Data Mining HW 2
E. Falcon
2/17/2020
Problem 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
“KNN classifier first identifies the neighbors K points in the training data that are closest to \(x_{0}\), represented by N. It then estimates the conditional probability for class j as the fraction of points in N…KNN applies Bayes rule and classifies the test observation \(x_{0}\) to the class with the largest probability.” “Given a value for K and a prediction point \(x_{0}\), KNN regression first identifies the K training observations that are closest to \(x_{0}\), represented by N. It then estimates \(f(x_{0})\) using the average of all the training responses in N.”
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().
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.
lm.fit <- lm(mpg ~ . -name, data = Auto)
summary(lm.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-16Comment on the output.
i. Is there a relationship between the predictors and the response?
- Looking at the results of the multiple regression model we will see that based on the F test the overall model is significant. The F-statistic is 252.4, since this is far larger than 1, it provides compelling evidence against the null hypothesis. In other words, the large F-statistic suggests that at least one of the predictors be related to mpg.
ii. Which predictors appear to have a statistically significant relationship to the response?
- Yes, all predictors are significant excluding, “cylinders,” “horsepower,” and “acceleration.”
iii. What does the coefficient for the year variable suggest?
- We would expect a increase of about 0.750773 in mpg for every one year increase.
(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(lm.fit)
Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
- In regards to the Residuals vs Fitted, the residuals have a slight parabola shape. This suggests that the assumption that the relationship in linear is not reasonable.
- The Q-Q plot, the residuals seem to normally distributed. However, there are several observations that might be a potential problem. Like observations number 323 and 326.
- The Scale-Location shows that the assumption of equal variance is met.
- The Residuals vs Leverage plot we see that observation 14 is far beyond Cook’s distance, indicating that we have a influential case in our data. Removing the observation will alter our regression results.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.fit2 <- lm(mpg ~ displacement*weight + year*origin, data = Auto)
summary(lm.fit2)##
## Call:
## lm(formula = mpg ~ displacement * weight + year * origin, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5758 -1.6211 -0.0537 1.3264 13.3266
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.793e+01 8.044e+00 2.229 0.026394 *
## displacement -7.519e-02 9.091e-03 -8.271 2.19e-15 ***
## weight -1.035e-02 6.450e-04 -16.053 < 2e-16 ***
## year 4.864e-01 1.017e-01 4.782 2.47e-06 ***
## origin -1.503e+01 4.232e+00 -3.551 0.000432 ***
## displacement:weight 2.098e-05 2.179e-06 9.625 < 2e-16 ***
## year:origin 1.980e-01 5.436e-02 3.642 0.000308 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.969 on 385 degrees of freedom
## Multiple R-squared: 0.8575, Adjusted R-squared: 0.8553
## F-statistic: 386.2 on 6 and 385 DF, p-value: < 2.2e-16Using only the significant predictors given in the first model, such as “displacement,” “weight,” “year,” and “origin.” The displacement and weight interaction is significant and the year and origin interaction is also significant with both having a p-value of <0.001.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
lm.fit3 <- lm(mpg ~ weight + I(weight^2) + sqrt(displacement)+ displacement + year + origin + log(acceleration) + acceleration, data = Auto)
summary(lm.fit3)##
## Call:
## lm(formula = mpg ~ weight + I(weight^2) + sqrt(displacement) +
## displacement + year + origin + log(acceleration) + acceleration,
## data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5393 -1.8394 0.0469 1.6850 12.6360
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.956e+01 1.290e+01 3.842 0.000143 ***
## weight -2.046e-02 2.518e-03 -8.127 6.14e-15 ***
## I(weight^2) 2.249e-06 3.673e-07 6.125 2.25e-09 ***
## sqrt(displacement) 1.042e-01 8.148e-01 0.128 0.898265
## displacement -5.956e-03 2.773e-02 -0.215 0.830067
## year 8.359e-01 4.437e-02 18.840 < 2e-16 ***
## origin 4.923e-01 2.640e-01 1.865 0.063011 .
## log(acceleration) -3.048e+01 7.652e+00 -3.984 8.12e-05 ***
## acceleration 2.050e+00 4.741e-01 4.324 1.95e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.936 on 383 degrees of freedom
## Multiple R-squared: 0.8614, Adjusted R-squared: 0.8585
## F-statistic: 297.5 on 8 and 383 DF, p-value: < 2.2e-16I included “weight” and I squared it in the lm.fit3 model, both are statistically significant with p-value < 0.001. I also included displacement and took the square root of displacement, both are not significant. I left year as it is, and it was significant with a p-value < 0.001. I also left “origin” as is, and it was it was not significant. Lastly, I included both acceleration and I took the log of acceleration, both of which were significant with a p-value < 0.001.
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.
fit1 <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(fit1)##
## 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!
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 \(H_{0} : β_{j} = 0\)?
- Price
- 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.
fit2 <- lm(Sales~ Price + US, data = Carseats)
summary(fit2)##
## 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?
They fit terribly, these models only account for 23% of the variance in Sales.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit2)## 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)?
par(mfrow=c(2,2))
plot(fit2)
summary(influence.measures(fit2))## 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 0.00outyling.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,]
fit3<-lm(Sales~Price+US,data=Carseats.small)
summary(fit3)##
## 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-16“With 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 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? \[\sum_{n=1}^{\infty} x^{2} = \sum_{n=1}^{\infty} y^{2}\] (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 <- 100:1
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
## -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-08summary(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