1. Carefully explain the differences between the KNN classifier and KNN regression methods. KNN classifier is used for qualitative things and KNN regression is used for quantitative things.

  2. 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.
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
library(MASS)
plot(Auto)
  1. Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
Auto1<-Auto
Auto1$name=NULL
cor(Auto1)
## 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
  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?
y1<-lm(mpg~ .-name,data=Auto)
summary(y1)
## 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 *
  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(y1)
plot(predict(y1),rstudent(y1))
plot(hatvalues(y1))
which.max(hatvalues(y1))
  1. Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
y2<-lm(mpg~.:.,Auto1)
summary(y2)
## Coefficients:
##                             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)                3.548e+01  5.314e+01   0.668  0.50475   
## cylinders                  6.989e+00  8.248e+00   0.847  0.39738   
## displacement              -4.785e-01  1.894e-01  -2.527  0.01192 * 
## horsepower                 5.034e-01  3.470e-01   1.451  0.14769   
## weight                     4.133e-03  1.759e-02   0.235  0.81442   
## acceleration              -5.859e+00  2.174e+00  -2.696  0.00735 **
## year                       6.974e-01  6.097e-01   1.144  0.25340   
## origin                    -2.090e+01  7.097e+00  -2.944  0.00345 **
## cylinders:displacement    -3.383e-03  6.455e-03  -0.524  0.60051   
## cylinders:horsepower       1.161e-02  2.420e-02   0.480  0.63157   
## cylinders:weight           3.575e-04  8.955e-04   0.399  0.69000   
## cylinders:acceleration     2.779e-01  1.664e-01   1.670  0.09584 . 
## cylinders:year            -1.741e-01  9.714e-02  -1.793  0.07389 . 
## cylinders:origin           4.022e-01  4.926e-01   0.816  0.41482   
## displacement:horsepower   -8.491e-05  2.885e-04  -0.294  0.76867   
## displacement:weight        2.472e-05  1.470e-05   1.682  0.09342 . 
## displacement:acceleration -3.479e-03  3.342e-03  -1.041  0.29853   
## displacement:year          5.934e-03  2.391e-03   2.482  0.01352 * 
## displacement:origin        2.398e-02  1.947e-02   1.232  0.21875   
## horsepower:weight         -1.968e-05  2.924e-05  -0.673  0.50124   
## horsepower:acceleration   -7.213e-03  3.719e-03  -1.939  0.05325 . 
## horsepower:year           -5.838e-03  3.938e-03  -1.482  0.13916   
## horsepower:origin          2.233e-03  2.930e-02   0.076  0.93931   
## weight:acceleration        2.346e-04  2.289e-04   1.025  0.30596   
## weight:year               -2.245e-04  2.127e-04  -1.056  0.29182   
## weight:origin             -5.789e-04  1.591e-03  -0.364  0.71623   
## acceleration:year          5.562e-02  2.558e-02   2.174  0.03033 * 
## acceleration:origin        4.583e-01  1.567e-01   2.926  0.00365 
## acceleration              -5.859e+00  2.174e+00  -2.696  0.00735 **
## year                       6.974e-01  6.097e-01   1.144  0.25340   
## origin                    -2.090e+01  7.097e+00  -2.944  0.00345 **
## cylinders:displacement    -3.383e-03  6.455e-03  -0.524  0.60051   
## cylinders:horsepower       1.161e-02  2.420e-02   0.480  0.63157   
## cylinders:weight           3.575e-04  8.955e-04   0.399  0.69000   
## cylinders:acceleration     2.779e-01  1.664e-01   1.670  0.09584 . 
## cylinders:year            -1.741e-01  9.714e-02  -1.793  0.07389 . 
## cylinders:origin           4.022e-01  4.926e-01   0.816  0.41482   
## displacement:horsepower   -8.491e-05  2.885e-04  -0.294  0.76867   
## displacement:weight        2.472e-05  1.470e-05   1.682  0.09342 . 
## displacement:acceleration -3.479e-03  3.342e-03  -1.041  0.29853   
## displacement:year          5.934e-03  2.391e-03   2.482  0.01352 * 
## displacement:origin        2.398e-02  1.947e-02   1.232  0.21875   
## horsepower:weight         -1.968e-05  2.924e-05  -0.673  0.50124   
## horsepower:acceleration   -7.213e-03  3.719e-03  -1.939  0.05325 . 
## horsepower:year           -5.838e-03  3.938e-03  -1.482  0.13916   
## horsepower:origin          2.233e-03  2.930e-02   0.076  0.93931   
## weight:acceleration        2.346e-04  2.289e-04   1.025  0.30596   
## weight:year               -2.245e-04  2.127e-04  -1.056  0.29182   
## weight:origin             -5.789e-04  1.591e-03  -0.364  0.71623
  1. Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings.
y4<-lm(mpg~weight+I((weight)^2),Auto)
summary(y4)
## lm(formula = mpg ~ weight + I((weight)^2), data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -12.6246  -2.7134  -0.3485   1.8267  16.0866 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    6.226e+01  2.993e+00  20.800  < 2e-16 ***
## weight        -1.850e-02  1.972e-03  -9.379  < 2e-16 ***
## I((weight)^2)  1.697e-06  3.059e-07   5.545 5.43e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. This question should be answered using the Carseats data set.
  1. Fit a multiple regression model to predict Sales using Price, Urban, and US.
sales<-lm(Sales~Price+Urban+US,data=Carseats)
summary(sales)
## lm(formula = Sales ~ Price + Urban + US, data = Carseats)
## 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 ***
  1. For which of the predictors can you reject the null hypothesis H0 : βj = 0? 0.05
  2. 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.
sales_var<-lm(Sales~Price+US,data=Carseats)
summary(sales_var)
## 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 ***
  1. How well do the models in (a) and (e) fit the data?
anova(sales,sales_var)
  1. Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).?
confint(sales_var)
plot(predict(sales_var),rstudent(sales_var))
  1. This problem involves simple linear regression without an intercept.

  1. Recall that the coefficient estimate βˆ 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? x^2 y^2

  2. 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=rbinom(100,2,0.3)
eg<-lm(y~x+0)
summary(eg)

eg1<-lm(x~y+0)
summary(eg1)
## lm(formula = x ~ y + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.7685 -0.6248 -0.1147  0.6004  2.4687 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)
## y  0.09873    0.10400   0.949    0.345
## 
## Residual standard error: 0.9645 on 99 degrees of freedom
## Multiple R-squared:  0.00902,    Adjusted R-squared:  -0.0009897 
## F-statistic: 0.9011 on 1 and 99 DF,  p-value: 0.3448
  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=1:100
y=100:1
eg3<-lm(y~x+0)
summary(eg3)
eg4<-lm(x~y+0)
summary(eg4)
## Call:
## lm(formula = x ~ y + 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|)    
## y   0.5075     0.0866    5.86 6.09e-08 ***