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.)

Load Data

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

Visualization of the data

plot(cars$speed, cars$dist, xlab='Speed in MPH', ylab='Stopping Distance in FT',main='Speed vs. Stopping Distance')

Linear Regression model

model <- lm(dist ~ speed, cars)
plot(cars$speed, cars$dist, xlab='Speed (mph)', ylab='Stopping Distance (ft)', 
     main='Stopping Distance vs. Speed')
abline(model)

Quality Evaluation of the model

summary(model)
## 
## Call:
## lm(formula = dist ~ speed, data = cars)
## 
## 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 *  
## 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
#Error: larger ratio indicates little variability in the slope estimate
err<-3.9324/0.4155
res_err<-15.38/9.215
res_err2<-15.38/-9.525

In a good fit model residuals will have a median value near to zero and Min/Max values roughly of the same magnitude. We can see 1st/3rd quartile values are roughly of same magnitude.

Standard error is smaller than corresponding coefficient: So we can conclude that Probability that speed is not relevant in this model

Normal distribution of 1st/3rd quartiles are about 1.5x the residual standard error

R^2: The model explains 65% of data’s variation.

Residual Analysis

plot(fitted(model), residuals(model), xlab="Fitted", ylab="Residuals")
abline(h=0);

The residuals look like they are uniformly scattered above/below zero.

#checking for normal distribution
qqnorm(model$residuals)
qqline(model$residuals)

The plot looks almost normal, except at the tails.So we may conclude model is an OK fit.