Assignment #2
2024-02-21
QUESTION 2
Carefully explain the differences between the KNN classifier and KNN regression methods.
KNN classifier: identifies nearest K points of the input data and outputs a class label via majority vote of neighbors.
KNN regression: identifies nearest K points of the input data and outputs a continuous value via average of neighbors.
QUESTION 9
This question involves the use of multiple linear regression on the Auto data set.
- Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto, cex = 0.2)
- Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative.
auto_noname <- subset(Auto, select = -name)
cor(auto_noname)## 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- 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:
auto_reg <- lm(mpg ~ ., data = auto_noname)
summary(auto_reg)##
## Call:
## lm(formula = mpg ~ ., data = auto_noname)
##
## 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 response? Yes, there is a relationship
Which predictors appear to have a statistically significant relationship to the response? displacement(+), weight(-), year(+), and origin(+)
What does the coefficient for the year variable suggest? Coefficient for year is 0.751. It suggests mpg increases on average by 0.751 for every year increase
- 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(auto_reg, cex = 0.5)
According to the diagnostic plots, there may be some non-linearity with the data. There could be some outliers and observation 14 has high leverage.
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(mpg~.+year:cylinders, data = auto_noname))##
## Call:
## lm(formula = mpg ~ . + year:cylinders, data = auto_noname)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.3346 -2.0533 -0.1177 1.6257 12.0879
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -9.290e+01 1.256e+01 -7.399 8.75e-13 ***
## cylinders 1.469e+01 2.378e+00 6.178 1.66e-09 ***
## displacement 1.175e-02 7.259e-03 1.619 0.1063
## horsepower -2.765e-02 1.322e-02 -2.092 0.0371 *
## weight -6.242e-03 6.212e-04 -10.049 < 2e-16 ***
## acceleration 1.213e-01 9.423e-02 1.287 0.1988
## year 1.734e+00 1.601e-01 10.826 < 2e-16 ***
## origin 1.375e+00 2.647e-01 5.197 3.30e-07 ***
## cylinders:year -1.949e-01 3.027e-02 -6.439 3.60e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.165 on 383 degrees of freedom
## Multiple R-squared: 0.8389, Adjusted R-squared: 0.8356
## F-statistic: 249.3 on 8 and 383 DF, p-value: < 2.2e-16summary(lm(mpg~.+acceleration:horsepower, data = auto_noname))##
## Call:
## lm(formula = mpg ~ . + acceleration:horsepower, data = auto_noname)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.0329 -1.8177 -0.1183 1.7247 12.4870
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -32.499820 4.923380 -6.601 1.36e-10 ***
## cylinders 0.083489 0.316913 0.263 0.792350
## displacement -0.007649 0.008161 -0.937 0.349244
## horsepower 0.127188 0.024746 5.140 4.40e-07 ***
## weight -0.003976 0.000716 -5.552 5.27e-08 ***
## acceleration 0.983282 0.161513 6.088 2.78e-09 ***
## year 0.755919 0.048179 15.690 < 2e-16 ***
## origin 1.035733 0.268962 3.851 0.000138 ***
## horsepower:acceleration -0.012139 0.001772 -6.851 2.93e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.145 on 383 degrees of freedom
## Multiple R-squared: 0.841, Adjusted R-squared: 0.8376
## F-statistic: 253.2 on 8 and 383 DF, p-value: < 2.2e-16There are interaction effects between year:cylinders and also between acceleration:horsepower. (statistically significant p-values)
- Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
par(mfrow = c(2, 2))
plot(Auto$weight, Auto$mpg, cex = 0.5)
plot(sqrt(Auto$weight), Auto$mpg, cex = 0.5)
plot(Auto$weight ^ 2, Auto$mpg, cex = 0.5)
plot(log(Auto$weight), Auto$mpg, cex = 0.5)
Out of the three other transformations, log(weight) appears to have the most linear relationship with mpg. It will be fitted to our model below:
auto_noname2 <- auto_noname
auto_noname2$weight <- log(auto_noname2$weight)
auto_reg_log <- lm(mpg ~ ., data = auto_noname)
summary(auto_reg_log)##
## Call:
## lm(formula = mpg ~ ., data = auto_noname)
##
## 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-16Unlike before, the significant variables are: displacement(+), weight(-), year(+), and origin(+). However, the R^2 value was actually lower than the initial fit. (0.841 > 0.8215)
QUESTION 10
This question should be answered using the Carseats data set.
- Fit a multiple regression model to predict Sales using Price, Urban, and US.
carseat_reg <- lm(Sales~ Price + Urban + US, data=Carseats)
summary(carseat_reg)##
## 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- Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
Price(-): Sales will fall by 54.46 for every dollar increase in price. UrbanYes: Not considered statistically significant. USYes: If the store is in the US, Sales will increase by 1,200.
- Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales = 13.04 - 0.54(Price) - 0.02(Urban:Yes = 1, No = 0) + 1.20(US:Yes = 1, No = 0)
- For which of the predictors can you reject the null hypothesis H0 :βj =0?
Price and US. Urban was not statistically significant.
- 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.
carseat_reg_sigonly <- lm(Sales~ Price + US, data=Carseats)
summary(carseat_reg_sigonly)##
## 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- How well do the models in (a) and (e) fit the data?
All variables from model (e) were considered significant. However, the R^2 values from both models were similar (0.2393) suggesting they have similar goodness-of-fit.
- Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(carseat_reg_sigonly, level = 0.95)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632- Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow = c(2, 2))
plot(carseat_reg_sigonly, cex = 0.5)
There is evidence of some outliers.
QUESTION 11
In this problem we will investigate the t-statistic for the null hypothesis H0 : β = 0 in simple linear regression without an intercept. To begin, we generate a predictor x and a response y as follows.
set.seed(1)
x <- rnorm(100)
y <- 2 * x + rnorm(100)- Perform a simple linear regression of y onto x, without an intercept. Report the coefficient estimate βˆ, the standard error of this coefficient estimate, and the t-statistic and p-value associated with the null hypothesis H0 : β = 0. Comment on these results. (You can perform regression without an intercept using the command lm(y∼x+0).)
lm.yx = lm(y ~ x + 0)
summary(lm.yx)##
## 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-16Variable x is statistically significant. For every unit increase in x, y is expected to increase by 1.99 on average. The standard error for x is 0.11, t-stat: 18.73, and p-value: <2e-16.
- Now perform a simple linear regression of x onto y without an intercept, and report the coefficient estimate, its standard error, and the corresponding t-statistic and p-values associated with the null hypothesis H0 : β = 0. Comment on these results.
lm.xy = lm(x ~ y + 0)
summary(lm.xy)##
## 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-16Variable y is statistically significant. For every unit increase in y, x is expected to increase by 0.39 on average. The standard error for y is 0.02, t-stat: 18.73, and p-value: <2e-16.
- What is the relationship between the results obtained in (a) and (b)?
The t-stats and p-values from both models are the same. The axis has changed, causing inversion.
- For the regression of Y onto X without an intercept, the t-statistic for H0 : β = 0 takes the form βˆ/SE(βˆ), where βˆ is given by (3.38), and where: [equation in book]
Show algebraically, and confirm numerically in R, that the t-statistic can be written as: [equation in book]
n = length(x)
t = sqrt(n - 1) * (x %*% y)/sqrt(sum(x^2) * sum(y^2) - (x %*% y)^2)
as.numeric(t)## [1] 18.72593- Using the results from (d), argue that the t-statistic for the regression of y onto x is the same as the t-statistic for the regression of x onto y.
Replacing x with y, or y with x, will give the same t-statistic.
- In R, show that when regression is performed with an intercept, the t-statistic for H0 : β1 = 0 is the same for the regression of y onto x as it is for the regression of x onto y.
coef(summary(lm(y ~ x)))## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.03769261 0.09698729 -0.3886346 6.983896e-01
## x 1.99893961 0.10772703 18.5555993 7.723851e-34coef(summary(lm(x ~ y)))## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.03880394 0.04266144 0.9095787 3.652764e-01
## y 0.38942451 0.02098690 18.5555993 7.723851e-34Same t-stat values are observed in both cases.