#Exercise 13, Ch08 (Wooldridge)
Use the data in FERTIL2 to answer these questions.
children = b0 +b1age + b2age^2 + b3educ +b4electric+ b5urban + u
children=number of living children age=age in years educ=years of education electric = 1 if has electricity urban=1 if live in urban area
and report the usual and heteroskedasticity-robust standard errors. Are the robust standard errors always bigger than the nonrobust ones?
library(wooldridge)
library(sandwich)
library(lmtest)
## Loading required package: zoo
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric
data=fertil2
Modelxd=lm(children~age+age^2+educ+electric+urban,data=fertil2)
summary(Modelxd)
##
## Call:
## lm(formula = children ~ age + age^2 + educ + electric + urban,
## data = fertil2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.6563 -0.6734 -0.0760 0.7540 7.0503
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.999655 0.096612 -20.698 < 2e-16 ***
## age 0.176652 0.002727 64.780 < 2e-16 ***
## educ -0.075611 0.006369 -11.872 < 2e-16 ***
## electric -0.303370 0.069795 -4.347 1.41e-05 ***
## urban -0.171306 0.046953 -3.648 0.000267 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.469 on 4353 degrees of freedom
## (3 observations deleted due to missingness)
## Multiple R-squared: 0.5634, Adjusted R-squared: 0.563
## F-statistic: 1404 on 4 and 4353 DF, p-value: < 2.2e-16
Algunos son más pequeños, que sus respectivos errores estandares
bptest(Modelxd)
##
## studentized Breusch-Pagan test
##
## data: Modelxd
## BP = 1041.5, df = 4, p-value < 2.2e-16
#Ho: No hay heteroskedasticity H1: Hay heteroskedasticity
El valor de p, es menor que el valor de alfa, con cl (95%), por ende, no hay evidencia suficiente, para rechazar ho, por lo tanto, hayheteroskedasticity
#Ho: no hay heteroskedasticity H1: Hay heteroskedasticity
usqr=(summary(Modelxd)$residual)^2
yhat=Modelxd$fitted.values
yhatsqr=Modelxd$fitted.values^2
ModelxdWT=summary(lm(usqr~yhat+yhatsqr,data=fertil2))$r.squared
ModelxdWT*4361
## [1] 1093.183
1-pchisq(1093.183,2)
## [1] 0
El valor de p.value, es menor que alfa, con un 95% de confianza, por lo tanto, no se puede rechazar HO, por ende, hay heteroskedasticity
No, ya que la diferencia, de los datos, es sumamante pequeña, por ende, no existe una importancia en el metodo que uses, por ende, puede esciger el que se te ha más comodo