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

The KNN regression method seeks to predict Y for a given value X by considering the K nearest neighbors to X in the training data and taking the average of those responses.

The KNN classifier seeks to predict Y for a given X by classifying an observation to the class with the highest probability based on its closest neighbors.

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

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.1.2
data(Auto)

(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.

names(Auto)
## [1] "mpg"          "cylinders"    "displacement" "horsepower"   "weight"      
## [6] "acceleration" "year"         "origin"       "name"
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.

lm.fit=lm(mpg~.-name, data=Auto)
summary(lm.fit)
## 
## Call:
## lm(formula = mpg ~ . - name, 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-16

Comment on the output. For instance:

  1. Is there a relationship between the predictors and the response?

Yes, the R-squared statistic suggests that around 82% of the variation in the response can be explained by the predictors.

  1. Which predictors appear to have a statistically significant relationship to the response?

The predictors with significant P values are displacement, weight, year, and origin.

  1. What does the coefficient for the year variable suggest?

For each increase in year, there is a 0.75 increase in mpg. Fuel efficiency increases with each year.

(d) Use the plot() function to produce diagnostic plots of the linear regression fit.

par(mfrow=c(2,2))
plot(lm.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?

The observations at 323, 326, and 327 have the highest standardized residuals, and it appears the data is slightly skewed right.

The observation at 14 has unusually high leverage, but does not cross over the dashed line of Cook’s distance, so it is not an influential point.

(e) Use the * and : symbols to fit linear regression models with interaction effects.

summary(lm(mpg~cylinders*displacement, data = Auto))
## 
## Call:
## lm(formula = mpg ~ cylinders * displacement, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.0432  -2.4308  -0.2263   2.2048  20.9051 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            48.22040    2.34712  20.545  < 2e-16 ***
## cylinders              -2.41838    0.53456  -4.524 8.08e-06 ***
## displacement           -0.13436    0.01615  -8.321 1.50e-15 ***
## cylinders:displacement  0.01182    0.00207   5.711 2.24e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.454 on 388 degrees of freedom
## Multiple R-squared:  0.6769, Adjusted R-squared:  0.6744 
## F-statistic:   271 on 3 and 388 DF,  p-value: < 2.2e-16
summary(lm(mpg~displacement*weight, data = Auto))
## 
## Call:
## lm(formula = mpg ~ displacement * weight, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.8664  -2.4801  -0.3355   1.8071  17.9429 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)          5.372e+01  1.940e+00  27.697  < 2e-16 ***
## displacement        -7.831e-02  1.131e-02  -6.922 1.85e-11 ***
## weight              -8.931e-03  8.474e-04 -10.539  < 2e-16 ***
## displacement:weight  1.744e-05  2.789e-06   6.253 1.06e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.097 on 388 degrees of freedom
## Multiple R-squared:  0.7265, Adjusted R-squared:  0.7244 
## F-statistic: 343.6 on 3 and 388 DF,  p-value: < 2.2e-16
summary(lm(mpg~horsepower*weight, data = Auto))
## 
## Call:
## lm(formula = mpg ~ horsepower * weight, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.7725  -2.2074  -0.2708   1.9973  14.7314 
## 
## Coefficients:
##                     Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        6.356e+01  2.343e+00  27.127  < 2e-16 ***
## horsepower        -2.508e-01  2.728e-02  -9.195  < 2e-16 ***
## weight            -1.077e-02  7.738e-04 -13.921  < 2e-16 ***
## horsepower:weight  5.355e-05  6.649e-06   8.054 9.93e-15 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.93 on 388 degrees of freedom
## Multiple R-squared:  0.7484, Adjusted R-squared:  0.7465 
## F-statistic: 384.8 on 3 and 388 DF,  p-value: < 2.2e-16

Do any interactions appear to be statistically significant?

All of the interactions tested are significantly statistic according to their P values.

(f) Try a few different transformations of the variables, such as log(X), √X, X^2. Comment on your findings.

lm.fit2=lm(mpg~horsepower+log(horsepower), data=Auto)
lm.fit3=lm(mpg~horsepower+I(horsepower^2), data=Auto)
lm.fit4=lm(mpg~horsepower+sqrt(horsepower), data=Auto)
summary(lm.fit2)
## 
## Call:
## lm(formula = mpg ~ horsepower + log(horsepower), data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -14.5118  -2.5018  -0.2533   2.4446  15.3102 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     156.04057   12.08267  12.914  < 2e-16 ***
## horsepower        0.11846    0.02929   4.044 6.34e-05 ***
## log(horsepower) -31.59815    3.28363  -9.623  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.415 on 389 degrees of freedom
## Multiple R-squared:  0.6817, Adjusted R-squared:  0.6801 
## F-statistic: 416.6 on 2 and 389 DF,  p-value: < 2.2e-16
summary(lm.fit3)
## 
## 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-16
summary(lm.fit4)
## 
## Call:
## lm(formula = mpg ~ horsepower + sqrt(horsepower), data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -14.5479  -2.5677  -0.2663   2.2998  15.5098 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      105.31581    6.64657  15.845  < 2e-16 ***
## horsepower         0.41913    0.05867   7.144 4.49e-12 ***
## sqrt(horsepower) -12.48574    1.26337  -9.883  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.392 on 389 degrees of freedom
## Multiple R-squared:  0.685,  Adjusted R-squared:  0.6834 
## F-statistic:   423 on 2 and 389 DF,  p-value: < 2.2e-16

Transforming the horsepower variable did not change its significance.

10. This question should be answered using the Carseats data set.

data(Carseats)
attach(Carseats)

(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.

fitA=lm(Sales~Price+Urban+US)
summary(fitA)
## 
## Call:
## lm(formula = Sales ~ Price + Urban + US)
## 
## 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!

The observations with significance are Price and US. Urban is not significant related to Sales. For each $1 increase in Price, Sales go down $54. Sales inside the US are $1200 more than outside the US.

(c) Write out the model in equation form, being careful to handle the qualitative variables properly.

\(Sales = 13.043469 − 0.054459Price − 0.021916Urban_{Yes} + 1.200573 x US_{Yes}\)

(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?

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.

fitB=lm(Sales~Price+US)
summary(fitB)
## 
## Call:
## lm(formula = Sales ~ Price + US)
## 
## 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?

Neither fits very well as only around 24% of the variance in Sales can be explained by the predictors according to the R-square statistic.

(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).

confint(fitB)
##                   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)?

par(mfrow=c(2,2))
plot(fitB)

library(MASS)
stud_res=studres(fitB)
plot(stud_res)

There does not seem to be evidence of outliers or high leverage points that would be influential. The studentized residuals plot shows that no values have an absolute value greater than 3, so there are no clear outliers. The leverage plot shows that no observations cross the Cook’s distance line.

12. This problem involves simple linear regression without an intercept.

(a) 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?

They are the same if \(\sum_{j=1}^{n} x^2_j\) is equal to \(\sum_{j=1}^{n} y^2_j\)

(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(1)
x=rnorm(100)
y=rnorm(100)
sum(x^2)
## [1] 81.05509
sum(y^2)
## [1] 90.97864
fit1=lm(x~y+0)
fit2=lm(y~x+0)
summary(fit1)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.2182 -0.4973  0.1162  0.6898  2.4039 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)
## y -0.005456   0.094863  -0.058    0.954
## 
## Residual standard error: 0.9048 on 99 degrees of freedom
## Multiple R-squared:  3.341e-05,  Adjusted R-squared:  -0.01007 
## F-statistic: 0.003308 on 1 and 99 DF,  p-value: 0.9543
summary(fit2)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.9154 -0.6472 -0.1771  0.5056  2.3109 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)
## x -0.006124   0.106477  -0.058    0.954
## 
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared:  3.341e-05,  Adjusted R-squared:  -0.01007 
## F-statistic: 0.003308 on 1 and 99 DF,  p-value: 0.9543

(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.

x1=1:100
y1=100:1
sum(x1^2)
## [1] 338350
sum(y1^2)
## [1] 338350
fit3=lm(x1~y1+0)
fit4=lm(y1~x1+0)
summary(fit3)
## 
## Call:
## lm(formula = x1 ~ y1 + 0)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -49.75 -12.44  24.87  62.18  99.49 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)    
## y1   0.5075     0.0866    5.86 6.09e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared:  0.2575, Adjusted R-squared:   0.25 
## F-statistic: 34.34 on 1 and 99 DF,  p-value: 6.094e-08
summary(fit4)
## 
## Call:
## lm(formula = y1 ~ x1 + 0)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -49.75 -12.44  24.87  62.18  99.49 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)    
## x1   0.5075     0.0866    5.86 6.09e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared:  0.2575, Adjusted R-squared:   0.25 
## F-statistic: 34.34 on 1 and 99 DF,  p-value: 6.094e-08