Assignment_2
Jesus Gonzalez
2022-05-05
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.
library(ISLR)
data(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 [,-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:
Auto$origin <- factor(Auto$origin, labels = c("American", "European", "Japanese"))
mpg_lm <- lm(mpg ~ . - name, data = Auto)
summary(mpg_lm)##
## Call:
## lm(formula = mpg ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.0095 -2.0785 -0.0982 1.9856 13.3608
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.795e+01 4.677e+00 -3.839 0.000145 ***
## cylinders -4.897e-01 3.212e-01 -1.524 0.128215
## displacement 2.398e-02 7.653e-03 3.133 0.001863 **
## horsepower -1.818e-02 1.371e-02 -1.326 0.185488
## weight -6.710e-03 6.551e-04 -10.243 < 2e-16 ***
## acceleration 7.910e-02 9.822e-02 0.805 0.421101
## year 7.770e-01 5.178e-02 15.005 < 2e-16 ***
## originEuropean 2.630e+00 5.664e-01 4.643 4.72e-06 ***
## originJapanese 2.853e+00 5.527e-01 5.162 3.93e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.307 on 383 degrees of freedom
## Multiple R-squared: 0.8242, Adjusted R-squared: 0.8205
## F-statistic: 224.5 on 8 and 383 DF, p-value: < 2.2e-16coef(mpg_lm)[7]## year
## 0.7770269i. Is there a relationship between the predictors and the response?
Based on the p-value we reject the null hypothesis, so there is a relationship between the predictors and mpg.
ii. Which predictors appear to have a statistically significant relationship to the response?
With the exception of cylinders, horsepower, and acceleration, all variables are significant because they all have a p-value lower than .05
iii. What does the coefficient for the year variable suggest?
Having a coefficient of .77 means that when increasing year mpg increases by .77. This would mean that as the years go the efficiency of the cars will increase.
(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(mpg_lm)
There might be some nonlinearity that the residuals graph is not showing. There also is high leverage and high residuals which suggests that there are uneven predictions.
Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(formula = mpg ~ . * ., data = Auto[, -9]))##
## Call:
## lm(formula = mpg ~ . * ., data = Auto[, -9])
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6008 -1.2863 0.0813 1.2082 12.0382
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.401e+01 5.147e+01 0.855 0.393048
## cylinders 3.302e+00 8.187e+00 0.403 0.686976
## displacement -3.529e-01 1.974e-01 -1.788 0.074638 .
## horsepower 5.312e-01 3.390e-01 1.567 0.117970
## weight -3.259e-03 1.820e-02 -0.179 0.857980
## acceleration -6.048e+00 2.147e+00 -2.818 0.005109 **
## year 4.833e-01 5.923e-01 0.816 0.415119
## originEuropean -3.517e+01 1.260e+01 -2.790 0.005547 **
## originJapanese -3.765e+01 1.426e+01 -2.640 0.008661 **
## cylinders:displacement -6.316e-03 7.106e-03 -0.889 0.374707
## cylinders:horsepower 1.452e-02 2.457e-02 0.591 0.555109
## cylinders:weight 5.703e-04 9.044e-04 0.631 0.528709
## cylinders:acceleration 3.658e-01 1.671e-01 2.189 0.029261 *
## cylinders:year -1.447e-01 9.652e-02 -1.499 0.134846
## cylinders:originEuropean -7.210e-01 1.088e+00 -0.662 0.508100
## cylinders:originJapanese 1.226e+00 1.007e+00 1.217 0.224379
## displacement:horsepower -5.407e-05 2.861e-04 -0.189 0.850212
## displacement:weight 2.659e-05 1.455e-05 1.828 0.068435 .
## displacement:acceleration -2.547e-03 3.356e-03 -0.759 0.448415
## displacement:year 4.547e-03 2.446e-03 1.859 0.063842 .
## displacement:originEuropean -3.364e-02 4.220e-02 -0.797 0.425902
## displacement:originJapanese 5.375e-02 4.145e-02 1.297 0.195527
## horsepower:weight -3.407e-05 2.955e-05 -1.153 0.249743
## horsepower:acceleration -3.445e-03 3.937e-03 -0.875 0.382122
## horsepower:year -6.427e-03 3.891e-03 -1.652 0.099487 .
## horsepower:originEuropean -4.869e-03 5.061e-02 -0.096 0.923408
## horsepower:originJapanese 2.289e-02 6.252e-02 0.366 0.714533
## weight:acceleration -6.851e-05 2.385e-04 -0.287 0.774061
## weight:year -8.065e-05 2.184e-04 -0.369 0.712223
## weight:originEuropean 2.277e-03 2.685e-03 0.848 0.397037
## weight:originJapanese -4.498e-03 3.481e-03 -1.292 0.197101
## acceleration:year 6.141e-02 2.547e-02 2.412 0.016390 *
## acceleration:originEuropean 9.234e-01 2.641e-01 3.496 0.000531 ***
## acceleration:originJapanese 7.159e-01 3.258e-01 2.198 0.028614 *
## year:originEuropean 2.932e-01 1.444e-01 2.031 0.043005 *
## year:originJapanese 3.139e-01 1.483e-01 2.116 0.035034 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.628 on 356 degrees of freedom
## Multiple R-squared: 0.8967, Adjusted R-squared: 0.8866
## F-statistic: 88.34 on 35 and 356 DF, p-value: < 2.2e-16There are 6 interactions that seem to be significant.
cylinders:acceleration
acceleration:year
acceleration:originEuropean
acceleration:originJapanese
year:originEuropean
year:originJapanese#9.f
Problem 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. sales_lm <- lm(Sales ~ Price + Urban + US, data = Carseats)
sales_lm <- lm(Sales ~ Price + Urban + US, data = Carseats)(b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
summary(sales_lm)##
## 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(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales=13.043469-.054459*Price-.21916*Urban+1.00573*US(d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
There isn’t enough evidence to reject the null hypothesis for Urban = 0. However, we reject the null hypothesis for 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.
sales_lm_2 <- lm(Sales ~ Price + US, data = Carseats)
summary(sales_lm_2)##
## 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?
Model (a) has a higher R2 but a lower adjusted R2.
(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(sales_lm_2, level = 0.95)## 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)?
round(((2 + 1) / nrow(Carseats)), 10) == round(mean(hatvalues(sales_lm_2)), 10)## [1] TRUEProblem 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?
The coefficient remains the same.
(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(1)
x=rnorm(100)
y=x+rnorm(100)
head(data.frame(y,x))## y x
## 1 -1.2468205 -0.6264538
## 2 0.2257592 0.1836433
## 3 -1.7465503 -0.8356286
## 4 1.7533096 1.5952808
## 5 -0.3250769 0.3295078
## 6 0.9468189 -0.8204684lm.fit <- lm(y~x+0)
summary(lm.fit)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9154 -0.6472 -0.1771 0.5056 2.3109
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.9939 0.1065 9.334 3.1e-15 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared: 0.4681, Adjusted R-squared: 0.4627
## F-statistic: 87.13 on 1 and 99 DF, p-value: 3.1e-15lm.fit <- lm(x~y+0)
summary(lm.fit)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.35410 -0.37468 0.09974 0.48799 1.55406
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.47099 0.05046 9.334 3.1e-15 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.6599 on 99 degrees of freedom
## Multiple R-squared: 0.4681, Adjusted R-squared: 0.4627
## F-statistic: 87.13 on 1 and 99 DF, p-value: 3.1e-15(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(2)
x=rnorm(100)
y=abs(x)
head(data.frame(y,x))## y x
## 1 0.89691455 -0.89691455
## 2 0.18484918 0.18484918
## 3 1.58784533 1.58784533
## 4 1.13037567 -1.13037567
## 5 0.08025176 -0.08025176
## 6 0.13242028 0.13242028lm.fit <- lm(y~x+0)
summary(lm.fit)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## 0.00488 0.39563 0.86041 1.45172 2.47908
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.01116 0.10050 0.111 0.912
##
## Residual standard error: 1.161 on 99 degrees of freedom
## Multiple R-squared: 0.0001246, Adjusted R-squared: -0.009975
## F-statistic: 0.01234 on 1 and 99 DF, p-value: 0.9118lm.fit <- lm(x~y+0)
summary(lm.fit)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.4791 -0.8654 -0.1418 0.8047 2.0675
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.01116 0.10050 0.111 0.912
##
## Residual standard error: 1.161 on 99 degrees of freedom
## Multiple R-squared: 0.0001246, Adjusted R-squared: -0.009975
## F-statistic: 0.01234 on 1 and 99 DF, p-value: 0.9118