Chapter 2 HW
Tyler Crawford
2022-09-08
Problem 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
- KNN regression tries to predict the value of the output variable by using a local average. KNN classification attempts to predict the class to which the output variable belongs by computing the local probability.
Problem 9
This question involves the use of multiple linear regression on the Auto data set.
library(ISLR2)
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, cor() which is qualitative.
cor(Auto[-9])## 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:
lm1=lm(mpg~.,data=Auto[-9])
summary(lm1)##
## Call:
## lm(formula = mpg ~ ., data = Auto[-9])
##
## 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-16c1. Is there a relationship between the predictors and the response?
- Yes, there is a relationship between the response variable of MPG and it’s predictors. Our R-squared is 0.8215 which means that approximately 82.15% of our response variable can be explained by the different predictors in our model. Further, our p-value of 2.2e-16 is much less than the level of significance 0.05, meaning that the overall model is statistically significant.
c2. Which predictors appear to have a statistically significant relationship to the response?
- To determine this, we can compare the p-value of the predictors to the significance level of 0.05. displacement, weight, year, and origin are all statistically significant to MPG.
c3. What does the coefficient for the year variable suggest?
- The coefficient for the year variable is 0.7507 which means that for every increment that mpg increases by 1, year increases by 0.75. Essentially, every 3 years mpg increases by 4.
(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(lm1)
Residuals vs Fitted: The U-shape pattern in our residuals indicates that the data is non-linear. This chart also shows that the variance is not constant as shown by the funnel shape and variance in data points on the right side of the chart.
Normal Q-Q: This chart shows that our data is normally distributed, except for a few data points towards the upper-right, namely, 326, 327, and 323.
Scale-Location: This chart is used to determine if there are any outliers to the data. Anything outside of three standard deviations from the mean will be defined as an outlier. In this case, our data ranges from 0 to 2 which means that there are no statistical outliers.
Residuals vs Leverage: The leverage plot does not indicate of high leverage in our data.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(mpg~.-name+horsepower*displacement,data=Auto))##
## Call:
## lm(formula = mpg ~ . - name + horsepower * displacement, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.7010 -1.6009 -0.0967 1.4119 12.6734
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.894e+00 4.302e+00 -0.440 0.66007
## cylinders 6.466e-01 3.017e-01 2.143 0.03275 *
## displacement -7.487e-02 1.092e-02 -6.859 2.80e-11 ***
## horsepower -1.975e-01 2.052e-02 -9.624 < 2e-16 ***
## weight -3.147e-03 6.475e-04 -4.861 1.71e-06 ***
## acceleration -2.131e-01 9.062e-02 -2.351 0.01921 *
## year 7.379e-01 4.463e-02 16.534 < 2e-16 ***
## origin 6.891e-01 2.527e-01 2.727 0.00668 **
## displacement:horsepower 5.236e-04 4.813e-05 10.878 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.912 on 383 degrees of freedom
## Multiple R-squared: 0.8636, Adjusted R-squared: 0.8608
## F-statistic: 303.1 on 8 and 383 DF, p-value: < 2.2e-16- One test I did was to fit a linear regression model to mpg with the interaction effect of horsepower*displacement. This fit was more statistically significant than our original model through the product of horsepower and displacement. The predictors cylinders and acceleration came into relationships with mpg as well as horsepower and displacement.
(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-16summary(lm(mpg~.-name+(acceleration^2),data=Auto))##
## Call:
## lm(formula = mpg ~ . - name + (acceleration^2), 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- log(acceleration) has strong significance to mpg, but not as strong as acceleration.
- Acceleration-squared has a strong significance to mpg, but not as strong as year or weight.
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.
lm1=lm(Sales~Price+Urban+US,data=Carseats)
summary(lm1)##
## 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 in my price, sales decrease by $54. Sales inside 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.
\(Sales=13.043469 - 0.054459Price - 0.021916UrbanYes + 1.200573USYes\)
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
- 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.
lm2=lm(Sales~Price+US,data=Carseats)
summary(lm2)##
## 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?
- Both of our models have an R-Squared of 23.93% (variance in sales), so neither of them fit particularly well.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(lm2)## 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(lm2)
- We can see on the Residuals vs Leverage plot that all data points fall between -3:3, meaning that there is no evidence of any outliers of high leverage observations.
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?
- Both X and Y have equal tendency to be away from the mean. They should equal when the coefficient is 1.
(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.
x=rnorm(100)
y=0.5*x+rnorm(100)
coefficients(lm(x~y+0))## y
## 0.3106775coefficients(lm(y~x+0))## x
## 0.5104772(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=rnorm(100)
y=1*x
coefficients(lm(x~y+0))## y
## 1coefficients(lm(y~x+0))## x
## 1