Assignment #2
Tien Vo
2023-02-17
library(ISLR2)Exercise 2
Carefully explain the differences between the KNN classifier and KNN regression methods
The difference between the KNN classifier and KNN regression methods is that the classifier is used in situations where the response variable is categorical (qualitative). THe KNN regression methods is used for solve regression problem (quantitative response) by identifying the neighborhood of x0 and then estimating f(x0) as the average of all the training responses in the neighborhood.
Exercise 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.
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(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
lm.fit=lm(mpg~.-name, data=Auto)
summary(lm.fit)##
## 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-16i. Is there a relationship between the predictors and the response?
Yes, There is a relationship between predictor and the response variable by a p-value < 0.05
ii. Which predictors appear to have a statistically significant relationship to the response?
Displacement, weight, year and origin have a statistically significant relationship with mpg.
iii. What does the coefficient for the year variable
suggest?
The year coefficient suggests a positive correlation between increasing 1 year, then the mpg is increased 0.75.
(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 highleverage?
par(mfrow = c(2, 2))
plot(lm.fit)
which.max(hatvalues(lm.fit))## 14
## 14The plot of residuals versus fitted values indicates the non linearity in the data. The plot of standardized residuals versus leverage indicate the outliers (higher than 2 or lower than -2) and one high leverage point 14.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.fit1 <- lm(mpg ~ +cylinders : displacement+displacement : weight, data = Auto[, 1:8])
summary(lm.fit1)##
## Call:
## lm(formula = mpg ~ +cylinders:displacement + displacement:weight,
## data = Auto[, 1:8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.5522 -3.3109 -0.6592 2.7231 17.4894
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.126e+01 3.884e-01 80.47 < 2e-16 ***
## cylinders:displacement -7.481e-04 1.335e-03 -0.56 0.576
## displacement:weight -1.041e-05 2.546e-06 -4.09 5.25e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.767 on 389 degrees of freedom
## Multiple R-squared: 0.6288, Adjusted R-squared: 0.6269
## F-statistic: 329.5 on 2 and 389 DF, p-value: < 2.2e-16lm.fit2 <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(lm.fit2)##
## 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-16The interaction between displacement and weight is statistically significant (p-value < 0.05), while the interaction between cylinders and displacement is not (p-value > 0.05)
(f) Try a few different transformations of the variables, such aslog(X), √X, X2. Comment on your findings.
lm.fit3<-lm(log(mpg) ~ . - name, data = Auto)
summary(lm.fit3)##
## Call:
## lm(formula = log(mpg) ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.40955 -0.06533 0.00079 0.06785 0.33925
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.751e+00 1.662e-01 10.533 < 2e-16 ***
## cylinders -2.795e-02 1.157e-02 -2.415 0.01619 *
## displacement 6.362e-04 2.690e-04 2.365 0.01852 *
## horsepower -1.475e-03 4.935e-04 -2.989 0.00298 **
## weight -2.551e-04 2.334e-05 -10.931 < 2e-16 ***
## acceleration -1.348e-03 3.538e-03 -0.381 0.70339
## year 2.958e-02 1.824e-03 16.211 < 2e-16 ***
## origin 4.071e-02 9.955e-03 4.089 5.28e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1191 on 384 degrees of freedom
## Multiple R-squared: 0.8795, Adjusted R-squared: 0.8773
## F-statistic: 400.4 on 7 and 384 DF, p-value: < 2.2e-16par(mfrow=c(2,2))
plot(lm.fit3)
All predictors are statistically significant except acceleration
Exercise 10
This question should be answered using the Carseats
data set
attach(Carseats)(a) Fit a multiple regression model to predict Sales
using Price, Urban, and US.
mult_regression <- lm(Sales~Price+Urban+US)
summary(mult_regression)##
## 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 interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
Urban has no effect on Sales. So, Price and US are two variables used for Sales predictions.Increase $1 in Price, Sales go down by $54.4. Sales inside the US are $1,200 higher than outside the US.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales = 13.043469−0.054459XPrice − 0.021916XUrbanYes + 1.200573XUSYes
(d) For which of the predictors can you reject the null hypothesis
Price and US because they have p-value <
0.05
(e) On the basis of your response to the previous question, fit asmaller model that only uses the predictors for which there is evidence of association with the outcome
mult_regression <- lm(Sales~Price+US)
summary(mult_regression)##
## 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?
not good
(g) Using the model from (e), obtain 95 % confidence intervals forthe coefficient(s).
confint(mult_regression)## 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)?
par(mfrow=c(2,2))
plot(mult_regression)
Exercise 12
This problem involves simple linear regression without an intercept
(a) 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?
It will be equal if the summation of Xi^2 equals with the summation of Yi^2
(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 (1)
x <- rnorm (100)
y <- 2 * x + rnorm (100)
sum(x^2)## [1] 81.05509sum(y^2)## [1] 413.2135fit.X <- lm(x ~ y + 0)
summary(fit.X)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.8699 -0.2368 0.1030 0.2858 0.8938
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.39111 0.02089 18.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4246 on 99 degrees of freedom
## Multiple R-squared: 0.7798, Adjusted R-squared: 0.7776
## F-statistic: 350.7 on 1 and 99 DF, p-value: < 2.2e-16fit.Y <- lm(y ~ x + 0)
summary(fit.Y)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9154 -0.6472 -0.1771 0.5056 2.3109
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 1.9939 0.1065 18.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared: 0.7798, Adjusted R-squared: 0.7776
## F-statistic: 350.7 on 1 and 99 DF, p-value: < 2.2e-16(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 (1)
x <- rnorm (100)
y <- 1 * x
sum(x^2)## [1] 81.05509sum(y^2)## [1] 81.05509fit.X <- lm(x ~ y + 0)
summary(fit.X)## Warning in summary.lm(fit.X): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.888e-16 -1.689e-17 1.339e-18 3.057e-17 2.552e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 1.000e+00 6.479e-18 1.543e+17 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.833e-17 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 2.382e+34 on 1 and 99 DF, p-value: < 2.2e-16fit.Y <- lm(y ~ x + 0)
summary(fit.Y)## Warning in summary.lm(fit.Y): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.888e-16 -1.689e-17 1.339e-18 3.057e-17 2.552e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 1.000e+00 6.479e-18 1.543e+17 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.833e-17 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 2.382e+34 on 1 and 99 DF, p-value: < 2.2e-16