Assignment 2
Phillip Cisneros
2022-09-16
Problem 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
The KNN classifier method tries to estimate the distribution of Y given X, and then classifies that test observation into qualitative groups. The KNN regression method attempts to predict the output variable by using the average of all training responses.
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, 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:
Based on the p-values from the data, there appears to be a relationship between the predictors and the responses.Displacement, weight, year, origin are the most statistically significant relationships because they have the p-values closest to 0. The year variable has a coefficient of 0.750773 which means that for every year that passed, the mpg increases by that coefficient. This positive correlation is logical, since technology and the push for more fuel efficient vehicles have increased.
fit.auto <- lm(mpg ~ . -name, data = Auto)
summary(fit.auto)##
## 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(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?
Looking at the residual plots, they suggest that there are unusually large outliers and leverage points in both the ‘Residual vs Fitted’ and the ‘Residuals vs Leverage’ graphs.
par(mfrow = c(2, 2))
plot(fit.auto)
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
Using interaction effects, I found that weight:displacement and cylinders:year appear to be statistically significant.
fit.auto1 <- lm(mpg ~ . -name + weight * displacement + cylinders * horsepower + year * cylinders , data = Auto)
summary(fit.auto1)##
## Call:
## lm(formula = mpg ~ . - name + weight * displacement + cylinders *
## horsepower + year * cylinders, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.5716 -1.5438 -0.0587 1.2227 12.6501
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4.738e+01 1.432e+01 -3.309 0.001027 **
## cylinders 8.190e+00 2.758e+00 2.969 0.003172 **
## displacement -4.966e-02 1.317e-02 -3.770 0.000189 ***
## horsepower -1.415e-01 4.656e-02 -3.039 0.002535 **
## weight -8.076e-03 1.135e-03 -7.118 5.49e-12 ***
## acceleration 7.576e-03 9.404e-02 0.081 0.935836
## year 1.394e+00 1.624e-01 8.585 2.35e-16 ***
## origin 6.497e-01 2.524e-01 2.574 0.010433 *
## displacement:weight 1.458e-05 3.514e-06 4.149 4.12e-05 ***
## cylinders:horsepower 1.448e-02 6.415e-03 2.257 0.024585 *
## cylinders:year -1.241e-01 3.058e-02 -4.057 6.03e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.838 on 381 degrees of freedom
## Multiple R-squared: 0.8712, Adjusted R-squared: 0.8678
## F-statistic: 257.6 on 10 and 381 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.
After testing a few transformations, I found that ‘log(displacement)’ and ‘log(horsepower)’ made them more significant.
fit.auto2 <- lm(mpg ~ . -name + sqrt(cylinders) + log(displacement) + log(horsepower) + weight + I(acceleration^2) , data = Auto)
summary(fit.auto2)##
## Call:
## lm(formula = mpg ~ . - name + sqrt(cylinders) + log(displacement) +
## log(horsepower) + weight + I(acceleration^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.7764 -1.5944 -0.0246 1.5107 11.8156
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 77.368033 16.132641 4.796 2.33e-06 ***
## cylinders -2.849421 2.895641 -0.984 0.32572
## displacement 0.030908 0.014520 2.129 0.03392 *
## horsepower 0.091823 0.032669 2.811 0.00520 **
## weight -0.002993 0.000666 -4.494 9.31e-06 ***
## acceleration -1.329879 0.548162 -2.426 0.01573 *
## year 0.764716 0.045392 16.847 < 2e-16 ***
## origin 0.563242 0.272105 2.070 0.03913 *
## sqrt(cylinders) 14.059123 13.795237 1.019 0.30879
## log(displacement) -8.390157 2.859876 -2.934 0.00355 **
## log(horsepower) -17.583513 3.707430 -4.743 2.99e-06 ***
## I(acceleration^2) 0.031506 0.016095 1.958 0.05102 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.922 on 380 degrees of freedom
## Multiple R-squared: 0.8638, Adjusted R-squared: 0.8598
## F-statistic: 219 on 11 and 380 DF, p-value: < 2.2e-16Problem 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.
fit <- lm(Sales ~ Price + Urban + US)
summary(fit)##
## 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!
Based on the coefficients of each variable in the model above, we can conclude that Price and US are significant predictors of ‘Sales’ while ‘Urban’ is not a significant predictor. For every $1 increase in my price, sales decreases by $54. Sales made in the US are $1,200 higher than sales outside the US
(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?
We can reject the null hypothesis for both ‘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.
fit <- lm(Sales ~ Price + US)
summary(fit)##
## 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?
The models in (a) and (e) do not fit the data that well because they both have an R-squared of about .23. This means that it only accounts for about 23% of the variance in sales.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit)## 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)?
The ‘Residual vs Fitted’ and the ‘Residual vs Leverage’ graphs both show evidence of outliers in the model we fit from (e).
par(mfrow = c(2, 2))
plot(fit)
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?
They should be the same if the \(∑x^2=∑y^2\) where there is no variability or noise.
(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