1. Carefully explain the differences between the KNN classifier, and the KNN regression methods.

The KNN classifier is used to solve classifiction problems with a qualitative response by identifying the neighborhood of Xo and then extimating the conditional probability P(Y=j|X=Xo) for class j as the fraction of points in the neighborhood whose respnse values equals j. The KNN regression method is used to solve regression problems with a quantitative response by again identifying the neighborhood of Xo and then estimating f(Xo) as the average of all the training responses in the neighborhood.

  1. This question involves the use of multiple linear regression of the Auto data set.
library(ISLR)
data("Auto")
  1. Produce a scatter plot 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
  1. Use the lm() function to perform a multiple lieaner regression with mpg as the respnse and all the other variables except name as the rpedictors. Use the summary() function to print the results. Comment the output. For instance: i. Is there a relationsshiopt between the predictors and the response?
fit2 <- lm(mpg ~ . - name, data = Auto)
summary(fit2)
## 
## 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
We can answer this question by again testing the hypothesis H0:βi=0 ∀i. The p-value corresponding  to the F-statistics is 2.037105910^{-139}, this indicates a clear evidence of a relationship between “mpg” and the other predictors.

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

We can answer this question by checking the p-values associated with each predictor’s t-statistic. We may conclude that all predictors are statistically significant except “cylinders”, “horsepower” and “acceleration”.

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

The coefficient ot the “year” variable suggests that the average effect of an increase of 1 year is an increase of 0.7507727 in “mpg” (all other predictors remaining constant). In other words, cars become more fuel efficient every year by almost 1 mpg / year.
  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 plots identify any observations with unusually high leverages ?
par(mfrow = c(2, 2))
plot(fit2)

As before, the plot of residuals versus fitted values indicates the presence of mild non linearity in the data. The plot of standardized residuals versus leverage indicates the presence of a few outliers (higher than 2 or lower than -2) and one high leverage point (point 14)

e. Use the * and : symbols to fit linear regression models with interaction effects. Do any interactinos appear to be statistically significant?

From the correlation matrix, we obtained the two highest correlated pairs and used them in picking interaction effects.
fit3 <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(fit3)
## 
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement * 
##     weight, data = Auto[, 1:8])
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.2934  -2.5184  -0.3476   1.8399  17.7723 
## 
## Coefficients:
##                          Estimate Std. Error t value Pr(>|t|)    
## (Intercept)             5.262e+01  2.237e+00  23.519  < 2e-16 ***
## cylinders               7.606e-01  7.669e-01   0.992    0.322    
## displacement           -7.351e-02  1.669e-02  -4.403 1.38e-05 ***
## weight                 -9.888e-03  1.329e-03  -7.438 6.69e-13 ***
## cylinders:displacement -2.986e-03  3.426e-03  -0.872    0.384    
## displacement:weight     2.128e-05  5.002e-06   4.254 2.64e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.103 on 386 degrees of freedom
## Multiple R-squared:  0.7272, Adjusted R-squared:  0.7237 
## F-statistic: 205.8 on 5 and 386 DF,  p-value: < 2.2e-16
  1. Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)

We limit ourselves to examining “horsepower” as sole predictor. It seems that the log transformation gives the most linear looking plot.

  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”
data("Carseats")
fit3
## 
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement * 
##     weight, data = Auto[, 1:8])
## 
## Coefficients:
##            (Intercept)               cylinders            displacement  
##              5.262e+01               7.606e-01              -7.351e-02  
##                 weight  cylinders:displacement     displacement:weight  
##             -9.888e-03              -2.986e-03               2.128e-05

  1. Provide an interpretation of each coefficient in the model. Becareful some of the variables in the model are qualitative! The coefficient of the “Price” variable may be interpreted by saying that the average effect of a price increase of 1 dollar is a decrease of 54.4588492 units in sales all other predictors remaining fixed. The coefficient of the “Urban” variable may be interpreted by saying that on average the unit sales in urban location are 21.9161508 units less than in rural location all other predictors remaining fixed. The coefficient of the “US” variable may be interpreted by saying that on average the unit sales in a US store are 1200.5726978 units more than in a non US store all other predictors remaining fixed.

  2. Write out the model in equation form, being careful to handle the qualitative variables properly. The model may be written as Sales=13.0434689+(−0.0544588)×Price+(−0.0219162)×Urban+(1.2005727)×US+ε with Urban = 1 If the store is in an urban location and 0 if not, and US = 1 if the store is in the US and 0 if not.

  3. For which of the predictors can you reject the null hypothesis H0 :βj =0? We can reject the null hypothesis for the “Price” and “US” variables.

  4. 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.

fit4 <- lm(Sales ~ Price + US, data = Carseats)
summary(fit4)
## 
## 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

  1. How well do the models in (a) and (e) fit the data? The R2 for the smaller model is marginally better than for the bigger model. Essentially about 23.9262888% of the variability is explained by the model.

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

confint(fit4)
##                   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(fit4)

The plot of standardized residuals versus leverage indicates the presence of a few outliers (higher than 2 or lower than -2) and some leverage points as some points exceed (p+1)/n(0.01).

  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? The coefficient estimate for the regression of Y onto X is β̂ =∑ixiyi∑jx2j; The coefficient estimate for the regression of X onto Y is β′ =∑ixiyi∑jy2j. The coefficients are the same iff ∑jx2j=∑jy2j.

  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.

set.seed(1)
x <- 1:100
sum(x^2)
## [1] 338350
y <- 2 * x + rnorm(100, sd = 0.1)
sum(y^2)
## [1] 1353606
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.223590 -0.062560  0.004426  0.058507  0.230926 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)    
## x 2.0001514  0.0001548   12920   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.09005 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 1.669e+08 on 1 and 99 DF,  p-value: < 2.2e-16
summary(fit.X)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.115418 -0.029231 -0.002186  0.031322  0.111795 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## y 5.00e-01   3.87e-05   12920   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.04502 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 1.669e+08 on 1 and 99 DF,  p-value: < 2.2e-16
  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
sum(x^2)
## [1] 338350
y <- 100:1
sum(y^2)
## [1] 338350
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
## 
## Call:
## lm(formula = y ~ x + 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|)    
## x   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(fit.X)
## 
## 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 ***
## ---
## 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