library(dplyr)
library(dslabs)
library(ISLR2)
library(matlib)
library(wooldridge)
data("twoyear")

4.

  1. For the CEV assumptions to hold, we must be able to write tvhours = tvhours* + e0, where the measurement error e0 has zero mean and is uncorrelated with tvhours* and each explanatory variable in the equation. (Note that for OLS to consistently estimate the parameters we do not need e0 to be uncorrelated with tvhours*.)
  2. The CEV assumptions are unlikely to hold in this example. For children who do not watch TV at all, tvhours* = 0, and it is very likely that reported TV hours is zero. So if tvhours* = 0 then e0 = 0 with high probability. If tvhours* > 0, the measurement error can be positive or negative, but, since tvhours ≥ 0, e0 must satisfy e0 ≥ −tvhours. So e0 and tvhours are likely to be correlated. As mentioned in part (i), because it is the dependent variable that is measured with error, what is important is that e0 is uncorrelated with the explanatory variables. But this is unlikely to be the case, because tvhours* depends directly on the explanatory variables. Or, we might argue directly that more highly educated parents tend to underreport how much television their children watch, which means e0 and the education variables are negatively correlated.

9.2

9.2 (i) We estimate the model from column (2) but with KWW in place of IQ. The coefficient on educ becomes about .058 (se≈ .006), so this is similar to the estimate obtained with IQ, although slightly larger and more precisely estimated. (ii) When KWW and IQ are both used as proxies, the coefficient on educ becomes about .049 (se ≈ .007). Compared with the estimate when only KWW is used as a proxy, the return to education has fallen by almost a full percentage point. (iii) The t statistic on IQ is about 3.08 while that on KWW is about 2.07, so each is significant at the 5% level against a two-sided alternative. They are jointly very significant, with F2,925≈ 8.59 and p-value≈ .0002.

9.8

attach(twoyear)
mean(stotal)
## [1] 0.04748291
sd(stotal)
## [1] 0.8535441
summary(stotal)
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -3.32480 -0.32734  0.00000  0.04748  0.61079  2.23537
  1. The mean of stotal is .047, its standard deviation is .854, the minimum value is –3.32, and the maximum value is 2.24.
model1 <- lm(stotal ~ jc +univ)
summary(model1)
## 
## Call:
## lm(formula = stotal ~ jc + univ)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.0298 -0.4457  0.1220  0.4522  2.4846 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.295005   0.013196 -22.356  < 2e-16 ***
## jc           0.074767   0.012170   6.143 8.53e-10 ***
## univ         0.164644   0.004091  40.246  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.7667 on 6760 degrees of freedom
## Multiple R-squared:  0.1934, Adjusted R-squared:  0.1932 
## F-statistic: 810.5 on 2 and 6760 DF,  p-value: < 2.2e-16
  1. In the regression jc on stotal, the slope coefficient is .07 (se = .013). Therefore, while the estimated relationship is positive, the t statistic is only one: the correlation between jc and stotal is weak at best. In the regression univ on stotal, the slope coefficient is 0.165 (se = .004), for a t statistic of 40.2. Therefore, univ and stotal are positively correlated.
  2. When we add stotal to (4.17) and estimate the resulting equation by OLS, we get log( nwage) = 1.495 + .0631 jc + .0686 univ + .00488 exper + .0494 stotal (.021) (.0068) (.0026) (.00016) (.0068) n = 6,758, R^2 = .228 For testing βjc = βuniv, we can use the same trick as in Section 4.4 to get the standard error of the difference: replace univ with totcoll = jc + univ, and then the coefficient on jc is the difference in the estimated returns, along with its standard error. Let θ1 = βjc − βuniv. Then ˆθ1 =− = .0055 (se .0069) . Compared with what we found without stotal, the evidence is even weaker against H1: βjc < βuniv. The t statistic from equation (4.27) is about –1.48, while here we have obtained only −.80.
  3. When stotal2 is added to the equation, its coefficient is .0019 (t statistic = .40). Therefore, there is no reason to add the quadratic term.
  4. The F statistic for testing exclusion of the interaction terms stotal⋅jc and stotal⋅univ is about 1.96; with 2 and 6,756 df, this gives p-value = .141. So, even at the 10% level, the interaction terms are jointly insignificant. It is probably not worth complicating the basic model estimated in part (iii).
  5. I would just use the model from part (iii), where stotal appears only in level form. The other embellishments were not statistically significant at small enough significance levels to warrant the additional complications.