Homework3
Dario Bonaretti
9/11/2017
- 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.
Auto %>% select(mpg : origin) %>%
ggpairs()
- Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
Auto %>% select(mpg : origin) %>%
ggcorr(., palette = "RdBu", label = TRUE)
- 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.
Auto %>% select(mpg : origin) -> dt
fit <- lm(mpg ~ ., data = dt)
summary(fit)##
## Call:
## lm(formula = mpg ~ ., data = dt)
##
## 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-16Comment on the output. For instance:
i. Is there a relationship between the predictors and the response?
Only between some the predictors depending on the observations we have.
ii. Which predictors appear to have a statistically significant relationship to the response?
Displacement, weight, year and origin are significant, thus we can claim that there is a significant relationship.
iii. What does the coefficient for the year variable suggest? That the newer the car, the higher 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. Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
library(broom)
model <- broom::augment(fit)
residual <- function(model) {ggplot(model, aes(.fitted, .resid)) +
geom_point() +
geom_hline(yintercept = 0) +
geom_smooth(se = FALSE)}
stdResidual <- function(model) {ggplot(model, aes(.fitted, .std.resid)) +
geom_point() +
geom_hline(yintercept = 0) +
geom_smooth(se = FALSE)}
ggqqplot <- function(model) {ggplot(model) +
stat_qq(aes(sample = .std.resid)) +
geom_abline()}
sqrtResidual <- function(model) {ggplot(model, aes(.fitted, sqrt(abs(.std.resid)))) +
geom_point() +
geom_smooth(se = FALSE)}
cooksD <- function(model) { ggplot(model, aes(seq_along(.cooksd), .cooksd)) +
geom_col() }
hatPlot <- function(model) { ggplot(model, aes(.hat, .std.resid)) +
geom_vline(size = 2, colour = "white", xintercept = 0) +
geom_hline(size = 2, colour = "white", yintercept = 0) +
geom_point() + geom_smooth(se = FALSE) }
hatCooksD <- function(model) { ggplot(model, aes(.hat, .std.resid)) +
geom_point(aes(size = .cooksd)) +
geom_smooth(se = FALSE, size = 0.5) }
hatCooksD2 <- function(model) { ggplot(model, aes(.hat, .cooksd)) +
geom_vline(xintercept = 0, colour = NA) +
geom_abline(slope = seq(0, 3, by = 0.5), colour = "white") +
geom_smooth(se = FALSE) +
geom_point() }
purrr::invoke_map(.f = list(residual, stdResidual, ggqqplot, sqrtResidual, cooksD, hatPlot, hatCooksD, hatCooksD2), .x = list(list(model))) %>%
gridExtra::grid.arrange(grobs = ., ncol = 2,
top = paste("Diagnostic plots for", as.expression(fit$call)))
There is a trend in the variance of the residuals. In fact the model fits more worse on higher values of the predicted variable. That’s probably due to the effect of an influencial too as we can see from hat and Cook’s D value.
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
summary(lm(mpg ~ .*., dt))##
## Call:
## lm(formula = mpg ~ . * ., data = dt)
##
## 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-16All the interactions (:) marked with at least * are significant
- Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
fit2 <- lm(mpg ~ cylinders + log(displacement) + log(horsepower) + log(weight) + acceleration + year + origin, dt)map(list(fit, fit2), summary) %>%
map('adj.r.squared')## [[1]]
## [1] 0.8182238
##
## [[2]]
## [1] 0.8450547If I look at the partial residual plots, it seems like the correlation between variables such as displacement, horsepower and weight is not actually linear. Thus we can consider a log transformation given the pattern of the residuals.
Problem 2: Work on the problem 10 in Section 4.7 (Exercises) on page 171. You are required to work on part (a)~(d). But I encourage you to try the part(e)~(i) if you are familiar with KNN, LDA and QDA.
- This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
summary(Weeklyd)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## Today Direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260ggpairs(Weeklyd)
ggcorr(Weeklyd, palette = "RdBu", label = TRUE)
It seems like there is a very high correlation between year and volume.
(b) Use the full data set to perform a logistic regression with. Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
fit <- glm(Direction ~ . - Today - Year, family = binomial, data = Weeklyd)
summary(fit)##
## Call:
## glm(formula = Direction ~ . - Today - Year, family = binomial,
## data = Weeklyd)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4Only lag2 appears significant.
- Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
#1 are Up
augment(fit) %>% mutate(observed = as.numeric(Direction)-1) %>%
select(observed, .fitted) %>% mutate(fitted = ifelse(.fitted > .5, '1', '0')) %>%
select(-.fitted) %>%
count(observed, fitted) %>%
spread(fitted, n)## # A tibble: 2 x 3
## observed `0` `1`
## * <dbl> <int> <int>
## 1 0 465 19
## 2 1 563 42The model does well in predicting true negative but performs poorly on the rest.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
fit2 <- glm(Direction ~ Lag2, family = binomial, data = filter(Weeklyd, Year <= 2008) )
augment(fit2) %>% mutate(observed = as.numeric(Direction)-1) %>%
select(observed, .fitted) %>% mutate(fitted = ifelse(.fitted > .5, '1', '0')) %>%
select(-.fitted) %>%
count(observed, fitted) %>%
spread(fitted, n)## # A tibble: 2 x 3
## observed `0` `1`
## * <dbl> <int> <int>
## 1 0 439 2
## 2 1 533 11select(filter(Weeklyd, Year > 2008), Direction) %>%
mutate(fitted = ifelse(predict.glm(fit2, filter(Weeklyd, Year > 2008), type = 'response') > .5, 1, 0), observed = as.numeric(Direction)-1) %>%
count(observed, fitted) %>%
mutate(prop = n/sum(n)) %>%
select(-n) %>%
spread(fitted, prop)## # A tibble: 2 x 3
## observed `0` `1`
## * <dbl> <dbl> <dbl>
## 1 0 0.08653846 0.3269231
## 2 1 0.04807692 0.5384615The confusions matrix says that we are underperforming a random guess.
- Repeat (d) using LDA.
library(MASS)
fitLDA <- lda(formula(fit2), data = filter(Weeklyd, Year <= 2008))
predict(fitLDA, filter(Weeklyd, Year > 2008))$class## [1] Up Up Down Down Up Up Up Down Down Down Down Up Up Up
## [15] Up Up Up Up Up Up Down Up Up Up Up Up Up Up
## [29] Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [43] Up Up Down Up Up Up Up Up Up Up Up Up Up Up
## [57] Down Up Up Up Up Up Up Up Up Up Up Up Up Up
## [71] Up Down Up Down Up Up Up Up Down Down Up Up Up Up
## [85] Up Down Up Up Up Up Up Up Up Up Up Up Up Up
## [99] Up Up Up Up Up Up
## Levels: Down Up- Repeat (d) using QDA.
fitQDA <- lda(formula(fit2), data = filter(Weeklyd, Year <= 2008))
predict(fitQDA, filter(Weeklyd, Year > 2008))$class## [1] Up Up Down Down Up Up Up Down Down Down Down Up Up Up
## [15] Up Up Up Up Up Up Down Up Up Up Up Up Up Up
## [29] Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [43] Up Up Down Up Up Up Up Up Up Up Up Up Up Up
## [57] Down Up Up Up Up Up Up Up Up Up Up Up Up Up
## [71] Up Down Up Down Up Up Up Up Down Down Up Up Up Up
## [85] Up Down Up Up Up Up Up Up Up Up Up Up Up Up
## [99] Up Up Up Up Up Up
## Levels: Down Up- Repeat (d) using KNN with K = 1.
detach("package:MASS", unload=TRUE)
library(class)
train <- sample(1:nrow(Weekly), 900)
train.X <- cbind(Weekly$Lag1 ,Weekly$Lag2)[train ,]
test.X <- cbind(Weekly$Lag1 ,Weekly$Lag2)[-train,]
train.Direction <- Weekly$Direction[train]
test.Direction <- Weekly$Direction[-train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction, k=1)
table(knn.pred , test.Direction)## test.Direction
## knn.pred Down Up
## Down 44 37
## Up 37 71knn.pred <- knn(train.X, test.X, train.Direction, k=3)
table(knn.pred , test.Direction)## test.Direction
## knn.pred Down Up
## Down 34 39
## Up 47 69- Which of these methods appears to provide the best results on this data?
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier