Linear Regression
Richard Rivas
2022-09-17
knitr::opts_chunk$set(echo = TRUE, message=FALSE, warning = FALSE)Problem 2
2. Carefully explain the differences between the KNN classifier and KNN regression methods. KNN Classifier uses dependent variables that are qualitative and KNN regression method uses quanitative. Regression models tend to be more flexible than a classifier model.
Problem 9
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(ISLR)
attach(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.
auto_num =subset(Auto, select=-name)
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:
m1=lm(mpg ~ ., data=auto_num)
summary(m1)##
## Call:
## lm(formula = mpg ~ ., data = auto_num)
##
## 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 the R-squared shows a corrolation of
82.15% between dependant variables. ii. Which
predictors appear to have a statistically significant relationship to
the response? Origin, year,
weight, and displacement all have a
significant relationship based on their p-value. iii. What does
the coefficient for the year variable suggest? The coefficient
year shows a 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. Do the residual plots
suggest any unusually large outliers? Does the leverage plot identify
any observations with unusually high leverage?
plot(m1)
(e) Use the and : symbols to fit linear regression models with
interaction effects. Do any interactions appear to be statistically
significant?
m2=lm(mpg~origin+year+displacement*weight, data=auto_num)
summary(m2)##
## Call:
## lm(formula = mpg ~ origin + year + displacement * weight, data = auto_num)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.6119 -1.7290 -0.0115 1.5609 12.5584
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -8.007e+00 3.798e+00 -2.108 0.0357 *
## origin 3.567e-01 2.574e-01 1.386 0.1666
## year 8.194e-01 4.518e-02 18.136 < 2e-16 ***
## displacement -7.148e-02 9.176e-03 -7.790 6.27e-14 ***
## weight -1.054e-02 6.530e-04 -16.146 < 2e-16 ***
## displacement:weight 2.104e-05 2.214e-06 9.506 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.016 on 386 degrees of freedom
## Multiple R-squared: 0.8526, Adjusted R-squared: 0.8507
## F-statistic: 446.5 on 5 and 386 DF, p-value: < 2.2e-16(f) Try a few different transformations of the variables, such as log(X), √, X, X2. Comment on your findings.
m3=lm(mpg~I(origin^2)+log(year)+displacement*weight, data=auto_num)
summary(m3)##
## Call:
## lm(formula = mpg ~ I(origin^2) + log(year) + displacement * weight,
## data = auto_num)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.6529 -1.7786 -0.0351 1.6398 12.6164
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.121e+02 1.491e+01 -14.221 < 2e-16 ***
## I(origin^2) 7.084e-02 6.399e-02 1.107 0.269
## log(year) 6.167e+01 3.454e+00 17.855 < 2e-16 ***
## displacement -7.268e-02 9.104e-03 -7.983 1.65e-14 ***
## weight -1.063e-02 6.598e-04 -16.104 < 2e-16 ***
## displacement:weight 2.130e-05 2.218e-06 9.605 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.036 on 386 degrees of freedom
## Multiple R-squared: 0.8506, Adjusted R-squared: 0.8487
## F-statistic: 439.7 on 5 and 386 DF, p-value: < 2.2e-16##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.
f= lm(Sales~Price+Urban+US)
summary(f)##
## 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! Price and US are significant. (c) Write out the model in equation form, being careful to handle the qualitative variables properly. Sales= 13.043469-0.054459Price-.021916Uarban +1.200573 US (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.
f2 =lm(Sales~Price+US)
summary(f2)##
## 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? Each model Explains around 24% of variance in sales. (g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).**
confint(f2)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632##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?
(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=1:100
y=x^2
fx=lm(x~y)
fy=lm(y~x)
summary(fy)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -833 -677 -208 573 1617
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1717.000 151.683 -11.32 <2e-16 ***
## x 101.000 2.608 38.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 752.7 on 98 degrees of freedom
## Multiple R-squared: 0.9387, Adjusted R-squared: 0.9381
## F-statistic: 1500 on 1 and 98 DF, p-value: < 2.2e-16summary(fx)##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.064 -5.121 2.036 6.357 7.845
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 19.054254 1.086528 17.54 <2e-16 ***
## y 0.009294 0.000240 38.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.221 on 98 degrees of freedom
## Multiple R-squared: 0.9387, Adjusted R-squared: 0.9381
## F-statistic: 1500 on 1 and 98 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
y=x
fxg=lm(x~y)
fyg=lm(y~x)
summary(fyg)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.680e-13 -4.300e-16 2.850e-15 5.302e-15 3.575e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.684e-14 5.598e-15 -1.015e+01 <2e-16 ***
## x 1.000e+00 9.624e-17 1.039e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.778e-14 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.08e+32 on 1 and 98 DF, p-value: < 2.2e-16summary(fxg)##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.680e-13 -4.300e-16 2.850e-15 5.302e-15 3.575e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.684e-14 5.598e-15 -1.015e+01 <2e-16 ***
## y 1.000e+00 9.624e-17 1.039e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.778e-14 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.08e+32 on 1 and 98 DF, p-value: < 2.2e-16