| title | author | date | output | ||||
|---|---|---|---|---|---|---|---|
Regression-model-for-automobile-industry |
luis |
11/7/2020 |
|
This project consists on a regression model for analysis of a data collection of automobile industry. The main objectivo is to explore the relationship between a set of variables and miles per gallon (MPG). Particularly in the answer of the following questions:
- Is an automatic or manual transmission better for MPG
- Quantify the MPG difference between automatic and manual transmissions
The data contains a set of 32 observations on 11 different variables
- mpg: Miles/(US) gallon
- cyl: Number of cylinders
- disp: Displacement (cu.in.)
- hp: Gross horsepower
- drat: Rear axle ratio
- wt: Weight (1000 lbs)
- qsec: 1/4 mile time
- vs: Engine (0 = V-shaped, 1 = straight)
- am: Transmission (0 = automatic, 1 = manual)
- gear: Number of forward gears
- carb: Number of carburetors
The libraries and set of data are loaded
library(ggplot2)
library(dplyr)
data("mtcars")A first aproximation is made by the boxplot that differentiates the automatic and manual transmissions.
boxplot(mpg ~ am, data = mtcars, xlab = "Transmission (Automatic = 0, Manual = 1)", ylab = "Miles/gallon", main = "Miles/gallon for manual and automatic transmissions")
The boxplot shows the manual transmission has in general greater values (mean = 24.39) for Miles/Gallon (mpg) than automatic transmissions (mean = 17.14). Nevertheless, the range of values is bigger for the manual transmission.
For the model selection, a first guess is made by fitting all the variables in the dataframe in order to look the diagnostics and decide whicho ones are needed.
allData <- mtcars %>% mutate(cyl = as.factor(cyl), vs = as.factor(vs), am = as.factor(am), gear = as.factor(gear), carb = as.factor(carb))
fitAllData <- lm(mpg ~ ., data = allData)
summary(fitAllData)##
## Call:
## lm(formula = mpg ~ ., data = allData)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.5087 -1.3584 -0.0948 0.7745 4.6251
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 23.87913 20.06582 1.190 0.2525
## cyl6 -2.64870 3.04089 -0.871 0.3975
## cyl8 -0.33616 7.15954 -0.047 0.9632
## disp 0.03555 0.03190 1.114 0.2827
## hp -0.07051 0.03943 -1.788 0.0939 .
## drat 1.18283 2.48348 0.476 0.6407
## wt -4.52978 2.53875 -1.784 0.0946 .
## qsec 0.36784 0.93540 0.393 0.6997
## vs1 1.93085 2.87126 0.672 0.5115
## am1 1.21212 3.21355 0.377 0.7113
## gear4 1.11435 3.79952 0.293 0.7733
## gear5 2.52840 3.73636 0.677 0.5089
## carb2 -0.97935 2.31797 -0.423 0.6787
## carb3 2.99964 4.29355 0.699 0.4955
## carb4 1.09142 4.44962 0.245 0.8096
## carb6 4.47757 6.38406 0.701 0.4938
## carb8 7.25041 8.36057 0.867 0.3995
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.833 on 15 degrees of freedom
## Multiple R-squared: 0.8931, Adjusted R-squared: 0.779
## F-statistic: 7.83 on 16 and 15 DF, p-value: 0.000124
The model shows a residual standard error of 2.833 on 15 degrees of freedom. The R-squared value 0.8931 indicates 89.3% of the data is explained by this model. Nevertheless, none of the variables show significant values less than 5%.
The followed procedure was to find the most insignificant variable in data set and remove it. The line "which.max(summary(model)$coef[, 4])" can find the least significant variable. The process was repeated until all the variables in the model resulted significant.
dataBetter <- allData %>% select(-cyl); fitBetter <- lm(mpg ~ ., data = allData)
dataBetter <- dataBetter %>% select(-carb); fitBetter <- lm(mpg ~ ., data = dataBetter)
dataBetter <- dataBetter %>% select(-gear); fitBetter <- lm(mpg ~ ., data = dataBetter)
dataBetter <- dataBetter %>% select(-vs); fitBetter <- lm(mpg ~ ., data = dataBetter)
dataBetter <- dataBetter %>% select(-drat); fitBetter <- lm(mpg ~ ., data = dataBetter)
dataBetter <- dataBetter %>% select(-disp); fitBetter <- lm(mpg ~ ., data = dataBetter)
dataBetter <- dataBetter %>% select(-hp); fitBetter <- lm(mpg ~ ., data = dataBetter)
summary(fitBetter)##
## Call:
## lm(formula = mpg ~ ., data = dataBetter)
##
## 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 ***
## am1 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
After the process is concluded, the final model shows a residual standard error of 2.459 on 28 degrees of freedom. Even though we removed 8 variables of the data out of 11 initial variables, the R-squared value is 0.8497 so the 84.97% of the data variance is still explained with all the variables show significance values less than 5%. The p-value indicates that we fail to reject the null hypothesis.
par(mfrow = c(2, 2))
plot(fitBetter)
In the QQ plot we can see there is no special pattern for the residuals, also the standarized and theoretical residuals have a good correlation.
This model states that if given the other vairables constant (qsec: 1/4 mile time, wt: Weight(1000 lbs)), the Miles/gallon (mpg) for the transmissions is:
- Automatic: 9.6178 + 0*(2.9358) = 9.6178 Miles/Gallon
- Manual: 9.6178 + 1*(2.9358) = 12.5536 Miles/Gallon Concluding, the manual transmission is better for Miles/Gallon, with a 2.9358 Miles/Gallon difference with a constant weight and 1/4 mile time.