Fuel efficiency of 1970s’ cars: manual transmission vs automatic
Anton Kasenkov
27 01 2020
knitr::opts_chunk$set(echo = TRUE,
fig.width = 12,
fig.height = 8,
warning = FALSE,
message = FALSE)
Sys.setlocale(locale = "English")
pacman::p_load(tidyverse, car, psych, DT)1. Executive summary
In this document a fuel efficiency of 1970’s automobiles is the main subject of interest. In particular, there are 2 major questions of the research:
- Is an automatic transmission better than manual in regard to US miles per gallon (MPG) travelled?
- And what is the actual MPG difference between automatic and manual transmissions?
Judging by the results of the study, it appears that there is in fact a significant difference in MPG between the 2 types of transmissions. Unfortunately, the data for this study does not provide satisfactory basis for more or less reliable quantification of the difference in fuel economy between 2 groups.
2. Analysis
First, we have to prepare the data:
m_df <-
as_tibble(mtcars, rownames = "models") %>%
mutate_at("vs", ~factor(., labels = c("V", "S"))) %>%
mutate_at("am", ~factor(., labels = c("auto", "manual"))) %>%
mutate_at(c("cyl", "gear", "carb"), ~factor(.))2.1. Data exploration
The basic overview of the data:
str(m_df)## Classes 'tbl_df', 'tbl' and 'data.frame': 32 obs. of 12 variables:
## $ models: chr "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
## $ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
## $ cyl : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ...
## $ 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 : Factor w/ 2 levels "V","S": 1 1 2 2 1 2 1 2 2 2 ...
## $ am : Factor w/ 2 levels "auto","manual": 2 2 2 1 1 1 1 1 1 1 ...
## $ gear : Factor w/ 3 levels "3","4","5": 2 2 2 1 1 1 1 2 2 2 ...
## $ carb : Factor w/ 6 levels "1","2","3","4",..: 4 4 1 1 2 1 4 2 2 4 ...The MPG difference can be visualised with the simple violin plots:
ggplot(m_df, aes(y = mpg, x = am, fill = am))+
geom_violin(draw_quantiles = 0.5, lwd = 1.5)+
labs(title = "MPG by types of transmission",
x = "type of transmission",
y = "miles per gallon")+
theme_minimal()
It seems like there is a difference between the 2 groups. An attempt can be made to extrapolate this colnclusion beyond the sample and to quantify this difference. But to accomplish the latter, we have to first explore the relationships between the possible predictors of MPG. To do this, we can look at the correlations of continuous variables, namely, “mpg”, “disp” (displacement (cu.in.)), “drat” (rear axle ratio), “wt” (weight in 1000 lbs), “qsec” (1/4 mile time). The corresponding plot can be found here.
Out of 5 mentioned variables (excluding MPG) 3 are highly linearly correlated with each other. This means possible problems for estimation of variable contribution to the difference in MPG (multicollinearity):
cor.plot(m_df[c("mpg", "hp", "disp", "drat", "wt", "qsec")])
We can also compare variables on their pairwise relationship with MPG. This can give us clues on the possible order of predictor inclusion for the final model. The strongest magnitude of linear relationship can be noticed in the weight variable (\(\rho \approx -0.87\)).
2.2. Is an automatic or manual transmission better for MPG?
We can test our “mpg” variable for normality with Shapiro-Wilk test of normality. From here and forward on we would use significance level (\(\alpha\)) = 5%. results are here With this data we cannot reject the hypothesis that in the overall “population” of the 1970s cars miles per gallon are distributed normally. Of course, we have only 32 samples, so perhaps collecting more data would show otherwise. By assuming the normality of mpg we can perform a t-test to compare both groups of observations - automatic and manual transmission.
The output of t-test allows us to conclude that there is in fact a statistically significant difference between the 2 types of transmissions in terms of their fuel economy: automobiles with manual transmission tend to travel longer distances on the same amount of petrol. In the following section we will attempt to estimate this difference.
2.3. What is the MPG difference between automatic and manual transmissions?
2.3.1. Model fit
We begin with the basic model: \(mpg = \beta_0 + \beta_1 * am + \epsilon\). The interpretation of this model is that on average, cars with automatic transmission tend to travel ~17.1 miles per gallon of petrol, while an “average” car with the manual transmission will travel ~7.2 miles longer. However, the quality of the model is not very good (adjusted \(R^2 \approx 0.34\)), we can do better:
fit0 <- lm(mpg ~ am, data = m_df)
summary(fit0)##
## Call:
## lm(formula = mpg ~ am, data = m_df)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.3923 -3.0923 -0.2974 3.2439 9.5077
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 17.147 1.125 15.247 1.13e-15 ***
## ammanual 7.245 1.764 4.106 0.000285 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.902 on 30 degrees of freedom
## Multiple R-squared: 0.3598, Adjusted R-squared: 0.3385
## F-statistic: 16.86 on 1 and 30 DF, p-value: 0.0002852.3.2. Model selection
We can include additional covariates into our model, buliding several models. As was mentioned before, we should probably start with the weight (“wt”) and then gradually include more variables, leaving previous features untouched (so-called “nested” approach). We can then use ANOVA to test the significance of “model improvement”. The entire procedure can be found in the corresponding section of Appendix.
Note that we recentered our continuous variables for better interpretability of coefficients 1. Thus, the final model (“prefin”) can be interpreted in the following terms: if we compare average models (those that have the average weight, displacement, gross horsepower and 1/4 mile time), models with manual transmission tend to travel ~3.5 miles per gallon of petrol longer than models with automatic transmission. The inference here is that with 95% certainty we can say that the mentioned cars are travelling from ~0.42 to ~6.52 miles longer on a gallon of petrol:
sc <- summary(prefin)$coef
c_int <- sc[2, 1] + c(-1, 1) * qt(.975, df = prefin$df) * sc[2, 2]
cat("CI 95% for transmission coefficient is from", round(c_int, 2)[1], "to", round(c_int, 2)[2])## CI 95% for transmission coefficient is from 0.42 to 6.52Preliminary diagnostics. Here are the residuals plots:
par(mfrow = c(2,2))
plot(prefin)
On the \(1^{st}\) plot (fitted values vs residuals) we can notice that there is a certain “hockey stick”-like pattern, which basically means that our model fit is not optimal.
The second plot in the series shows us that the standardised residuals are not really normally distributed. Overall, observations 17, 18 and 20 (Chrysler Imperial, Fiat 128 and Toyota Corolla) break the linear pattern slightly. This might be suggestive of the need for non-linear relationship model.
One of the simplest variable transformation is the log-transformation. This can improve the quality of our model without loss in the interpretability. There are 3 main options for each of the model specification: we can either take the log of the response variable, or take the log of the predictors, or do both (for simplicity we would not consider options where only part of the continuous predictors will be transformed). In the exploratory phase of the analysis it was shown that perhaps the log transformation of the predictors is the best option we have.
Now, conducting the same procedure as before, we are arrriving at the following specification: \[mpg = \beta_0 + am + log(wt) + log(disp) + log(hp) + \epsilon\] The details of selection process can be found in the according section of Appendix
Summary of the model:
summary(finfit)##
## Call:
## lm(formula = mpg ~ am + log(wt) + log(disp) + log(hp), data = m_df)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.5697 -1.5426 -0.5873 1.1312 4.5337
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 59.7479 5.5792 10.709 3.2e-11 ***
## ammanual 0.4928 1.2937 0.381 0.70623
## log(wt) -9.4190 3.0550 -3.083 0.00468 **
## log(disp) -1.1203 2.1957 -0.510 0.61403
## log(hp) -4.7873 1.8899 -2.533 0.01742 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.135 on 27 degrees of freedom
## Multiple R-squared: 0.8907, Adjusted R-squared: 0.8745
## F-statistic: 55 on 4 and 27 DF, p-value: 1.372e-122.3.3. Model diagnostics
par(mfrow = c(2,2))
plot(finfit)
Now the explained variance of the model has improved. But as is shown on the residuals plots, some problems with fit are still present. And a couple of coefficients, transmission type - one of them, became insignificantly different from zero. Thus, we cannot draw inference from those. Apparently, the task of explanation of MPG variance among cars is not that strongly dependent on the type of transmission. Further investigation of the subject can be done with a bigger dataset.
3. Appendix
Data exploration (scatter plot)
ggplot(m_df, aes(x = wt, y = mpg, color = am))+
geom_point(cex = 5, alpha = .5)+
geom_label(aes(label = models), label.size = 0.2)+
labs(title = "MPG and weight",
color = "transmission",
x = "weight in 1000 lbs",
y = "miles per gallon")+
geom_smooth(method = lm, se = FALSE)+
theme_minimal()
Normailty test for MPG
shapiro.test(m_df$mpg)##
## Shapiro-Wilk normality test
##
## data: m_df$mpg
## W = 0.94756, p-value = 0.1229T-test
t.test(mpg ~ am, m_df)##
## Welch Two Sample t-test
##
## data: mpg by 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 auto mean in group manual
## 17.14737 24.39231Model selection 1
fit1 <- lm(mpg ~ am + I(wt - mean(wt)), data = m_df)
fit2 <- lm(mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)), data = m_df)
fit3 <- lm(mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp)), data = m_df)
fit4 <- lm(mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp)) +
I(drat - mean(drat)), data = m_df)
anova(fit0, fit1, fit2, fit3, fit4)## Analysis of Variance Table
##
## Model 1: mpg ~ am
## Model 2: mpg ~ am + I(wt - mean(wt))
## Model 3: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp))
## Model 4: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp))
## Model 5: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp)) +
## I(drat - mean(drat))
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 30 720.90
## 2 29 278.32 1 442.58 65.5044 1.422e-08 ***
## 3 28 246.56 1 31.76 4.7012 0.039478 *
## 4 27 179.91 1 66.65 9.8645 0.004171 **
## 5 26 175.67 1 4.24 0.6275 0.435428
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1apparently, fit4 does not improve our model. We replace “drat” variable with “qsec” and start
prefin <- lm(mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp)) +
I(qsec - mean(qsec)), data = m_df)
anova(fit0, fit1, fit2, fit3, prefin)## Analysis of Variance Table
##
## Model 1: mpg ~ am
## Model 2: mpg ~ am + I(wt - mean(wt))
## Model 3: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp))
## Model 4: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp))
## Model 5: mpg ~ am + I(wt - mean(wt)) + I(disp - mean(disp)) + I(hp - mean(hp)) +
## I(qsec - mean(qsec))
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 30 720.90
## 2 29 278.32 1 442.58 74.9946 3.877e-09 ***
## 3 28 246.56 1 31.76 5.3823 0.028457 *
## 4 27 179.91 1 66.65 11.2936 0.002413 **
## 5 26 153.44 1 26.47 4.4853 0.043908 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Model selection 2
fit1a <- lm(mpg ~ am + log(wt), data = m_df)
fit2a <- lm(mpg ~ am + log(wt) + log(disp), data = m_df)
finfit <- lm(mpg ~ am + log(wt) + log(disp) + log(hp), data = m_df)
anova(fit0, fit1a, fit2a, finfit)## Analysis of Variance Table
##
## Model 1: mpg ~ am
## Model 2: mpg ~ am + log(wt)
## Model 3: mpg ~ am + log(wt) + log(disp)
## Model 4: mpg ~ am + log(wt) + log(disp) + log(hp)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 30 720.90
## 2 29 207.97 1 512.93 112.5023 3.954e-11 ***
## 3 28 152.35 1 55.62 12.1987 0.001665 **
## 4 27 123.10 1 29.26 6.4166 0.017424 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1we can also test inclusion of qsec variable, but it seems that this is redundant feature.
fit4d <- lm(mpg ~ am + log(wt) + log(disp) + log(hp) + log(qsec), data = m_df)
anova(fit0, fit1a, fit2a, finfit, fit4d)## Analysis of Variance Table
##
## Model 1: mpg ~ am
## Model 2: mpg ~ am + log(wt)
## Model 3: mpg ~ am + log(wt) + log(disp)
## Model 4: mpg ~ am + log(wt) + log(disp) + log(hp)
## Model 5: mpg ~ am + log(wt) + log(disp) + log(hp) + log(qsec)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 30 720.90
## 2 29 207.97 1 512.93 113.0995 5.796e-11 ***
## 3 28 152.35 1 55.62 12.2635 0.001688 **
## 4 27 123.10 1 29.26 6.4507 0.017410 *
## 5 26 117.91 1 5.19 1.1433 0.294779
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1One can notice from the summary of our final model that displacement and gross horsepower are not statistically significant. This happend most likely due to the high correlation between them. The decision to include them in the model was based on 2 facts: (1) removing them would lead to biased coefficients and (2) the only important coefficients for the interpretation are intercept and dummy variable “am” (transmission).↩