Assignment 2
Adrianne Kristianto
2/18/2021
Question 2
Carefully explain the differences between the KNN classifier and KNN regression methods
Although these two are quite similar, KNN classifier is used to solve classification problems (qualitative response) using the most common group among the K nearest neighbor. KNN regression is used to make a quantitative estimate by averaging the result of the K nearest neighbors.
“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 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.
9A: Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto)
9B: Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative
cor(Auto[, names(Auto) !="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 1.00000009C: 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.
lm.auto <- lm(mpg ~. -name, 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-16- Is there a relationship between the predictors and the response?
- Yes, there’s a significant relationship between the predictors and
mpg, the response. However, predictors likecylinders,horsepower, andaccelerationaren’t statistically significant (>0.05), meaning that they don’t have a significant effect onmpg.
- Yes, there’s a significant relationship between the predictors and
- Which predictors appear to have a statistically significant relationship to the response?
displacement,weight,year, andorigin
- What does the coefficient for the
yearvariable suggest?- The regression coefficient for the variable
yearis 0.750773. This means that with with every one year increase in car making/modeling,mpgincreases by 0.750773 (newer cars are fuel efficient)
- The regression coefficient for the variable
9D: 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)
Comments:
- As seen in the residual plot, there appears to be a solid curve. This indicates that there are issues with the fit and there is a non-linear relationship between predictors and response
- In the Normal Q-Q plot, we can see that residuals are normally distributed, and that it’s right skewed
- The sqrt standardized residual model shows that constant variance of error assumption is not true for this model
- Last but not least, the standardized residual model indiciates that observation 14 has high leverage
9E: Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.auto1 <- lm(mpg ~. -name + weight:acceleration, Auto)
summary(lm.auto1)##
## Call:
## lm(formula = mpg ~ . - name + weight:acceleration, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.247 -2.048 -0.045 1.619 12.193
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4.364e+01 5.811e+00 -7.511 4.18e-13 ***
## cylinders -2.141e-01 3.078e-01 -0.696 0.487117
## displacement 3.138e-03 7.495e-03 0.419 0.675622
## horsepower -4.141e-02 1.348e-02 -3.071 0.002287 **
## weight 4.027e-03 1.636e-03 2.462 0.014268 *
## acceleration 1.629e+00 2.422e-01 6.726 6.36e-11 ***
## year 7.821e-01 4.833e-02 16.184 < 2e-16 ***
## origin 1.033e+00 2.686e-01 3.846 0.000141 ***
## weight:acceleration -5.826e-04 8.408e-05 -6.928 1.81e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.141 on 383 degrees of freedom
## Multiple R-squared: 0.8414, Adjusted R-squared: 0.838
## F-statistic: 253.9 on 8 and 383 DF, p-value: < 2.2e-16Comment: For this model, we can see that weight:acceleration appear to be statistically significant
lm.auto2 <- lm(mpg ~. -name + weight:acceleration + weight:displacement + acceleration:horsepower, Auto)
summary(lm.auto2)##
## Call:
## lm(formula = mpg ~ . - name + weight:acceleration + weight:displacement +
## acceleration:horsepower, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.0240 -1.5736 0.0379 1.5342 12.2525
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -6.799e+00 6.962e+00 -0.977 0.32939
## cylinders 4.841e-01 2.991e-01 1.619 0.10630
## displacement -8.372e-02 1.242e-02 -6.740 5.89e-11 ***
## horsepower 1.031e-01 3.449e-02 2.990 0.00297 **
## weight -1.318e-02 2.555e-03 -5.158 4.02e-07 ***
## acceleration 1.490e-01 2.847e-01 0.523 0.60111
## year 7.751e-01 4.462e-02 17.373 < 2e-16 ***
## origin 4.097e-01 2.582e-01 1.587 0.11331
## weight:acceleration 2.631e-04 1.362e-04 1.932 0.05415 .
## displacement:weight 2.246e-05 2.841e-06 7.905 2.91e-14 ***
## horsepower:acceleration -1.050e-02 2.407e-03 -4.363 1.66e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.892 on 381 degrees of freedom
## Multiple R-squared: 0.8663, Adjusted R-squared: 0.8627
## F-statistic: 246.8 on 10 and 381 DF, p-value: < 2.2e-16Comments: Adding two extra interaction effects, however, we see that weight:acceleration appears to be insignificant (0.5415), leaving displacement:weight and horsepwoer:acceleration statistically significant
9F: Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
lm.auto3 <- lm(mpg ~ . - name + log(weight) + sqrt(horsepower) + I(displacement^2) + I(cylinders^2), Auto)
summary(lm.auto3)##
## Call:
## lm(formula = mpg ~ . - name + log(weight) + sqrt(horsepower) +
## I(displacement^2) + I(cylinders^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.2216 -1.4972 -0.1142 1.4184 11.9541
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.228e+02 4.381e+01 2.803 0.00532 **
## cylinders 3.360e-01 1.451e+00 0.232 0.81697
## displacement -3.765e-02 2.175e-02 -1.731 0.08421 .
## horsepower 2.197e-01 6.684e-02 3.287 0.00111 **
## weight 1.181e-03 2.074e-03 0.569 0.56949
## acceleration -2.036e-01 1.004e-01 -2.028 0.04323 *
## year 7.654e-01 4.526e-02 16.911 < 2e-16 ***
## origin 5.497e-01 2.679e-01 2.052 0.04088 *
## log(weight) -1.493e+01 6.714e+00 -2.223 0.02678 *
## sqrt(horsepower) -5.998e+00 1.493e+00 -4.018 7.06e-05 ***
## I(displacement^2) 6.788e-05 3.773e-05 1.799 0.07279 .
## I(cylinders^2) -1.067e-02 1.164e-01 -0.092 0.92702
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.904 on 380 degrees of freedom
## Multiple R-squared: 0.8654, Adjusted R-squared: 0.8615
## F-statistic: 222.2 on 11 and 380 DF, p-value: < 2.2e-16Comments: We can see that by adding transformations, the Adjusted R-squared increases
Question 10
This question should be answered using the Carseats data set
data(Carseats)10A: Fit a multiple regression model to predict Sales using Price, Urban, and US
lm.carseats <- lm(Sales ~ Price+Urban+US, Carseats)
summary(lm.carseats)##
## 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-1610B: Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
- When price increases by $1,000 and other variables held constant,
Salesdecrease by 54,459 - Store location (urban/rural) does not have any affect on sales (>0.05)
- A US store has 1,200 more carseats sold than a store located abroad
10C: Write out the model in equation form, being careful to handle the qualitative variables properly.
- SALES = 13.04 + PRICE(-0.05) + URBANYes(-0.02) + USYes(1.2)
10D: For which of the predictors can you reject the null hypothesis H0 : βj = 0? - Price and USYes (p-value > 0.05)
10E: 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.carseats1 <- lm(Sales ~ Price+US, Carseats)
summary(lm.carseats1)##
## 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-1610F: How well do the models in (a) and (e) fit the data?
- They both fit similarly, with higher RSE and Adjusted R-squared values. This means that the latter model fits the data slightly better
10G: Using the model from (e), obtain 95 % confidence intervals for the coefficient(s)
confint(lm.carseats1)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.7077663210H: Is there evidence of outliers or high leverage observations in the model from (e)?
plot(predict(lm.carseats1), rstudent(lm.carseats1))
All studentized residuals appear between -3 and 3, so no potential outliers are suggested
par(mfrow=c(2,2))
plot(lm.carseats1)
The leverage-statistic plot that suggest that the corresponding points have high leverage
Question 12
This problem involves simple linear regression without an intercept
Q12A: Recall that the coefficient estimate ˆ β 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?
- sum of squares of observed y values = sum of squares of observed x values
Q12B: 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-16Q12C: 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