Assignment2

Problem 2

Carefully explain the differences between the KNN classifies and KNN regressions methods

• The KNN Classifier and Regression have the same core idea, using the same neighbor-finding algorithm. They differ depending on what they are trying to predict. The Classifier is used to predict a qualitative variable but the Regression is used to predict a quantitative variable. Classifier basically holds a vote among the neighbors and assigns whichever category wins (Majority rules) whole the regression takes the average of the neighbors to make its prediction. Classification picks a category and regression predicts a number, both relying on the same idea.

Problem 9

(a) Produce a scatter-plot matrix which includes all of the variables in the data set.

library(ISLR2)
pairs(Auto[,-9])

(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(Auto[,-9])

                    mpg  cylinders displacement horsepower     weight acceleration       year     origin
mpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442    0.4233285  0.5805410  0.5652088
cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273   -0.5046834 -0.3456474 -0.5689316
displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944   -0.5438005 -0.3698552 -0.6145351
horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377   -0.6891955 -0.4163615 -0.4551715
weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000   -0.4168392 -0.3091199 -0.5850054
acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392    1.0000000  0.2903161  0.2127458
year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199    0.2903161  1.0000000  0.1815277
origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054    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.

fit_auto <- lm(mpg~ . - name , Auto)
summary(fit_auto)

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

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

- Yes, there are several variables that reject the null hypothesis.

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

- displacement, weight, year, and origin

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

- A positive coefficient for `year` tells us that newer cars get better miles per gallon.

(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?

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

• The plot shows us a non-linear pattern, showing us potential outliers towards the end of the Q-Q Residuals plot. Number 14 on the Leverage plot has a really high leverage compared to the rest.

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

lm_inter <- lm(mpg~horsepower*weight+displacement:weight, Auto)
summary(lm_inter)

Call:
lm(formula = mpg ~ horsepower * weight + displacement:weight, 
    data = Auto)

Residuals:
     Min       1Q   Median       3Q      Max 
-10.6474  -2.2039  -0.3157   2.0342  14.8543 

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)    
(Intercept)          6.313e+01  2.402e+00  26.278  < 2e-16 ***
horsepower          -2.571e-01  2.837e-02  -9.062  < 2e-16 ***
weight              -1.046e-02  8.644e-04 -12.102  < 2e-16 ***
horsepower:weight    5.692e-05  7.845e-06   7.255 2.21e-12 ***
weight:displacement -1.491e-06  1.842e-06  -0.809    0.419    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.932 on 387 degrees of freedom
Multiple R-squared:  0.7489,    Adjusted R-squared:  0.7463 
F-statistic: 288.5 on 4 and 387 DF,  p-value: < 2.2e-16

• The horsepower:weight interaction is significant but the weight:displacment interaction is non-significant.

(f)Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.

lm_nolog <- lm(mpg~horsepower + weight, Auto)
lm_log <- lm(mpg~log(horsepower) + sqrt(weight), Auto)
summary(lm_nolog)

Call:
lm(formula = mpg ~ horsepower + weight, data = Auto)

Residuals:
     Min       1Q   Median       3Q      Max 
-11.0762  -2.7340  -0.3312   2.1752  16.2601 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) 45.6402108  0.7931958  57.540  < 2e-16 ***
horsepower  -0.0473029  0.0110851  -4.267 2.49e-05 ***
weight      -0.0057942  0.0005023 -11.535  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.24 on 389 degrees of freedom
Multiple R-squared:  0.7064,    Adjusted R-squared:  0.7049 
F-statistic: 467.9 on 2 and 389 DF,  p-value: < 2.2e-16

summary(lm_log)

Call:
lm(formula = mpg ~ log(horsepower) + sqrt(weight), data = Auto)

Residuals:
     Min       1Q   Median       3Q      Max 
-10.4597  -2.4807  -0.3574   2.2158  15.4697 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)      89.2761     3.3768  26.438  < 2e-16 ***
log(horsepower)  -7.9526     1.2365  -6.431 3.71e-10 ***
sqrt(weight)     -0.5431     0.0554  -9.803  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.035 on 389 degrees of freedom
Multiple R-squared:  0.734, Adjusted R-squared:  0.7327 
F-statistic: 536.8 on 2 and 389 DF,  p-value: < 2.2e-16

• The log transformation on horsepower made R-squared jump up my 3%. From looking at the scatter-plolts in the beginning we could tell horsepower was slightly skewed but with the help of the log transformation it made it more normal and predicted mpg better.

Problem 10

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

library(ISLR2)
attach(Carseats)
fit <- lm(Sales~ Price+Urban+US)
summary(fit)

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 interpritation of each coefficient in the model. Be careful some of the variables in the model are qualitative!

From the table above, price and us are significant predictors of Sales, for every \(1 increase my price. my sales go down\) 54. Sales inside of the US are $1,200 higher than sales outside of the us. Urban has little to no effect on sales

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

• Sales = 13 ;- .054Price-.021UrbanYes+1.200USYes

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

• Price and US can reject the null hypothesis.

(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_better <- lm(Sales~ Price+US)
summary(fit_better)

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?

• Both perform poorly with an R-squared being around 23%, there was not much of a difference in removing Urban.

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

confint(fit_better)

                  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)?

#Leverage: # of predictors + 1/n (cook.d)
summary(influence.measures(fit_better))

Potentially influential observations of
     lm(formula = Sales ~ Price + US) :

    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   0.00  

Problem 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 estimates are only equal when X and Y vary by the same amount overall. If X is much more spread out than Y, or vice versa, you’ll get different answers

(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(42)
n <- 100
x1 <- rnorm(n,mean=0,sd=1)
y1 <- rnorm(n,mean=0,sd=2)

sum(x1*y1)/ sum(x1^2)

[1] 0.04897101

sum(x1*y1)/ sum(y1^2)

[1] 0.01610329

(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(42)
n <- 100
x2 <- rnorm(n,mean=0,sd=1)
y_raw <- rnorm(n,mean=0,sd=1)
y2 <- y_raw*sqrt(sum(x2^2)/sum(y_raw^2))

sum(x2*y2)/ sum(x2^2)

[1] 0.02808193

sum(x2*y2)/ sum(y2^2)

[1] 0.02808193