Regression Model Project (2014.11.23)
Questions
Take the mtcars data set and write up an analysis to answer the questions below using regression models and exploratory data analyses.
- Is an automatic or manual transmission better for MPG?
- Quantify the MPG difference between automatic and manual transmissions
Summary
The first analysis is to analyze the relationship between MPG performance and transmission by dataset mtcars. The result shows the relation is significant. The cars with manual transmissions are statistically outperformed those with automatic transmissions.
In the second analysis, several regression models with different combinations of variables are compared. With the practice of less number of regressors and high \(R^2\) value, we choose mpg~wt+qsec+am as the best fitness.
- Is an automatic or manual transmission better for MPG?
Load data
data(mtcars)Boxplot: MPG with automatic
am=0& manualam=1(defined in?mtcars)g = ggplot(mtcars, aes(factor(am), mpg, fill=factor(am))) g = g + geom_boxplot() g = g + geom_jitter(position=position_jitter(width=.1, height=0)) g = g + scale_colour_discrete(name = "Type") g = g + scale_fill_discrete(name="Type", breaks=c("0", "1"), labels=c("Automatic", "Manual")) g = g + scale_x_discrete(breaks=c("0", "1"), labels=c("Automatic", "Manual")) g = g + xlab("") g
T-Test: compare group #1 with
am=1& group #2 witham=0.
\(H_0\): mpg of manual is less than mpg of automatic
\(H_A\): mpg of manual is greater than mpg of automaticg1 <- subset(mtcars, mtcars$am==0) g2 <- subset(mtcars, mtcars$am==1) amt <- t.test(g1, g2, alternative="greater", paired=F)The p-value is 0.0307 which is less than 0.05. So, the mpg of
am=1group is significant larger thanam=0group.
ANS: The manual transmission has better performance in miles per gallon (mpg).
- Quantify the MPG difference between automatic and manual transmissions
From the previous analysis, we know transmission is an important variable. How about other models? Use stepwise-selected model
stepin R to find better fitness model.s_model <- step(lm(data = mtcars, mpg~.), trace=0) summary(s_model)
With weight## ## 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 *** ## am 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-11wt, 1/4 mile timeqsecand transmissionam, we can build the a model with best \(R^2\) fitness.Get the variables which have the high corraltions
var_cor <- round(cor(mtcars)[1,], 2) name_cor <- names(sort(abs(var_cor),decreasing=T)) ## the top four variables: wt, cyl, disp, hpcor(
mpg,wt) = -0.87
cor(mpg,cyl) = -0.85
cor(mpg,disp) = -0.85
cor(mpg,hp) = -0.78Use
mpg ~ wt + qsec + amas the formula of linear regression model.fit1 <- lm(data=mtcars, mpg~wt+qsec+am) ## coefficient summary(fit1)$coefficient## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 9.618 6.9596 1.382 1.779e-01 ## wt -3.917 0.7112 -5.507 6.953e-06 ## qsec 1.226 0.2887 4.247 2.162e-04 ## am 2.936 1.4109 2.081 4.672e-02Compare with other models
Stepwise-selected modelstepin R:mpg~wt+qsec+am
All variables:mpg~.
Weight:mpg~wt(with the largest value of correlation)
Four high-correlated variables:mpg~wt+cyl+disp+hpfit2 <- lm(data=mtcars, mpg~.) fit3 <- lm(data=mtcars, mpg~wt) fit4 <- lm(data=mtcars, mpg~wt+cyl+disp+hp) ## using anova to compare models anova(fit1, fit2, fit3, fit4)## Analysis of Variance Table ## ## Model 1: mpg ~ wt + qsec + am ## Model 2: mpg ~ cyl + disp + hp + drat + wt + qsec + vs + am + gear + carb ## Model 3: mpg ~ wt ## Model 4: mpg ~ wt + cyl + disp + hp ## Res.Df RSS Df Sum of Sq F Pr(>F) ## 1 28 169 ## 2 21 148 7 21.8 0.44 0.8636 ## 3 30 278 -9 -130.8 2.07 0.0816 . ## 4 27 170 3 107.9 5.12 0.0082 ** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1In the ANOVA comparison
mpg~wt+cyl+disp+hpis significant different frommpg~wt+qsec+am. We may find which one has better \(R^2\) value:
Modelmpg~wt+qsec+am= 0.8497
Modelmpg~wt+cyl+disp+hp= 0.8486Compare reesiduals of the models (
mpg~wt+qsec+amandmpg~wt+cyl+disp+hp)x <- mtcars$wt res1 <- resid(fit1) e <- res1 n <- length(e) plot(x, e, main="mpg~wt+qsec+am", xlab="wt", ylab="Residuals", bg="lightblue", col="black", cex = 2, pch = 21,frame = FALSE) abline(h = 0, lwd = 2) for (i in 1 : n) lines(c(x[i], x[i]), c(e[i], 0), col = "red" , lwd = 2)
res4 <- resid(fit4) e <- res4 n <- length(e) plot(x, e, main="mpg~wt+cyl+disp+hp", xlab="wt", ylab="Residuals", bg="lightblue", col="black", cex = 2, pch = 21,frame = FALSE) abline(h = 0, lwd = 2) for (i in 1 : n) lines(c(x[i], x[i]), c(e[i], 0), col = "red" , lwd = 2)
The variables in both models can strongly explain the relationship between mpg and car-samples! Howerver, it recomemeds the model with less number of regressors –
mpg~wt+qsec+am.Explanation of the MPG difference between automatic and manual transmissions
Let’s use two variable and their interactionmpg~wt+am+am*wtmodel to explain the contribution ofamwithwt.fit <- lm(data=mtcars, mpg~wt+am+wt*am) g1 <- subset(mtcars, mtcars$am==0) g2 <- subset(mtcars, mtcars$am==1) plot(g1$wt, g1$mpg, col="lightblue", pch=19, cex=2, xlab="Weight", ylab="mpg") points(g2$wt, g2$mpg, col="salmon", pch=19, cex=2) ## am=0; abline(c(fit$coeff[1], fit$coeff[2]), col="lightblue", lwd=3, lty=2) ## am=1 abline(c(fit$coeff[1]+fit$coeff[3], fit$coeff[2]+fit$coeff[4]), col="salmon", lwd=3, lty=2) legend("topright", pch=19, col=c("lightblue", "salmon"), legend=c("Manual (am=1)", "Automatic (am=0)"))
The number of samples with
am=0is less than number ofam=1. With the regression lines, the mpg payoff ofam=1is better ifwt > 2.8. We still need more data to prove and increase the reliability.