Assignment 2
ikq392
2022-09-18
Problem 2.
Carefully explain the differences between the KNN classifier and KNN regression methods.
The KNN classifier method uses information from the k neighbors to predict a categorical class assignment (qualitative). The KNN regression method uses information from k neighbors to predict a numerical value (quantitative).
Problem 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.
library(ISLR2)
attach(Auto)
View(Auto)
pairs(Auto)
(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[,!(names(Auto) %in% c("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(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:
i. Is there a relationship between the predictors and the
response?
ii. Which predictors appear to have a statistically significant
relationship to the response?
iii. What does the coefficient for the year variable
suggest?
mpglr<-lm(mpg~.-name, data=Auto)
summary(mpglr)##
## 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- There is a relationship between the predictors and the response. The
f-statistic is high at 252.4 with a low p-value meaning that the null
hypothesis of no relationship should be rejected.
- The predictors with statistically significant relationships can be
determined by identifying which predictors have p-values lower than the
typical threshold of 0.05: displacement, weight, year, and origin.
- The coefficient for the year variable, 0.750773, quantifies the relationship between year and mpg. For every year, mpg is improved by roughly 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 high leverage?
plot(mpglr)
summary(influence.measures(mpglr))## Potentially influential observations of
## lm(formula = mpg ~ . - name, data = Auto) :
##
## dfb.1_ dfb.cyln dfb.dspl dfb.hrsp dfb.wght dfb.accl dfb.year dfb.orgn
## 9 -0.01 -0.15 0.16 0.14 -0.14 0.09 0.00 0.03
## 13 0.01 -0.01 0.02 -0.01 -0.01 -0.01 -0.01 0.01
## 14 0.11 0.17 -0.42 -0.43 0.68 -0.36 -0.08 0.00
## 26 -0.04 0.01 -0.06 0.15 -0.02 0.10 -0.01 -0.02
## 27 -0.03 0.05 -0.08 0.11 -0.01 0.07 -0.02 -0.03
## 28 -0.05 0.07 -0.13 0.19 -0.02 0.10 -0.02 -0.05
## 29 -0.06 0.08 -0.14 0.18 0.02 0.15 -0.03 -0.04
## 112 -0.22 0.16 0.03 0.02 -0.14 0.19 0.18 -0.16
## 167 -0.01 -0.22 -0.02 0.03 0.21 0.06 -0.03 0.04
## 245 -0.10 0.05 0.12 0.05 -0.23 0.33 0.02 0.03
## 271 0.03 0.15 -0.11 0.03 -0.04 0.06 -0.06 -0.24
## 300 -0.03 -0.03 0.02 0.01 0.01 0.06 0.01 0.02
## 301 -0.04 0.05 -0.01 0.01 -0.03 0.05 0.02 0.00
## 310 -0.04 0.01 -0.04 -0.02 -0.01 -0.13 0.13 -0.02
## 323 -0.19 0.06 0.05 -0.01 -0.06 0.09 0.16 0.31
## 326 -0.20 0.05 0.10 0.06 -0.19 0.34 0.12 0.04
## 327 -0.28 0.04 0.06 0.11 -0.13 0.49 0.12 0.05
## 328 -0.14 0.14 -0.17 -0.07 0.19 0.06 0.08 0.06
## 330 -0.02 0.02 0.09 -0.15 -0.05 -0.21 0.12 0.22
## 387 -0.16 -0.16 0.40 -0.19 -0.19 0.05 0.26 0.04
## 394 -0.38 0.03 0.14 0.25 -0.32 0.58 0.23 0.03
## dffit cov.r cook.d hat
## 9 0.30 1.06 0.01 0.06_*
## 13 0.02 1.07_* 0.00 0.05
## 14 -0.79_* 1.19_* 0.08 0.19_*
## 26 0.18 1.07_* 0.00 0.06
## 27 0.13 1.09_* 0.00 0.07_*
## 28 0.22 1.08_* 0.01 0.07_*
## 29 0.25 1.11_* 0.01 0.09_*
## 112 -0.46_* 0.87_* 0.03 0.02
## 167 -0.39 0.93_* 0.02 0.03
## 245 0.48_* 0.82_* 0.03 0.02
## 271 -0.33 0.90_* 0.01 0.02
## 300 0.09 1.07_* 0.00 0.05
## 301 0.08 1.08_* 0.00 0.06
## 310 0.29 0.86_* 0.01 0.01
## 323 0.47_* 0.74_* 0.03 0.01
## 326 0.49_* 0.81_* 0.03 0.02
## 327 0.63_* 0.79_* 0.05 0.03
## 328 0.40 0.88_* 0.02 0.02
## 330 0.43 0.87_* 0.02 0.02
## 387 0.55_* 0.88_* 0.04 0.03
## 394 0.68_* 0.90_* 0.06 0.05The scale-location plot suggests no outliers because all points are within the -3 and 3 range.
The leverage plot does show point 14 at high leverage.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
interact.fit=lm(mpg~.-name+horsepower*displacement, data=Auto)
origin.hp=lm(mpg~.-name+horsepower*origin, data=Auto)
summary(origin.hp)##
## Call:
## lm(formula = mpg ~ . - name + horsepower * origin, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.277 -1.875 -0.225 1.570 12.080
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.196e+01 4.396e+00 -4.996 8.94e-07 ***
## cylinders -5.275e-01 3.028e-01 -1.742 0.0823 .
## displacement -1.486e-03 7.607e-03 -0.195 0.8452
## horsepower 8.173e-02 1.856e-02 4.404 1.38e-05 ***
## weight -4.710e-03 6.555e-04 -7.186 3.52e-12 ***
## acceleration -1.124e-01 9.617e-02 -1.168 0.2434
## year 7.327e-01 4.780e-02 15.328 < 2e-16 ***
## origin 7.695e+00 8.858e-01 8.687 < 2e-16 ***
## horsepower:origin -7.955e-02 1.074e-02 -7.405 8.44e-13 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.116 on 383 degrees of freedom
## Multiple R-squared: 0.8438, Adjusted R-squared: 0.8406
## F-statistic: 258.7 on 8 and 383 DF, p-value: < 2.2e-16Significant interactions include displacement and horsepower, and horsepower and origin.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
summary(lm(mpg~.-name+log(acceleration),data=Auto))##
## Call:
## lm(formula = mpg ~ . - name + log(acceleration), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.7931 -2.0052 -0.1279 1.9299 13.1085
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.552e+01 1.479e+01 3.077 0.00224 **
## cylinders -2.796e-01 3.193e-01 -0.876 0.38172
## displacement 8.042e-03 7.805e-03 1.030 0.30344
## horsepower -3.434e-02 1.401e-02 -2.450 0.01473 *
## weight -5.343e-03 6.854e-04 -7.795 6.15e-14 ***
## acceleration 2.167e+00 4.782e-01 4.532 7.82e-06 ***
## year 7.560e-01 4.978e-02 15.186 < 2e-16 ***
## origin 1.329e+00 2.724e-01 4.877 1.58e-06 ***
## log(acceleration) -3.513e+01 7.886e+00 -4.455 1.10e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.249 on 383 degrees of freedom
## Multiple R-squared: 0.8303, Adjusted R-squared: 0.8267
## F-statistic: 234.2 on 8 and 383 DF, p-value: < 2.2e-16Log(acceleration) is less significant than acceleration.
summary(lm(mpg~.-name+I(horsepower^2),data=Auto))##
## Call:
## lm(formula = mpg ~ . - name + I(horsepower^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.5497 -1.7311 -0.2236 1.5877 11.9955
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.3236564 4.6247696 0.286 0.774872
## cylinders 0.3489063 0.3048310 1.145 0.253094
## displacement -0.0075649 0.0073733 -1.026 0.305550
## horsepower -0.3194633 0.0343447 -9.302 < 2e-16 ***
## weight -0.0032712 0.0006787 -4.820 2.07e-06 ***
## acceleration -0.3305981 0.0991849 -3.333 0.000942 ***
## year 0.7353414 0.0459918 15.989 < 2e-16 ***
## origin 1.0144130 0.2545545 3.985 8.08e-05 ***
## I(horsepower^2) 0.0010060 0.0001065 9.449 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.001 on 383 degrees of freedom
## Multiple R-squared: 0.8552, Adjusted R-squared: 0.8522
## F-statistic: 282.8 on 8 and 383 DF, p-value: < 2.2e-16There is no change in significance when horsepower is squared.
Problem 10.
This question should be answered using the Carseats data set.
library(ISLR2)
attach(Carseats)(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
fit<-lm(Sales~Price+Urban+US)
summary(fit)##
## 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!
The Price coefficient has a statistically significant
effect on Sales. When price increases by $1000, the number
of car seats sold decreases by 54 units. The UrbanYes
coefficient is not statistically significant. Sales are not
impacted by store location. The USYes coefficient is
statistically significant. A US store sells 1200 more carseats than a
foreign store.
(c) Write out the model in equation form, being careful to
handle the qualitative variables properly.
\(Sales=13.043469 - 0.054459Price -
0.021916Urban_{Yes} + 1.200573US_{Yes}\)
(d) For which of the predictors can you reject the null
hypothesis \(H_0 : /beta_j =
0\)?
The null hypothesis can be rejected for the Price and
USYes predictors
(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.
fit<-lm(Sales~Price+US)
summary(fit)##
## 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?
The models do not fit the data well. The r-squared values in both
summaries suggest that only 24% of the change is explained.
(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(fit)## 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)?
The average leverage \(/frac{(p+1)}{n}\) is \(/frac{(2+1)}{400} = 0.0075\)
par(mfrow=c(2,2))
plot(fit)
summary(influence.measures(fit))## Potentially influential observations of
## lm(formula = Sales ~ Price + US) :
##
## 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 0.00There are several outlying observations as listed above. To remove and run a new regression:
outlying.obs<-c(26,29,43,50,51,58,69,126,160,166,172,175,210,270,298,314,353,357,368,377,384,387,396)
Carseats.small<-Carseats[-outlying.obs,]
fit2<-lm(Sales~Price+US,data=Carseats.small)
summary(fit2)##
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats.small)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.263 -1.605 -0.039 1.590 5.428
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 12.925232 0.665259 19.429 < 2e-16 ***
## Price -0.053973 0.005511 -9.794 < 2e-16 ***
## USYes 1.255018 0.248856 5.043 7.15e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.29 on 374 degrees of freedom
## Multiple R-squared: 0.2387, Adjusted R-squared: 0.2347
## F-statistic: 58.64 on 2 and 374 DF, p-value: < 2.2e-16When outliers are removed, there is very little change.
Problem 12.
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate ˆ β 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 coefficient estimate for the regression of X onto Y is the same as
the coefficient estimate for the regression of Y onto X if both the
numerator and denominator terms are the same.
(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
sum(x^2)## [1] 338350y<-2*x+rnorm(100,sd=0.1)
sum(y^2)## [1] 1353606fit.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-16summary(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) 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
sum(x^2)## [1] 338350y<-1:100
sum(y^2)## [1] 338350fit.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
## -3.082e-13 -2.094e-15 2.900e-17 2.218e-15 1.294e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 1.000e+00 5.379e-17 1.859e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.129e-14 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 3.457e+32 on 1 and 99 DF, p-value: < 2.2e-16fit.Y<-lm(y~x+0)
fit.X<-lm(x~y+0)
summary(fit.X)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.082e-13 -2.094e-15 2.900e-17 2.218e-15 1.294e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 1.000e+00 5.379e-17 1.859e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.129e-14 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 3.457e+32 on 1 and 99 DF, p-value: < 2.2e-16