Question

Using the “cars” dataset in R, build a linear model for stopping distance as a function of speed and replicate the analysis of your textbook chapter 3 (visualization, quality evaluation of the model, and residual analysis.)

Solution

Load required libraries

library(tidyverse)

Take a glimpse at the dataset

glimpse(cars)
## Rows: 50
## Columns: 2
## $ speed <dbl> 4, 4, 7, 7, 8, 9, 10, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13…
## $ dist  <dbl> 2, 10, 4, 22, 16, 10, 18, 26, 34, 17, 28, 14, 20, 24, 28, 26, 34…

We can see that the dataset has 50 rows and 2 columns (speed and dist).

Summary for the dataset

summary(cars)
##      speed           dist       
##  Min.   : 4.0   Min.   :  2.00  
##  1st Qu.:12.0   1st Qu.: 26.00  
##  Median :15.0   Median : 36.00  
##  Mean   :15.4   Mean   : 42.98  
##  3rd Qu.:19.0   3rd Qu.: 56.00  
##  Max.   :25.0   Max.   :120.00

Visualize the data 

p = ggplot(cars, aes(x=speed, y=dist)) + geom_point() + theme(panel.grid.major = element_line(colour = "lemonchiffon3"),
    panel.grid.minor = element_line(colour = "lemonchiffon3"),
    axis.title = element_text(size = 13),
    axis.text = element_text(size = 11),
    axis.text.x = element_text(family = "sans",
        size = 11), axis.text.y = element_text(family = "sans",
        size = 11), plot.title = element_text(size = 15,
        hjust = 0.5), panel.background = element_rect(fill = "gray85"),
    plot.background = element_rect(fill = "antiquewhite")) +labs(title = "Stopping Distance vs Speed",
    x = "Speed", y = "Stopping Distance")
p

From the scatter plot, as the speed increases, the stopping distance tend to increase as well.

Build the Model 

lm_cars = lm(cars$dist ~ cars$speed)
lm_cars
## 
## Call:
## lm(formula = cars$dist ~ cars$speed)
## 
## Coefficients:
## (Intercept)   cars$speed  
##     -17.579        3.932

From the output of the model, the linear function is given by:
\(dist = 3.932 * speed - 17.579\)

Evaluate the Model - Quality Evaluation

# Check the summary of the model
summary(lm_cars)
## 
## Call:
## lm(formula = cars$dist ~ cars$speed)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -29.069  -9.525  -2.272   9.215  43.201 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -17.5791     6.7584  -2.601   0.0123 *  
## cars$speed    3.9324     0.4155   9.464 1.49e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 15.38 on 48 degrees of freedom
## Multiple R-squared:  0.6511, Adjusted R-squared:  0.6438 
## F-statistic: 89.57 on 1 and 48 DF,  p-value: 1.49e-12

The values from the summary show that there is little variability in the data and the result is statistically significant.

Residual Analysis
Residual vs Fitted plot

plot(fitted(lm_cars),resid(lm_cars), main="Residuals vs Fitted", xlab = "Fitted", ylab = "Residuals")
abline(0, 0)

From this plot, we can say that there seems to be constant variance for the residuals and the data is not heteroscedacity.

QQ Plot

qqnorm(resid(lm_cars))
qqline(resid(lm_cars))

From the Q-Q plot, we can see that the residuals follow a nearly normal distribution.

Entire plot

par(mfrow=c(2,2))
plot(lm_cars)

From the residual analysis, we can see that the conditions for linear regression are satisfied for this data and it can be fitted by using a linear model approach. Although I am not satisfied with the R-squared value of about 0.65, acceptable values for the R-squared value is highly dependent on the use case of the model.