Assignment_2
Ashley Windham
6/15/2021
Question #2
Carefully explain the difference between the KNN classifier and the KNN regression methods.
KNN classifier is best used for qualitative responses. The KNN classifier methods, for a given value for K and a prediction point Xo, identifies the K training observations that are closest to xo (represented by No). It then estimates the conditional probability for class j as the fraction of points in No whose response values equal j.
KNN regression is best used for quantitative responses. The KNN regression methods, for a given value for K and a prediction point Xo, identifies the K training observations that are closest to xo (represented by No). It then estimates the F(xo) using the average of all the training responses in No.
Question 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.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5data(Auto)
pairs(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.
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 1.0000000(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:
i. Is there a relationship between the predictors and the response?
- yes there is a relationship between the predictors and the response. The F-statistic is far from one and the p-value is small (less than the usual critical value 0.05). This is evidence against the null hypothesis that all the regression coefficients are zero.
ii. Which predictors appear to have a statistically significant relationship to the response?
- The predictors that appear to have a statistically significant relationship to the response are displacement, weight, year and origin. The p-values for these predictor’s t-statistic are all less than 0.01. Therefore, we cannot reject the null hypothesis. Cylinders, horsepower and acceleration do not have a statistically significant relationship.
iii. What does the coefficient for the year variable suggest?
- The coefficient for the year variable is 0.750773. This suggests that for every one year, the mpg increases by 0.75. In other words, the cars become more efficient by increases mpg for each year.
lm.auto1 = lm(mpg~.-name, data=Auto)
summary(lm.auto1)##
## 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-16(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?
- The residuals vs fitted plot shows somewhat of a pattern indicates mild non-linearity of the data.
- The residuals vs leverage plot shows some outliers in the data (above 2 and below -2). Also the studentized residuals show values greater than three, indicating outliers. There is one high leverage point (point 14).
par(mfrow=c(2,2))
plot(lm.auto1)
plot(predict(lm.auto1), rstudent(lm.auto1))
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
lm.auto2 = lm(mpg~displacement*cylinders+displacement*weight, data = Auto)
summary(lm.auto2)##
## Call:
## lm(formula = mpg ~ displacement * cylinders + displacement *
## weight, data = Auto)
##
## 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 ***
## displacement -7.351e-02 1.669e-02 -4.403 1.38e-05 ***
## cylinders 7.606e-01 7.669e-01 0.992 0.322
## weight -9.888e-03 1.329e-03 -7.438 6.69e-13 ***
## displacement:cylinders -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-16- From the correlation matrix, two high correlation pairs are displacement and cylinders as well as displacement and weight. When assessing the interaction effects, we see that the interaction between displacement and weight is statistically significant (small p value) while the interaction between displacement and cylinders is not (large p value).
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
par(mfrow = c(2, 2))
plot(Auto$displacement, Auto$mpg)
plot(log(Auto$displacement), Auto$mpg)
plot(sqrt(Auto$displacement), Auto$mpg)
plot((Auto$displacement)^2, Auto$mpg)
- When looking at transformations of the displacement variable, it appears that the log transformation may give a better linear plot.
Question 10
This question should be answered using the Carseats data set.
(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
data("Carseats")
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!
- The coefficient for the price variable is negative which suggests that when price increases, sales decrease. Also, the p-value for the t-statistic is low and statistically significant, suggesting that there is a relationship between price and sales. The unit of sales is in the thousands. Therefore, for every $1 increase in price, there would be a decrease of approximately $54 in sales.
- The coefficient for the Urban variable has a high p-value for the t-statistic suggesting that there is no relationship between the Urban variable and Sales.
- The coefficient for the US variable has a low p-value that is statistically significant which suggests that there is a relationship between US and Sales. This variable indicates whether the store is in the US or not. This variable USYes has a positive relationship suggesting that sales within the US are higher compared to those outside of the US (approximately $1,200 higher).
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
- Sales = 13.043469 + (-0.054459 x Price) + (-0.021916 x UrbanYes) + (1.200573 X USYes)
(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
- We can reject the null hypothesis for predictors Price and USYes because the p-value for the t-statistic is low and statistically significant (less than 0.01).
(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~ US + Price, data = Carseats)
summary(lm.carseats2)##
## Call:
## lm(formula = Sales ~ US + Price, 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 ***
## USYes 1.19964 0.25846 4.641 4.71e-06 ***
## Price -0.05448 0.00523 -10.416 < 2e-16 ***
## ---
## 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?
- The RSE for the first and second model are 2.472 and 2.469, respectively. The R^2 for the first and second model are 23.93% and 23.54% respectively. Therefore, the fits for these models are similar with a slightly better fit for the second model. However, neither are very good fits.
(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
## USYes 0.69151957 1.70776632
## Price -0.06475984 -0.04419543(h) Is there evidence of outliers or high leverage observations in the model from (e)?
plot(predict(lm.carseats2), rstudent(lm.carseats2))
- The studentized residuals for the observations appear to be within -3 and 3 which suggests that there are no potential outliers present.
par(mfrow=c(2,2))
plot(lm.carseats2)
- There are few points on the residuals vs levarage plot that suggest high leverage.
Question 12
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate ˆ 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?
- This occurs when sum of squares of the observed y-values are the same as the sum of squares for the observed x-values.
(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)
y = 5 * x
lm.fit = lm(y~x+0)
lm.fit2 = lm(x~y+0)
summary(lm.fit)## Warning in summary.lm(lm.fit): essentially perfect fit: summary may be
## unreliable##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.653e-14 -7.210e-17 9.100e-18 1.694e-16 1.207e-15
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 5.000e+00 1.726e-16 2.896e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.718e-15 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 8.387e+32 on 1 and 99 DF, p-value: < 2.2e-16summary(lm.fit2)## Warning in summary.lm(lm.fit2): essentially perfect fit: summary may be
## unreliable##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.269e-15 -2.240e-17 8.600e-19 2.493e-17 2.611e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 2.000e-01 4.847e-18 4.126e+16 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.412e-16 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.703e+33 on 1 and 99 DF, p-value: < 2.2e-16- The regression coefficients are different for each linear regression.
(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.
x2 = rnorm(100)
y2 = sample(x2, 100)
sum(x2^2)## [1] 87.7259sum(y2^2)## [1] 87.7259lm.fit3 = lm(y2~x2+0)
lm.fit4 = lm(x2~y2+0)
summary(lm.fit3)##
## Call:
## lm(formula = y2 ~ x2 + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2485 -0.7843 -0.1972 0.4646 2.3715
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x2 0.08146 0.10017 0.813 0.418
##
## Residual standard error: 0.9382 on 99 degrees of freedom
## Multiple R-squared: 0.006635, Adjusted R-squared: -0.003399
## F-statistic: 0.6613 on 1 and 99 DF, p-value: 0.4181summary(lm.fit4)##
## Call:
## lm(formula = x2 ~ y2 + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2754 -0.7432 -0.1574 0.5095 2.2362
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y2 0.08146 0.10017 0.813 0.418
##
## Residual standard error: 0.9382 on 99 degrees of freedom
## Multiple R-squared: 0.006635, Adjusted R-squared: -0.003399
## F-statistic: 0.6613 on 1 and 99 DF, p-value: 0.4181- The regression coefficients are the same for both linear regressions. This is because the sum of x^2 equals the sum of y^2.