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”

0. Preprocessing

data(mtcars)
names(mtcars)
##  [1] "mpg"  "cyl"  "disp" "hp"   "drat" "wt"   "qsec" "vs"   "am"   "gear"
## [11] "carb"
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

1. Analysis

As we can see, there are 11 variables in the dataset. We are interested in the relationship between mpg and other variables, so first we check the correlation between mpg and other variables by using the cor() function

cor(mtcars$mpg,mtcars[,-1])
##            cyl       disp         hp      drat         wt     qsec        vs
## [1,] -0.852162 -0.8475514 -0.7761684 0.6811719 -0.8676594 0.418684 0.6640389
##             am      gear       carb
## [1,] 0.5998324 0.4802848 -0.5509251

From the correlation results, we can see, that the cyl, hp, wt and carb are negatively correlated with mpg.

2. Automatic or manual transmission? Checking at the help of the dataset (?mtcars), we can identify the transmission type as:

0 = automatic 1 = manual We need to make some data conversions.

mtcars$am <- as.factor(mtcars$am)
levels(mtcars$am) <-c("Automatic", "Manual")

A boxplot can be seen in the appendix, with the relationship between mpg and am type (Appendix 1). It seems like automatic car has better mpg compared with manual cars.

We perform a statistical analysis to support this hypothesis with a t-test.

t.test(mtcars$mpg~mtcars$am,conf.level=0.95)
## 
##  Welch Two Sample t-test
## 
## data:  mtcars$mpg by mtcars$am
## t = -3.7671, df = 18.332, p-value = 0.001374
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -11.280194  -3.209684
## sample estimates:
## mean in group Automatic    mean in group Manual 
##                17.14737                24.39231

The p-value is 0.001374, we may reject the null hypothesis and conclude, that automatic transmission cars have lower mpg compared with manual transmission cars - but this assumption is based on all other characteristics of automatic transmission cars and manual transmission cars are same (e.g: both have same weight distribution). This needs to be further explored in a multiple linear regression analysis.

3. Quantifying mpg difference Here we can adopt a stepwise algorithm, to choose the best model. We are using the step() function.

stepmodel = step(lm(data = mtcars, mpg ~ .),trace=0,steps=10000)
summary(stepmodel)
## 
## Call:
## lm(formula = mpg ~ wt + qsec + am, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.4811 -1.5555 -0.7257  1.4110  4.6610 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   9.6178     6.9596   1.382 0.177915    
## wt           -3.9165     0.7112  -5.507 6.95e-06 ***
## qsec          1.2259     0.2887   4.247 0.000216 ***
## amManual      2.9358     1.4109   2.081 0.046716 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.459 on 28 degrees of freedom
## Multiple R-squared:  0.8497, Adjusted R-squared:  0.8336 
## F-statistic: 52.75 on 3 and 28 DF,  p-value: 1.21e-11

At this point we have a model, which includes 3 variables

wt qsec am This model has a 0.85 of total variance. To further optimize the model, we can examine mpg ~ wt + qsec correlation with am.

model <- lm(mpg~ factor(am):wt + factor(am):qsec,data=mtcars)
summary(model)
## 
## Call:
## lm(formula = mpg ~ factor(am):wt + factor(am):qsec, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.9361 -1.4017 -0.1551  1.2695  3.8862 
## 
## Coefficients:
##                          Estimate Std. Error t value Pr(>|t|)    
## (Intercept)               13.9692     5.7756   2.419  0.02259 *  
## factor(am)Automatic:wt    -3.1759     0.6362  -4.992 3.11e-05 ***
## factor(am)Manual:wt       -6.0992     0.9685  -6.297 9.70e-07 ***
## factor(am)Automatic:qsec   0.8338     0.2602   3.205  0.00346 ** 
## factor(am)Manual:qsec      1.4464     0.2692   5.373 1.12e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.097 on 27 degrees of freedom
## Multiple R-squared:  0.8946, Adjusted R-squared:  0.879 
## F-statistic: 57.28 on 4 and 27 DF,  p-value: 8.424e-13

4. Summary Interpreting the results, we can see this model has a 89.5% total variance with an adjusted variance of 0.879. By adding the coefficients, we have the following conclusions:

when the weight increased by 1000 lbs, the mpg decreased by -3.176 for automatic transmission cars, and -6.09 for manual transmission cars so with increasing car weight we should choose manual transmission cars when the acceleration speed dropped, and 1/4 mile time increased (by 1 sec), the mpg factor increased by 0.834 miles for automatic transmission cars, and 1.446 miles for manual transmission cars so with lower acceleration speed, but same weight, manual transmission cars are better for mpg

Main conclusion The mpg is largely determined by the interplay between weight, acceleration and transmission. Given the above analysis, the original question (automatic transmission vs manual transmission) is not really answered, and should be considered in the context of weight and acceleration speed.

Appendix Appendix 1. Boxplot of mpg vs transmission type

boxplot(mtcars$mpg ~ mtcars$am, data = mtcars, outpch = 19, ylab="mpg:miles per 
gallon",xlab="transmission type",main="mpg vs transmission type", col="blue")

Appendix 2. Residual check and diagnostics plot

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

Appendix 3. Further plots

plot(model)