Assignment 2
Lauren Wilson
1/19/2021
Q2
- Carefully explain the differences between the KNN classifier and KNN regression methods.
KNN Regression starts by identifying the observations that are closest to the point being predicted, the test observation. It then estimates the value of this point using the average of all the closest observations values. KNN Classifier starts in the same way, identifying the observations that are closest to the test observation but, instead of the average values it uses the estimated probability. From this probability, a decision boundary is drawn separating the values by their probability of occurrence. The point being predicted will fall on one side of this decision boundary.
Q9
- This question involves the use of multiple linear regression on the
Autodata set.
a)
Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto)
There is some clear correlation between variables such as horsepower and displacement which also seem to have a linear relationship with mpg.
b)
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[,c(-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 1.0000000Both displacement and weight seem to strongly correlate with the response variable mpg
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?
ii. Which predictors appear to have a statistically significant relationship to the response?
iii. What does the coefficient for the year variable suggest?
The output shows a small p-value suggesting the model is significant, meaning there is a relationship between at least one of the predictors and the response variable, mpg. Looking at the significance of the individual terms, displacement, weight,year, and origin are significant while the remaining terms cylinders,horsepower, and acceleration are not significant. Some variables may need to be removed from the model. The year variable suggests that for every 1 unit increase in year there is a .75 increase in mpg.
#lm.auto <- lm(mpg ~ .,auto[,c(-9)])
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-16d)
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?
The residual plot appears to violate the linearity assumption with the quadratic shape of the residuals vs fitted plot. Multiple points appear to be outliers laying higher than 2 or lower than -2 in the Residuals vs Leverage plot. This plot shows unusually high leverage in point 14.
par(mfrow=c(2,3))
plot(lm.auto, which=c(1:6))
e)
Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
From the correlation matrix we find the most highly correlated terms (over 90%) and will use these to test for any interaction effects
lm.auto <- lm(mpg~cylinders:displacement+ cylinders + displacement + displacement * weight, Auto[,c(-9)])
summary(lm.auto)##
## Call:
## lm(formula = mpg ~ cylinders:displacement + cylinders + displacement +
## displacement * weight, data = Auto[, c(-9)])
##
## 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-16The interaction between displacement and weight is significant while the other interaction term is not. This means mpg gets an extra boost from the inclusion of the interaction effect between displacment and weight
f)
Try a few different transformations of the variables, such as logX, X‾‾√, X2. Comment on your findings.
par(mfrow=c(2,2))
plot(log(Auto$weight), Auto$mpg)
plot(sqrt(Auto$weight), Auto$mpg)
plot(I(Auto$weight^2), Auto$mpg)
Focusing on the weight variable, the sqrt and log transformations seem to plot the most linear looking data.
Q10
This question should be answered using the built-in Carseats data set from the ISLR package.
a)
Fit a multiple regression model to predict Sales using Price, Urban, and US.
summary(lm(Sales~Price+Urban+US, 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-16?Carseatsb)
Provide an interpretation of each coefficient in the model. Be careful some of the variables in the model are qualitative!
Price can be interpreted as, for every 1000 units, if I increase the Price by 1 dollar Sales are set to decrease by 54 units on average, all other predictors remaining fixed. In the regression output, UrbanYes is negative indicating an urban area leads to lower sales than UrbanNo. Specifically, Sales in urban locations have average sales about 21 units less than rural locations, all other predictors remaining fixed. USYes is positive indicating US stores have higher sales in this dataset. Specifically, sales in US stores average 1200 units more than sales outside the US.
c)
Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.043469 -0.054459 *Price -0.021916 *Urban{Yes}+1.200573*US{Yes} + ε\)
d)
For which of the predictors can you reject the null hypothesis H0 :βj =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.carseats = lm(Sales~Price+US, Carseats)
summary(lm.carseats)##
## 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-16f)
How well do the models in (a) and (e) fit the data?
Removing the Urban variable from the model in question (e) did not change the R2 value. Even after the removal both models maintained a prediction power of only 23%. The adjusted R^2 ,which takes into account the multiple predictors, does slightly increase in this smaller model.
g)
Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
?confint
confint(lm.carseats, level = 0.95)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632h)
Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow=c(3,2))
plot(lm.carseats, which = c(1:6))
There appears to be a few outliers higher than 2 and lower than -2, but no high leverage points that exceed the average leverage of \(\frac{(p+1)}{n}\) which in this case is \(\frac{(2+1)}{400} = 0.0075\)
Q12
This problem involves simple linear regression without an intercept.
a)
Recall that the coefficient estimate 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?
The coeffients are the same if the summation of the squared jith values of x = the summation of the squared jith values of y
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 <- (x ^ 2) + rnorm(100)
sum(y^2)## [1] 2050395816fit.y <- lm(y~x)
fit.x <- lm(x~y)
summary(fit.y)##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -832.7 -676.6 -208.0 572.3 1616.4
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1716.868 151.678 -11.32 <2e-16 ***
## x 101.000 2.608 38.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 752.7 on 98 degrees of freedom
## Multiple R-squared: 0.9387, Adjusted R-squared: 0.9381
## F-statistic: 1500 on 1 and 98 DF, p-value: < 2.2e-16summary(fit.x)##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.056 -5.116 2.039 6.359 7.856
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.905e+01 1.087e+00 17.54 <2e-16 ***
## y 9.294e-03 2.399e-04 38.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.221 on 98 degrees of freedom
## Multiple R-squared: 0.9387, Adjusted R-squared: 0.9381
## F-statistic: 1500 on 1 and 98 DF, p-value: < 2.2e-16c)
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 <- 1:100
sum(y^2)## [1] 338350fit.y <- lm(y~x)
fit.x <- lm(x~y)
summary(fit.y)## Warning in summary.lm(fit.y): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.680e-13 -4.300e-16 2.850e-15 5.302e-15 3.575e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.684e-14 5.598e-15 -1.015e+01 <2e-16 ***
## x 1.000e+00 9.624e-17 1.039e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.778e-14 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.08e+32 on 1 and 98 DF, p-value: < 2.2e-16summary(fit.x)## Warning in summary.lm(fit.x): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.680e-13 -4.300e-16 2.850e-15 5.302e-15 3.575e-14
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.684e-14 5.598e-15 -1.015e+01 <2e-16 ***
## y 1.000e+00 9.624e-17 1.039e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.778e-14 on 98 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.08e+32 on 1 and 98 DF, p-value: < 2.2e-16