Question

tions in this section. Question 1 (20%) You regress the logarithm of monthly earnings (lwage) on years of education (educ) and both the mother's and father's education (meduc and feduc). You obtain the following R output: Call: lm(formula = lwage ~ educ + meduc + feduc) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 5.92864 0.09462 62.66 < 2e-16 * educ 0.04701 (-1-) 6.36 3.7e-10 * meduc 0.01280 0.00647 1.98 0.048 * feduc 0.00884 0.00569 (-2-) (-3-) Signif. codes: 0 * 0.001 0.01 * 0.05 . 0.1 1 Residual standard error: (-4-) on 718 degrees of freedom Multiple R-squared: (-5-), Adjusted R-squared: 0.109 F-statistic: (-6-) on 3 and (-7-) DF, p-value: <2e-16 Analysis of Variance Table Response: lwage Df Sum Sq Mean Sq F value Pr(>F) educ 1 12.4 12.39 79.10 <2e-16 * meduc 1 1.6 1.58 10.06 0.0016 feduc 1 0.4 0.38 2.42 0.1205 Residuals 718 112.5 0.16 a) Calculate the missing values (-1-) to (-7-). b) Derive a 90% confidence interval for the coefficient on meduc.

          tions in this section.
Question 1 (20%)
You regress the logarithm of monthly earnings (lwage) on years of education
(educ) and both the mother's and father's education (meduc and feduc). You
obtain the following R output:
Call:
lm(formula = lwage ~ educ + meduc + feduc)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.92864 0.09462 62.66 < 2e-16 ***
educ 0.04701 (-1-) 6.36 3.7e-10 ***
meduc 0.01280 0.00647 1.98 0.048 *
feduc 0.00884 0.00569 (-2-) (-3-)
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Residual standard error: (-4-) on 718 degrees of freedom
Multiple R-squared: (-5-), Adjusted R-squared: 0.109
F-statistic: (-6-) on 3 and (-7-) DF, p-value: <2e-16
Analysis of Variance Table
Response: lwage
Df Sum Sq Mean Sq F value Pr(>F)
educ 1 12.4 12.39 79.10 <2e-16 ***
meduc 1 1.6 1.58 10.06 0.0016 **
feduc 1 0.4 0.38 2.42 0.1205
Residuals 718 112.5 0.16
a) Calculate the missing values (-1-) to (-7-).
b) Derive a 90% confidence interval for the coefficient on meduc.

tions in this section.
Question 1 (20%)
You regress the logarithm of monthly earnings (lwage) on years of education
(educ) and both the mother's and father's education (meduc and feduc). You
obtain the following R output:
Call:
lm(formula = lwage educ + meduc + feduc)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.92864 0.09462 62.66 < 2e-16 ***
educ 0.04701 (-1-) 6.36 3.7e-10 ***
meduc 0.01280 0.00647 1.98 0.048 *
feduc 0.00884 0.00569 (-2-) (-3-)
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Residual standard error: (-4-) on 718 degrees of freedom
Multiple R-squared: (-5-), Adjusted R-squared: 0.109
F-statistic: (-6-) on 3 and (-7-) DF, p-value: <2e-16
Analysis of Variance Table
Response: lwage
Df Sum Sq Mean Sq F value Pr(>F)
educ 1 12.4 12.39 79.10 <2e-16 ***
meduc 1 1.6 1.58 10.06 0.0016 **
feduc 1 0.4 0.38 2.42 0.1205
Residuals 718 112.5 0.16
a) Calculate the missing values (-1-) to (-7-).
b) Derive a 90% confidence interval for the coefficient on meduc.

Added by Francisco J.

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