Data Mining Homework 2
John Thomas
February 11, 2019
Homework Problems 2, 9, 10, 12
Problem 2:
We use KNN classifier to solve problems with qualitative responses. This is done by identifying a neighborhood of xo and then estimating a conditional probability of P(Y=c | X=x0 ) for the class c as the portion of the neighborhood if the response value is c. We use KNN Regression method to solve problems with quantitative responses. This is done by analyzing data around xo and then estimating f(xo) as the average of all training responses in the neighborhood around xo.
Problem: 9
This question involves the use of multiple linear regression on the Auto data set.
Part A:
Produce a scatterplot matrix which includes all of the variables in the data set.
Auto <- read.csv("Auto.csv")
pairs(Auto)
Part 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.
Auto <- read.csv("Auto.csv")
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower"
## [5] "weight" "acceleration" "year" "origin"
## [9] "name"Auto$horsepower <- as.numeric(as.character(Auto$horsepower))## Warning: NAs introduced by coercionAuto <- na.omit(Auto)
cor(Auto[1:8])## 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 1.0000000Part 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:
i. Is there a relationship between the predictors and the response?
Auto <- read.csv("Auto.csv")
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower"
## [5] "weight" "acceleration" "year" "origin"
## [9] "name"Auto$horsepower <- as.numeric(as.character(Auto$horsepower))## Warning: NAs introduced by coercionAuto <- na.omit(Auto)
fit2 <- lm(mpg ~ . - name, data = Auto)
summary(fit2)##
## 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-16With the data shown above we can see there is a relation between MPG and the other predictors of the data set “Auto”.
ii. Which predictors appear to have a statistically significant relationship to the response?
We also can observe that the predictors with statistical significant relationship are, MPG, displacement, weight, year and origin.
iii. What does the coefficient for the year variable suggest?
The coefficient of “Year” suggests that the increase of 1 Year on a vehicle would increase MPG by .77 if all other predictors are constant. We can say that every year cars are more fuel efficient by .77 MPG.
Part 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 plots identify any observations with unusually high leverages ?
par(mfrow = c(2, 2))
plot(fit2)
Reviewing the graphs above we can see that the fit of the data is somewhat non-linear. With observations on the Residuals VS Leverage graph below and above -2 and 2 respectively, the point 14 is a high leverage point.
Part E
Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
fit3 <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(fit3)##
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement *
## weight, data = Auto[, 1:8])
##
## 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-16We can see from our summary of our latest fit, that weight and displacement are statistically significant but cylinders and displacement is not
Part F
Try a few different transformations of the variables, such as log(X), ???X, X^2. Comment on your findings.
par(mfrow = c(2, 2))
plot(log(Auto$weight), Auto$mpg)
plot(sqrt(Auto$weight), Auto$mpg)
plot((Auto$weight)^2, Auto$mpg)
Isolating MPG and the Wieght of the vehicle, we can see that the log function has a more linear plot.
Problem 10
This question should be answered using the Carseats data set.
Part A
Fit a multiple regression model to predict Sales using Price, Urban, and US.
library(ISLR)##
## Attaching package: 'ISLR'## The following object is masked _by_ '.GlobalEnv':
##
## Autodata(Carseats)
fit3 <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(fit3)##
## 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-16Part B
Provide an interpretation of each coefficient in the model. Be careful - some of the variables in the model are qualitative.
We can interpret the coefficient of the “Price” variable as the average effect of price given that there is an increase a dollar given that there’s a decrease in about 54.46 carseats in sales with all other predictors fixed. The coefficient for “Urban” variable represents the difference of sales between urban locations and rural locations. We can expect roughly 21.92 less carseats to be sold in urban than rural given that all other predictors remain unchanged. On average we can also see that average sales in the U.S. is 1200.57 more than a Non-U.S. store as long as other predictors stay constant.
Part C
Write out the model in equation form, being careful to handle the qualitative variables properly.
The model may be written as:
Sales = 13.04 + (???0.054)×Price + (???0.02)×Urban + (1.20)×US + errorWhere Urban and US are 1 if they are located in Urban or US respectively, 0 if they are in rural or Non-U.S.
Part D
For which of the parameters can you reject the null hypothesis Ho : Bj = 0?
We can reject that null hypothesis for “Price” and “U.S.” since both are statistically significant.
Part 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.
fit4 <- lm(Sales ~ Price + US, data = Carseats)
summary(fit4)##
## 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-16Part F
How well do the models in (a) and (e) fit the data?
The R^2 value for the previous model is better than the first model. Roughly 23.93% of the variability is explained in model (e).
Part G
Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(fit4)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632Part H
Is there evidence of outliers or high leverage observations in the model from (e) ?
par(mfrow = c(2, 2))
plot(fit4)
The plot of standardized residuals versus leverage shows us a few outliers (higher than 2 or lower than -2) and some leverage points reaching towards .03 and .04.
Problem 12
This problem involves simple linear regression without an intercept.
Part A
Recall that the coefficient estimate ??^ 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 regression of Y onto X is \(\hat{\beta}\) = \(\Sigma_i[(x_i)(y_i)]\over\Sigma_j[(xj)^{2}]\)
The coefficient estimate for the regression of X onto Y is \(\hat{\beta}\) = \((\Sigma_i[(x_i)(y_i)])\over(\Sigma_j[(yj)^{2}])\)
The coefficients are the same if and only if \(\Sigma_j(x_j)^2 = \Sigma_j(y_j)^2\).
Part 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
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-16Part 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
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