When u is low in figure 3 less than zero lnwagerun is low below the solid line

# When u is low in figure 3 less than zero lnwagerun is

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Note the correspondence between figure three and figure four. When u is low in figure 3 (less than zero), lnwagerun is low (below the solid line) in figure four. When u is high (greater than zero) in figure three, lnwagerun is high (above the solid line) in figure four. reg lnwageran education Source | SS df MS Number of obs = 19 -------------+------------------------------ F( 1, 17) = 1.44 Model | 1.54899602 1 1.54899602 Prob > F = 0.2470 Residual | 18.3173689 17 1.07749229 R-squared = 0.0780 -------------+------------------------------ Adj R-squared = 0.0237 Total | 19.866365 18 1.10368694 Root MSE = 1.038 ------------------------------------------------------------------------------ lnwageran | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- education | .05213 .043478 1.20 0.247 -.0396006 .1438606 _cons | 4.906743 .4957256 9.90 0.000 3.860854 5.952633 The slope for this line is .05213, which implies that the return to education is 5.213% (see Wooldridge, p. 49, line 3. Now let’s introduce a new value for the random component: gen u2 = invnormal(uniform()) Note that the values of u and u2 are not equal u u2 Subscribe to view the full document.

.1952169 -.9470641 -1.23091 -.5522633 1.953314 .3275578 -.7849584 1.255616 -.6400359 -.036478 -.9740305 -.7476848 .7082468 1.534772 -.3281993 1.252978 -.3066868 1.081464 1.85042 -.8742716 .2078537 .3231856 .8249497 .6407216 -1.335853 .1043822 .9023688 -1.358528 -.6263756 -.2949743 -.1184899 -1.684372 -1.121178 .8746579 -.7535513 -.2219705 -1.787513 .0124836 Furthermore, they are not correlated: . correlate u u2 (obs=19) | u u2 -------------+------------------ u | 1.0000 u2 | -0.0847 1.0000 Generate a new variable that incorporates the values of u2: gen lnwageran2 = lnwage + u2 reg lnwageran2 education Source | SS df MS Number of obs = 19 -------------+------------------------------ F( 1, 17) = 3.90 Model | 3.4676034 1 3.4676034 Prob > F = 0.0648 Residual | 15.1145584 17 .889091669 R-squared = 0.1866 -------------+------------------------------ Adj R-squared = 0.1388 Total | 18.5821618 18 1.03234232 Root MSE = .94292 ------------------------------------------------------------------------------ lnwageran2 | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- education | .0779969 .0394944 1.97 0.065 -.0053291 .1613229 _cons | 4.861528 .4503058 10.80 0.000 3.911466 5.81159 Here we observe a point estimate of .0779969*100 = 7.7999% as the rate-of-return for an additional year of education. Note too that the confidence interval estimate for education extends from -.005 to .1613229. This range includes what we know to be the true value, .1. Therefore our confidence interval estimate for 1 includes the true value of 1 , .1. If we were to repeat this process 100 times, that is generate u3, u4…u100, and then run the regression where the dependent variable is the sum of the error + lnwage, I would expect that 95 of the 100 confidence interval estimates for 1 would include the true value, .1. • Spring '08
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