library(ISLR2)Warning: package 'ISLR2' was built under R version 4.4.2data(Auto)
pairs(Auto[, 1:8])
Assignment #2
2. Carefully explain the differences between the KNN classifier and KNN regression methods.
- The KNN classifier predicts the most common class among the k nearest neighbor
- The KNN regression method predicts the average response of the k nearest neighbor
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.
(b) Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
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. For instance:
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-16i. Is there a relationship between the predictors and the response?
- Yes, the model is significant (p < 2.2e-16). There is a relationship
ii. Which predictors appear to have a statistically significant relationship to the response?
- Significant predictors: displacement, weight, year, origin
iii. What does the coefficient for the year variable suggest?
- The year coefficient means mpg increases by about 0.75 for each year
(d) Use the plot() function to produce diagnostic plots of the linear regression ft. Comment on any problems you see with the ft. 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(fit2)
- Residuals show mild non-linearity and a few large outliers
- One point has high leverage
(e) Use the * and : symbols to ft linear regression models with interaction effects. Do any interactions appear to be statistically significant?
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- Displacement:weight is significant
- Cylinders:displacement is not
(f) 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)
- Log(horsepower) looks most linear
- Square root is okay, squared is worst
10. This question should be answered using the Carseats data set. (a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
library(ISLR2)
attach(Carseats)
fit <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(fit)
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(b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
- Sales decrease by about 10.39 units for each $1 increase in Price
- US stores have 1.20 higher Sales on average than non-US stores
- Urban location does not affect Sales
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
- Sales = 13.04 - 0.10 * Price - 0.02 * UrbanYes + 1.20 * USYes
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
- Price and US are significant (p < 0.05)
- Urban is not significant
(e) On the basis of your response to the previous question, ft a smaller model that only uses the predictors for which there is evidence of association with the outcome.
fit_sm <- lm(Sales ~ Price + US, data = Carseats)
summary(fit_sm)
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(f) How well do the models in (a) and (e) fit the data?
- The smaller model is better because it is simpler and performs slightly better after removing Urban
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit_sm) 2.5 % 97.5 %
(Intercept) 11.79032020 14.27126531
Price -0.06475984 -0.04419543
USYes 0.69151957 1.70776632- Price: confidence interval is from -0.0648 to -0.0442
- USYes: confidence interval is from 0.6915 to 1.7077
- Since both intervals do not include 0, the predictors are statistically significant.
(h) Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow = c(2, 2))
plot(fit_sm)
summary(influence.measures(fit_sm))Potentially influential observations of
lm(formula = Sales ~ Price + US, data = Carseats) :
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 - No strong outliers or high leverage points
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?
The slopes are the same only when the total of X squared equals the total of Y squared.
(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 <- 1:100
y <- 2 * x + rnorm(100, sd = 0.1)
fit_y <- lm(y ~ x + 0)
fit_x <- lm(x ~ y + 0)
summary(fit_y)$coefficients[1][1] 2.000151summary(fit_x)$coefficients[1][1] 0.4999619(c) Generate an example in R with n = 100 observations in which the coefficient estimate for the regression of X onto Y is the same as the coefficient estimate for the regression of Y onto X.
x <- 1:100
y <- 100:1 # reversed x
sum(x^2) # 338350[1] 338350sum(y^2) # 338350[1] 338350fit_y <- lm(y ~ x + 0)
fit_x <- lm(x ~ y + 0)
summary(fit_y)$coefficients[1][1] 0.5074627summary(fit_x)$coefficients[1][1] 0.5074627