Assignment2
Nicholas Mata jnm705
2025-02-17
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## 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.
KNN regression uses average of the nearest points to predict a value, where KNN classifier uses the highest majority category among its neighbors to predict a class.
Problem 9
## The following object is masked from package:lubridate:
##
## origin## The following object is masked from package:ggplot2:
##
## mpgThis 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.
plot(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 1.0000000(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:
lm_auto = lm(mpg ~ . -name, data = Auto)
summary(lm_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-16i. Is there a relationship between the predictors and the response?
There is a relationship between the response and at least one of the predictors.
ii. Which predictors appear to have a statistically significant relationship to the response?
displacement, weight, year,
and origin have statistically significant relationships to
mpg.
iii. What does the coefficient for the year variable suggest?
The coefficient for year is 0.750773 which means that a
vehicle model has, on average, 0.75 mpg more than the previous year’s
model.
(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(lm_auto)
The residual plot shows some outliers around 30. The data is normally distributed according to the Q-Q plot, except for the end values. There are no values with high leverage according to the Leverage plot. No points appear above the Cook’s distance dotted line of 0.5.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm_auto_interact = lm(mpg ~ acceleration*year + cylinders*displacement + horsepower*weight + cylinders*origin, data = Auto)
summary(lm_auto_interact)##
## Call:
## lm(formula = mpg ~ acceleration * year + cylinders * displacement +
## horsepower * weight + cylinders * origin, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.6734 -1.5452 -0.0555 1.2971 11.3844
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.145e+02 1.850e+01 6.188 1.58e-09 ***
## acceleration -6.815e+00 1.149e+00 -5.929 6.84e-09 ***
## year -6.300e-01 2.395e-01 -2.630 0.00888 **
## cylinders -1.088e+00 7.457e-01 -1.459 0.14534
## displacement -2.203e-02 1.574e-02 -1.400 0.16232
## horsepower -2.266e-01 2.590e-02 -8.749 < 2e-16 ***
## weight -9.916e-03 9.050e-04 -10.957 < 2e-16 ***
## origin -1.670e+00 1.293e+00 -1.292 0.19713
## acceleration:year 8.773e-02 1.490e-02 5.888 8.62e-09 ***
## cylinders:displacement 3.223e-03 2.310e-03 1.395 0.16381
## horsepower:weight 4.927e-05 6.771e-06 7.276 1.98e-12 ***
## cylinders:origin 5.450e-01 2.991e-01 1.822 0.06917 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.79 on 380 degrees of freedom
## Multiple R-squared: 0.8759, Adjusted R-squared: 0.8723
## F-statistic: 243.7 on 11 and 380 DF, p-value: < 2.2e-16acceleration:year and horsepower:weight are
statistically significant.
lm_auto_interact = lm(mpg ~ displacement:weight + displacement:year + displacement:origin + weight:year + weight:origin + year:origin, data = Auto)
summary(lm_auto_interact)##
## Call:
## lm(formula = mpg ~ displacement:weight + displacement:year +
## displacement:origin + weight:year + weight:origin + year:origin,
## data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5226 -2.2594 -0.5171 1.4368 16.3439
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.650e+01 1.846e+00 8.936 < 2e-16 ***
## displacement:weight -2.034e-06 1.853e-06 -1.098 0.273048
## displacement:year -1.090e-03 2.576e-04 -4.234 2.88e-05 ***
## displacement:origin 4.913e-02 1.460e-02 3.364 0.000845 ***
## weight:year 1.078e-04 1.564e-05 6.892 2.25e-11 ***
## weight:origin -8.805e-03 8.707e-04 -10.112 < 2e-16 ***
## year:origin 2.099e-01 1.254e-02 16.735 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.576 on 385 degrees of freedom
## Multiple R-squared: 0.7934, Adjusted R-squared: 0.7901
## F-statistic: 246.4 on 6 and 385 DF, p-value: < 2.2e-16Auto2 = Auto |>
select(-name)lm_auto_interact = lm(mpg ~ . * ., data = Auto2)
summary(lm_auto_interact)##
## Call:
## lm(formula = mpg ~ . * ., data = Auto2)
##
## 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-16(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
lm_auto_different = lm(mpg ~ weight + I(weight^2), data = Auto2)
summary(lm_auto_different)##
## Call:
## lm(formula = mpg ~ weight + I(weight^2), data = Auto2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.6246 -2.7134 -0.3485 1.8267 16.0866
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.226e+01 2.993e+00 20.800 < 2e-16 ***
## weight -1.850e-02 1.972e-03 -9.379 < 2e-16 ***
## I(weight^2) 1.697e-06 3.059e-07 5.545 5.43e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.176 on 389 degrees of freedom
## Multiple R-squared: 0.7151, Adjusted R-squared: 0.7137
## F-statistic: 488.3 on 2 and 389 DF, p-value: < 2.2e-16lm_auto_different = lm(mpg ~ log(weight), data = Auto2)
summary(lm_auto_different)##
## Call:
## lm(formula = mpg ~ log(weight), data = Auto2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.4315 -2.6752 -0.2888 1.9429 16.0136
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 209.9433 6.0002 34.99 <2e-16 ***
## log(weight) -23.4317 0.7534 -31.10 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.189 on 390 degrees of freedom
## Multiple R-squared: 0.7127, Adjusted R-squared: 0.7119
## F-statistic: 967.3 on 1 and 390 DF, p-value: < 2.2e-16lm_auto_different = lm(mpg ~ weight + I(weight^(1/2)), data = Auto2)
summary(lm_auto_different)##
## Call:
## lm(formula = mpg ~ weight + I(weight^(1/2)), data = Auto2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.5660 -2.6552 -0.4161 1.7373 16.1001
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 109.218284 11.573797 9.437 < 2e-16 ***
## weight 0.013191 0.003828 3.446 0.000631 ***
## I(weight^(1/2)) -2.314535 0.424250 -5.456 8.7e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.181 on 389 degrees of freedom
## Multiple R-squared: 0.7145, Adjusted R-squared: 0.713
## F-statistic: 486.7 on 2 and 389 DF, p-value: < 2.2e-16The p values of the squared and square root of weight
were not as close to zero as weight on its own so they were
not improvements on the regression.
Problem 10
## Sales CompPrice Income Advertising Population Price ShelveLoc Age Education
## 1 9.50 138 73 11 276 120 Bad 42 17
## 2 11.22 111 48 16 260 83 Good 65 10
## 3 10.06 113 35 10 269 80 Medium 59 12
## 4 7.40 117 100 4 466 97 Medium 55 14
## 5 4.15 141 64 3 340 128 Bad 38 13
## 6 10.81 124 113 13 501 72 Bad 78 16
## Urban US
## 1 Yes Yes
## 2 Yes Yes
## 3 Yes Yes
## 4 Yes Yes
## 5 Yes No
## 6 No YesThis question should be answered using the Carseats data set.
(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-16(b) Provide an interpretation of each coefficient 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 a carseat, the store’s sales
decrease by $54 on average.
The coefficient for US = Yes is 1.200573 which means, on
average, US stores sell $1,200 more in carseats compared to stores
outside the US.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.04 - 0.05Price - 0.02Urban + 1.2US\)
(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 thus 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-16(f) How well do the models in (a) and (e) fit the data?
They fit terribly.
Adjusted R-squared is 0.2335 for part (a)
Adjusted R-squared is 0.2354 for part (e)
(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)?
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 0.00There are high leverage observations.
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?
The coefficients will be the same when the sum of the squares of x and y are equal (they are the denominators of the coefficient equations)
(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 = rnorm(100)
y = 2 * x + rnorm(100)sum(x^2)## [1] 81.05509sum(y^2)## [1] 413.2135lm_y = lm(y~x + 0)
lm_x = lm(x~y + 0)
summary(lm_y)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9154 -0.6472 -0.1771 0.5056 2.3109
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 1.9939 0.1065 18.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared: 0.7798, Adjusted R-squared: 0.7776
## F-statistic: 350.7 on 1 and 99 DF, p-value: < 2.2e-16summary(lm_x)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.8699 -0.2368 0.1030 0.2858 0.8938
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.39111 0.02089 18.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4246 on 99 degrees of freedom
## Multiple R-squared: 0.7798, Adjusted R-squared: 0.7776
## F-statistic: 350.7 on 1 and 99 DF, p-value: < 2.2e-16The coefficients are different because the sum of the squares of x and y are different.
(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.
set.seed(1)
x = 1:100
y = 100:1
sum(x^2)## [1] 338350sum(y^2)## [1] 338350lm_y = lm(y~x + 0)
lm_x = lm(x~y + 0)
summary(lm_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(lm_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