Assignment 2
R Markdown document by Diana L Cisneros
- Carefully explain the differences between the KNN classifier and KNN regression methods.
KNN algorithms use data and classify new data points based on similarity measures (e.g. distance function). Classification is done by a majority vote to its neighbors. KNN regression is a non-parametric method that, in an intuitive manner, approximates the association between independent variables and the continuous outcome by averaging the observations in the same neighborhood. If k is small KNN is much more flexible than linear regression.
- This question involves the use of multiple linear regression on the Auto data set.
- Produce a scatterplot matrix which includes all of the variables in the data set.
library("ISLR")
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:
mod_1=lm(mpg~. -name, data=Auto)
summary(mod_1)##
## 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? ## Some of the predictors are not statisticcally signifficant on the response variable, like cylinders, horsepower and acceleration. The rest of the predictors are signifficant. overall there is a relationship between them with 82% of variance can be explained by the independent variables.
- Which predictors appear to have a statistically significant relationship to the response? ## displacement, weight, year and origin
- What does the coefficient for the year variable suggest? ## When every other predictor held constant, the mpg value increases with each year that passes by 0.7507
- 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(mod_1)
Do the residual plots suggest any unusually large outliers? ## No outliers fund outside of the range -3,3
NOn linear relationship between Residual vs Fitted graph, instead there is a funnel shape that suggest heteroscedasticity (no constant variance).
Does the leverage plot identify any observations with unusually high leverage?
No high leverage points found.
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
mod_2 = lm(mpg ~ displacement * weight + origin * year + displacement * origin, data = Auto[, 1:8])
summary(mod_2)##
## Call:
## lm(formula = mpg ~ displacement * weight + origin * year + displacement *
## origin, data = Auto[, 1:8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.4232 -1.5709 -0.0384 1.3298 13.3393
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.844e+01 8.195e+00 2.250 0.025045 *
## displacement -7.873e-02 1.399e-02 -5.629 3.51e-08 ***
## weight -1.048e-02 7.425e-04 -14.110 < 2e-16 ***
## origin -1.521e+01 4.273e+00 -3.560 0.000418 ***
## year 4.878e-01 1.019e-01 4.787 2.42e-06 ***
## displacement:weight 2.133e-05 2.420e-06 8.811 < 2e-16 ***
## origin:year 1.966e-01 5.458e-02 3.602 0.000357 ***
## displacement:origin 2.583e-03 7.758e-03 0.333 0.739343
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.973 on 384 degrees of freedom
## Multiple R-squared: 0.8576, Adjusted R-squared: 0.855
## F-statistic: 330.2 on 7 and 384 DF, p-value: < 2.2e-16There are significant differences between the interactions “displacement * weight” and “origin * weight”, no signifficant difference in the interaction “displacement * origin”. R-Squares is about 86% of the variance.
- Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
mod_log = lm(data = Auto[,1:8], mpg ~ log(displacement)*log(weight)*log(origin))
summary(mod_log)##
## Call:
## lm(formula = mpg ~ log(displacement) * log(weight) * log(origin),
## data = Auto[, 1:8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.0334 -2.3867 -0.4177 1.7467 18.4256
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 244.746 110.085 2.223 0.0268
## log(displacement) -25.990 19.751 -1.316 0.1890
## log(weight) -23.378 14.114 -1.656 0.0985
## log(origin) 143.021 362.386 0.395 0.6933
## log(displacement):log(weight) 2.395 2.490 0.962 0.3366
## log(displacement):log(origin) -5.578 77.487 -0.072 0.9427
## log(weight):log(origin) -25.732 46.601 -0.552 0.5811
## log(displacement):log(weight):log(origin) 2.275 9.916 0.229 0.8187
##
## (Intercept) *
## log(displacement)
## log(weight) .
## log(origin)
## log(displacement):log(weight)
## log(displacement):log(origin)
## log(weight):log(origin)
## log(displacement):log(weight):log(origin)
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.074 on 384 degrees of freedom
## Multiple R-squared: 0.7325, Adjusted R-squared: 0.7276
## F-statistic: 150.2 on 7 and 384 DF, p-value: < 2.2e-16None of the log variables are signifficant, and the R-squared is 73% of the variance.
- 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.
model_1=lm(Sales ~ Price + Urban + US, data=Carseats)
summary(model_1)##
## 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!
The coefficient “Price” variable is interpreted as the average effect of a price increase of 1 dollar is a decrease of 54.4588492 units in sales all other predictors remaining fixed. The coefficient “Urban” variable is interpreted as on average the unit sales in urban location yes are 21.9161508 units less than in rural location all other predictors remaining fixed. The coefficient “US” variable can be interpreted as on average the unit sales in a US store are 1200.5726978 units more than in a non US store all other predictors remaining fixed.
- Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales= 13.0434689 − 0.0544588Price − 0.0219162Urban + 1.2005727*US + ε
Urban=1 if the store is in an urban location and 0 if not, US=1 if the store is in the US and 0 if not.
- For which of the predictors can you reject the null hypothesis H0 : βj = 0?
“Price” and “US”
- 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.
sig_mod=lm(Sales ~ Price + US, data = Carseats)
summary(sig_mod)##
## 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?
#Both of the models are about 24% variance explained by the predictor variables, they are not good models because the low R-Squared.
- Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(sig_mod)## 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)?
par(mfrow=c(2,2))
plot(sig_mod)
In the Residuals vs Leverage graph notes a very high leverage observation, (#43).
hatvalues(sig_mod)[order(hatvalues(sig_mod), decreasing = T)][1]## 43
## 0.04333766- This problem involves simple linear regression without an intercept.
- Recall that the coefficient estimate βˆ 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 sum of the squares of the “Y” variable must be the same as the sum of the “X” variable that is squared.
∑yx = ∑xy so: ∑ y^2 = ∑ x^2
- 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=rnorm(100)sum(x^2)## [1] 85.86661sum(y^2)## [1] 102.1562summary(lm(y~x))##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.4107 -0.5898 0.0080 0.6190 2.5394
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.03658 0.10212 0.358 0.721
## x -0.06057 0.11021 -0.550 0.584
##
## Residual standard error: 1.019 on 98 degrees of freedom
## Multiple R-squared: 0.003073, Adjusted R-squared: -0.007099
## F-statistic: 0.3021 on 1 and 98 DF, p-value: 0.5838summary(lm(y~x-1))##
## Call:
## lm(formula = y ~ x - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.3707 -0.5542 0.0436 0.6540 2.5744
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x -0.0634 0.1094 -0.579 0.564
##
## Residual standard error: 1.014 on 99 degrees of freedom
## Multiple R-squared: 0.003378, Adjusted R-squared: -0.006688
## F-statistic: 0.3356 on 1 and 99 DF, p-value: 0.5637summary(lm(x~y))##
## Call:
## lm(formula = x ~ y)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.7088 -0.6920 -0.1723 0.7280 2.7741
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.06424 0.09330 -0.689 0.493
## y -0.05074 0.09231 -0.550 0.584
##
## Residual standard error: 0.9322 on 98 degrees of freedom
## Multiple R-squared: 0.003073, Adjusted R-squared: -0.007099
## F-statistic: 0.3021 on 1 and 98 DF, p-value: 0.5838summary(lm(x~y-1))##
## Call:
## lm(formula = x ~ y - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.7738 -0.7564 -0.2377 0.6658 2.7099
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y -0.05329 0.09199 -0.579 0.564
##
## Residual standard error: 0.9297 on 99 degrees of freedom
## Multiple R-squared: 0.003378, Adjusted R-squared: -0.006688
## F-statistic: 0.3356 on 1 and 99 DF, p-value: 0.5637- 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=-sample(x)
head(x)## [1] -0.6788538 -1.2351503 -1.0936383 0.9673890 1.6168616 -0.4077981head(x)## [1] -0.6788538 -1.2351503 -1.0936383 0.9673890 1.6168616 -0.4077981head(y)## [1] 0.2537250 -2.2477557 -0.2200013 -0.1415669 -0.4047063 -0.1539842sum(x^2)## [1] 115.4475sum(y^2)## [1] 115.4475