[ECON2206_UNSW]Econometrics_Assignment2_problem1
Jay Kang
2019 8 19
ECON2206 Assignment2 2018 2nd Sem
(Problem1) * Originally, STATA assignment * Adpated to R
The file beauty.dta collects information on wages hourly wage, years of experience and gender for 1; 260 individuals. We also have information on a subjective index of beauty (looks) measured from 1 to 5.
\[ lwage = \beta_0 + exper + exper^2 + \beta_3 looks + \beta_4female + u \ (1) \]
- Required packages
library(tidyverse) #data wrangling
library(foreign) #read.dta()
library(modelr) #modelling
library(car) #linearHypothesis()
library(MASS) #robust regression
- Estimate the parameter of the econometric model in equation (1). What is the estimated effect of experience on wages?
Importing STATA data file(.dta) into R
1) open STATA
2) open the corresponding data file(.dta)
3) type “saveold (file_name).dta, version(12)”
4) import the saved file into R using read.dta function
beauty <- read.dta("C:/Users/Kang/Desktop/UNSW/S2/ECON2206/STATA and Assignments/beauty(1).dta")
colnames(beauty)## [1] "wage" "lwage" "belavg" "abvavg" "exper" "looks"
## [7] "union" "goodhlth" "black" "female" "married" "south"
## [13] "bigcity" "smllcity" "service" "expersq" "educ"beauty %>% head()## wage lwage belavg abvavg exper looks union goodhlth black female
## 1 5.73 1.745715 0 1 30 4 0 1 0 1
## 2 4.28 1.453953 0 0 28 3 0 1 0 1
## 3 7.96 2.074429 0 1 35 4 0 1 0 1
## 4 11.57 2.448416 0 0 38 3 0 1 0 0
## 5 11.42 2.435366 0 0 27 3 0 1 0 0
## 6 3.91 1.363537 0 0 20 3 0 0 0 1
## married south bigcity smllcity service expersq educ
## 1 1 0 0 1 1 900 14
## 2 1 1 0 1 0 784 12
## 3 0 0 0 1 0 1225 10
## 4 1 0 1 0 1 1444 16
## 5 1 0 0 1 0 729 16
## 6 1 0 1 0 0 400 12A. estimated parameters
mod1 <- lm(lwage ~ exper + expersq + looks + female, data = beauty)
mod1##
## Call:
## lm(formula = lwage ~ exper + expersq + looks + female, data = beauty)
##
## Coefficients:
## (Intercept) exper expersq looks female
## 1.0783855 0.0462051 -0.0008054 0.0887511 -0.4664837B. estimated effect of experience on avg
impact_exper <- function(current_exper) {
coef(mod1)[2] + 2*coef(mod1)[3] * current_exper #coef(mod1)[2]: exper regressor
} #coef(mod1)[3]: expersq regressor
impact_exper(5) #ex) estimated effect on log(wage) of one more yr of exper for a person with 5 yrs of exper## exper
## 0.03815152
- What can we say about the effect of an increase of one point in the looks scale on wages?
coef(mod1)[4]## looks
## 0.08875113
- Is there any evidence of functional form misspecification for equation (1)?
\[ \text{RESET test} \\ \begin{align} &lwage = \gamma_1exper+\gamma_2exper^2 + \gamma_3looks + \gamma_4female + \gamma_5fitted^2 + \gamma_6fitted^3 \ \text{(unrestricted model)} \\ &lwage = \beta_0 + exper + exper^2 + \beta_3 looks + \beta_4female + u \ (1) \ \text{(restricted model)} \\ \end{align} \\ \] \[ \begin{align} &H_0: \gamma_5 = \gamma_6 = 0 \ &\text{(no misspecification)}\\ &H_1: \text{~} H_0 \ &\text{(misspecifcation)} \end{align} \]
1) create fitted^2, fitted^3 values (column)
beauty$fitted1 <- predict(mod1)
beauty$fitted2 <- predict(mod1)^2
beauty$fitted3 <- predict(mod1)^32) run the unrestricted model
mod1_1 <- lm(lwage ~ exper + expersq + looks + female + fitted2 + fitted3, data = beauty)3) joint F-test
linearHypothesis(mod1_1, c("fitted2=0", "fitted3=0")) #H0: cannot be rejected## Linear hypothesis test
##
## Hypothesis:
## fitted2 = 0
## fitted3 = 0
##
## Model 1: restricted model
## Model 2: lwage ~ exper + expersq + looks + female + fitted2 + fitted3
##
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 1255 321.21
## 2 1253 320.39 2 0.81863 1.6008 0.2022 #No significant misspecification
- How would you test the hypothesis that beauty has the same effect on wages both for males and females? Provide a p-value for such test.
1) create an interaction term * between female and looks
* if this term is significant, different effects of looks on each different gender confirmed
beauty <-
beauty %>%
mutate(femlooks = female*looks)2) model estimates + hypothesis testing
mod3 <- lm(lwage ~ exper + expersq + looks + female + femlooks, data = beauty)
summary(mod3) ##
## Call:
## lm(formula = lwage ~ exper + expersq + looks + female + femlooks,
## data = beauty)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9626 -0.3068 0.0111 0.3068 3.0112
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.1347492 0.0977813 11.605 < 2e-16 ***
## exper 0.0466609 0.0046682 9.995 < 2e-16 ***
## expersq -0.0008166 0.0001039 -7.862 8.12e-15 ***
## looks 0.0700875 0.0267244 2.623 0.00883 **
## female -0.6228310 0.1410797 -4.415 1.10e-05 ***
## femlooks 0.0489631 0.0431046 1.136 0.25621
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5059 on 1254 degrees of freedom
## Multiple R-squared: 0.2789, Adjusted R-squared: 0.276
## F-statistic: 96.99 on 5 and 1254 DF, p-value: < 2.2e-16#not significant (p-value: 25.6%)
- Is there any evidence of heteroskedasticity in equation (1)? Provide an heteroskedas- ticity robust estimate of the parameters in equation (1).
A. BP test
[ \[\begin{align} \text{Theoratical background}\\ &Var(u|X) = E(u^2|X) - [E(u|X)]^2 = E(u^2|X) \ \text{(zero conditional mean)} \\ \text{In case of homoskedasticity,} \\ &Var(u|X) = Var(u) = \sigma^2 \\ \text{and accordingly} \\ &E(u^2|X) = E(u^2) = \sigma^2 \\ \text{Therefore,} \\ &H_0: u^2\ \text{has no linear relationship with respect to x (homoskedasticity)} \\ &H_1: \text{~} H_0 \end{align}\] ]
beauty$resid <- resid(mod1)
beauty$resid2 <- resid(mod1)^2
hetero1 <- lm(resid2 ~ exper + expersq + looks + female, data = beauty)
summary(hetero1)##
## Call:
## lm(formula = resid2 ~ exper + expersq + looks + female, data = beauty)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.3082 -0.2241 -0.1507 0.0353 8.9580
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.218e-01 7.911e-02 1.539 0.12399
## exper 1.133e-02 4.367e-03 2.594 0.00961 **
## expersq -2.250e-04 9.709e-05 -2.317 0.02064 *
## looks 8.844e-03 1.979e-02 0.447 0.65505
## female 1.595e-02 2.907e-02 0.549 0.58338
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.475 on 1255 degrees of freedom
## Multiple R-squared: 0.005741, Adjusted R-squared: 0.002572
## F-statistic: 1.812 on 4 and 1255 DF, p-value: 0.1242#no heteroskedasticity by BP test (F-statistic)B. White test
\[ u^2 = \delta_0 + \delta_1\hat{y} + \delta_2\hat{y}^2 + \epsilon \\ H_0: \delta_1 = \delta_2 =0 \]
hetero2 <- lm(resid2 ~ fitted1 + fitted2, data = beauty)
summary(hetero2) ##
## Call:
## lm(formula = resid2 ~ fitted1 + fitted2, data = beauty)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.2994 -0.2261 -0.1586 0.0351 8.9843
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.4714 0.3467 1.360 0.174
## fitted1 -0.3549 0.4521 -0.785 0.433
## fitted2 0.1306 0.1430 0.913 0.361
##
## Residual standard error: 0.4755 on 1257 degrees of freedom
## Multiple R-squared: 0.002035, Adjusted R-squared: 0.000447
## F-statistic: 1.282 on 2 and 1257 DF, p-value: 0.278#no heteroskedasticity by White test (F-statistic)C. Robust Regression
mod4 <- rlm(lwage ~ exper + expersq + looks + female, data = beauty)
- Bonus: Residual Analysis
par(mfrow = c(2,2))
plot(mod1) #No significant problem observed
par(mfrow = c(2,2))
plot(mod4) #robust model