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.)
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
# Compare two variables speed and dist
plot(cars$speed, cars$dist, xlab='Speed (mph)', ylab='Stopping Distance (ft)', main='Stopping Distance vs. Speed',col = c("red", "blue"))
cars_linearmodel<- lm(cars$dist ~ cars$speed)
cars_linearmodel
##
## Call:
## lm(formula = cars$dist ~ cars$speed)
##
## Coefficients:
## (Intercept) cars$speed
## -17.579 3.932
plot(cars$speed, cars$dist, xlab='Speed (mph)', ylab='Stopping Distance (ft)',
main='Stopping Distance vs. Speed', col = "red")
abline(cars_linearmodel, col="yellow")
summary(cars_linearmodel)
##
## Call:
## lm(formula = cars$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 residuals distribution suggests that the distribution is normal.
The standard error for the speed coefficient is ~ 9.4 (3.93/.42) times the coefficient value, which is good. From the book it says “For a good model, we typically would like to see a standard error that is at least five to ten times smaller than the corresponding coefficient”.
The probability that the speed coefficient is not relevant in the model is 1.49e-12 (p-value), which means that speed is very relevant in modeling stopping distiance.
The p-value of the intercept is 0.0123, which means the intercept is pretty relevant in the model.
The multiple R-squared is 0.6511, which means that this model explains 65.11% of the data’s variation.
plot(cars_linearmodel$fitted.values, cars_linearmodel$residuals, xlab='Fitted Values', ylab='Residuals',col = "green")
abline(0,0, col="yellow4")
qqnorm(resid(cars_linearmodel))
qqline(resid(cars_linearmodel))