Assignment 2
Arwa N
6/17/2020
Chapter 03 (page 120): 2, 9, 10, 12
- Carefully explain the differences between the KNN classifier and KNN regression methods.
To predict Y for a given value of X, consider k closest points to X in training data and take the average of the responses. The KNN classifier is used for classification and The KNN regression method is used for regression problems.
- This question involves the use of multiple linear regression on the Auto data set.
library(ISLR)## Warning: package 'ISLR' was built under R version 3.6.3data(Auto)- Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto )
- 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[, names(Auto) !="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- 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:
Lm_model = lm(mpg ~. -name, data = Auto)
summary(Lm_model)##
## 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- Is there a relationship between the predictors and the response?
This model is significant with small p values and four of the predictors are highly significant.
- Which predictors appear to have a statistically significant relationship to the response?
displacement, weight, year, origin.
- What does the coefficient for the year variable suggest?
With one unit increase in year of the car, mpg are increased by 0.75 by gallon
- 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?
Residuals vs Fitted graph: evidence of a non-linear relationship between the response and the predictors. QQ plot: slighly right skewed. Residuals vs leverage: no major leverage points except #14 on the graph
par(mfrow = c(2, 2))
plot(Lm_model)
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
Displacement and weight are statistically significant.
Lm_model2 = lm(mpg ~ displacement*weight+weight*cylinders, data = Auto)summary(Lm_model2)##
## Call:
## lm(formula = mpg ~ displacement * weight + weight * cylinders,
## data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.3698 -2.5514 -0.3861 1.7206 18.0838
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.798e+01 6.440e+00 7.451 6.15e-13 ***
## displacement -1.065e-01 3.066e-02 -3.473 0.000573 ***
## weight -7.232e-03 2.165e-03 -3.341 0.000916 ***
## cylinders 1.993e+00 2.055e+00 0.970 0.332710
## displacement:weight 2.457e-05 8.205e-06 2.995 0.002924 **
## weight:cylinders -5.380e-04 6.016e-04 -0.894 0.371771
## ---
## 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: 0.7273, Adjusted R-squared: 0.7237
## F-statistic: 205.8 on 5 and 386 DF, p-value: < 2.2e-16- Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
I did a log transformation. All the predictors are significant. The Residuals vs Fitted graph doesn;t have that curve visible, QQ plot doesn’t support normal distribution.
lmlog = lm(mpg ~ log(horsepower) + log(weight) + log(displacement), data = Auto)
summary(lmlog)##
## Call:
## lm(formula = mpg ~ log(horsepower) + log(weight) + log(displacement),
## data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.6539 -2.4232 -0.3147 2.0760 15.2014
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 161.495 11.898 13.574 < 2e-16 ***
## log(horsepower) -6.929 1.263 -5.487 7.38e-08 ***
## log(weight) -11.835 2.264 -5.228 2.81e-07 ***
## log(displacement) -2.353 1.187 -1.982 0.0482 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.978 on 388 degrees of freedom
## Multiple R-squared: 0.7422, Adjusted R-squared: 0.7402
## F-statistic: 372.3 on 3 and 388 DF, p-value: < 2.2e-16par(mfrow=c(2,2))
plot(lmlog)
- This question should be answered using the Carseats data set.
data(Carseats)- Fit a multiple regression model to predict Sales using Price, Urban, and US.
Lm_model3 = lm(Sales ~ Price+Urban+US, data = Carseats)
summary(Lm_model3)##
## 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- Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
Price and US are significant predictors. One $1 increase in price and the sales dcrease by $54. Stores in US sell $1200 more in car seats than non US stores.
- Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales=13.043469−0.054459Price−0.021916Urban_{Yes}+1.200573XUS_{Yes}\)
- For which of the predictors can you reject the null hypothesis H0 : βj = 0?
‘Price’ and ‘USYes’ because the p value is significant
- 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_model4 = lm(Sales ~ Price+US, data = Carseats)
summary(Lm_model4)##
## 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- How well do the models in (a) and (e) fit the data?
only 23.54% of the variance in Sales can be explained by the model. Not that great.
- Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(Lm_model4)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632- Is there evidence of outliers or high leverage observations in the model from (e)?
Residuals vs Fitted graph and QQplot support normal distribution Rresiduals vs. leverage plot indicates there are outliers and some high leverage points.
par(mfrow=c(2,2))
plot(Lm_model4)
- This problem involves simple linear regression without an intercept.
- 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?
sum of squares of observed y values = sum of squares of observed x values
- 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 <- x^2
coefficients(lm(x ~ y))## (Intercept) y
## 0.1708051 -0.2072981coefficients(lm(y ~ x))## (Intercept) x
## 1.1243761 -0.4287084- 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)
y <- x
coefficients(lm(x ~ y))## (Intercept) y
## -1.110223e-17 1.000000e+00coefficients(lm(y ~ x))## (Intercept) x
## -1.110223e-17 1.000000e+00