Homework 2
Sadiyah Hotakey
2/18/2022
##2## Carefully explain the differences between the KNN classifier and KNN regression methods. The KNN classifier classifies a result in a qualitative manner based on using the most common group found among the K nearest neighbors The regression method makes a quantitative estimate by taking the average of the K nearest neighbors.
##9## This question involves the use of multiple linear regression on the Auto data set.
library(ISLR)
attach(Auto)- Produce a scatterplot matrix which includes all of the variables in the data set.
plot(Auto)
- Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "name"cor(Auto[1:8])## 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.
fit<-lm(mpg~., data = Auto[, 1:8])
summary(fit)##
## Call:
## lm(formula = mpg ~ ., data = Auto[, 1:8])
##
## 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-16For instance: i. Is there a relationship between the predictors and the response? There’s a relationship between some of the predictors and the response because the F-statistic is < 2.2e-16. ii. Which predictors appear to have a statistically significant relationship to the response? Displacement, weight, year, and origin. iii. What does the coefficient for the year variable suggest? The average effect of the year variable increasing by one unit equals to an increase of .750773 units in mpg. The model seems to be a good fit because the R-squared is 82.15% of the variability can be explained by the model.
- 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 indiciates that there is non-linearity, presence of utliers, and high leverage obs > .05.
par(mfrow = c(2, 2))
plot(fit)
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant? Interaction between horsepower & weight and then between acceleration & year are significant.
fit2 <- lm(mpg ~ cylinders * displacement + horsepower * weight + acceleration * year, data = Auto[, 1:8])
summary(fit2)##
## Call:
## lm(formula = mpg ~ cylinders * displacement + horsepower * weight +
## acceleration * year, data = Auto[, 1:8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.3265 -1.5779 0.0389 1.3483 11.6961
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.162e+02 1.853e+01 6.274 9.53e-10 ***
## cylinders -1.803e-01 4.776e-01 -0.377 0.7061
## displacement -2.867e-02 1.425e-02 -2.013 0.0449 *
## horsepower -2.261e-01 2.609e-02 -8.664 < 2e-16 ***
## weight -1.019e-02 9.020e-04 -11.296 < 2e-16 ***
## acceleration -7.081e+00 1.158e+00 -6.113 2.41e-09 ***
## year -6.719e-01 2.417e-01 -2.780 0.0057 **
## cylinders:displacement 2.790e-03 2.067e-03 1.350 0.1779
## horsepower:weight 5.154e-05 6.727e-06 7.661 1.53e-13 ***
## acceleration:year 9.113e-02 1.502e-02 6.069 3.10e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.819 on 382 degrees of freedom
## Multiple R-squared: 0.8726, Adjusted R-squared: 0.8696
## F-statistic: 290.6 on 9 and 382 DF, p-value: < 2.2e-16- Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings. Horsepower seems the most linear than other variations when looking at mpg from original model, horspower, acceleration, & cylinders.
par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)
par(mfrow = c(2, 2))
plot(log(Auto$acceleration), Auto$mpg)
plot(sqrt(Auto$acceleration), Auto$mpg)
plot((Auto$acceleration)^2, Auto$mpg)
par(mfrow = c(2, 2))
plot(log(Auto$cylinders), Auto$mpg)
plot(sqrt(Auto$cylinders), Auto$mpg)
plot((Auto$cylinders)^2, Auto$mpg)
##10## This question should be answered using the Carseats data set.
library(ISLR)
attach(Carseats)- Fit a multiple regression model to predict Sales using Price, Urban, and US.
fit3<-lm(Sales~Price+Urban+US)
summary(fit3)##
## 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-16Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative! Price coefficient indiciates the average effect of price increase of $1 equals a decrease of $54.459 in sales. Urban’s coefficient indicates no relationship between Sales & Urban due to large p-value. US coefficient indiciates that on average sales in a US store is $1200.573 more than in a non-US store.
Write out the model in equation form, being careful to handle the qualitative variables properly. Sales=13.043469−0.054459Price−0.021916UrbanYes+1.200573USYes where UrbanYes=1 for Urban and 0 for not Urban and USYes=1 for U.S store and 0 for non U.S stores.
For which of the predictors can you reject the null hypothesis H0 : βj = 0? Reject null for Price & US because both have an effect on Sales.
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.
fit4<-lm(Sales~Price+US)
summary(fit4)##
## 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-16How well do the models in (a) and (e) fit the data? The model doesn’t seem to be a good fit because the R-squared for both models shows 23.93% of the variability can be explained by the model.
Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit4)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632- Is there evidence of outliers or high leverage observations in the model from (e)? The plots indicate linearity and the Residuals vs Leverage plot shows outliers as well as high leverage obs > .01.
par(mfrow = c(2, 2))
plot(fit4)
##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? Using (c)’s example, the coefficient estimates for the regressions are the same when sum of x^2 is equal to sum of y^2.
- 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 <- x * -153
sum(y^2)## [1] 7920435150fit.X<-lm(y~x + 0)
fit.Y<-lm(x~y + 0)summary(fit.Y)## Warning in summary.lm(fit.Y): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.493e-13 -1.826e-15 4.100e-17 1.549e-15 1.140e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y -6.536e-03 2.846e-19 -2.297e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.532e-14 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 5.276e+32 on 1 and 99 DF, p-value: < 2.2e-16summary(fit.X)## Warning in summary.lm(fit.X): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.926e-12 -2.460e-13 -1.000e-15 2.110e-13 3.282e-11
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x -1.530e+02 5.745e-15 -2.663e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.342e-12 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 7.093e+32 on 1 and 99 DF, p-value: < 2.2e-16- Generate an example in R with n = 100 observations in whichthe 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