Assignment2
Eduardo Ramirez"
2/17/2020
Problem 2
Carefully explain the differences between the KNN classifier and KNN regression methods
KNN regression predicts the quatntitave value for f(X) while KNN classifier is the classification output for y.
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.
#install.packages('ISLR')
library(ISLR)## Warning: package 'ISLR' was built under R version 3.6.2attach(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.
library(ISLR)
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(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:
library(ISLR)
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-16i. Is there a relationship between the predictors and the response?
Yes, there are multiple predictors that are statistically signifigant
ii. Which predictors appear to have a statistically significant relationship to the response?
displacement, weight, year, and origin
iii. What does the coefficient for the year variable suggest?
for every increase in year there is an increase of mpg by about .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?
par(mfrow = c(2,2))
plot(lm.fit1)
summary(influence.measures(lm.fit1))## 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 0.05There is an outliar but there is no certain unussaly high leverage on the plots
(e) Use the and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
library('ISLR')
attach(Auto)## The following objects are masked from Auto (pos = 3):
##
## acceleration, cylinders, displacement, horsepower, mpg, name,
## origin, weight, yearlm.fit2<-lm(mpg~cylinders*displacement+displacement*weight)
summary(lm.fit2)##
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement *
## weight)
##
## 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-16To see if there was an interaction, you use the two most highly correlated predictors and using the p values, we can see an interaction between displacemnet and wieght.
Problem 10
This question should be answered using the Carseats data set.
library(ISLR)
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!
From the table above, price and US are significant predictors of Sales, for every $1 increase my price, my sales go down by $54. Sales inside of the US are $1,200 higher than sales outside of the US. Urban has no effect on Sales.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.043469 -0.054459Price-0.021916Urban_{Yes}+1.200573XUS_{Yes}\)
(d) For which of the predictors can you reject the null hypothesis.
Price and US
(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?
Terrible, each model explains around 23% of the variance in Sales.
(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)? R has built in functions to that can help us identify influential points using various statistics with one simple command. Researchers have suggested several cutoff levels or upper limits as to what is the acceptable influence an observation should have before being considered an outlier. For example, the average leverage \(\frac{(p+1)}{n}\) which for us 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 0.00R points out a few observations that violate various rules for each influence measure. Typically, one can demonstrate these statistics and report both a regression with all data included and one with the outliers removed and compare.
outyling.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[-outyling.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-16With these potential outliers or influential observations removed, very little changes from the linear model fit to the full data set. The confidence interval for the coefficient estimates produced by the linear model fit to the full data set contain the estimates of the coefficients for the estimates of the model with the outliers removed. It’s safe to include all of the data points in our model.
##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 coefficient estimate for x onto y is the same as y onto x when the sum of the squares of the observed y-values are equal to the sum of the squares of the observed x-values
(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
lm.fit = lm(y~x+0)
lm.fit2 = lm(x~y+0)
summary(lm.fit)## Warning in summary.lm(lm.fit): essentially perfect fit: summary may be
## unreliable##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.776e-16 -3.378e-17 2.680e-18 6.113e-17 5.105e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 2.000e+00 1.296e-17 1.543e+17 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.167e-16 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-16summary(lm.fit2)## Warning in summary.lm(lm.fit2): 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 5.00e-01 3.24e-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(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 <- -sample(x, 100)
sum(x^2)## [1] 81.05509sum(y^2)## [1] 81.05509lm.fit <- lm(y~x+0)
lm.fit2 <- lm(x~y+0)
summary(lm.fit)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2833 -0.6945 -0.1140 0.4995 2.1665
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.07768 0.10020 0.775 0.44
##
## Residual standard error: 0.9021 on 99 degrees of freedom
## Multiple R-squared: 0.006034, Adjusted R-squared: -0.004006
## F-statistic: 0.601 on 1 and 99 DF, p-value: 0.4401summary(lm.fit2)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2182 -0.4969 0.1595 0.6782 2.4017
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.07768 0.10020 0.775 0.44
##
## Residual standard error: 0.9021 on 99 degrees of freedom
## Multiple R-squared: 0.006034, Adjusted R-squared: -0.004006
## F-statistic: 0.601 on 1 and 99 DF, p-value: 0.4401