This is a regression analysis that tries to respond to the following 2 questions:
Based on mtcars small dataset analysis we can conclude:
Loading mtcars dataset produces:
data(mtcars)
str(mtcars)
## 'data.frame': 32 obs. of 11 variables:
## $ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
## $ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
## $ disp: num 160 160 108 258 360 ...
## $ hp : num 110 110 93 110 175 105 245 62 95 123 ...
## $ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
## $ wt : num 2.62 2.88 2.32 3.21 3.44 ...
## $ qsec: num 16.5 17 18.6 19.4 17 ...
## $ vs : num 0 0 1 1 0 1 0 1 1 1 ...
## $ am : num 1 1 1 0 0 0 0 0 0 0 ...
## $ gear: num 4 4 4 3 3 3 3 4 4 4 ...
## $ carb: num 4 4 1 1 2 1 4 2 2 4 ...
where features have the following meanings:
Let's convert am feature into a categorical one:
mtcars$am <- as.factor(mtcars$am)
levels(mtcars$am) <-c("AT", "MT")
So, am feature / predictor has 2 categories:
First, we need to check if mpg has a distribution close to the gaussian distribution so we can apply regression.
The mpg histogram is presented in Appendix.1 and in spite of the reduced observations in mtcars definitely it resembles a normal distribution.
Next, the means and ranges of mpg for AT versus MT are presented in Appendix.2 boxplot
Now we can address the first question:
“Is an automatic or manual transmission better for MPG ?”
To get exact values and confidence interval for fuel consumption by AT vs. MT vehicles, we split the dataset by AT vs. MT
and apply the t-test:
mpg.at <- mtcars[mtcars$am == "AT",]$mpg
mpg.mt <- mtcars[mtcars$am == "MT",]$mpg
t.test(mpg.at, mpg.mt)
##
## Welch Two Sample t-test
##
## data: mpg.at and mpg.mt
## t = -3.767, df = 18.33, p-value = 0.001374
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -11.28 -3.21
## sample estimates:
## mean of x mean of y
## 17.15 24.39
As the p-value is 0.001374 well bellow 5% or 1%, the alternative hypothesis is true: the difference in means is not equal to 0.
So, the mean mileage of automatic transmission is 17.15 mpg and the manual transmission is 24.39 mpg.
The 95% confidence interval of the difference in mean gas mileage is between 3.21 and 11.28 mpg.
The first question was answered.
Let's start with the simplest regression model for mpg with only one predictor: am.
model1 <- lm(mpg ~ am, data = mtcars)
summary(model1)
##
## Call:
## lm(formula = mpg ~ am, data = mtcars)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.392 -3.092 -0.297 3.244 9.508
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 17.15 1.12 15.25 1.1e-15 ***
## amMT 7.24 1.76 4.11 0.00029 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.9 on 30 degrees of freedom
## Multiple R-squared: 0.36, Adjusted R-squared: 0.338
## F-statistic: 16.9 on 1 and 30 DF, p-value: 0.000285
The null hypothesis is rejected by p-value = 0.000285 but the regression model covers only 36% of the variance.
So, the model coefficients are:
Next, we'll use a more complex regression model containing all independent variables as predictors:
model.all <- lm(mpg ~ ., data=mtcars)
and go backwards to the fitest model using step() function:
model.best <- step(model.all, trace=0)
summary(model.best)
##
## Call:
## lm(formula = mpg ~ wt + qsec + am, data = mtcars)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.481 -1.556 -0.726 1.411 4.661
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.618 6.960 1.38 0.17792
## wt -3.917 0.711 -5.51 7e-06 ***
## qsec 1.226 0.289 4.25 0.00022 ***
## amMT 2.936 1.411 2.08 0.04672 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.46 on 28 degrees of freedom
## Multiple R-squared: 0.85, Adjusted R-squared: 0.834
## F-statistic: 52.7 on 3 and 28 DF, p-value: 1.21e-11
The “Occam's razor” model explains 85% of mpg variance and contains only 3 predictors:
formula = mpg ~ wt + qsec + am
amMT estimated coefficient equals now to 2.9358 and represents the adjusted estimate for the expected change in mpg comparing AT versus MT for this new model containing 2 other predictors besides am.
amMT estimated coefficient is the answer to the second question.
Best model residuals are depicted in Appendix.3
First graphic, “Residuals vs. Fitted values” is not quite a straight line, proof of some outliers.
hist(mtcars$mpg, breaks=12, xlab="miles per gallon", main="mpg histogram", col="blue")
boxplot(mpg ~ am, data=mtcars, xlab="Transmission Type", ylab="Miles per Gallon",
main="Automatic versus Manual Transmission MPG")
par(mfrow = c(2,2))
plot(model.best)
