+------------------------------ Adj R-squared = 0.5731 Total | 24.1598154 24 1.00665898 Root MSE =.65553 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g |.4846863.0840907 5.764 0.000.3107315.6586411 _cons | -.5208643.2707298 -1.924 0.067 -1.080912.039183 ------------------------------------------------------------------------------ 4 The intercept indicates that if there were no output growth, employment would shrink at the rate of 0.52 percent per year, again as a consequence of increasing productivity. EXERCISE 4.5

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. reg e g Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 1, 23) = 33.22 Model | 14.2762167 1 14.2762167 Prob > F = 0.0000 Residual | 9.88359869 23.429721682 R-squared = 0.5909 ---------+------------------------------ Adj R-squared = 0.5731 Total | 24.1598154 24 1.00665898 Root MSE =.65553 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g |.4846863.0840907 5.764 0.000.3107315.6586411 _cons | -.5208643.2707298 -1.924 0.067 -1.080912.039183 ------------------------------------------------------------------------------ 5 R 2 is quite high for a simple regression using cross-section data. EXERCISE 4.5

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6 However, it is evident that the relationship is nonlinear. A double-logarithmic specification is ruled out by the negative values of e in the sample and the fact that the relationship clearly does not pass through the origin. EXERCISE 4.5

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. g gsq = g*g. reg e g gsq Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 2, 22) = 20.00 Model | 15.58633 2 7.793165 Prob > F = 0.0000 Residual | 8.57348541 22.389703882 R-squared = 0.6451 ---------+------------------------------ Adj R-squared = 0.6129 Total | 24.1598154 24 1.00665898 Root MSE =.62426 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g | 1.171194.3828871 3.059 0.006.3771345 1.965253 gsq | -.0810133.0441844 -1.834 0.080 -.1726461.0106196 _cons | -1.614902.6500016 -2.484 0.021 -2.962922 -.2668809 ------------------------------------------------------------------------------ 7 The quadratic form is an alternative that sometimes fits simple curves well. R 2 rises from 0.59 to 0.65. EXERCISE 4.5

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8 Visually the quadratic specification is a large improvement on the linear one. The most obvious defect is that it becomes downward-sloping for high values of g. EXERCISE 4.5

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. g gsq = g*g. reg e g gsq Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 2, 22) = 20.00 Model | 15.58633 2 7.793165 Prob > F = 0.0000 Residual | 8.57348541 22.389703882 R-squared = 0.6451 ---------+------------------------------ Adj R-squared = 0.6129 Total | 24.1598154 24 1.00665898 Root MSE =.62426 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g | 1.171194.3828871 3.059 0.006.3771345 1.965253 gsq | -.0810133.0441844 -1.834 0.080 -.1726461.0106196 _cons | -1.614902.6500016 -2.484 0.021 -2.962922 -.2668809 ------------------------------------------------------------------------------ 9 Note that the t statistic for the squared term is rather low. It is only just significant at the 5 percent level, using a one-sided test. (A one-sided test is justified here because we would not expect the sensitivity of e to g to increase with g.) EXERCISE 4.5

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. g gsq = g*g. reg e g gsq Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 2, 22) = 20.00 Model | 15.58633 2 7.793165 Prob > F = 0.0000 Residual | 8.57348541 22.389703882 R-squared = 0.6451 ---------+------------------------------ Adj R-squared = 0.6129 Total | 24.1598154 24 1.00665898 Root MSE =.62426 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- g | 1.171194.3828871 3.059 0.006.3771345 1.965253 gsq | -.0810133.0441844 -1.834 0.080 -.1726461.0106196 _cons | -1.614902.6500016 -2.484 0.021 -2.962922 -.2668809 ------------------------------------------------------------------------------ 10. cor g gsq (obs=26) | gdp gdpsq --------+------------------ g| 1.0000 gsq| 0.9775 1.0000 However, a low t statistic is not surprising, given that g and gsq are highly correlated. EXERCISE 4.5

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. g lgg=ln(g). reg e lgg Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 1, 23) = 38.44 Model | 15.116284 1 15.116284 Prob > F = 0.0000 Residual | 9.04353146 23.39319702 R-squared = 0.6257 ---------+------------------------------ Adj R-squared = 0.6094 Total | 24.1598154 24 1.00665898 Root MSE =.62705 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- lgg | 1.709096.2756444 6.200 0.000 1.138883 2.27931 _cons | -.728669.2830097 -2.575 0.017 -1.314119 -.1432188 ------------------------------------------------------------------------------ 11 The semilogarithmic specification, with the explanatory variable logarithmic, is another possible specification. EXERCISE 4.5

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12 Comparing it with the quadratic specification, it does have the advantage of not being downward-sloping for high values of g, but it exhibits excessive sensitivity to g for low values. R 2 is about the same as for the quadratic specification. EXERCISE 4.5

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. g Z=1/g. reg e Z Source | SS df MS Number of obs = 25 ---------+------------------------------ F( 1, 23) = 25.82 Model | 12.7770171 1 12.7770171 Prob > F = 0.0000 Residual | 11.3827983 23.494904274 R-squared = 0.5289 ---------+------------------------------ Adj R-squared = 0.5084 Total | 24.1598154 24 1.00665898 Root MSE =.70349 ------------------------------------------------------------------------------ e | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- Z | -3.997462.7867382 -5.081 0.000 -5.624954 -2.36997 _cons | 2.592225.3716506 6.975 0.000 1.823407 3.361043 ------------------------------------------------------------------------------ 13 A further possibility is a hyperbolic function, with e being regressed on the reciprocal of g. EXERCISE 4.5

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14 However R 2 is lower than in the previous specifications. e is systematically underestimated for high values of g and the specification exhibits excessive sensitivity to g for low values. EXERCISE 4.5

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15 One reason for the poor performance of the hyperbolic function is that it is constrained to be asymptotically tangential to the vertical axis. EXERCISE 4.5

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============================================================ Dependent Variable: e Method: Least Squares Sample: 1 25 Included observations: 25 Convergence achieved after 18 iterations e=C(1)+C(2)/(g+C(3)) ============================================================ CoefficientStd. Errort-Statistic Prob. ============================================================ C(1) 5.467457 2.826812 1.934142 0.0661 C(2) -31.07502 41.79895 -0.743440 0.4651 C(3) 4.148426 4.871805 0.851517 0.4037 ============================================================ R-squared 0.635231 Mean dependent var 0.833600 Adjusted R-squared 0.602070 S.D. dependent var 1.014519 S.E. of regression 0.639975 Akaike info criteri2.057393 Sum squared resid 9.010510 Schwarz criterion 2.203658 Log likelihood -22.71741 F-statistic 19.15609 Durbin-Watson stat 0.770997 Prob(F-statistic) 0.000015 ============================================================ 16 If we relax that assumption, we obtain a much better fit. Note that there is no way of linearizing this nonlinear specification. We have to use nonlinear least squares. EXERCISE 4.5

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============================================================ Dependent Variable: e Method: Least Squares Sample: 1 25 Included observations: 25 Convergence achieved after 18 iterations e=C(1)+C(2)/(g+C(3)) ============================================================ CoefficientStd. Errort-Statistic Prob. ============================================================ C(1) 5.467457 2.826812 1.934142 0.0661 C(2) -31.07502 41.79895 -0.743440 0.4651 C(3) 4.148426 4.871805 0.851517 0.4037 ============================================================ R-squared 0.635231 Mean dependent var 0.833600 Adjusted R-squared 0.602070 S.D. dependent var 1.014519 S.E. of regression 0.639975 Akaike info criteri2.057393 Sum squared resid 9.010510 Schwarz criterion 2.203658 Log likelihood -22.71741 F-statistic 19.15609 Durbin-Watson stat 0.770997 Prob(F-statistic) 0.000015 ============================================================ 17 EViews was used instead of Stata for this purpose because its nonlinear least squares facility is much easier to use. To fit a nonlinear least squares specification in EViews, you write it as an explicit mathematical equation, with C(1), C(2), etc being the parameters. EXERCISE 4.5

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18 This specification is compared with the semilogarithmic one. There is very little to choose between them, and indeed the quadratic specification is virtually as good, at least for the data range in the sample. EXERCISE 4.5

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Copyright Christopher Dougherty 2001–2006. This slideshow may be freely copied for personal use. 22.06.06

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