HW2
Lily He
2/3/2022
library(dplyr)
library(ISLR)Problem 2.
Carefully explain the differences between the KNN classifier and KNN regression methods
The KNN classifier is typically used to solve classification problems (those with a qualitative response) by identifying the neighborhood of x0 and then estimating the conditional probability P(Y=j|X=x0) for class j as the fraction of points in the neighborhood whose response values equal j. The KNN regression method is used to solve regression problems (those with a quantitative response) by again identifying the neighborhood of x0 and then estimating f(x0) as the average of all the training responses in the neighborhood.
Problem 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.
data=Autopairs(Auto)
(b) Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "name"cor(Auto[,-9])## 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.fit <- lm(mpg ~. -name, data = Auto)
summary(lm.fit)##
## 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 response? Yes, because there is a very small p-value.
Which predictors appear to have a statistically significance relationship to the response? If we used a significant level of 0.05 to determine which predictors were significant in this regression model, we’d say that
displacement,weight,yearandoriginare statistically significant predictors.What does the coefficient for the year variable suggest? Mpg is predicted to have an increase of 0.750773 when all other predictors are fixed.
(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.fit)
From Normal Q-Q plot, there are many values away from the same direction, means that the data skewed, and do not follow the normality assumption. From Residuals vs Leverage plot, there are many standardized residual values with absolute value greater than 3. They are outliers.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(mpg ~ . -name + displacement*weight, data = Auto))##
## Call:
## lm(formula = mpg ~ . - name + displacement * weight, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.9027 -1.8092 -0.0946 1.5549 12.1687
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.389e+00 4.301e+00 -1.253 0.2109
## cylinders 1.175e-01 2.943e-01 0.399 0.6899
## displacement -6.837e-02 1.104e-02 -6.193 1.52e-09 ***
## horsepower -3.280e-02 1.238e-02 -2.649 0.0084 **
## weight -1.064e-02 7.136e-04 -14.915 < 2e-16 ***
## acceleration 6.724e-02 8.805e-02 0.764 0.4455
## year 7.852e-01 4.553e-02 17.246 < 2e-16 ***
## origin 5.610e-01 2.622e-01 2.139 0.0331 *
## displacement:weight 2.269e-05 2.257e-06 10.054 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.964 on 383 degrees of freedom
## Multiple R-squared: 0.8588, Adjusted R-squared: 0.8558
## F-statistic: 291.1 on 8 and 383 DF, p-value: < 2.2e-16summary(lm(mpg ~. -name +displacement:weight, data = Auto))##
## Call:
## lm(formula = mpg ~ . - name + displacement:weight, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.9027 -1.8092 -0.0946 1.5549 12.1687
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.389e+00 4.301e+00 -1.253 0.2109
## cylinders 1.175e-01 2.943e-01 0.399 0.6899
## displacement -6.837e-02 1.104e-02 -6.193 1.52e-09 ***
## horsepower -3.280e-02 1.238e-02 -2.649 0.0084 **
## weight -1.064e-02 7.136e-04 -14.915 < 2e-16 ***
## acceleration 6.724e-02 8.805e-02 0.764 0.4455
## year 7.852e-01 4.553e-02 17.246 < 2e-16 ***
## origin 5.610e-01 2.622e-01 2.139 0.0331 *
## displacement:weight 2.269e-05 2.257e-06 10.054 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.964 on 383 degrees of freedom
## Multiple R-squared: 0.8588, Adjusted R-squared: 0.8558
## F-statistic: 291.1 on 8 and 383 DF, p-value: < 2.2e-16Use displacement * weight and displacement:weight in the lm model, the results are same, the interaction is significant.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
lm.cyld = lm(mpg ~ log(cylinders), data = Auto)
summary(lm.cyld)##
## Call:
## lm(formula = mpg ~ log(cylinders), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.5678 -2.7969 -0.7969 2.3783 17.8031
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 56.6059 1.3837 40.91 <2e-16 ***
## log(cylinders) -20.0600 0.8234 -24.36 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.921 on 390 degrees of freedom
## Multiple R-squared: 0.6034, Adjusted R-squared: 0.6024
## F-statistic: 593.5 on 1 and 390 DF, p-value: < 2.2e-16par(mfrow= c(2,2))
plot(lm.cyld)
lm.cyld2 = lm(mpg ~ sqrt(cylinders), data = Auto)
summary(lm.cyld2)##
## Call:
## lm(formula = mpg ~ sqrt(cylinders), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.3194 -3.0285 -0.6513 2.3607 17.8430
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 62.8112 1.6268 38.61 <2e-16 ***
## sqrt(cylinders) -17.0271 0.6955 -24.48 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.906 on 390 degrees of freedom
## Multiple R-squared: 0.6058, Adjusted R-squared: 0.6048
## F-statistic: 599.4 on 1 and 390 DF, p-value: < 2.2e-16par(mfrow= c(2,2))
plot(lm.cyld2)
lm.cyld3 = lm(mpg ~ (cylinders^2), data = Auto)
summary(lm.cyld3)##
## Call:
## lm(formula = mpg ~ (cylinders^2), data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.2413 -3.1832 -0.6332 2.5491 17.9168
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 42.9155 0.8349 51.40 <2e-16 ***
## cylinders -3.5581 0.1457 -24.43 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.914 on 390 degrees of freedom
## Multiple R-squared: 0.6047, Adjusted R-squared: 0.6037
## F-statistic: 596.6 on 1 and 390 DF, p-value: < 2.2e-16par(mfrow= c(2,2))
plot(lm.cyld3)
After apply log to predictor, looks like there is subtle change, the lines a little bit smoother when apply square operation.
Problem 10.
This question should be answered using the Carseats data set.
library(ISLR)
attach(Carseats)(a) Fit a multiple regression model to predict Sales using Price,Urban, and US.
lm.carfit <- lm(Sales ~ Price+Urban +US)
summary(lm.carfit)##
## Call:
## lm(formula = Sales ~ Price + Urban + US)
##
## 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. Becareful—some of the variables in the model are qualitative!
From the table above, price and US are significant predictors of Sales, for every $1 increase in my price, sales descrease by $54. Sales inside the US are $1200 higher than sales outside of the US. Urban has no effect on Sales.
(c) Write out the model in equation form, being careful to handlethe qualitative variables properly.
\(Sales = 13.043469-0.05445Price + -0.021916Urban_{Yes} + 1.200573US_{Yes}\)
(d) For which of the predictors can you reject the null hypothesis \(H0 : \beta_j = 0\)?
Price and US
(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.carfit <- lm(Sales ~ Price +US)
summary(lm.carfit)##
## Call:
## lm(formula = Sales ~ Price + US)
##
## 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?
Not well, each model explains around 23% of the varience in Sales.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(lm.carfit)## 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)?
R has built in functions that can help us ifulential points various statistics with one simple command. Researchers have suggested several cutoff levels or upper limits as to what is the acceptable influence an observation should have before being considered an outlier. For example, the average leverage \(fract{p+1}{n}\) which for us is \(\frac{(2+1)}{400} = 0.0075\).
par(mfrow= c(2,2))
plot(lm.carfit)
summary(influence.measures(lm.carfit))## Potentially influential observations of
## lm(formula = Sales ~ Price + US) :
##
## 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 0.00R points out a few observations that violate various rules for each influence measure. Typically, one can demonstrate these statistics and report both a regression with all data included and one with the outliers removed and compare.
outyling.obs<-c(26,29,43,50,51,58,69,126,160,166,172,175,210,270,298,314,353,357,368,377,384,387,396)
Carseats.small<-Carseats[-outyling.obs,]
fit2<-lm(Sales~Price+US,data=Carseats.small)
summary(fit2)##
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats.small)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.263 -1.605 -0.039 1.590 5.428
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 12.925232 0.665259 19.429 < 2e-16 ***
## Price -0.053973 0.005511 -9.794 < 2e-16 ***
## USYes 1.255018 0.248856 5.043 7.15e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.29 on 374 degrees of freedom
## Multiple R-squared: 0.2387, Adjusted R-squared: 0.2347
## F-statistic: 58.64 on 2 and 374 DF, p-value: < 2.2e-16With these potential outliers or influential observations removed, very little changes from the linear model fit to the full data set. The confidence interval for the coefficient estimates produced by the linear model fit to the full data set contain the estimates of the coefficients for the estimates of the model with the outliers removed. It’s safe to include all of the data points in our model.
Problem 12.
This problem involves simple linear regression without an intercept.
data("Carseats")(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 numerator term of the equation is equal for both regressions. For the coefficients be the same the denominator term must be equal too.
(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 <- 1:100
y <- 2 * x + rnorm(100, sd = 0.1)fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
cbind(XonY = coef(fit.Y),YonX=coef(fit.X))[1,]## XonY YonX
## 2.0001514 0.4999619(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
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
cbind(XonY = coef(fit.Y),YonX=coef(fit.X))[1,]## XonY YonX
## 0.5074627 0.5074627