Assignment 2
Karim I
2022-9-10
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Carefully explain the differences between the KNN classifier and KNN regression methods. K-nearest neighbors regression is a nonparametric regression technique. KNN regression first determines the K training observations that are most similar to x0, after which it uses the mean of all the training responses to estimate the average for f(x0). In contrast, a KNN classifier divides a result into qualitative (categorical) groups based on the most prevalent group identified among the K closest neighbors. The test observation x0 is classified to the class with the highest likelihood by first identifying the K positive integer training data points that are closest to x0, then estimating the conditional probability for class j as the proportion of points.
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)
Compute the matrix of correlations between the variables using the
function cor(). You will need to exclude the name variable, which is
qualitative
cor(subset(Auto, select = -name))## 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 1.0000000Use 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.fit1 <- lm(mpg ~ . - name, data = Auto)
summary(lm.fit1)##
## 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-16Is there a relationship between the predictors and the re- sponse? Yes, the link between the predictors and the answer may be shown by testing the hypothesis to determine if all regression coefficients are zero. Which predictors appear to have a statistically significant relationship to the response? We can observe from the graphs that there is a substantial association between weight, year, displacement, and origin, as evidenced by their p-values being less than 0.5 f. We can observe that there is no statistical correlation between acceleration, cylinders, and horsepower. What does the coefficient for the year variable suggest? If all other variables remain constant and the coefficient for the ‘year’ variable is positive, it signifies that the mpg grows by the coefficient for every year. We may deduce that automobiles become approximately 1% more fuel-efficient annually since we can identify the link between them by looking at their p-values.
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.fit1)
plot(predict(lm.fit1), rstudent(lm.fit1))
By the conclusion of the Normal Q-Q plot, we can observe some outliers
from (2,2) to (3,4). Additionally, we may see outliers by (30,1.5) to
the scale location plot (35,2.0). Outliers may be seen in the residuals
vs. Leverage plot from the beginning (0.025,2) to (0.05,4), to answer
the other question: Yes, point 14 looks to have an atypically high
leverage, as shown on the Residuals vs. Leverage figure.
Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant? Yes, a statistically significant interaction between weight and displacement appears to exist.
lm.fit2 <- lm(mpg ~ cylinders * displacement + displacement * weight, data = Auto)
summary(lm.fit2)##
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement *
## weight, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.2934 -2.5184 -0.3476 1.8399 17.7723
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.262e+01 2.237e+00 23.519 < 2e-16 ***
## cylinders 7.606e-01 7.669e-01 0.992 0.322
## displacement -7.351e-02 1.669e-02 -4.403 1.38e-05 ***
## weight -9.888e-03 1.329e-03 -7.438 6.69e-13 ***
## cylinders:displacement -2.986e-03 3.426e-03 -0.872 0.384
## displacement:weight 2.128e-05 5.002e-06 4.254 2.64e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.103 on 386 degrees of freedom
## Multiple R-squared: 0.7272, Adjusted R-squared: 0.7237
## F-statistic: 205.8 on 5 and 386 DF, p-value: < 2.2e-16Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings. The Residuals versus Leverage plot contains no points inside its boundaries, indicating that there are no important points that might affect the slope coefficient, as seen in the preceding plots.
y1<-lm(mpg~weight+I((weight)^2),Auto)
summary(y1)##
## Call:
## lm(formula = mpg ~ weight + I((weight)^2), data = Auto)
##
## 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-16This question should be answered using the Carseats data set. (a) Fit a multiple regression model to predict Sales using Price,Urban, and US.
data("Carseats", package = "ISLR2")
head(Carseats)## 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 Yesstr(Carseats)## 'data.frame': 400 obs. of 11 variables:
## $ Sales : num 9.5 11.22 10.06 7.4 4.15 ...
## $ CompPrice : num 138 111 113 117 141 124 115 136 132 132 ...
## $ Income : num 73 48 35 100 64 113 105 81 110 113 ...
## $ Advertising: num 11 16 10 4 3 13 0 15 0 0 ...
## $ Population : num 276 260 269 466 340 501 45 425 108 131 ...
## $ Price : num 120 83 80 97 128 72 108 120 124 124 ...
## $ ShelveLoc : Factor w/ 3 levels "Bad","Good","Medium": 1 2 3 3 1 1 3 2 3 3 ...
## $ Age : num 42 65 59 55 38 78 71 67 76 76 ...
## $ Education : num 17 10 12 14 13 16 15 10 10 17 ...
## $ Urban : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 1 2 2 1 1 ...
## $ US : Factor w/ 2 levels "No","Yes": 2 2 2 2 1 2 1 2 1 2 ...lm.fit = lm(Sales ~ Price+Urban+US, data= Carseats)
summary(lm.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-16Provide an interpretation of each coefficient in the model. Becareful—some of the variables in the model are qualitative! The model contains three coefficients. The “Price” coefficient may be interpreted as the average of car seat price increases of one dollar affecting the r abs(summary(fit3)coef[2, 1]) * 1000 of units for sale if all predictors remain constant. If all other variables remain constant, the coefficient “UrbanYes” means that on average, urban location unit sales are r abs(summary(fit3)coef[3, 1]) * 1000)’ less than rural location unit sales. If all other variables stay constant, the “USYes” coefficient may be understood as meaning that US store unit sales are r abs(summary(fit3)$coef[4, 1]) * 1000’ units higher than non-US store unit sales.
Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales=13.04−0.05∗Price−0.02∗UrbanYes+1.2∗USYes
- For which of the predictors can you reject the null hypothesis H0 :βj =0?
predictor ‘Urban’ and ‘Price’ can be rejected as it presents higher P-values.
- 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.
lm.fit2 = lm(Sales ~ Price+US, data= Carseats)
summary(lm.fit2)##
## 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-16How well do the models in (a) and (e) fit the data? With R-square values of 24% and 24%, respectively, I would argue that (a) and (e) do not match the data very well. We can also see that the smaller model is somewhat better for the R2.
Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(lm.fit2)## 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(lm.fit2)
- This problem involves simple linear regression without an intercept.
- 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? β̂ =∑ixiyi/∑jx2j
coeffcient estimate for regression of X and Y
β̂ ′=∑ixiyi/∑jy2j.
The coefficients are the same iff ∑jx2j=∑jy2j.
- 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- 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