Assignment 2
Connor Allison fbu535
2025-02-26
knitr::opts_chunk$set(echo = TRUE)
library(ISLR2)
library(olsrr)##
## Attaching package: 'olsrr'## The following object is masked from 'package:datasets':
##
## riverslibrary(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.5.1 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.1
## ✔ purrr 1.0.2## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsProblem 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
The KNN classifier attempts to estimate the conditional distribution of Y given X, and then classify the observation of interest based on the highest estimated probability. It does this but taking a group of known points \(\mathbb{N}_{0}\) similar to X and estimating the conditional probability for class j as the fraction of the points in \(\mathbb{N}_{0}\), whose response is j. The test observation is then classified by the largest probability.
The KNN regression model is similar. The key difference is that the KNN regression model estimates the response of the test observation by taking average of all training responses in \(\mathbb{N}_{0}\).
Problem 9
a) Produce a scatterplot matrix which includes all of the variables in the data set.
auto = read.csv("auto1.csv", na.strings = "?", stringsAsFactors = T)
dim(auto)## [1] 397 9auto = na.omit(auto)
dim(auto)## [1] 392 9plot(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.
auto %>%
select(-name) %>%
cor()## 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 1.0000000c) 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.
m1 = lm(mpg ~ .-name, data = auto)
summary(m1)##
## 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-16coef(m1)[7]## year
## 0.7507727There is a relationship between the predictors and the response.
Displacement, weight, year and origin seem to have a significant
relationship with mpg. The coefficient for Year is 0.75
which means that for every 1 increase in year the mpg increases by
0.75
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?
par(mfrow=c(2,2))
plot(m1)
summary(influence.measures(m1))## Potentially influential observations of
## lm(formula = mpg ~ . - name, data = auto) :
##
## dfb.1_ dfb.cyln dfb.dspl dfb.hrsp dfb.wght dfb.accl dfb.year dfb.orgn
## 9 -0.01 -0.15 0.16 0.14 -0.14 0.09 0.00 0.03
## 13 0.01 -0.01 0.02 -0.01 -0.01 -0.01 -0.01 0.01
## 14 0.11 0.17 -0.42 -0.43 0.68 -0.36 -0.08 0.00
## 26 -0.04 0.01 -0.06 0.15 -0.02 0.10 -0.01 -0.02
## 27 -0.03 0.05 -0.08 0.11 -0.01 0.07 -0.02 -0.03
## 28 -0.05 0.07 -0.13 0.19 -0.02 0.10 -0.02 -0.05
## 29 -0.06 0.08 -0.14 0.18 0.02 0.15 -0.03 -0.04
## 112 -0.22 0.16 0.03 0.02 -0.14 0.19 0.18 -0.16
## 167 -0.01 -0.22 -0.02 0.03 0.21 0.06 -0.03 0.04
## 245 -0.10 0.05 0.12 0.05 -0.23 0.33 0.02 0.03
## 271 0.03 0.15 -0.11 0.03 -0.04 0.06 -0.06 -0.24
## 300 -0.03 -0.03 0.02 0.01 0.01 0.06 0.01 0.02
## 301 -0.04 0.05 -0.01 0.01 -0.03 0.05 0.02 0.00
## 310 -0.04 0.01 -0.04 -0.02 -0.01 -0.13 0.13 -0.02
## 323 -0.19 0.06 0.05 -0.01 -0.06 0.09 0.16 0.31
## 326 -0.20 0.05 0.10 0.06 -0.19 0.34 0.12 0.04
## 327 -0.28 0.04 0.06 0.11 -0.13 0.49 0.12 0.05
## 328 -0.14 0.14 -0.17 -0.07 0.19 0.06 0.08 0.06
## 330 -0.02 0.02 0.09 -0.15 -0.05 -0.21 0.12 0.22
## 387 -0.16 -0.16 0.40 -0.19 -0.19 0.05 0.26 0.04
## 394 -0.38 0.03 0.14 0.25 -0.32 0.58 0.23 0.03
## dffit cov.r cook.d hat
## 9 0.30 1.06 0.01 0.06_*
## 13 0.02 1.07_* 0.00 0.05
## 14 -0.79_* 1.19_* 0.08 0.19_*
## 26 0.18 1.07_* 0.00 0.06
## 27 0.13 1.09_* 0.00 0.07_*
## 28 0.22 1.08_* 0.01 0.07_*
## 29 0.25 1.11_* 0.01 0.09_*
## 112 -0.46_* 0.87_* 0.03 0.02
## 167 -0.39 0.93_* 0.02 0.03
## 245 0.48_* 0.82_* 0.03 0.02
## 271 -0.33 0.90_* 0.01 0.02
## 300 0.09 1.07_* 0.00 0.05
## 301 0.08 1.08_* 0.00 0.06
## 310 0.29 0.86_* 0.01 0.01
## 323 0.47_* 0.74_* 0.03 0.01
## 326 0.49_* 0.81_* 0.03 0.02
## 327 0.63_* 0.79_* 0.05 0.03
## 328 0.40 0.88_* 0.02 0.02
## 330 0.43 0.87_* 0.02 0.02
## 387 0.55_* 0.88_* 0.04 0.03
## 394 0.68_* 0.90_* 0.06 0.05The residual plots indict lack of fit. Additionally, the leverage plot shows several outliers and we also see that observation 14 has very high leverage.
e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
m2 = lm(mpg ~ weight * year * origin, data = auto)
summary(m2)##
## Call:
## lm(formula = mpg ~ weight * year * origin, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.7880 -1.9187 -0.1022 1.4576 12.1862
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.170e+02 3.551e+01 -6.111 2.43e-09 ***
## weight 7.198e-02 1.334e-02 5.398 1.18e-07 ***
## year 3.331e+00 4.660e-01 7.147 4.50e-12 ***
## origin 9.961e+01 2.508e+01 3.972 8.51e-05 ***
## weight:year -1.005e-03 1.749e-04 -5.749 1.83e-08 ***
## weight:origin -4.313e-02 1.080e-02 -3.995 7.75e-05 ***
## year:origin -1.236e+00 3.254e-01 -3.798 0.000170 ***
## weight:year:origin 5.402e-04 1.399e-04 3.861 0.000132 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.055 on 384 degrees of freedom
## Multiple R-squared: 0.8495, Adjusted R-squared: 0.8468
## F-statistic: 309.7 on 7 and 384 DF, p-value: < 2.2e-16After testing multiple combinations of interactions using the variables displacement, weight, year, and origin, I found that a model with weight, year, origin and all combinations of interactions of the three, produced the highest adj r-square.
f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
m3 = lm(mpg ~ log(weight) * year * origin, data = auto)
summary(m3)##
## Call:
## lm(formula = mpg ~ log(weight) * year * origin, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.3341 -1.6334 0.0224 1.2993 12.3905
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1285.5968 266.8732 -4.817 2.10e-06 ***
## log(weight) 160.0439 33.8676 4.726 3.23e-06 ***
## year 19.1901 3.5125 5.463 8.41e-08 ***
## origin 692.8308 187.9104 3.687 0.000260 ***
## log(weight):year -2.3543 0.4458 -5.281 2.15e-07 ***
## log(weight):origin -89.7194 24.2803 -3.695 0.000252 ***
## year:origin -8.9417 2.4437 -3.659 0.000289 ***
## log(weight):year:origin 1.1589 0.3157 3.671 0.000276 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.951 on 384 degrees of freedom
## Multiple R-squared: 0.8596, Adjusted R-squared: 0.857
## F-statistic: 335.8 on 7 and 384 DF, p-value: < 2.2e-16m4 = lm(mpg ~ weight * year^2 * origin, data = auto)
summary(m4)##
## Call:
## lm(formula = mpg ~ weight * year^2 * origin, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.7880 -1.9187 -0.1022 1.4576 12.1862
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.170e+02 3.551e+01 -6.111 2.43e-09 ***
## weight 7.198e-02 1.334e-02 5.398 1.18e-07 ***
## year 3.331e+00 4.660e-01 7.147 4.50e-12 ***
## origin 9.961e+01 2.508e+01 3.972 8.51e-05 ***
## weight:year -1.005e-03 1.749e-04 -5.749 1.83e-08 ***
## weight:origin -4.313e-02 1.080e-02 -3.995 7.75e-05 ***
## year:origin -1.236e+00 3.254e-01 -3.798 0.000170 ***
## weight:year:origin 5.402e-04 1.399e-04 3.861 0.000132 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.055 on 384 degrees of freedom
## Multiple R-squared: 0.8495, Adjusted R-squared: 0.8468
## F-statistic: 309.7 on 7 and 384 DF, p-value: < 2.2e-16Log weight gave less power to the weight variable and actually improved the model a bit. Year squared gave more power to the year variable and did not have a significant effect on our model.
Problem 10
a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
fit = lm(Sales ~ Price + Urban + US, data = Carseats)
summary(fit)##
## 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-16coef(fit)[2]## Price
## -0.05445885coef(fit)[4]## USYes
## 1.200573b) Provide an interpretation of each coeffcient in the model. Be careful—some of the variables in the model are qualitative!
The coefficient for Price is -0.054459 which means that
for every dollar increase in the price of my carseat, my store’s sales
decrease by $54
The coefficient for US = Yes is 1.200573 which means, on
average, our carseats sell $1,200 more compared to stores outside the
US
Urban is not significant.
c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.04 - 0.054Price - 0.022Urban + 1.2US\)
d) For which of the predictors can you reject the null hypothesis \(\H_0 : \beta_j = 0\)?
See part (b) for interpretation, but Price and
US_Yes are significant so we can reject the null hypothesis
\(H_0 : \beta_j = 0\)
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, data = Carseats)
summary(fit)##
## 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-16f) How well do the models in (a) and (e) ft the data?
Terrible Adj R-Squared is 0.2335 for part (a) and 0.2354 for part (e)
g) Using the model from (e), obtain 95 % confdence intervals for the coeffcient(s).
confint(fit)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632h) Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow=c(2,2))
plot(fit)
summary(influence.measures(fit))## Potentially influential observations of
## lm(formula = Sales ~ Price + US, data = Carseats) :
##
## dfb.1_ dfb.Pric dfb.USYs dffit cov.r cook.d hat
## 26 0.24 -0.18 -0.17 0.28_* 0.97_* 0.03 0.01
## 29 -0.10 0.10 -0.10 -0.18 0.97_* 0.01 0.01
## 43 -0.11 0.10 0.03 -0.11 1.05_* 0.00 0.04_*
## 50 -0.10 0.17 -0.17 0.26_* 0.98 0.02 0.01
## 51 -0.05 0.05 -0.11 -0.18 0.95_* 0.01 0.00
## 58 -0.05 -0.02 0.16 -0.20 0.97_* 0.01 0.01
## 69 -0.09 0.10 0.09 0.19 0.96_* 0.01 0.01
## 126 -0.07 0.06 0.03 -0.07 1.03_* 0.00 0.03_*
## 160 0.00 0.00 0.00 0.01 1.02_* 0.00 0.02
## 166 0.21 -0.23 -0.04 -0.24 1.02 0.02 0.03_*
## 172 0.06 -0.07 0.02 0.08 1.03_* 0.00 0.02
## 175 0.14 -0.19 0.09 -0.21 1.03_* 0.02 0.03_*
## 210 -0.14 0.15 -0.10 -0.22 0.97_* 0.02 0.01
## 270 -0.03 0.05 -0.03 0.06 1.03_* 0.00 0.02
## 298 -0.06 0.06 -0.09 -0.15 0.97_* 0.01 0.00
## 314 -0.05 0.04 0.02 -0.05 1.03_* 0.00 0.02_*
## 353 -0.02 0.03 0.09 0.15 0.97_* 0.01 0.00
## 357 0.02 -0.02 0.02 -0.03 1.03_* 0.00 0.02
## 368 0.26 -0.23 -0.11 0.27_* 1.01 0.02 0.02_*
## 377 0.14 -0.15 0.12 0.24 0.95_* 0.02 0.01
## 384 0.00 0.00 0.00 0.00 1.02_* 0.00 0.02
## 387 -0.03 0.04 -0.03 0.05 1.02_* 0.00 0.02
## 396 -0.05 0.05 0.08 0.14 0.98_* 0.01 0.00Yes there is.
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?
Regression of Y on X without intercept, \(\hat{\beta} = (X'X)^{-1}X'Y = \frac{\sum x_iy_i}{\sum x_i^2}\)
Regression of X on Y without intercept, \(\hat{\beta} = (Y'Y)^{-1}Y'X = \frac{\sum x_iy_i}{\sum y_i^2}\)
These are equal when \(\frac{\sum x_iy_i}{\sum x_i^2} = \frac{\sum x_iy_i}{\sum y_i^2}\)
Therefore \(\sum x_i^2 = \sum y_i^2\), when the sum of squared X values equals the sum of squared Y values
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(123)
x_b = rnorm(100, mean = 0, sd = 1)
y_b = rnorm(100, mean = 0, sd = 2)xy_fit_b = lm(y_b ~ x_b - 1)
yx_fit_b = lm(x_b ~ y_b - 1)coef(xy_fit_b)## x_b
## -0.1272558coef(yx_fit_b)## y_b
## -0.02827693c) 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.
set.seed(23)
x_c = rnorm(100, mean = 0, sd = 1)
y_c = rnorm(100, mean = 0, sd = 1)xy_fit_c = lm(y_c ~ x_c - 1)
yx_fit_c = lm(x_c ~ y_c - 1)coef(xy_fit_c)## x_c
## -0.02207371coef(yx_fit_c)## y_c
## -0.02283415