Assignment 2
STA-6543 Predictive Modeling — Chapter 3: An Introduction to Statistical Learning with Applications in R
Jerimey Simons
June 21, 2026
Question 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
The difference between KNN Classification and KNN Regression lies in the type of target variable being predicted and the method used to generate the prediction. While both techniques use the same underlying approach of calculating distances between observations to identify the k nearest neighbors, they differ in how the final output is determined.
KNN Classification is used when the response variable is categorical. After identifying the k nearest neighbors, the algorithm predicts the class of a new observation based on the majority class among those neighbors. In contrast, KNN Regression is used when the response variable is continuous and numerical. Rather than predicting a class, it estimates the output value by taking the average of the response values of the k nearest neighbors.
Question 9
This question involves the use of multiple linear regression on the Auto Data set.
(a) Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto)
(b) Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative. cor()
cor(Auto[1:8])## 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(c) 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.
model <- lm(mpg ~ cylinders + displacement + horsepower + weight + acceleration + year +
origin, data = Auto )
summary(model)##
## Call:
## lm(formula = mpg ~ cylinders + displacement + horsepower + weight +
## acceleration + year + origin, data = Auto)
##
## 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-16For instance: i. Is there a relationship between the predictors and the response?
Based on the regression model summary for the Auto dataset, several
variables appear to be significantly related to a vehicle’s miles per
gallon mpg.
ii. Which predictors appear to have a statistically significant relationship to the response?
In particular, displacement, weight,
year, and origin have statistically
significant relationships with mpg.
iii. What does the coefficient for the year variable suggest?
Holding all other variables constant, a one-year increase in a vehicle’s model year is associated with an average increase of approximately 0.75 miles per gallon (mpg).
(d) 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?
plot(model)
# Leverage values
min(hatvalues(model))## [1] 0.005334609# Leverage values
max(hatvalues(model))## [1] 0.1899129Based on the diagnostic plots, several observations appear to be potential outliers. The leverage values range from 0.0053 to 0.1899, and the maximum leverage exceeds the common cutoff value of 0.041 (2(p+1)/n). This suggests that one or more observations have unusual predictor values and may exert substantial influence on the regression model.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
interaction_model <- lm(mpg ~ horsepower * weight + year * origin + displacement * weight,
data = Auto )
summary(interaction_model)##
## Call:
## lm(formula = mpg ~ horsepower * weight + year * origin + displacement *
## weight, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.4474 -1.6780 -0.0525 1.4634 11.8047
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.841e+01 7.914e+00 2.326 0.02056 *
## horsepower -1.357e-01 4.204e-02 -3.229 0.00135 **
## weight -1.094e-02 6.996e-04 -15.641 < 2e-16 ***
## year 5.288e-01 1.021e-01 5.177 3.65e-07 ***
## origin -1.091e+01 4.319e+00 -2.525 0.01198 *
## displacement -3.282e-02 1.798e-02 -1.825 0.06874 .
## horsepower:weight 2.936e-05 1.216e-05 2.415 0.01622 *
## year:origin 1.491e-01 5.528e-02 2.698 0.00729 **
## weight:displacement 1.102e-05 5.187e-06 2.125 0.03421 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.899 on 383 degrees of freedom
## Multiple R-squared: 0.8648, Adjusted R-squared: 0.862
## F-statistic: 306.3 on 8 and 383 DF, p-value: < 2.2e-16The interaction terms horsepower:weight, year:origin, and weight:displacement are all statistically significant, with p-values less than 0.05.
interaction_model <- lm(mpg ~ horsepower:weight + year:origin + displacement:weight,
data = Auto )
summary(interaction_model)##
## Call:
## lm(formula = mpg ~ horsepower:weight + year:origin + displacement:weight,
## data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.4144 -3.0707 -0.3628 2.2426 14.6139
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.728e+01 8.180e-01 33.354 < 2e-16 ***
## horsepower:weight -1.815e-05 3.924e-06 -4.625 5.10e-06 ***
## year:origin 3.167e-02 4.479e-03 7.071 7.19e-12 ***
## weight:displacement -2.252e-06 1.744e-06 -1.291 0.197
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.447 on 388 degrees of freedom
## Multiple R-squared: 0.6779, Adjusted R-squared: 0.6754
## F-statistic: 272.2 on 3 and 388 DF, p-value: < 2.2e-16A regression model containing only the interaction terms horsepower:weight, year:origin, and weight:displacement was fit to predict mpg. The interactions horsepower:weight (p=5.10×10 −6) and year:origin (p=7.19×10 −12) are statistically significant, indicating that the combined effects of these variables are associated with fuel efficiency. However, the interaction weight:displacement is not statistically significant (p=0.197), suggesting insufficient evidence that the combined effect of weight and displacement is related to mpg after accounting for the other interaction terms.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
fit_log <- lm(mpg ~ log(horsepower) + log(weight), data = Auto)
summary(fit_log)##
## Call:
## lm(formula = mpg ~ log(horsepower) + log(weight), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.6665 -2.4028 -0.3842 2.1558 15.3359
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 179.973 7.420 24.25 < 2e-16 ***
## log(horsepower) -7.672 1.210 -6.34 6.36e-10 ***
## log(weight) -15.244 1.478 -10.32 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.993 on 389 degrees of freedom
## Multiple R-squared: 0.7396, Adjusted R-squared: 0.7382
## F-statistic: 552.4 on 2 and 389 DF, p-value: < 2.2e-16A linear regression model was fit using the log-transformed
predictors horsepower and weight. Both log(horsepower) and
log(weight) were highly statistically significant
predictors of mpg (p < 0.001), indicating that horsepower and vehicle
weight have strong nonlinear relationships with fuel efficiency. The
negative coefficients suggest that increases in horsepower and weight
are associated with decreases in miles per gallon.
fit_sqrt <- lm(mpg ~ sqrt(horsepower) + sqrt(weight), data = Auto)
summary(fit_sqrt)##
## Call:
## lm(formula = mpg ~ sqrt(horsepower) + sqrt(weight), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.9211 -2.6240 -0.3587 2.2098 15.7776
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 68.49726 1.49097 45.941 < 2e-16 ***
## sqrt(horsepower) -1.27083 0.23678 -5.367 1.38e-07 ***
## sqrt(weight) -0.59713 0.05522 -10.813 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.096 on 389 degrees of freedom
## Multiple R-squared: 0.726, Adjusted R-squared: 0.7246
## F-statistic: 515.5 on 2 and 389 DF, p-value: < 2.2e-16A linear regression model was fit using sqrt(horsepower)
and sqrt(weight) as predictors of mpg. Both transformed
variables were highly statistically significant (p < 0.001),
indicating that horsepower and vehicle weight remain important
predictors of fuel efficiency after transformation. The negative
coefficients suggest that increases in horsepower and weight are
associated with lower miles per gallon.
fit_quad <- lm(mpg ~ horsepower + I(horsepower^2), data = Auto)
summary(fit_quad)##
## Call:
## lm(formula = mpg ~ horsepower + I(horsepower^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.7135 -2.5943 -0.0859 2.2868 15.8961
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 56.9000997 1.8004268 31.60 <2e-16 ***
## horsepower -0.4661896 0.0311246 -14.98 <2e-16 ***
## I(horsepower^2) 0.0012305 0.0001221 10.08 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.374 on 389 degrees of freedom
## Multiple R-squared: 0.6876, Adjusted R-squared: 0.686
## F-statistic: 428 on 2 and 389 DF, p-value: < 2.2e-16A quadratic regression model was fit using horsepower and horsepower² as predictors of mpg. Both the linear term (horsepower) and the quadratic term (horsepower²) were highly statistically significant (p < 0.001). The significance of the squared term provides strong evidence that the relationship between horsepower and fuel efficiency is nonlinear rather than purely linear.
The model explains approximately 68.8% of the variation in
mpg (\(R^2 = 0.6876\)),
with an adjusted \(R^2\) of 0.6860.
Although the quadratic term is statistically significant, the model
explains less variation than the logarithmic transformation model (\(R^2 = 0.7396\)) and substantially less than
the interaction model (\(R^2 =
0.8648\)). The residual standard error of 4.374 is also larger
than that of the other transformed models, indicating a poorer overall
fit.
Question 10
This question should be answered using the Carseats data set.
library(ISLR)
attach(Carseats)(a) Fit a multiple regression model to predict
Sales using Price, Urban, and
US.
fit <- lm(Sales ~ Price + Urban + US, data = Carseats)
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(b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
From the summary from the linear regression model Price
and US are significant predictors of Sales.
While holding all other variable fixed, every $1.00 increase in
Price, sales decrease by 54 units sold. Additionally,
Sales inside the US sale 1.2 thousand more
units than stores outside the US.
However, the model \(R^2\) = 0.2393
means that 24% of the variation in Sales is explained by
the predictors Price, Urban, and
US, suggesting that other factors not included in the model
also influence sales.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.043469 - 0.054459(\text{Price}) - 0.021916(\text{UrbanYes})+ 1.200573(\text{USYes})\)
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
We reject the null hypothesis for Price and
US
(e) 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.
fit <- lm(Sales ~ Price + US, data = Carseats)
summary(fit)##
## 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(f) How well do the models in (a) and (e) fit the data?
Models in (A) and (E) only explain on 23% of the variance in
Sales.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632(h) Is there evidence of outliers or high leverage observations in the model from (e)?
plot(fit)
Based of the above graphs there are several outliers. Using the common cutoff of \(\frac{(p+1)}{n}\) or \(\frac{(2+1)}{400}=0.0075\), where p=2 predictors and n=400 observations, several observations exceed the threshold for high leverage.
par(mfrow=c(2,2))
summary(influence.measures(fit))## Potentially influential observations of
## lm(formula = Sales ~ Price + US, data = Carseats) :
##
## dfb.1_ dfb.Pric dfb.USYs dffit cov.r cook.d hat
## 26 0.24 -0.18 -0.17 0.28_* 0.97_* 0.03 0.01
## 29 -0.10 0.10 -0.10 -0.18 0.97_* 0.01 0.01
## 43 -0.11 0.10 0.03 -0.11 1.05_* 0.00 0.04_*
## 50 -0.10 0.17 -0.17 0.26_* 0.98 0.02 0.01
## 51 -0.05 0.05 -0.11 -0.18 0.95_* 0.01 0.00
## 58 -0.05 -0.02 0.16 -0.20 0.97_* 0.01 0.01
## 69 -0.09 0.10 0.09 0.19 0.96_* 0.01 0.01
## 126 -0.07 0.06 0.03 -0.07 1.03_* 0.00 0.03_*
## 160 0.00 0.00 0.00 0.01 1.02_* 0.00 0.02
## 166 0.21 -0.23 -0.04 -0.24 1.02 0.02 0.03_*
## 172 0.06 -0.07 0.02 0.08 1.03_* 0.00 0.02
## 175 0.14 -0.19 0.09 -0.21 1.03_* 0.02 0.03_*
## 210 -0.14 0.15 -0.10 -0.22 0.97_* 0.02 0.01
## 270 -0.03 0.05 -0.03 0.06 1.03_* 0.00 0.02
## 298 -0.06 0.06 -0.09 -0.15 0.97_* 0.01 0.00
## 314 -0.05 0.04 0.02 -0.05 1.03_* 0.00 0.02_*
## 353 -0.02 0.03 0.09 0.15 0.97_* 0.01 0.00
## 357 0.02 -0.02 0.02 -0.03 1.03_* 0.00 0.02
## 368 0.26 -0.23 -0.11 0.27_* 1.01 0.02 0.02_*
## 377 0.14 -0.15 0.12 0.24 0.95_* 0.02 0.01
## 384 0.00 0.00 0.00 0.00 1.02_* 0.00 0.02
## 387 -0.03 0.04 -0.03 0.05 1.02_* 0.00 0.02
## 396 -0.05 0.05 0.08 0.14 0.98_* 0.01 0.00outlying_obs <- c(26, 29, 43, 50, 51, 58, 69, 126, 160, 166, 172, 175, 210, 270, 298, 314, 353, 357, 368, 377, 384, 387, 396)
model_removed <- lm(Sales ~ Price + US, data = Carseats[-outlying_obs, ])
summary(model_removed)##
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats[-outlying_obs,
## ])
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.263 -1.605 -0.039 1.590 5.428
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 12.925232 0.665259 19.429 < 2e-16 ***
## Price -0.053973 0.005511 -9.794 < 2e-16 ***
## USYes 1.255018 0.248856 5.043 7.15e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.29 on 374 degrees of freedom
## Multiple R-squared: 0.2387, Adjusted R-squared: 0.2347
## F-statistic: 58.64 on 2 and 374 DF, p-value: < 2.2e-16Removing the potentially influential observations resulted in only
minor changes to the coefficient estimates and model fit statistics. The
coefficients for Price and US remained highly
significant and retained similar magnitudes. The model’s \(R^2\) value remained virtually unchanged
(0.2393 versus 0.2387), while the residual standard error decreased
slightly from 2.469 to 2.290. Therefore, there is little evidence that
the identified observations substantially influenced the overall
conclusions of the regression model.
Question 12
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate ˆβ 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?
The coefficient estimates are the same when the sum of squares of X equals the sum of squares of Y.
(b) 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.
set.seed(123)
n <- 100
# Generate X
X <- rnorm(n, mean = 0, sd = 1)
# Generate Y with a different scale
Y <- 2 * X + rnorm(n, mean = 0, sd = 1)
# Regression of Y onto X (no intercept)
fit_yx <- lm(Y ~ X - 1)
# Regression of X onto Y (no intercept)
fit_xy <- lm(X ~ Y - 1)
coef(fit_yx)## X
## 1.936372coef(fit_xy)## Y
## 0.3975655(c) 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.
set.seed(123)
n <- 100
# Generate X
X <- rnorm(n, mean = 0, sd = 1)
# Set Y equal to X
Y <- X
# Regression of Y onto X (no intercept)
fit_yx <- lm(Y ~ X + 0)
# Regression of X onto Y (no intercept)
fit_xy <- lm(X ~ Y + 0)
coef(fit_yx)## X
## 1coef(fit_xy)## Y
## 1