Introduction

This document presents answers to several exercises from Chapter 3 (“Linear Regression”) in An Introduction to Statistical Learning. In particular, we address:

For all analyses, we use the ISLR2 package (which contains these data sets), along with ggplot2 and dplyr for data manipulation and visualization.

Setup

Load the necessary libraries and data sets.

# Load libraries
library(ISLR2)    # Contains Auto and Boston data sets
library(ggplot2)  # For plotting
library(dplyr)    # For data manipulation

# Set a consistent theme for ggplot2
theme_set(theme_minimal())
data(Auto)
# Display the first few rows of the dataset
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)  # Display variable types
## '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        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
##  - attr(*, "na.action")= 'omit' Named int [1:5] 33 127 331 337 355
##   ..- attr(*, "names")= chr [1:5] "33" "127" "331" "337" ...
### Question 2

> Carefully explain the differences between the KNN classifier and KNN
> regression methods.

The KNN classifier is categorical and assigns a value based on the most
frequent observed category among $K$ nearest neighbors, whereas KNN regression
assigns a continuous variable, the average of the response variables for
the $K$ nearest neighbors.

Question 9

This question involves the use of multiple linear regression on the Auto data set.

  1. Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto, cex = 0.2)

  1. Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, name which is qualitative.
x <- subset(Auto, select = -name)
cor(x)
##                     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. Use the lm() function to perform a multiple linear regression with mpg as the response and all other variables except name as the predictors. Use the summary() function to print the results. Comment on the output. For instance:
    1. Is there a relationship between the predictors and the response?
    2. Which predictors appear to have a statistically significant relationship to the response?
    3. What does the coefficient for the year variable suggest?
fit <- lm(mpg ~ ., data = x)
summary(fit)
## 
## Call:
## lm(formula = mpg ~ ., data = x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.5903 -2.1565 -0.1169  1.8690 13.0604 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -17.218435   4.644294  -3.707  0.00024 ***
## cylinders     -0.493376   0.323282  -1.526  0.12780    
## displacement   0.019896   0.007515   2.647  0.00844 ** 
## horsepower    -0.016951   0.013787  -1.230  0.21963    
## weight        -0.006474   0.000652  -9.929  < 2e-16 ***
## acceleration   0.080576   0.098845   0.815  0.41548    
## year           0.750773   0.050973  14.729  < 2e-16 ***
## origin         1.426141   0.278136   5.127 4.67e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.328 on 384 degrees of freedom
## Multiple R-squared:  0.8215, Adjusted R-squared:  0.8182 
## F-statistic: 252.4 on 7 and 384 DF,  p-value: < 2.2e-16

Yes, there is a relationship between some predictors and response, such as “displacement”, “weight”, and “year”

The coefficient for “year” suggests that “mpg” increases every year, on average.

  1. Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit. Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
par(mfrow = c(2, 2))
plot(fit, cex = 0.2)

One point has high leverage, the residuals also show a trend with fitted values.

  1. Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(mpg ~ . + weight:horsepower, data = x))
## 
## Call:
## lm(formula = mpg ~ . + weight:horsepower, data = x)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -8.589 -1.617 -0.184  1.541 12.001 
## 
## Coefficients:
##                     Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        2.876e+00  4.511e+00   0.638 0.524147    
## cylinders         -2.955e-02  2.881e-01  -0.103 0.918363    
## displacement       5.950e-03  6.750e-03   0.881 0.378610    
## horsepower        -2.313e-01  2.363e-02  -9.791  < 2e-16 ***
## weight            -1.121e-02  7.285e-04 -15.393  < 2e-16 ***
## acceleration      -9.019e-02  8.855e-02  -1.019 0.309081    
## year               7.695e-01  4.494e-02  17.124  < 2e-16 ***
## origin             8.344e-01  2.513e-01   3.320 0.000986 ***
## horsepower:weight  5.529e-05  5.227e-06  10.577  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.931 on 383 degrees of freedom
## Multiple R-squared:  0.8618, Adjusted R-squared:  0.859 
## F-statistic: 298.6 on 8 and 383 DF,  p-value: < 2.2e-16
summary(lm(mpg ~ . + acceleration:horsepower, data = x))
## 
## Call:
## lm(formula = mpg ~ . + acceleration:horsepower, data = x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.0329 -1.8177 -0.1183  1.7247 12.4870 
## 
## Coefficients:
##                           Estimate Std. Error t value Pr(>|t|)    
## (Intercept)             -32.499820   4.923380  -6.601 1.36e-10 ***
## cylinders                 0.083489   0.316913   0.263 0.792350    
## displacement             -0.007649   0.008161  -0.937 0.349244    
## horsepower                0.127188   0.024746   5.140 4.40e-07 ***
## weight                   -0.003976   0.000716  -5.552 5.27e-08 ***
## acceleration              0.983282   0.161513   6.088 2.78e-09 ***
## year                      0.755919   0.048179  15.690  < 2e-16 ***
## origin                    1.035733   0.268962   3.851 0.000138 ***
## horsepower:acceleration  -0.012139   0.001772  -6.851 2.93e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.145 on 383 degrees of freedom
## Multiple R-squared:  0.841,  Adjusted R-squared:  0.8376 
## F-statistic: 253.2 on 8 and 383 DF,  p-value: < 2.2e-16
summary(lm(mpg ~ . + cylinders:weight, data = x))
## 
## Call:
## lm(formula = mpg ~ . + cylinders:weight, data = x)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.9484  -1.7133  -0.1809   1.4530  12.4137 
## 
## Coefficients:
##                    Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       7.3143478  5.0076737   1.461  0.14494    
## cylinders        -5.0347425  0.5795767  -8.687  < 2e-16 ***
## displacement      0.0156444  0.0068409   2.287  0.02275 *  
## horsepower       -0.0314213  0.0126216  -2.489  0.01322 *  
## weight           -0.0150329  0.0011125 -13.513  < 2e-16 ***
## acceleration      0.1006438  0.0897944   1.121  0.26306    
## year              0.7813453  0.0464139  16.834  < 2e-16 ***
## origin            0.8030154  0.2617333   3.068  0.00231 ** 
## cylinders:weight  0.0015058  0.0001657   9.088  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.022 on 383 degrees of freedom
## Multiple R-squared:  0.8531, Adjusted R-squared:  0.8501 
## F-statistic: 278.1 on 8 and 383 DF,  p-value: < 2.2e-16

There are a few coefficients that appear to be highly significant.

  1. Try a few different transformations of the variables, such as \(log(X)\), \(\sqrt{X}\), \(X^2\). Comment on your findings.

Below are transformations for horsepower:

par(mfrow = c(2, 2))
plot(Auto$horsepower, Auto$mpg, cex = 0.2)
plot(log(Auto$horsepower), Auto$mpg, cex = 0.2)
plot(sqrt(Auto$horsepower), Auto$mpg, cex = 0.2)
plot(Auto$horsepower^2, Auto$mpg, cex = 0.2)

x <- subset(Auto, select = -name)
x$horsepower <- log(x$horsepower)
fit <- lm(mpg ~ ., data = x)
summary(fit)
## 
## Call:
## lm(formula = mpg ~ ., data = x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.3115 -2.0041 -0.1726  1.8393 12.6579 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  27.254005   8.589614   3.173  0.00163 ** 
## cylinders    -0.486206   0.306692  -1.585  0.11372    
## displacement  0.019456   0.006876   2.830  0.00491 ** 
## horsepower   -9.506436   1.539619  -6.175 1.69e-09 ***
## weight       -0.004266   0.000694  -6.148 1.97e-09 ***
## acceleration -0.292088   0.103804  -2.814  0.00515 ** 
## year          0.705329   0.048456  14.556  < 2e-16 ***
## origin        1.482435   0.259347   5.716 2.19e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.18 on 384 degrees of freedom
## Multiple R-squared:  0.837,  Adjusted R-squared:  0.834 
## F-statistic: 281.6 on 7 and 384 DF,  p-value: < 2.2e-16
par(mfrow = c(2, 2))
plot(fit, cex = 0.2)

horsepower appears to give a linear relationship with mpg.

Question 10

This question should be answered using the Carseats data set.

  1. Fit a multiple regression model to predict Sales using Price, Urban, and US.
fit <- lm(Sales ~ Price + Urban + US, data = Carseats)
  1. Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
summary(fit)
## 
## Call:
## lm(formula = Sales ~ Price + Urban + US, data = Carseats)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -6.9206 -1.6220 -0.0564  1.5786  7.0581 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 13.043469   0.651012  20.036  < 2e-16 ***
## Price       -0.054459   0.005242 -10.389  < 2e-16 ***
## UrbanYes    -0.021916   0.271650  -0.081    0.936    
## USYes        1.200573   0.259042   4.635 4.86e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.472 on 396 degrees of freedom
## Multiple R-squared:  0.2393, Adjusted R-squared:  0.2335 
## F-statistic: 41.52 on 3 and 396 DF,  p-value: < 2.2e-16
  1. Write out the model in equation form, being careful to handle the qualitative variables properly.

\[ \textit{Sales} = 13 + -0.054 \times \textit{Price} + \begin{cases} -0.022, & \text{if $\textit{Urban}$ is Yes, $\textit{US}$ is No} \\ 1.20, & \text{if $\textit{Urban}$ is No, $\textit{US}$ is Yes} \\ 1.18, & \text{if $\textit{Urban}$ and $\textit{US}$ is Yes} \\ 0, & \text{Otherwise} \end{cases} \]

  1. For which of the predictors can you reject the null hypothesis \(H_0 : \beta_j = 0\)?

Price and US

  1. On the basis of your response to the previous question, fit a smaller model that only uses the predictors for which there is evidence of association with the outcome.
fit2 <- lm(Sales ~ Price + US, data = Carseats)
  1. How well do the models in (a) and (e) fit the data?
summary(fit)
## 
## Call:
## lm(formula = Sales ~ Price + Urban + US, data = Carseats)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -6.9206 -1.6220 -0.0564  1.5786  7.0581 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 13.043469   0.651012  20.036  < 2e-16 ***
## Price       -0.054459   0.005242 -10.389  < 2e-16 ***
## UrbanYes    -0.021916   0.271650  -0.081    0.936    
## USYes        1.200573   0.259042   4.635 4.86e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.472 on 396 degrees of freedom
## Multiple R-squared:  0.2393, Adjusted R-squared:  0.2335 
## F-statistic: 41.52 on 3 and 396 DF,  p-value: < 2.2e-16
summary(fit2)
## 
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -6.9269 -1.6286 -0.0574  1.5766  7.0515 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 13.03079    0.63098  20.652  < 2e-16 ***
## Price       -0.05448    0.00523 -10.416  < 2e-16 ***
## USYes        1.19964    0.25846   4.641 4.71e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.469 on 397 degrees of freedom
## Multiple R-squared:  0.2393, Adjusted R-squared:  0.2354 
## F-statistic: 62.43 on 2 and 397 DF,  p-value: < 2.2e-16
anova(fit, fit2)
## Analysis of Variance Table
## 
## Model 1: Sales ~ Price + Urban + US
## Model 2: Sales ~ Price + US
##   Res.Df    RSS Df Sum of Sq      F Pr(>F)
## 1    396 2420.8                           
## 2    397 2420.9 -1  -0.03979 0.0065 0.9357

They have similar \(R^2\) and the model containing the extra variable Urban is non-significantly better.

  1. Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(fit2)
##                   2.5 %      97.5 %
## (Intercept) 11.79032020 14.27126531
## Price       -0.06475984 -0.04419543
## USYes        0.69151957  1.70776632
  1. Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow = c(2, 2))
plot(fit2, cex = 0.2)

Yes, there are definitely some outliers, but most of the data is well fitted and grouped.

Question 12

This problem involves simple linear regression without an intercept.

  1. Recall that the coefficient estimate \(\hat{\beta}\) for the linear regression of \(Y\) onto \(X\) without an intercept is given by (3.38). Under what circumstance is the coefficient estimate for the regression of \(X\) onto \(Y\) the same as the coefficient estimate for the regression of \(Y\) onto \(X\)?

\[ \hat\beta = \sum_i x_iy_i / \sum_{i'} x_{i'}^2 \]

The coefficient for the regression of X onto Y swaps the \(x\) and \(y\) variables:

\[ \hat\beta = \sum_i x_iy_i / \sum_{i'} y_{i'}^2 \]

So they are the same when \(\sum_{i} x_{i}^2 = \sum_{i} y_{i}^2\)

  1. Generate an example in R with \(n = 100\) observations in which the coefficient estimate for the regression of \(X\) onto \(Y\) is different from the coefficient estimate for the regression of \(Y\) onto \(X\).
x <- rnorm(100)
y <- 2 * x + rnorm(100, 0, 0.1)
c(sum(x^2), sum(y^2))
## [1] 110.8061 445.4824
c(coef(lm(y ~ x))[2], coef(lm(x ~ y))[2])
##         x         y 
## 2.0038574 0.4978905
  1. Generate an example in R with \(n = 100\) observations in which the coefficient estimate for the regression of \(X\) onto \(Y\) is the same as the coefficient estimate for the regression of \(Y\) onto \(X\).
x <- rnorm(100)
y <- x + rnorm(100, 0, 0.1)
c(sum(x^2), sum(y^2))
## [1] 101.2825 102.0189
c(coef(lm(y ~ x))[2], coef(lm(x ~ y))[2])
##         x         y 
## 0.9997227 0.9923036