Examining the Difference of Manual and Automatic Cars : Miles per Gallon

Excutive Summary

This report investigated the difference of manual and automatic transmission cars by looking at their miles/gallon measures. Other variables are also investigated to find possible interactions and confounding effects. The data used is the mtcars taken from the package “datasets”.

Load Libraries

I used a number of libraries to help me with the visualizations and testing.

library(ggplot2)
library(datasets)
library(gridExtra)
library(corrplot)
library(car)
library(lmtest)
library(caret)

Exploratory Data Analysis

First, let us check the data structure.

dim(mtcars)
## [1] 32 11

The data has 32 observations with 11 variables.

Now let us convert some of the variables to factors.

mtcars[, c("cyl", "vs", "am", "gear", "carb")] <- lapply(mtcars[, c("cyl", "vs", "am", "gear", "carb")], as.factor)

Then let us examine the data by looking at the summary of am and mpg.

summary(mtcars$mpg)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   10.40   15.42   19.20   20.09   22.80   33.90
var(mtcars$mpg)
## [1] 36.3241
summary(mtcars$am)
##  0  1 
## 19 13

Looking at the mpg summary, the mean is 20.09 with variance of 36.3241. As for the variable am, there are 19 cars classified as automatic while 13 are manual.

To have an initial insight, it is best to visualize the data. We will focus first on mgp and am variables as this is our main concern on this problem. The result can be in Figure 1 below.

plot.1 <- ggplot(mtcars, aes(x = am, y = mpg, fill = am)) + geom_boxplot() + labs(title = "Figure 1")
plot.1 <- plot.1 + geom_point()
plot.1

Figure 1 shows the distribution of mpg broken down per transmission type. We can see that there is some sort of separation between 2 transmission types where manual transmission has higher median.

Another chart was created to examine the relationship of mpg to other categorical variables. The result is showed in the Appendix section as Figure 2. Looking at the figure, we can see that there is also some sort of separation between mpg vs. cyl, mpg vs. vs, and mpg vs. gear. As for mpg vs. carb, there is no clear separation to it.

We are done visualizing and examining mpg against the categorical variables. Next is we try to see if it also has some sort of relation with respect to the numerical variables. Let us see by looking at the correlation table below.

cor <- cor(mtcars[, c("mpg", "disp", "hp", "drat", "wt", "qsec")])
cor
##             mpg       disp         hp        drat         wt        qsec
## mpg   1.0000000 -0.8475514 -0.7761684  0.68117191 -0.8676594  0.41868403
## disp -0.8475514  1.0000000  0.7909486 -0.71021393  0.8879799 -0.43369788
## hp   -0.7761684  0.7909486  1.0000000 -0.44875912  0.6587479 -0.70822339
## drat  0.6811719 -0.7102139 -0.4487591  1.00000000 -0.7124406  0.09120476
## wt   -0.8676594  0.8879799  0.6587479 -0.71244065  1.0000000 -0.17471588
## qsec  0.4186840 -0.4336979 -0.7082234  0.09120476 -0.1747159  1.00000000

Looking at the correlation table, mpg has high negative correlation with disp, hp, and wt. On the other hand, drat has relatively high positive correlation with mpg, qsec meanwhile has the lowest correlation with mpg.

Inferential Analysis

Let us now examine the relationships by having an inferential analysis. The main objective is to see if there is an evidence of the difference of 2 types of transmission. We also want to quantify this effect by fitting a model.

We will use a t-test to determine if there is an evidence of difference in the mpg means for the 2 different types. Looking back at the graph, we have an assumption that automatic transmission has more mpg mean so we will have the hypothesis testing in this manner.
Ho : there is no difference of mpg means between 2 types of transmissions
Ha : automatic transmission has lesser mean than manual transmission

model.1 <- t.test(mpg ~ am, mtcars, alternative = "less")
model.1
## 
##  Welch Two Sample t-test
## 
## data:  mpg by am
## t = -3.7671, df = 18.332, p-value = 0.0006868
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##       -Inf -3.913256
## sample estimates:
## mean in group 0 mean in group 1 
##        17.14737        24.39231

The t-test result shows p-value = 0.00069 which very small. Hence we reject the null hypothesis and declare that there is a statistical difference between the 2 transmissions and that automatic transmission has lesser mpg mean than its manual counterpart. Ofcourse we dont want to stop the investigation at this point as we also wish to quantify the proven difference. We proceed by linear modelling.

model.1 <- lm(mpg ~ am, mtcars)
model.1$coefficients
## (Intercept)         am1 
##   17.147368    7.244939

The model shows the significance of the coefficient in am1. This means there is a 7.245 mpg difference between automatic transmission and manual transmission where manual is higher as per the previous result. However, we should also consider that there are other confounding variables that may also affect the this effect. Therefore, we will proceed on creating a model consisting of some of the variables that may also affect mpg.
We wish to build the best model by starting with our model.1 and nesting it by adding 1 variable at a time. The result is shown in the table below.

model.2 <- update(model.1, mpg ~ am + cyl, mtcars)
model.3 <- update(model.1, mpg ~ am + cyl + hp, mtcars)
model.4 <- update(model.1, mpg ~ am + cyl + hp + wt, mtcars)
anova(model.1, model.2, model.3, model.4)
## Analysis of Variance Table
## 
## Model 1: mpg ~ am
## Model 2: mpg ~ am + cyl
## Model 3: mpg ~ am + cyl + hp
## Model 4: mpg ~ am + cyl + hp + wt
##   Res.Df    RSS Df Sum of Sq      F    Pr(>F)    
## 1     30 720.90                                  
## 2     28 264.50  2    456.40 39.286 1.388e-08 ***
## 3     27 197.20  1     67.30 11.585  0.002164 ** 
## 4     26 151.03  1     46.17  7.949  0.009081 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

I stopped adding variables at model.4 as we don’t find any significant difference by adding the remaining variables. Let us examine model.4 then.

summary(model.4)
## 
## Call:
## lm(formula = mpg ~ am + cyl + hp + wt, 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 ***
## am1          1.80921    1.39630   1.296  0.20646    
## 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:  0.8659, Adjusted R-squared:  0.8401 
## F-statistic: 33.57 on 5 and 26 DF,  p-value: 1.506e-10
AIC(model.1, model.2, model.3, model.4)
##         df      AIC
## model.1  3 196.4844
## model.2  5 168.3989
## model.3  6 161.0033
## model.4  7 154.4669

The summary shows that the model explains 84.01% of the variance of mpg. The model also has the lowest Akaike Information Criterion (AIC) which means it is the best prediction model among the 4 considered.

The following is the residual plots to investigate the goodness of fit of model.4.

predicted <- predict.lm(model.4, mtcars[, -1])
resid <- model.4$residuals
resid.pred <- as.data.frame(cbind(resid, predicted))
plot.2 <- ggplot(resid.pred, aes(x = predicted, y = resid)) + geom_point() + geom_hline(yintercept = 0)
plot.2 <- plot.2 + labs(title = "Figure 3")
plot.2

The plots shows that the residuals are randomly scattered and distributed around 0 line. This is a good indication.

To make sure, the following diagnostic tests are also considered to see if the model is good enough.

dwtest(model.4)
## 
##  Durbin-Watson test
## 
## data:  model.4
## DW = 1.8088, p-value = 0.1567
## alternative hypothesis: true autocorrelation is greater than 0

The dwtest is the Durbin-Watson test which tests the existence of autocorrelation in the residuals. As observed, the p-value is greater than 0.05 which means there is no evidence that an autocorrelation exist.

Appendix

plot.cyl <- ggplot(mtcars, aes(x = mpg, fill = cyl)) + geom_histogram(bins = 32)
plot.vs <- ggplot(mtcars, aes(x = mpg, fill = vs)) + geom_histogram(bins = 32)
plot.gear <- ggplot(mtcars, aes(x = mpg, fill = gear)) + geom_histogram(bins = 32)
plot.carb <- ggplot(mtcars, aes(x = mpg, fill = carb)) + geom_histogram(bins = 32)
grid.arrange(plot.cyl, plot.vs, plot.gear, plot.carb, ncol = 2, top = "Figure 2")