Assignment 2

install.packages(c(“ISLR2”,“car”)) library(ISLR2) library(car)

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Q2

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

Answer:

KNN classifier deals with categorical targets and identifies the K closest neighbors, takes a majority vote to assign a class label.

whereas

KNN regression deals with continuous numerical targets, identifies the K closest neighbor and calculates the arithmetic average of their value to predice a specific number.

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#—————————————————- # Q9. # This question involves the use of multiple linear # regression on the Auto data set. #—————————————————-

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a. Produce a scatterplot matrix which includes all the variables in the data set.

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pairs(Auto[,-9], main ="Scatter Plot Matrix of Auto Dataset")

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b. Compute the matrix of correlations between the variables using the function cor(). Exclude the name variable,which is qualitativel

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cor(Auto[, -9])

                    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

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c. Use lm() function to perform a mutlitple linear regression with mpg as the response and all variables except name as the predictors.

Use summary() function to print the results.

i.Is there a relationship between predictors and the response

Answer: Yes

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

Answer: Displacement, Weight, Year and Origin

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

Answer: For every one year increase in model year, the vehicle’s fuel efficiency increases by 0.75 MPG.

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lr<-lm(mpg ~ . - name, data = Auto)
summary(lr)

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

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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?

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plot(lr)

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

Do any interactions appear to be statistically significant?

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lr2 <- lm(mpg ~ (.-name) ^2,data = Auto )
summary(lr2)

Call:
lm(formula = mpg ~ (. - name)^2, data = Auto)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.6303 -1.4481  0.0596  1.2739 11.1386 

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 **
year:origin                1.393e-01  7.399e-02   1.882  0.06062 . 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.695 on 363 degrees of freedom
Multiple R-squared:  0.8893,    Adjusted R-squared:  0.8808 
F-statistic: 104.2 on 28 and 363 DF,  p-value: < 2.2e-16

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f. Try a few different transformations of the variables, such a log(X), √X, X2.

Comment on your findings.

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lr_trans <- lm(mpg ~ . -name + log(horsepower) + I(horsepower ^2) + log (weight) + I(displacement^2), data = Auto)
summary(lr_trans)

Call:
lm(formula = mpg ~ . - name + log(horsepower) + I(horsepower^2) + 
    log(weight) + I(displacement^2), data = Auto)

Residuals:
   Min     1Q Median     3Q    Max 
-9.273 -1.497 -0.110  1.446 11.974 

Coefficients:
                    Estimate Std. Error t value Pr(>|t|)    
(Intercept)        1.415e+02  4.757e+01   2.976  0.00311 ** 
cylinders          1.732e-01  3.648e-01   0.475  0.63521    
displacement      -3.681e-02  1.994e-02  -1.846  0.06564 .  
horsepower         4.667e-02  1.714e-01   0.272  0.78556    
weight             1.078e-03  2.115e-03   0.510  0.61049    
acceleration      -2.018e-01  1.005e-01  -2.008  0.04533 *  
year               7.657e-01  4.514e-02  16.963  < 2e-16 ***
origin             5.465e-01  2.670e-01   2.046  0.04140 *  
log(horsepower)   -1.375e+01  9.530e+00  -1.442  0.15002    
I(horsepower^2)    6.682e-05  3.507e-04   0.191  0.84901    
log(weight)       -1.469e+01  6.803e+00  -2.159  0.03145 *  
I(displacement^2)  6.712e-05  3.436e-05   1.954  0.05148 .  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.905 on 380 degrees of freedom
Multiple R-squared:  0.8654,    Adjusted R-squared:  0.8615 
F-statistic: 222.1 on 11 and 380 DF,  p-value: < 2.2e-16

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Q10. This question should be answered using the Carseats data set.

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

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lr <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(lr)

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

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b. Provide an interpretation of each coefficient in the model.

Becareful—some of the variables in the model are qualitative!

Answer:

Price - for every increase in price, sales decrease

UrbanYes - results dont appear statistically significant

USYes - Stores in the US are expected to see higher sales on average compared to those outside of the US

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c. Write out the model in equation form, being careful to handle

the qualitative variables properly.

Answer: 𝑆𝑎𝑙𝑒𝑠=𝛽0+𝛽1(𝑃𝑟𝑖𝑐𝑒)+𝛽2(𝑈𝑟𝑏𝑎𝑛𝑌𝑒𝑠)+𝛽3(𝑈𝑆𝑌𝑒𝑠)+𝜖

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

Answer: We can reject null hypothesis for Price and US

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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.

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

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

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f. How well do the models in (a) and (e) fit the data?

Answer: second model appears to fit the model a generally better than the first model

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

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confint(lr)

                  2.5 %      97.5 %
(Intercept) 11.79032020 14.27126531
Price       -0.06475984 -0.04419543
USYes        0.69151957  1.70776632

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h. Is there evidence of outliers or high leverage observations in the model from (e)?

Answer: Yes, there are outliers in model from (e)

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Q12. This problem involves simple linear regression without an intercept.

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a. Recall that the coefficient estimateˆβ 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?

Answer: When the sum of squares of the X values equals the sum of squares of the Y values

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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.

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set.seed(42)
x<-rnorm(100, mean=0, sd =1)
y<-2 * x + rnorm(100, mean =0, sd =2)

x_fit <-lm(y ~ x - 1)
y_fit <-lm(x ~ y - 1)

coef(x_fit)

coef(y_fit)

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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.

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set.seed(42)
x <- rnorm(100, mean=0, sd = 1)
y<-sample(x)

x_fit <-lm(y~x-1)
y_fit <-lm(x~y-1)

coef(x_fit)

coef(y_fit)