Nicole Brown
HW 2
Algorithms II Summer 20
library(ISLR)Problem 2. Carefully explain the differences between the KNN classifier and KNN regression methods.
In KNN regresssions the algorithm will predict the output value by averaging out the nearest neighbor values based on the k setting. For KNN classifier, the algorithm will predict the class outcome based on which classes the nearest neighbors fall into and based on the k setting. So basically, the KNN regression is an averaged output while the KNN classifier is an output based on whichever class is the most highly represented near the given cluster.
attach(Auto)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.
pairs(Auto[,-9])
(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.
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:
lmfit<-lm(mpg~.-name, data=Auto)
summary(lmfit)##
## 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?
There is a relationship between some of the predictors and the response variable as evidenced by the significant p-values.
ii. Which predictors appear to have a statistically significant relationship to the response?
The following predictors have significant p-values:
displacement
weight
year
origin
iii. What does the coefficient for the year variable suggest?
For the year variable, every 1 year increase in the predictor variable will result in a .75 MPG increase. In other words, the newer the car the better the MPG.
(d) Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit.
par(mfrow=c(2,2))
plot(lmfit)
We can see from the residuals vs leverage output that there is a highly leveraged x data point, observation 14. There is also evidence of nonlinearty.
Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
Yes, we can tell that there are some outliers by the data points existing above the Cook’s distance cutoff. We can tell that observation 14 is highly leveraged by the placement in the far right corner of the Residuals vs Leverage plot. Also, the Residuals vs Fitted do show some abnormal data points in the top right corner.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm2<-lm(mpg~.*.-name*.+.-name,data=Auto)
summary(lm2)##
## Call:
## lm(formula = mpg ~ . * . - name * . + . - name, data = Auto)
##
## 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-16The following interaction effects show significant p-values:
acceleration:year
acceleration:origin
displacement:year
(f)Try a few different transformations of the variables. Comment on your findings.
lm3<-lm(log(mpg)~.-name,data=Auto)
summary(lm3)##
## Call:
## lm(formula = log(mpg) ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.40955 -0.06533 0.00079 0.06785 0.33925
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.751e+00 1.662e-01 10.533 < 2e-16 ***
## cylinders -2.795e-02 1.157e-02 -2.415 0.01619 *
## displacement 6.362e-04 2.690e-04 2.365 0.01852 *
## horsepower -1.475e-03 4.935e-04 -2.989 0.00298 **
## weight -2.551e-04 2.334e-05 -10.931 < 2e-16 ***
## acceleration -1.348e-03 3.538e-03 -0.381 0.70339
## year 2.958e-02 1.824e-03 16.211 < 2e-16 ***
## origin 4.071e-02 9.955e-03 4.089 5.28e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1191 on 384 degrees of freedom
## Multiple R-squared: 0.8795, Adjusted R-squared: 0.8773
## F-statistic: 400.4 on 7 and 384 DF, p-value: < 2.2e-16lm4<-lm((mpg*mpg)~.-name,data=Auto)
summary(lm4)##
## Call:
## lm(formula = (mpg * mpg) ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -483.45 -141.87 -19.62 103.58 1042.84
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.878e+03 2.928e+02 -6.412 4.22e-10 ***
## cylinders -1.436e+01 2.038e+01 -0.704 0.48157
## displacement 1.328e+00 4.738e-01 2.802 0.00534 **
## horsepower -3.587e-01 8.693e-01 -0.413 0.68009
## weight -3.522e-01 4.111e-02 -8.567 2.62e-16 ***
## acceleration 9.278e+00 6.232e+00 1.489 0.13740
## year 4.081e+01 3.214e+00 12.698 < 2e-16 ***
## origin 9.509e+01 1.754e+01 5.422 1.04e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 209.8 on 384 degrees of freedom
## Multiple R-squared: 0.7292, Adjusted R-squared: 0.7243
## F-statistic: 147.8 on 7 and 384 DF, p-value: < 2.2e-16lm5<-lm(sqrt(mpg)~.-name,data=Auto)
summary(lm5)##
## Call:
## lm(formula = sqrt(mpg) ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.98891 -0.18946 0.00505 0.16947 1.02581
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.075e+00 4.290e-01 2.506 0.0126 *
## cylinders -5.942e-02 2.986e-02 -1.990 0.0474 *
## displacement 1.752e-03 6.942e-04 2.524 0.0120 *
## horsepower -2.512e-03 1.274e-03 -1.972 0.0493 *
## weight -6.367e-04 6.024e-05 -10.570 < 2e-16 ***
## acceleration 2.738e-03 9.131e-03 0.300 0.7644
## year 7.381e-02 4.709e-03 15.675 < 2e-16 ***
## origin 1.217e-01 2.569e-02 4.735 3.09e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3074 on 384 degrees of freedom
## Multiple R-squared: 0.8561, Adjusted R-squared: 0.8535
## F-statistic: 326.3 on 7 and 384 DF, p-value: < 2.2e-16The log and square root transformations resulted in models with more predictive power than our initial model. The only transformation that didn’t seem to give the same kind of boost was the r square.
Problem 10.
This question should be answered using the [Carseats] https://rdrr.io/cran/ISLR/man/Carseats.html data set.
attach(Carseats)(a) Fit a multiple regression model to predict Sales using Price, Urban,and US.
fit<-lm(Sales~Price+Urban+US)
summary(fit)##
## 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. Be careful some of the variables in the model are qualitative!
Price and US can bee seen to be significant predictors of Sales per the table above. Thus, for every $1000 increase in price, sales will decrease by $54. Additionally, as USYes is identified as a baseline binary of 0 or NO, it can be said that sales within the US are $1,200 higher than sales that are outside of the US. Urban has no effect on Sales.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales=13.043469 - 0.054459(Price) - 0.021916(UrbanYes) + 1.200573(USYes)\)
(d) For which of the predictors can you reject the null hypothesis \(H_0 : \beta_j = 0\)?
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.
fit2<-lm(Sales~Price+US)
summary(fit2)##
## 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?
Neither model fits very well as both have r squares of ~23%
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(fit)## 2.5 % 97.5 %
## (Intercept) 11.76359670 14.32334118
## Price -0.06476419 -0.04415351
## UrbanYes -0.55597316 0.51214085
## USYes 0.69130419 1.70984121(h) Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow=c(2,2))
plot(fit2)
summary(influence.measures(fit))## Potentially influential observations of
## lm(formula = Sales ~ Price + Urban + US) :
##
## dfb.1_ dfb.Pric dfb.UrbY dfb.USYs dffit cov.r cook.d hat
## 26 0.28 -0.17 -0.19 -0.16 0.34_* 0.96_* 0.03 0.02
## 29 -0.07 0.11 -0.08 -0.09 -0.20 0.96_* 0.01 0.01
## 43 -0.10 0.10 -0.02 0.03 -0.12 1.06_* 0.00 0.05_*
## 50 -0.12 0.17 0.08 -0.17 0.27 0.97_* 0.02 0.01
## 51 -0.03 0.06 -0.09 -0.10 -0.20 0.94_* 0.01 0.01
## 58 -0.03 -0.01 -0.08 0.16 -0.21 0.96_* 0.01 0.01
## 69 -0.11 0.09 0.08 0.09 0.21 0.95_* 0.01 0.01
## 126 -0.07 0.06 0.03 0.02 -0.08 1.04_* 0.00 0.03_*
## 160 0.00 0.00 0.00 0.00 0.01 1.03_* 0.00 0.02
## 172 0.06 -0.07 0.02 0.02 0.08 1.03_* 0.00 0.02
## 175 0.11 -0.19 0.10 0.09 -0.24 1.03_* 0.01 0.04_*
## 210 -0.18 0.14 0.20 -0.11 -0.30 0.96_* 0.02 0.01
## 229 0.00 0.00 0.00 0.00 0.00 1.03_* 0.00 0.02
## 298 -0.04 0.06 -0.07 -0.08 -0.17 0.97_* 0.01 0.01
## 314 -0.04 0.04 -0.01 0.02 -0.05 1.03_* 0.00 0.02
## 353 -0.04 0.02 0.07 0.08 0.17 0.96_* 0.01 0.00
## 377 0.12 -0.16 0.10 0.11 0.26 0.94_* 0.02 0.01
## 396 -0.06 0.05 0.06 0.07 0.16 0.97_* 0.01 0.01The influence.measures output informs that there are quite a few data points that are seen as high leverage in the dataset. In order to truly identify if the potential outliers should be removed, one would need to run a new model without them and compare and contrast the full and minimized models. If the differences are minor enough, then it would be safe to keep all datapoints in the model.
Problem 12. This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate \(\beta\) Hat 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?
This will only occur when the data is perfectly correlated. i.e. when sx=sy and they equal 1. The correlation coefficient known as r is = 1.
(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.
x=rnorm(100, 10, 2)
y=rnorm(100, 7, 5)
lm(y~x)##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## 9.6689 -0.3068lm(x~y)##
## Call:
## lm(formula = x ~ y)
##
## Coefficients:
## (Intercept) y
## 10.27432 -0.04034(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=rnorm(100, 5, 3)
y=rnorm(100, 5, 3)
lm(y~x)##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept) x
## 5.41642 -0.05649lm(x~y)##
## Call:
## lm(formula = x ~ y)
##
## Coefficients:
## (Intercept) y
## 4.97708 -0.05942