You work for Motor Trend, a magazine about the automobile industry. Looking at a data set of a collection of cars, they are interested in exploring the relationship between a set of variables and miles per gallon (MPG) (outcome). They are particularly interested in the following two questions:
- Is an automatic or manual transmission better for MPG?
- Quantify the MPG difference between automatic and manual transmissions
Here we are loading the dataset.
data(mtcars)Plotting the MPG change to transmission modes.
boxplot(mpg ~ am, data = mtcars,xlab = "Type of transmission", ylab = "MPG count",
main = "mtcars data", varwidth = TRUE, col = "lightgray")
The basic view of our dataframe.
head(mtcars)## mpg cyl disp hp drat wt qsec vs am gear carb
## Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
## Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
## Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
## Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
## Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
## Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
Converting the specific variables from numeric as they are factor variables.
mtcars$cyl<-factor(mtcars$cyl)
mtcars$vs<-factor(mtcars$vs)
mtcars$am<-factor(mtcars$am)
mtcars$gear<-factor(mtcars$gear)
mtcars$carb<-factor(mtcars$carb)We will make a linear model of mpg to all the other variables. And let the step() function choose the best model with essential predictor variables.
mdl<-lm(mpg~.,mtcars)
mdl_final<-step(mdl,direction="both")Let`s have a look at the best model.
summary(mdl_final)##
## Call:
## lm(formula = mpg ~ cyl + hp + wt + am, data = mtcars)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.9387 -1.2560 -0.4013 1.1253 5.0513
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 33.70832 2.60489 12.940 7.73e-13 ***
## cyl6 -3.03134 1.40728 -2.154 0.04068 *
## cyl8 -2.16368 2.28425 -0.947 0.35225
## hp -0.03211 0.01369 -2.345 0.02693 *
## wt -2.49683 0.88559 -2.819 0.00908 **
## am1 1.80921 1.39630 1.296 0.20646
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.41 on 26 degrees of freedom
## Multiple R-squared: 0.8659, Adjusted R-squared: 0.8401
## F-statistic: 33.57 on 5 and 26 DF, p-value: 1.506e-10
We can see that the best predictor variables to include in the model are:
1: am
2: cyl
3: hp
4: wt
Let's plot some residuals and diagnostics parameters to check the validity of the model.
par(mfrow = c(2, 2))
plot(mdl_final)
Here we are building a linear model with zero intercept to have a look the coefficients.
mdl_final<-lm(mpg~am+cyl+hp+wt-1,mtcars)
summary(mdl_final)##
## Call:
## lm(formula = mpg ~ am + cyl + hp + wt - 1, data = mtcars)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.9387 -1.2560 -0.4013 1.1253 5.0513
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## am0 33.70832 2.60489 12.940 7.73e-13 ***
## am1 35.51754 2.03171 17.482 6.81e-16 ***
## cyl6 -3.03134 1.40728 -2.154 0.04068 *
## cyl8 -2.16368 2.28425 -0.947 0.35225
## hp -0.03211 0.01369 -2.345 0.02693 *
## wt -2.49683 0.88559 -2.819 0.00908 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.41 on 26 degrees of freedom
## Multiple R-squared: 0.9892, Adjusted R-squared: 0.9868
## F-statistic: 398.6 on 6 and 26 DF, p-value: < 2.2e-16
mdl_final$coef[[2]]-mdl_final$coef[[1]]## [1] 1.809211
Hence we can conclude that manual transmission is better than automatic in terms of mpg.
Manual transmission is better than automatic by a factor of 1.8.