To answer both questions below:

.“Is an automatic or manual transmission better for MPG”

.“Quantify the MPG difference between automatic and manual transmissions”

We need to evaluate mpg and am (as described below):

– mpg: Miles/(US) gallon

– am: Transmission (0 = automatic, 1 = manual)

– n =32 observations

– source: Henderson and Velleman (1981), Building multiplt regression models interactively. Biometrics, 37, 391-411.

With that in mind we will analyse only these two variables and see if any conclusions will be enough.

Exploratory Data Analyses

tapply(mtcars$mpg, mtcars$am,summary)
## $`0`
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   10.40   14.95   17.30   17.15   19.20   24.40 
## 
## $`1`
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.00   21.00   22.80   24.39   30.40   33.90

Regression Models

Linear Models

Only am:
model1 <- lm(mpg ~ as.factor(am), data=mtcars)
summary(model1)
## 
## Call:
## lm(formula = mpg ~ as.factor(am), data = mtcars)
## 
## 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 ***
## as.factor(am)1    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: 0.000285
anova(model1)
## Analysis of Variance Table
## 
## Response: mpg
##               Df Sum Sq Mean Sq F value   Pr(>F)    
## as.factor(am)  1 405.15  405.15   16.86 0.000285 ***
## Residuals     30 720.90   24.03                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
model1.res = resid(model1) 
plot(model1.res, 
      #ylab="Residuals", 
      #xlab="Transmissions", 
      main="Residuals") 
      abline(0, 0)   

qqnorm(model1.res)
qqline(model1.res)

Only am without intercept:
model2 <- lm(mpg ~ as.factor(am) - 1, data=mtcars)
summary(model2)
## 
## Call:
## lm(formula = mpg ~ as.factor(am) - 1, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.3923 -3.0923 -0.2974  3.2439  9.5077 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## as.factor(am)0   17.147      1.125   15.25 1.13e-15 ***
## as.factor(am)1   24.392      1.360   17.94  < 2e-16 ***
## ---
## 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:  0.9487, Adjusted R-squared:  0.9452 
## F-statistic: 277.2 on 2 and 30 DF,  p-value: < 2.2e-16
anova(model2)
## Analysis of Variance Table
## 
## Response: mpg
##               Df  Sum Sq Mean Sq F value    Pr(>F)    
## as.factor(am)  2 13321.4  6660.7  277.18 < 2.2e-16 ***
## Residuals     30   720.9    24.0                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
model2.res = resid(model2) 
plot(model2.res, 
      #ylab="Residuals", 
      #xlab="Transmissions", 
      main="Residuals") 
      abline(0, 0) 

qqnorm(model2.res)
qqline(model2.res)

As we can see both models have good residuals (diperse and normal) but the second one explain 94.5% of the mpg while the first one only 33% (R square).With a p-value less then 1% which means that for 100 cars less the 1 will have a mpg different then the one shown here.

Conclusions

Both models are able to show that manual transmission (1) in more effective because we can do more miles per gallon then with automatic ones (0): More specifically automatic transmissions do 7 miles per gallon less then manual:

Coefficients(model 1): Estimate
(Intercept) 17.147
as.factor(am)1 7.245

Coefficients (model 2): Estimate
as.factor(am)0 17.147 as.factor(am)1 24.392

Therefore manual transmissions are better for MPG and the difference between then are 7.25 miles per gallon more for manual.