Assignment 2
Aarthi Kapaleeswaran
2/24/2021
Question 2
Carefully explain the differences between between the KNN classifier and KNN regression methods.
KNN classifier : The goal of the KNN classifier is classification, where the response variable is categorical. Given a positive integer K and a test observation x0, the KNN classifier first identifies the K points in the training data that are closest to x0, represented by N0. It then estimates the conditional probability for class j as a fraction of the points in N0 whose response values equals j. Then, it applies the Bayes rule and classifies the test observation x0 to the class with the largest probability.
KNN regression: The goal of the KNN regression is where the response variable is quantitative and we try to find a relationship between the response variable and the predictors. Given a value for K and a prediction point x0, KNN regression first identifies the K training observations that are closest to x0, represented by N0. It then estimates the f(x0) using the average of all the training responses in N0.
Question 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.
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.
auto.num= Auto[-9] #Exclude name variable
auto.cor = cor(auto.num)
auto.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.
lm.auto = lm(mpg~cylinders+displacement+horsepower+weight+acceleration+year+origin,
data = Auto)
summary(lm.auto)##
## Call:
## lm(formula = mpg ~ cylinders + displacement + horsepower + weight +
## acceleration + year + origin, 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-16Comment on the output. For instance: i. Is there a relationship between the predictors and the response?
The p-value for the model is < 2.2e-16, the goodness-of-fit measured by R-squared value is 82.15% which means that 82% of the variance can be explained by the model. Hence, there is a relationship between the predictors and the response.
- Which predictors appear to have a statistically significant relationship to the response?
The predictors with p-value less than 0.05 which is the significance level are said to be statistically significant to the response. In our model, these predictors are displacement, weight, year and origin.
- What does the coefficient for the year variable suggest?
The year variable is statistically significant to the response since it has a p-value of < 2.2e-16. The coefficient is 0.750773 which suggests that for each year that progresses, the miles per gallon increases by 0.750773.
(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)
From the normal Q-Q plot, we can see that most of the data falls on the Q-Q line indicating that it follows normality. There are some outliers seen at the tail of the plot that are deviating from normality. The residuals plot has a curve that shows that there may be some non-linearity in the data. The leverage plot shows the influential point with unusually high leverage (between -2 and +2) is observation 14.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.inter1 = lm(mpg~-name+cylinders*displacement+origin+horsepower*weight*acceleration+year,data = Auto)
summary(lm.inter1)##
## Call:
## lm(formula = mpg ~ -name + cylinders * displacement + origin +
## horsepower * weight * acceleration + year, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.3693 -1.5354 0.0358 1.3874 11.9225
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -7.480e+00 1.117e+01 -0.670 0.50338
## cylinders -4.100e-01 5.011e-01 -0.818 0.41367
## displacement -2.079e-02 1.645e-02 -1.264 0.20692
## origin 6.491e-01 2.560e-01 2.536 0.01161 *
## horsepower 3.592e-02 1.040e-01 0.345 0.73007
## weight -9.580e-03 3.445e-03 -2.781 0.00569 **
## acceleration 6.872e-01 6.227e-01 1.104 0.27042
## year 7.631e-01 4.449e-02 17.154 < 2e-16 ***
## cylinders:displacement 2.521e-03 2.239e-03 1.126 0.26094
## horsepower:weight 1.115e-05 2.721e-05 0.410 0.68220
## horsepower:acceleration -1.832e-02 7.397e-03 -2.476 0.01371 *
## weight:acceleration 1.205e-05 2.099e-04 0.057 0.95425
## horsepower:weight:acceleration 2.700e-06 1.922e-06 1.404 0.16109
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.877 on 379 degrees of freedom
## Multiple R-squared: 0.8683, Adjusted R-squared: 0.8642
## F-statistic: 208.3 on 12 and 379 DF, p-value: < 2.2e-16The significant terms from the output are origin, weight, year and horsepower:acceleration.
lm.inter2 = lm(mpg~-name+displacement+origin+horsepower:weight+acceleration:cylinders+year,data = Auto)
summary(lm.inter2)##
## Call:
## lm(formula = mpg ~ -name + displacement + origin + horsepower:weight +
## acceleration:cylinders + year, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.2393 -2.1992 -0.2751 2.0135 13.7888
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.149e+01 4.464e+00 -4.815 2.12e-06 ***
## displacement -9.433e-03 6.433e-03 -1.466 0.143
## origin 1.545e+00 3.092e-01 4.997 8.85e-07 ***
## year 7.088e-01 5.581e-02 12.700 < 2e-16 ***
## horsepower:weight -1.289e-05 2.557e-06 -5.040 7.16e-07 ***
## acceleration:cylinders -6.221e-02 1.234e-02 -5.041 7.14e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.718 on 386 degrees of freedom
## Multiple R-squared: 0.776, Adjusted R-squared: 0.7731
## F-statistic: 267.5 on 5 and 386 DF, p-value: < 2.2e-16The significant terms from the output are origin, year, horsepower:weight and acceleration:cylinders.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
par(mfrow=c(2,2))
plot(log(horsepower), mpg)
plot(sqrt(horsepower), mpg)
plot(I(horsepower^2), mpg)
On examining the different transformations of horsepower, we see that the log transformation gives the most linear looking plot.
Question 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.
detach(Auto)
attach(Carseats)
str(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.car = lm(Sales ~ Price + Urban + US, data = Carseats)
summary(lm.car)##
## 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 variable Price has a p-value of < 2e-16, hence it is statistically significant to Sales. Since the coefficient is negative, we can say that for 1 unit increase in Price, the Sales decreases by 0.0544.
Urban is not significant to the response variable with a p-value 0.936, larger than 0.05
US is statistically significant with a p-value of 4.86e-06. Since the coefficient is positive, we can say that car seat stores in the US have more sales than those that are not in the US.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(y=13.043469 - 0.054459*Price - 0.021916*1 + 1.200573*1\)
In the above equation, Yes is coded as 1 and No as 0 for Urban and US.
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
The null hypothesis can be rejected for Price and US since their p-values are less than 0.05
(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.
lm.newfit = lm(Sales ~ Price+US, data = Carseats)
summary(lm.newfit)##
## 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?
anova(lm.car, lm.newfit)## Analysis of Variance Table
##
## Model 1: Sales ~ Price + Urban + US
## Model 2: Sales ~ Price + US
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 396 2420.8
## 2 397 2420.9 -1 -0.03979 0.0065 0.9357On looking at the outputs for lm.car and lm.newfit, we see that there is a slight increase in adjusted R-squared after removing the Urban variable. But the anova results show that there is not a big difference between both models.
(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(lm.newfit, level = 0.95)## 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.newfit)
which.max(hatvalues(lm.newfit))## 43
## 43From the normal Q-Q plot, we can see that most of the data falls on the normal Q-Q line. Hence, normality assumption is not violated. We can see in the Residuals vs Leverage plot that there are a few outliers scattered beyond -2 and +2. By using the hatvalues function, we can narrow down that observation 43 is an outlier.
Question 12
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate β-hat 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 estimate for the linear regression of Y onto X is \(\hat{\beta} = \frac{\sum_ix_iy_i}{\sum_jx_j^2}\)
The coefficient estimate for the linear regression of X onto Y is \(\hat{\beta}' = \frac{\sum_ix_iy_i}{\sum_jy_j^2}\)
The coefficients for both will be the same when \(\sum_jx_j^2 = \sum_jy_j^2\)
(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=rbinom(100,2,0.3)
example1<-lm(y~x+0)
summary(example1)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.25635 -0.01843 0.09642 0.96391 1.97561
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x -0.09083 0.06889 -1.319 0.19
##
## Residual standard error: 0.6757 on 99 degrees of freedom
## Multiple R-squared: 0.01726, Adjusted R-squared: 0.007333
## F-statistic: 1.739 on 1 and 99 DF, p-value: 0.1903example2<-lm(x~y+0)
summary(example2)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.82215 -0.75784 0.06094 0.59288 2.15707
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y -0.1900 0.1441 -1.319 0.19
##
## Residual standard error: 0.9773 on 99 degrees of freedom
## Multiple R-squared: 0.01726, Adjusted R-squared: 0.007333
## F-statistic: 1.739 on 1 and 99 DF, p-value: 0.1903(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
y=100:1
example3<-lm(y~x+0)
summary(example3)##
## 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-08example4<-lm(x~y+0)
summary(example4)##
## 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