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Question #2.) Carefully explain the differences between the KNN classifier and KNN regression methods. Answer: A KNN regression method is a nonparametric, K-nearest neighbors regression. KNN regression identifies first the K training observations that are closest to x0, then estimates the average for f(x0) using the average of all training responses. A KNN classifier differs in that it classifies a result into qualitative (categorical) group using the most common group found among the K nearest neighbors.Using a positive integer K points in the training data that are closest to x0, then estimates the conditional probability for class j as the fraction of points, then classifies the test observation x0 to the class with the alrgest possibility.

Question #9.)This question involves the use of multiple linear regression on the Auto data set. (a) Produce a scatterplot matrix which includes all of the variables in the data set. Answer:

library(ISLR2)
data("Auto", package = "ISLR2")
pairs(Auto)

Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.

Answer:

cor(subset(Auto, select = -name))

##                     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

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:

lm.fit1 <-  lm(mpg ~ . - name, data = Auto)
summary(lm.fit1)

## 
## 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

Is there a relationship between the predictors and the response? Answer: Yes, by testing the hypothesis to verify if all regression coefficients are zero, we can see there is a relationship between the predictors and the response.
Which predictors appear to have a statistically significant relationship to the response? Answer: From the graphs we can see that weight, year, and displacement and origin have a significant relationship shown by their p-value of less than 0.5 f. We can see that acceleration, cylinders and horsepower do not have any statistical relationship.
What does the coefficient for the year variable suggest? Answer: The coefficient for the ‘year’ variable means that for every one year and it is also positive if every other variable is constant then then it means that the mpg increases as well by the coefficient. We can conclude that cars become more efficient every year by nearly 1mpg/year which we can see the relationship between them by their p-values.

Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit.

View(Auto)
par(mfrow = c(2, 2))

plot(lm.fit1)

plot(predict(lm.fit1), rstudent(lm.fit1))

There seems to be a nonlinear pattern.

Do the residual plots suggest any unusually large outliers? Answer: The Normal Q-Q plot we can see some outliers by the end from (2,2) to (3,4). Also the scale location plot we can see outliers by (30,1.5) to (35,2.0). The residuals vs Leverage plot we can see outliers starting from the beginning (0.025,2) to (0.05,4).

Does the leverage plot identify any observations with unusually high leverage? Answer: Yes, we can observe on Residuals vs Leverage plot that point 14 appears to have an unusually high leverage.

Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant? Answer: Yes, there seems to be a statistically significant interaction between weight and displacement.

lm.fit2 <-  lm(mpg ~ cylinders * displacement + displacement * weight, data = Auto)
summary(lm.fit2)

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

Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings. Answer: We can see that the Residuals vs Leverage plot that it has no points within its bounds which means that there are no influential points that can cause the slope coefficient like those from the previous plots.

y4<-lm(mpg~weight+I((weight)^2),Auto)
summary(y4)

## 
## Call:
## 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
## 
## Residual standard error: 4.176 on 389 degrees of freedom
## Multiple R-squared:  0.7151, Adjusted R-squared:  0.7137 
## F-statistic: 488.3 on 2 and 389 DF,  p-value: < 2.2e-16

Homework

Sandy Diaz

2022-09-13

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Including Plots