Assignment 2
Kliner
6/7/2020
Chapter 03
Problem 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
KNN classifiers assign the test observation a qualitative class. This is done by using surrounding observations (neighbors) in a training data set to estimate the conditional probability for class assignment.
KNN Regression is a non-parametric method that is similar to KNN classification. However, instead of classifying the variable, the regression attempts to predict the value of the variable through a local average.
Problem 9
This question involves the use of multiple linear regression on the Auto data set.
library(ISLR)
attach(Auto)(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)
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:
lmAuto <- lm(mpg~.-name, data=Auto)
summary(lmAuto)##
## 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?
There is a relationship with four of the predictors, the remaining three are not statistically significant. I recall that “origin” should be removed from this lm as it should be a factor, the response 1, 2, & 3 represent American, European, & Japanese. I will change the variable to a factor and refit the model without it.
lapply(Auto, class)## $mpg
## [1] "numeric"
##
## $cylinders
## [1] "numeric"
##
## $displacement
## [1] "numeric"
##
## $horsepower
## [1] "numeric"
##
## $weight
## [1] "numeric"
##
## $acceleration
## [1] "numeric"
##
## $year
## [1] "numeric"
##
## $origin
## [1] "numeric"
##
## $name
## [1] "factor"Auto$origin <- as.factor(Auto$origin)
lapply(Auto, class)## $mpg
## [1] "numeric"
##
## $cylinders
## [1] "numeric"
##
## $displacement
## [1] "numeric"
##
## $horsepower
## [1] "numeric"
##
## $weight
## [1] "numeric"
##
## $acceleration
## [1] "numeric"
##
## $year
## [1] "numeric"
##
## $origin
## [1] "factor"
##
## $name
## [1] "factor"lmAuto2 <- lm(mpg~.-name, data=Auto)
summary(lmAuto2)##
## Call:
## lm(formula = mpg ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.0095 -2.0785 -0.0982 1.9856 13.3608
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.795e+01 4.677e+00 -3.839 0.000145 ***
## cylinders -4.897e-01 3.212e-01 -1.524 0.128215
## displacement 2.398e-02 7.653e-03 3.133 0.001863 **
## horsepower -1.818e-02 1.371e-02 -1.326 0.185488
## weight -6.710e-03 6.551e-04 -10.243 < 2e-16 ***
## acceleration 7.910e-02 9.822e-02 0.805 0.421101
## year 7.770e-01 5.178e-02 15.005 < 2e-16 ***
## origin2 2.630e+00 5.664e-01 4.643 4.72e-06 ***
## origin3 2.853e+00 5.527e-01 5.162 3.93e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.307 on 383 degrees of freedom
## Multiple R-squared: 0.8242, Adjusted R-squared: 0.8205
## F-statistic: 224.5 on 8 and 383 DF, p-value: < 2.2e-16I can now see that four variables are statistically significant.
ii. Which predictors appear to have a statistically significant relationship to the response?
The predictors weight, year, and origin have a statistically significant relationship to the response.Displacement also has a statistically significant relationship to the response, though slightly less in comparison.
iii. What does the coefficient for the year variable suggest?
The year coefficient is positive (.75). This suggests that when all other coefficient remain the same a one unit increase in year will cause a .75 increase in mpg.
(d) Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit.
par(mfrow = c(2,2))
plot(lmAuto2)
Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
The QQ plot show the data is relatively normal, though right-skewed. Cook’s D shows point 14 could produce unusually high leverage in the data. This suspicion is confirmed with which.max() below.
which.max(hatvalues(lmAuto2))## 14
## 14(e) Use the * and : symbols to fit linear regression models with interaction effects.
lmAuto3 <- lm(mpg~.-name +weight:horsepower +weight*acceleration +acceleration:horsepower, data=Auto)
summary(lmAuto3)##
## Call:
## lm(formula = mpg ~ . - name + weight:horsepower + weight * acceleration +
## acceleration:horsepower, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.5123 -1.6147 0.0547 1.4456 12.0577
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.254e-01 7.633e+00 0.069 0.94516
## cylinders 2.378e-01 2.951e-01 0.806 0.42096
## displacement -4.649e-03 7.982e-03 -0.582 0.56064
## horsepower -1.192e-01 4.025e-02 -2.962 0.00325 **
## weight -1.285e-02 2.540e-03 -5.061 6.5e-07 ***
## acceleration 5.326e-03 2.950e-01 0.018 0.98561
## year 7.706e-01 4.585e-02 16.806 < 2e-16 ***
## origin2 1.139e+00 5.308e-01 2.145 0.03257 *
## origin3 1.468e+00 4.998e-01 2.937 0.00352 **
## horsepower:weight 5.272e-05 6.634e-06 7.947 2.2e-14 ***
## weight:acceleration 1.897e-04 1.347e-04 1.409 0.15971
## horsepower:acceleration -7.967e-03 2.495e-03 -3.193 0.00153 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.885 on 380 degrees of freedom
## Multiple R-squared: 0.8672, Adjusted R-squared: 0.8634
## F-statistic: 225.6 on 11 and 380 DF, p-value: < 2.2e-16Do any interactions appear to be statistically significant?
I selected the following iterations; weight and horsepower, weight and acceleration, acceleration and horsepower. Of the selected iterations horsepower:weight and horsepower:acceleration suggest statistical significance.
(f) Try a few different transformations of the variables, such as log(X),√X, X2. Comment on your findings.
lmAuto4 <- lm(mpg~.-name -displacement -weight -horsepower +log(displacement) +sqrt(weight) +I(horsepower^2), data=Auto)
summary(lmAuto4)##
## Call:
## lm(formula = mpg ~ . - name - displacement - weight - horsepower +
## log(displacement) + sqrt(weight) + I(horsepower^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5245 -1.9833 0.0659 1.7781 12.9190
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.431e+00 6.363e+00 1.011 0.31281
## cylinders 5.424e-01 2.964e-01 1.830 0.06802 .
## acceleration 1.805e-01 8.398e-02 2.150 0.03220 *
## year 8.049e-01 4.930e-02 16.326 < 2e-16 ***
## origin2 1.217e+00 5.835e-01 2.085 0.03770 *
## origin3 1.151e+00 5.803e-01 1.984 0.04794 *
## log(displacement) -2.587e+00 1.472e+00 -1.758 0.07962 .
## sqrt(weight) -7.096e-01 7.162e-02 -9.908 < 2e-16 ***
## I(horsepower^2) 1.006e-04 3.625e-05 2.775 0.00579 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.174 on 383 degrees of freedom
## Multiple R-squared: 0.838, Adjusted R-squared: 0.8346
## F-statistic: 247.7 on 8 and 383 DF, p-value: < 2.2e-16These transformations made all the variables statistically significant, to one degree or another. It further illustrates the relationship of horsepower and weight have on mpg.
detach(Auto)Problem 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.
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 in price, 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.
Problem 12
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate ˆβ 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?
Both variables would need to be standardized for the variance to be equal before a regression of X to Y and Y to X can be 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(42)
x <- rnorm(100) #random number
y <- rbinom(100,5,.5) #random binomial distribution
xyset <- lm(x~y+0)
summary(xyset)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.06848 -0.65508 0.07654 0.62229 2.21125
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.02513 0.04095 0.614 0.541
##
## Residual standard error: 1.04 on 99 degrees of freedom
## Multiple R-squared: 0.00379, Adjusted R-squared: -0.006272
## F-statistic: 0.3767 on 1 and 99 DF, p-value: 0.5408yxset <- lm(y~x+0)
summary(yxset)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.1239 1.2680 2.0358 3.0913 5.0182
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.1508 0.2458 0.614 0.541
##
## Residual standard error: 2.548 on 99 degrees of freedom
## Multiple R-squared: 0.00379, Adjusted R-squared: -0.006272
## F-statistic: 0.3767 on 1 and 99 DF, p-value: 0.5408For X to Y, the coefficient estimate is 0.0251302
For Y to X, the coefficient estimate is 0.150832
(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
xyset2 <- lm(x~y+0)
summary(xyset2)##
## 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-08yxset2 <- lm(y~x+0)
summary(yxset2)##
## 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-08For X to Y, the coefficient estimate is 0.5074627
For Y to X, the coefficient estimate is 0.5074627