pdf('HW2.pdf')
##HW2 (CH3:2,9,10,12)
##Q2.
#The KNN classifier is typically used to solve classification problems (those with a qualitative response) by identifying the neighborhood of x0 and then estimating the conditional probability P(Y=j|X=x0) for class j as the fraction of points in the neighborhood whose response values equal j.
#The KNN regression method is used to solve regression problems (quantitative response) by again identifying the neighborhood of x0 and then estimating f(x0) as the average of all the training responses in the neighborhood.
##Q9.
#a
library(ISLR)
data(Auto)
pairs(Auto)
#b
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
#i.
fit <- lm(mpg ~ . - name, data = Auto)
summary(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-16
#Yes, there is a relationship.
#ii.
#displacement, weight, year, origin
#iii.
#The coefficient of the year variable suggests that when every other predictor held constant, then the mpg increases for each passing year.
#d
par(mfrow = c(2, 2))
plot(fit)
# The plots shows that there are no leverage points.
#e
fit2 <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(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-16
# By the p-values, the interaction between displacement and weight is statistically significant, while the interaction between cylinders and displacement is not.
#f
par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)
#It seems that the log transformation gives the most linear looking plot.
##Q10.
#a
data(Carseats)
fit3 <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(fit3)
##
## 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
# When price increase by 1000 and other variable are constant then sales also decrease;
# Store sale isnt affected by whether its in the Urban area
# A US store sells on average 1200 more carseats than others
#c
#Sales=13.0434689+(−0.0544588)×Price+(−0.0219162)×Urban+(1.2005727)×US+ε
#d
#We can reject the null hypothesis for the “Price” and “US” variables.
#e
fit4 <- lm(Sales ~ Price + US, data = Carseats)
summary(fit4)
##
## 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
#They both fit the data similarly, but linear regression from (e) fitting the data a bit better.
#g
confint(fit4)
## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632
#h
par(mfrow = c(2, 2))
plot(fit4)
#The plot of standardized residuals versus leverage shows the presence of a few outliers
#Q12.
#a
#The coefficients are the same if and only if ∑jx2j=∑jy2j.
#b
set.seed(1)
x <- 1:100
sum(x^2)
## [1] 338350
y <- 2 * x + rnorm(100, sd = 0.1)
sum(y^2)
## [1] 1353606
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.223590 -0.062560 0.004426 0.058507 0.230926
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 2.0001514 0.0001548 12920 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.09005 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.669e+08 on 1 and 99 DF, p-value: < 2.2e-16
summary(fit.X)
##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.115418 -0.029231 -0.002186 0.031322 0.111795
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 5.00e-01 3.87e-05 12920 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.04502 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.669e+08 on 1 and 99 DF, p-value: < 2.2e-16
#c
x <- 1:100
sum(x^2)
## [1] 338350
y <- 100:1
sum(y^2)
## [1] 338350
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -49.75 -12.44 24.87 62.18 99.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.5075 0.0866 5.86 6.09e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared: 0.2575, Adjusted R-squared: 0.25
## F-statistic: 34.34 on 1 and 99 DF, p-value: 6.094e-08
summary(fit.X)
##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -49.75 -12.44 24.87 62.18 99.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.5075 0.0866 5.86 6.09e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared: 0.2575, Adjusted R-squared: 0.25
## F-statistic: 34.34 on 1 and 99 DF, p-value: 6.094e-08
dev.off()
## quartz_off_screen
## 2