install.packages(‘knitr’)
The KNN classifier is used to classify a result into qualitative groups based on the most common group and the KNN regression makes a quantitative estimate by averaging the result.
library(ISLR) data(“Auto”)
Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto)
cor(subset(Auto, select = -name))
summary(my_lm)
Produce a scatterplot matrix which includes all of the variables in the data set. pairs(Auto)
Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative. 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 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: 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-16 i. Is there a relationship between the predictors and the response? Yes, there is a relatioship between the predictors and the response by testing the null hypothesis of whether all the regression coefficients are zero. The F-statistic is far from 1 (with a small p-value), indicating evidence against the null hypothesis.
Which predictors appear to have a statistically significant relationship to the response? Looking at the p-values associated with each predictor’s t-statistic, we see that displacement, weight, year, and origin have a statistically significant relationship, while cylinders, horsepower, and acceleration do not.
What does the coefficient for the year variable suggest? The regression coefficient for year, 0.7507727, suggests that for every one year, mpg increases by the coefficient. In other words, cars become more fuel efficient every year by almost 1 mpg / year.
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) Seems to be non-linear pattern, linear model not the best fit. From the leverage plot, point 14 appears to have high leverage, although not a high magnitude residual.
plot(predict(lm.fit1), rstudent(lm.fit1)) There are possible outliers as seen in the plot of studentized residuals because there are data with a value greater than 3.
Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant? lm.fit2 <- lm(mpg ~ cylinders * displacement + displacement * weight, data = Auto) summary(lm.fit2) ## ## Call: ## lm(formula = mpg ~ cylinders * displacement + displacement * ## weight, data = Auto) ## ## 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-16 Interaction between displacement and weight is statistically signifcant, while the interaction between cylinders and displacement is not.
