Instructions

This is the R portion of your mid-term exam. You will analyze the Auto dataset, which contains information about various car models (similar to mtcar). Follow the instructions carefully and write your R code in the provided chunks. You will be graded on the correctness of your code, the quality of your analysis, and your interpretation of the results.

Total points: 100 Time allowed: 45 minutes

Good luck!

1. Data Import and Exploration (20 points)

  1. Import the Auto dataset provided on Canvas, named it as Auto, and display the first few rows. (1 points)
Auto <- read.csv("Auto.csv")
head(Auto)
##   mpg cylinders displacement horsepower weight acceleration year origin
## 1  18         8          307        130   3504         12.0   70      1
## 2  15         8          350        165   3693         11.5   70      1
## 3  18         8          318        150   3436         11.0   70      1
## 4  16         8          304        150   3433         12.0   70      1
## 5  17         8          302        140   3449         10.5   70      1
## 6  15         8          429        198   4341         10.0   70      1
##                        name
## 1 chevrolet chevelle malibu
## 2         buick skylark 320
## 3        plymouth satellite
## 4             amc rebel sst
## 5               ford torino
## 6          ford galaxie 500
str(Auto)
## 'data.frame':    392 obs. of  9 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : int  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : int  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : int  3504 3693 3436 3433 3449 4341 4354 4312 4425 3850 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : int  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : int  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : chr  "chevrolet chevelle malibu" "buick skylark 320" "plymouth satellite" "amc rebel sst" ...
  1. Use appropriate R functions to display the structure of the dataset and report how many observations and variables are in the dataset? (10 points)
dim(Auto)
## [1] 392   9

392 rows, 9 variables

2. Data Preprocessing and Visualization (20 points)

  1. Create a correlation matrix for all numeric variables in the dataset. (10 points) (Optional) visualize the correlation matrix.
cor(Auto[sapply(Auto, is.numeric)])
##                     mpg  cylinders displacement horsepower     weight
## mpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442
## cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273
## displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944
## horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377
## weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000
## acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392
## year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199
## origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054
##              acceleration       year     origin
## mpg             0.4233285  0.5805410  0.5652088
## cylinders      -0.5046834 -0.3456474 -0.5689316
## displacement   -0.5438005 -0.3698552 -0.6145351
## horsepower     -0.6891955 -0.4163615 -0.4551715
## weight         -0.4168392 -0.3091199 -0.5850054
## acceleration    1.0000000  0.2903161  0.2127458
## year            0.2903161  1.0000000  0.1815277
## origin          0.2127458  0.1815277  1.0000000
  1. Create a scatter plot of ‘mpg’ vs ‘weight’ (you can use plot() or ggplot()). Add a title and proper axis labels. You don’t need to interpret the result here but you should know how. (10 points)
plot(Auto$weight, Auto$mpg,
     main = "MPG vs Weight",
     xlab = "Weight",
     ylab = "MPG")

3. Linear Regression Analysis (60 points)

  1. Fit a multiple linear regression model using ‘mpg’ as the response variable (Y) and ‘weight’, ‘horsepower’, and ‘year’ as predictors (X). Display the summary of the model. (10 points)
model <- lm(mpg ~ weight + horsepower + year, data = Auto)
summary(model)
## 
## Call:
## lm(formula = mpg ~ weight + horsepower + year, data = Auto)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.7911 -2.3220 -0.1753  2.0595 14.3527 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -1.372e+01  4.182e+00  -3.281  0.00113 ** 
## weight      -6.448e-03  4.089e-04 -15.768  < 2e-16 ***
## horsepower  -5.000e-03  9.439e-03  -0.530  0.59663    
## year         7.487e-01  5.212e-02  14.365  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.43 on 388 degrees of freedom
## Multiple R-squared:  0.8083, Adjusted R-squared:  0.8068 
## F-statistic: 545.4 on 3 and 388 DF,  p-value: < 2.2e-16
  1. Interpret the Intercept and coefficient of weight. What do they tell us about the relationship between the predictors and ‘mpg’? (10 points)

When weight, horsepower, and year are all equal to 0, the predicted MPG value is -13.719. Holding horsepower and year constant, a one-unit increase in weight is associated with an approximately 0.0064 decrease in predicted MPG.

  1. Create diagnostic plots for the model. Identify ONE potential issues you observe briefly. (10 points)
par(mfrow = c(2, 2))
plot(model)

The residuals show evidence of heteroscedasticity because the spread of the residuals is not constant across the fitted values.

  1. Obtain the R-squared and adjusted R-squared for the model. Interpret these values briefly. (10 points)
summary(model)$r.squared
## [1] 0.8083189
summary(model)$adj.r.squared
## [1] 0.8068368

r.squared: Approximately 80.8% of the variation in MPG is explained by weight, horsepower, and year.

Adj.r.squared: Approximately 80.7% of the variation in MPG is explained by the predictors after adjusting for the number of predictors in the model.

  1. Refit the model in 3a with one interaction term between weight and horsepower added to the ‘weight’, ‘horsepower’, and ‘year’ as predictors (X) and report the adjusted R-squared. (10 point)
model2 <- lm(mpg ~ weight + horsepower + year + weight:horsepower,
data = Auto)
summary(model2)$adj.r.squared
## [1] 0.8558773

[The adjusted R-squared for the interaction model is 0.856]

  1. Compare the R-squared and the adjusted R-squared from the interaction model and original model, does adding interaction improve the model? Provide your answer and reason below. (10 point)

Yes. The interaction model improves the model because both R-squared and adjusted R-squared increase substantially. R-squared increases from about 0.808 to 0.857, while adjusted R-squared increases from about 0.807 to 0.856. Since adjusted R-squared also increases, the improvement is not simply due to adding another predictor.

End of Exam. Please submit this RMD file along with a knitted HTML report. Failed to submit HTML will lead to 10pt deduction.