8/18/2019 Example LPM Handout 2016

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Example: LPM (percent correctly predicted) T. Monsod

Use the data set workwomen_practice.dta.

1.  What fraction of the women in the sample participated in the labor force in 1975?

2. 

Estimate a linear probability model explaining labor force participation in 1975 interms of husband’s earnings (non-wife income in thousands or nwifeinc); years ofeducation, years of past labor market experience, age, number of children less than 6years of age (kidslt6 ), number of kids between 6-18 (kidsge6 )

3.  Obtain the fitted values from the LPM estimated in (2). Are any fitted values negativeor greater than 1?

4.  Using the fitted values, define a new variable inlf_p =1 if the fitted value >=.5 andinlf_p=0 if the fitted value is F = 0.0000Residual | 135.919698 745 .182442547 R-squared = 0.2642

-------------+------------------------------ Adj R-squared = 0.2573Total | 184.727756 752 .245648611 Root MSE = .42713

------------------------------------------------------------------------------inlf | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+----------------------------------------------------------------nwifeinc | -.0034052 .0014485 -2.35 0.019 -.0062488 -.0005616

age | -.0160908 .0024847 -6.48 0.000 -.0209686 -.011213educ | .0379953 .007376 5.15 0.000 .023515 .0524756exper | .0394924 .0056727 6.96 0.000 .0283561 .0506287exper2 | -.0005963 .0001848 -3.23 0.001 -.0009591 -.0002335kidslt6 | -.2618105 .0335058 -7.81 0.000 -.3275875 -.1960335kidsge6 | .0130122 .013196 0.99 0.324 -.0128935 .0389179_cons | .5855192 .154178 3.80 0.000 .2828442 .8881943

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. predict y_hat

(option xb assumed; fitted values)

. sum y_hat

Variable | Obs Mean Std. Dev. Min Max-------------+--------------------------------------------------------

y_hat | 753 .5683931 .2547633 -.3451103 1.127151

8/18/2019 Example LPM Handout 2016

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. gen inlf_p=1 if y_hat>=.5(281 missing values generated)

. replace inlf_p=0 if y_hat