Econometrics 1.2 Midterm

Midterm R code

Task 4

library(wooldridge)
data("wage1")

model <- lm(log(wage) ~ educ + exper + tenure, data = wage1)
summary(model)

Call:
lm(formula = log(wage) ~ educ + exper + tenure, data = wage1)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.05802 -0.29645 -0.03265  0.28788  1.42809 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 0.284360   0.104190   2.729  0.00656 ** 
educ        0.092029   0.007330  12.555  < 2e-16 ***
exper       0.004121   0.001723   2.391  0.01714 *  
tenure      0.022067   0.003094   7.133 3.29e-12 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.4409 on 522 degrees of freedom
Multiple R-squared:  0.316, Adjusted R-squared:  0.3121 
F-statistic: 80.39 on 3 and 522 DF,  p-value: < 2.2e-16

we can see from F-statistic part that there are 522 degrees of freedom for the test, but additionally we can use

dfresidual<- df.residual(model)
dfresidual

[1] 522

Task 6

From the previous code chunk we see that coefficient are all positive, so A is true. P-value for tenure is 3.29e-12 which is much smaller than 0.05 so effect is statistically significant, so B is true as well. The adjusted r-squared is 0.3121 which confirms C, so D is not true.

Task 8

Again, from the model we see that p-value for the exper coefficient is 0.01714 which is smaller than 0.05 but bigger than 0.01, so the correct answer is 5% significance level.

Task 10

new_model <- lm(log(wage) ~ educ, data = wage1)
summary(new_model)

Call:
lm(formula = log(wage) ~ educ, data = wage1)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.21158 -0.36393 -0.07263  0.29712  1.52339 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 0.583773   0.097336   5.998 3.74e-09 ***
educ        0.082744   0.007567  10.935  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.4801 on 524 degrees of freedom
Multiple R-squared:  0.1858,    Adjusted R-squared:  0.1843 
F-statistic: 119.6 on 1 and 524 DF,  p-value: < 2.2e-16

The F-statistic for the model is 119.6, with 1 degree of freedom for the predictor (educ) and 524 degrees of freedom for the residuals. The p-value for the F-statistic is less than 2.2e-16, which suggests that the model is statistically significant overall. So the answer is D.

Task 13

confint(model, level = 0.95)

                   2.5 %     97.5 %
(Intercept) 0.0796755842 0.48904353
educ        0.0776292137 0.10642876
exper       0.0007356983 0.00750652
tenure      0.0159896850 0.02814475

For the education ourput is between 0.078 and 0.106 (option A).