Question 1

Part (a)

To estimate the model with wage on the LHS and all other variable on the RHS, we can use either:

mod1 <- lm(wage ~ ., dat=school)

or

mod1 <- lm(wage ~ ed + exp + iqscore + black + sinmom14 + momed + daded, data=school)

and then:

summary(mod1)

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -452.8176    64.3296  -7.039 2.72e-12 ***
ed            47.6204     3.3721  14.122  < 2e-16 ***
exp           28.1799     1.8580  15.167  < 2e-16 ***
iqscore        1.2859     0.4651   2.765  0.00575 ** 
blackyes     -81.2574    18.3346  -4.432 9.89e-06 ***
sinmom14yes   -7.6014    20.6584  -0.368  0.71295    
momed          2.7606     2.4434   1.130  0.25870    
daded          2.3158     2.2235   1.042  0.29777    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 239.1 on 1836 degrees of freedom
Multiple R-squared:   0.19,	Adjusted R-squared:  0.1869 
F-statistic: 61.51 on 7 and 1836 DF,  p-value: < 2.2e-16

Part (b)

The three variables: sinmom14yes, momed, and daded all have small t-statistics which indicate they are all individually insignificant. However, we know that dropping them all from the model, all at once, requires that all 3 variables be jointly insignificant. To justify dropping these 3 variables from the model, we need to use an F-test and test the null hypothesis:

$H_0: \beta_{sinmom14} = 0 \text{ and } \beta_{momed} = 0 \text{ and } \beta_{daded} = 0$

We can do this by estimating the model under the null hypothesis (the “restricted” model, where we drop the 3 variables):

mod2 <- lm(wage ~ ed + exp + iqscore + black, data=school)

and then calculating the F-statistic and p-value using the anova() function:

anova(mod1, mod2)

Model 1: wage ~ ed + exp + iqscore + black + sinmom14 + momed + daded
Model 2: wage ~ ed + exp + iqscore + black
  Res.Df       RSS Df Sum of Sq     F Pr(>F)
1   1836 104992112                          
2   1839 105237952 -3   -245840 1.433 0.2313

With an F-stat of 1.433 and a p-value of 0.2313, we fail to reject the null hypothesis that all 3 variables are jointly insignificant. We can drop them from the model.

You don’t have to use the anova() function for this question, you could also calculate the F-stat “by hand” and compare it to the critical value of 2.60.

Question 2

Download the data:

diam <- read.csv("https://rtgodwin.com/data/diamond.csv")

(a)

mod1 <- lm(price ~ carat + I(carat^2) + colour + clarity, data = diam)
summary(mod1)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   2028.6      224.2   9.049  < 2e-16 ***
carat         4351.3      687.6   6.328 9.16e-10 ***
I(carat^2)    6455.1      522.5  12.353  < 2e-16 ***
colourE      -1261.1      169.3  -7.448 1.04e-12 ***
colourF      -1679.2      158.9 -10.566  < 2e-16 ***
colourG      -2046.8      162.8 -12.574  < 2e-16 ***
colourH      -2543.0      164.8 -15.434  < 2e-16 ***
colourI      -3225.5      173.0 -18.646  < 2e-16 ***
clarityVS1   -1168.6      121.0  -9.661  < 2e-16 ***
clarityVS2   -1561.8      131.5 -11.875  < 2e-16 ***
clarityVVS1   -287.1      130.3  -2.203   0.0284 *  
clarityVVS2   -836.9      120.8  -6.927 2.69e-11 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 577.3 on 296 degrees of freedom
Multiple R-squared:  0.9723,	Adjusted R-squared:  0.9712 
F-statistic:   943 on 11 and 296 DF,  p-value: < 2.2e-16

colour and clarity are categorical variables - R has automatically created a system of dummy variables for us.

(b)

The effect of a 0.1 increase in carat will have a non-constant effect on price (that is the appeal of the polynomial model). In a polynomial model, we can interpret the estimated results by considering specific scenarios. For example, let’s get the predicted increase in price due to an increase in carat of 0.1, when the diamond has a size of 0.1:

predict(mod1, data.frame(carat = 0.2, colour = "D", clarity = "VS2")) - predict(
  mod1, data.frame(carat = 0.1, colour = "D", clarity = "VS2"))

628.7867

This has taken the difference in the predicted price of a diamond of size carat = 0.2 and carat = 0.1. The predicted effect is $628.79.

When using the predict function, we must choose values for all of the $X$ variables. I arbitrarily chose colour = "D" and clarity = "VS2". As we will see in the next part, these choices do not matter for the predicted effect.

The whole point of the non-linear model is that the effect of carat on price is non-constant. To illustrate that you understand this, you need to get the predicted effect of an increase of carats of 0.1, for a different starting value. For example, I will compare the effect of a 0.1 increase for when carat = 0.5:

predict(mod1, data.frame(carat = 0.6, colour = "D", clarity = "VS2")) - predict(
  mod1, data.frame(carat = 0.5, colour = "D", clarity = "VS2"))

1145.196

The same increase in size of 0.1 has almost double the effect when the diamond is 0.5 carats vs. 0.1 carats.

Question 3

Part (a)

cps <- read.csv("http://rtgodwin.com/data/cps1985.csv")
mod <- lm(log(wage) ~ education + gender + age + experience, data = cps)
summary(mod)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.89621    0.69229   1.295    0.196    
education    0.17746    0.11371   1.561    0.119    
gendermale   0.25736    0.03948   6.519 1.66e-10 ***
age         -0.07961    0.11365  -0.700    0.484    
experience   0.09234    0.11375   0.812    0.417    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4525 on 529 degrees of freedom
Multiple R-squared:  0.2703,	Adjusted R-squared:  0.2648 
F-statistic: 48.99 on 4 and 529 DF,  p-value: < 2.2e-16

Part (b)

The estimated coefficient of 0.17746 is interpreted as: an increase in education of 1 year is associated with an increase in wages of 17.8% on average.