Assignment 2
Simon Miller
2023-02-17
2. Carefully explain the differences between the KNN classifier and KNN regression methods.
The KNN classifier uses proximity to make predictions about the grouping of data and the output is categorical. The KNN regression gets the association between independent variables and the continuous outcomes by getting the means of the observations in the neighborhood and the output is numerical.
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(ISLR2)
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:
lmmpg <- lm(mpg ~ . - name, data = Auto)
summary(lmmpg)##
## 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- Is there a relationship between the predictors and the response? Yes, there is a relationship between the predictors and mpg.
- Which predictors appear to have a statistically significant relationship to the response? Displacement, weight, year, and origin.
- What does the coefficient for the year variable suggest? It suggests that mpg efficiency increases a little more than 75% every year.
(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(lmmpg)
Yes, the residual plots do have some large outliers and the leverage plot also has some points with high leverage.
(e) Use the * and : symbols to fit linear regression models with interaction effects.Do any interactions appear to be statistically significant?
summary(lm(formula = mpg ~ . * ., data = Auto[, -9]))##
## Call:
## lm(formula = mpg ~ . * ., data = Auto[, -9])
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6303 -1.4481 0.0596 1.2739 11.1386
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.548e+01 5.314e+01 0.668 0.50475
## cylinders 6.989e+00 8.248e+00 0.847 0.39738
## displacement -4.785e-01 1.894e-01 -2.527 0.01192 *
## horsepower 5.034e-01 3.470e-01 1.451 0.14769
## weight 4.133e-03 1.759e-02 0.235 0.81442
## acceleration -5.859e+00 2.174e+00 -2.696 0.00735 **
## year 6.974e-01 6.097e-01 1.144 0.25340
## origin -2.090e+01 7.097e+00 -2.944 0.00345 **
## cylinders:displacement -3.383e-03 6.455e-03 -0.524 0.60051
## cylinders:horsepower 1.161e-02 2.420e-02 0.480 0.63157
## cylinders:weight 3.575e-04 8.955e-04 0.399 0.69000
## cylinders:acceleration 2.779e-01 1.664e-01 1.670 0.09584 .
## cylinders:year -1.741e-01 9.714e-02 -1.793 0.07389 .
## cylinders:origin 4.022e-01 4.926e-01 0.816 0.41482
## displacement:horsepower -8.491e-05 2.885e-04 -0.294 0.76867
## displacement:weight 2.472e-05 1.470e-05 1.682 0.09342 .
## displacement:acceleration -3.479e-03 3.342e-03 -1.041 0.29853
## displacement:year 5.934e-03 2.391e-03 2.482 0.01352 *
## displacement:origin 2.398e-02 1.947e-02 1.232 0.21875
## horsepower:weight -1.968e-05 2.924e-05 -0.673 0.50124
## horsepower:acceleration -7.213e-03 3.719e-03 -1.939 0.05325 .
## horsepower:year -5.838e-03 3.938e-03 -1.482 0.13916
## horsepower:origin 2.233e-03 2.930e-02 0.076 0.93931
## weight:acceleration 2.346e-04 2.289e-04 1.025 0.30596
## weight:year -2.245e-04 2.127e-04 -1.056 0.29182
## weight:origin -5.789e-04 1.591e-03 -0.364 0.71623
## acceleration:year 5.562e-02 2.558e-02 2.174 0.03033 *
## acceleration:origin 4.583e-01 1.567e-01 2.926 0.00365 **
## year:origin 1.393e-01 7.399e-02 1.882 0.06062 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.695 on 363 degrees of freedom
## Multiple R-squared: 0.8893, Adjusted R-squared: 0.8808
## F-statistic: 104.2 on 28 and 363 DF, p-value: < 2.2e-16With a significance level of .01, acceleration:Origin With a significance level of .05, acceleration:year, displacement:year.
(f) Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings.
10. This question should be answered using the Carseats data set.
(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
lmsales <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(lmsales)##
## 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!
The Price variable with a 1 unit increase is a -.054 change in sales. The Urban variable is a change of -.022 in sales, but the p-value is so high it is possible to say that there is no relationship between the two. The store being in the US causes a change of 1.2 in sales.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales=13.043469−0.054459Price−0.021916Urban+1.200573*US
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
We can reject the null hypothesis for the Price and US variables.
(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.
lmsales2 <- lm(Sales ~ Price + US, data = Carseats)
summary(lmsales2)##
## 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?
There is a little difference in adjusted R^2 between the two, but both models explain a 23.9% of variance in sales.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(lmsales2, level = 0.95)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632The confidence interval is (-0.0648, -0.0442) for Price and (.6915, 1.7078) for US.
(h) Is there evidence of outliers or high leverage observations in the model from (e)?
There are not any outliers or high leverage observations.
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?
The coefficient estimates for which they would be the same is when both X and Y vectors are the same.
(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 <- 3*x + rnorm(100, sd = 2)
data <- data.frame(x, y)
lm0 <- lm(y ~ x)
lm1 <- lm(x ~ y)
summary(lm0)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.7269 -1.4255 -0.0902 1.2801 4.8632
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.3564 0.2101 -1.696 0.093 .
## x 2.7857 0.1921 14.499 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.092 on 98 degrees of freedom
## Multiple R-squared: 0.682, Adjusted R-squared: 0.6788
## F-statistic: 210.2 on 1 and 98 DF, p-value: < 2.2e-16summary(lm1)##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.69943 -0.39411 0.00651 0.41047 1.74045
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.05443 0.06295 0.865 0.389
## y 0.24483 0.01689 14.499 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.6201 on 98 degrees of freedom
## Multiple R-squared: 0.682, Adjusted R-squared: 0.6788
## F-statistic: 210.2 on 1 and 98 DF, p-value: < 2.2e-16This code will produce different estimates.
(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 <- x
data <- data.frame(x, y)
lm2 <- lm(y ~ x)
lm3 <- lm(x ~ y)
summary(lm2)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.802e-15 1.120e-17 3.950e-17 7.720e-17 4.714e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.110e-17 4.968e-17 2.230e-01 0.824
## x 1.000e+00 5.239e-17 1.909e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.967e-16 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 3.643e+32 on 1 and 98 DF, p-value: < 2.2e-16summary(lm3)##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.802e-15 1.120e-17 3.950e-17 7.720e-17 4.714e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.110e-17 4.968e-17 2.230e-01 0.824
## y 1.000e+00 5.239e-17 1.909e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.967e-16 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 3.643e+32 on 1 and 98 DF, p-value: < 2.2e-16For this code, both estimates are equal to one another