Assignment 2
Brett Stokes
2/26/2021
Q2
Carefully explain the differences between the KNN classifier and the KNN regression methods.
The KNN regression method is very similar to the KNN classifier. For a particular x0 and K, the KNN regression method will identify training observations with similar x values and will give an average continuous response value (y-hat). On the other hand, for a particular x0 and K, The KNN classification method will identify training observations with similar x values and give average probabilities that a particular x0 will belong to each of the categorical response classes, and then will assign it to the categorical response class with the highest probability.
Q9
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.
pairs(Auto)
(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(subset(Auto, select=-c(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 1.0000000(c) Using the lm() function to perform a multiple linear regression with mpg as the response and all other variables except name the predictors. Use the summary() function to print the results. Comment on the output.
lm.auto = lm(mpg~.-name, data=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-16i. Is there a relationship between the predictors and the response?
Yes, the p-value for the F statistic is small, therefore we reject the null can say that the predictors have an effect on the response.ii. Which predictors appear to have a statistically significant relationship to the response?
From the above table and corresponding p-values, we can see thatdisplacement,weight,year, andoriginall have a statistically significant relationship to the response.iii. What does the coefficient for the year variable suggest?
With a one unit increase in year, the mpg will increase by 0.75 assuming everything else stays the same.
(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.auto)
There appears to be some pattern in the residuals plot which may suggest non-linearity. There also appears to be a funnel-type shape on the residuals plot which would indicate heteroscedasticity. Yes, there appears to be a few outliers based on the residuals vs leverage graph. Yes, observation 14 looks like it has unusually high leverage compared to the other observations.
(e) Fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.auto2 = lm(mpg~.*., data=subset(Auto, select=-c(name)))
summary(lm.auto2)##
## Call:
## lm(formula = mpg ~ . * ., data = subset(Auto, select = -c(name)))
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6303 -1.4481 0.0596 1.2739 11.1386
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.548e+01 5.314e+01 0.668 0.50475
## cylinders 6.989e+00 8.248e+00 0.847 0.39738
## displacement -4.785e-01 1.894e-01 -2.527 0.01192 *
## horsepower 5.034e-01 3.470e-01 1.451 0.14769
## weight 4.133e-03 1.759e-02 0.235 0.81442
## acceleration -5.859e+00 2.174e+00 -2.696 0.00735 **
## year 6.974e-01 6.097e-01 1.144 0.25340
## origin -2.090e+01 7.097e+00 -2.944 0.00345 **
## cylinders:displacement -3.383e-03 6.455e-03 -0.524 0.60051
## cylinders:horsepower 1.161e-02 2.420e-02 0.480 0.63157
## cylinders:weight 3.575e-04 8.955e-04 0.399 0.69000
## cylinders:acceleration 2.779e-01 1.664e-01 1.670 0.09584 .
## cylinders:year -1.741e-01 9.714e-02 -1.793 0.07389 .
## cylinders:origin 4.022e-01 4.926e-01 0.816 0.41482
## displacement:horsepower -8.491e-05 2.885e-04 -0.294 0.76867
## displacement:weight 2.472e-05 1.470e-05 1.682 0.09342 .
## displacement:acceleration -3.479e-03 3.342e-03 -1.041 0.29853
## displacement:year 5.934e-03 2.391e-03 2.482 0.01352 *
## displacement:origin 2.398e-02 1.947e-02 1.232 0.21875
## horsepower:weight -1.968e-05 2.924e-05 -0.673 0.50124
## horsepower:acceleration -7.213e-03 3.719e-03 -1.939 0.05325 .
## horsepower:year -5.838e-03 3.938e-03 -1.482 0.13916
## horsepower:origin 2.233e-03 2.930e-02 0.076 0.93931
## weight:acceleration 2.346e-04 2.289e-04 1.025 0.30596
## weight:year -2.245e-04 2.127e-04 -1.056 0.29182
## weight:origin -5.789e-04 1.591e-03 -0.364 0.71623
## acceleration:year 5.562e-02 2.558e-02 2.174 0.03033 *
## acceleration:origin 4.583e-01 1.567e-01 2.926 0.00365 **
## year:origin 1.393e-01 7.399e-02 1.882 0.06062 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.695 on 363 degrees of freedom
## Multiple R-squared: 0.8893, Adjusted R-squared: 0.8808
## F-statistic: 104.2 on 28 and 363 DF, p-value: < 2.2e-16Yes, the interactions between displacement:year, acceleration:year, and acceleration:origin are all statistically significant.
(f) Try a few different transformations of the variables, such as \(log(X)\), \(sqrt(X)\), and \(X^2\). Comment on your findings.
lm.auto3 = lm(mpg~log(cylinders)+log(displacement)+log(horsepower)+log(weight)+log(acceleration)+log(year)+log(origin), data=subset(Auto, select=-c(name)))
summary(lm.auto3)##
## Call:
## lm(formula = mpg ~ log(cylinders) + log(displacement) + log(horsepower) +
## log(weight) + log(acceleration) + log(year) + log(origin),
## data = subset(Auto, select = -c(name)))
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5987 -1.8172 -0.0181 1.5906 12.8132
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -66.5643 17.5053 -3.803 0.000167 ***
## log(cylinders) 1.4818 1.6589 0.893 0.372273
## log(displacement) -1.0551 1.5385 -0.686 0.493230
## log(horsepower) -6.9657 1.5569 -4.474 1.01e-05 ***
## log(weight) -12.5728 2.2251 -5.650 3.12e-08 ***
## log(acceleration) -4.9831 1.6078 -3.099 0.002082 **
## log(year) 54.9857 3.5555 15.465 < 2e-16 ***
## log(origin) 1.5822 0.5083 3.113 0.001991 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.069 on 384 degrees of freedom
## Multiple R-squared: 0.8482, Adjusted R-squared: 0.8454
## F-statistic: 306.5 on 7 and 384 DF, p-value: < 2.2e-16par(mfrow=c(2,2))
plot(lm.auto3)
lm.auto4 = lm(mpg~(cylinders)^2+(displacement)^2+(horsepower)^2+(weight)^2+(acceleration)^2+(year)^2+(origin)^2, data=subset(Auto, select=-c(name)))
summary(lm.auto4)##
## Call:
## lm(formula = mpg ~ (cylinders)^2 + (displacement)^2 + (horsepower)^2 +
## (weight)^2 + (acceleration)^2 + (year)^2 + (origin)^2, data = subset(Auto,
## select = -c(name)))
##
## 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-16par(mfrow=c(2,2))
plot(lm.auto4)
lm.auto5 = lm(mpg~sqrt(cylinders)+sqrt(displacement)+sqrt(horsepower)+sqrt(weight)+sqrt(acceleration)+sqrt(year)+sqrt(origin), data=subset(Auto, select=-c(name)))
summary(lm.auto5)##
## Call:
## lm(formula = mpg ~ sqrt(cylinders) + sqrt(displacement) + sqrt(horsepower) +
## sqrt(weight) + sqrt(acceleration) + sqrt(year) + sqrt(origin),
## data = subset(Auto, select = -c(name)))
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5250 -1.9822 -0.1111 1.7347 13.0681
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -49.79814 9.17832 -5.426 1.02e-07 ***
## sqrt(cylinders) -0.23699 1.53753 -0.154 0.8776
## sqrt(displacement) 0.22580 0.22940 0.984 0.3256
## sqrt(horsepower) -0.77976 0.30788 -2.533 0.0117 *
## sqrt(weight) -0.62172 0.07898 -7.872 3.59e-14 ***
## sqrt(acceleration) -0.82529 0.83443 -0.989 0.3233
## sqrt(year) 12.79030 0.85891 14.891 < 2e-16 ***
## sqrt(origin) 3.26036 0.76767 4.247 2.72e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.21 on 384 degrees of freedom
## Multiple R-squared: 0.8338, Adjusted R-squared: 0.8308
## F-statistic: 275.3 on 7 and 384 DF, p-value: < 2.2e-16par(mfrow=c(2,2))
plot(lm.auto5)
The log transform and square root transform of the predictor variables seem to offer models with greater \(R^2\) values, and should probably be considered if we are interested in predicting mpg. However, we still seem to have the funnel shape indicating heteroscedasticity, log transformation on the response variable might help with this.
Q10
This Question should be answered using the Carseats data set.
(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
lm.carseats = lm(Sales~Price+Urban+US, data= 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-16(b) Provide an interpretation of each coefficient in the model. Be careful-some of the variables in the model are qualitative!
For every one unit increase in Price, Sales will decrease by 0.05. Stores in the US will have Sales that are 1.20 higher than stores not in the US. Urban or Rural has no effect on the response variable due to its large p-value.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales = -0.054459*Price + 1.200573*USYes + - 0.021916*UrbanYes + 13.043469
(d) For which of the predictors can you reject the null hypothesis {H_{0}: B_{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.carseats2 = lm(Sales~Price+US, data= Carseats)
summary(lm.carseats2)##
## 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(f) How well do the models in (a) and (e) fit the data?
Both models explain roughly 23% of the total variance of y, which is not very good.
(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(lm.carseats2)## 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)?
par(mfrow=c(2,2))
plot(lm.carseats2)
Based on the residuals vs leverage plot, there appears to be a few outliers, as well as one observation in particular with much higher leverage than the rest of the observations.
Q12
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate B for the linear regression of Y unto 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?
Coefficient estimate of Y onto X: \(\hat{\beta} = \frac{\sum_ix_iy_i}{\sum_jx_j^2}\);
Coefficient estimate of X onto Y: \(\hat{\beta}' = \frac{\sum_ix_iy_i}{\sum_jy_j^2}\).
These coefficient estimates will be equal when: \(\sum_jx_j^2 = \sum_jy_j^2\)
(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(254)
x=rnorm(100)
y=rnorm(100)
sum(I(x^2))## [1] 101.2619sum(I(y^2))## [1] 132.1775lm.y=lm(y~x+0)
lm.x=lm(x~y+0)
summary(lm.x)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2738 -0.5390 0.1106 0.8225 2.4104
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y -0.04466 0.08785 -0.508 0.612
##
## Residual standard error: 1.01 on 99 degrees of freedom
## Multiple R-squared: 0.002603, Adjusted R-squared: -0.007471
## F-statistic: 0.2584 on 1 and 99 DF, p-value: 0.6123summary(lm.y)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.6291 -0.5881 0.1003 0.7739 3.2874
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x -0.0583 0.1147 -0.508 0.612
##
## Residual standard error: 1.154 on 99 degrees of freedom
## Multiple R-squared: 0.002603, Adjusted R-squared: -0.007471
## F-statistic: 0.2584 on 1 and 99 DF, p-value: 0.6123From the above output, we can see that \(x^2\) and \(y^2\) are not equal and as expected the corresponding \(\beta\) coefficients are different for each model.
(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.
set.seed(254)
x=rnorm(100)
y=-x
sum(I(x^2))## [1] 101.2619sum(I(y^2))## [1] 101.2619lm.y=lm(y~x+0)
lm.x=lm(x~y+0)
summary(lm.x)## Warning in summary.lm(lm.x): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.036e-16 -2.710e-17 4.000e-19 2.900e-17 4.788e-15
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y -1.000e+00 4.831e-17 -2.07e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.862e-16 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 4.284e+32 on 1 and 99 DF, p-value: < 2.2e-16summary(lm.y)## Warning in summary.lm(lm.y): essentially perfect fit: summary may be unreliable##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.788e-15 -2.900e-17 -4.000e-19 2.710e-17 2.036e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x -1.000e+00 4.831e-17 -2.07e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.862e-16 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 4.284e+32 on 1 and 99 DF, p-value: < 2.2e-16From the above output, we can see that \(x^2\) equals \(y^2\) and the corresponding \(\beta\) coefficients are the same for both models.