Linear Regression
Justin Gould
library(tidyverse)
library(ISLR2)
library(gtsummary)
library(gt)Question 2
Carefully explain the differences between the KNN classifier and KNN regression methods. ***
Answer: KNN regression is a prediction method that uses prediction points to estimate what a value might be.Conversely, KNN classification is a classification method that uses the local values to predict which class a variable belongs to. KNN is classification is used for more broad research and determining which class a datapoint belongs to where KNN regression is more specific.
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.
plot(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.
auto1 = Auto |> select(mpg, cylinders, displacement, horsepower, weight, acceleration, year, origin)
cor(auto1)## 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? ii. Which predictors appear to have a statistically significant relationship to the response? iii. What does the coefficient for the year variable suggest
model = lm(mpg ~ ., data = auto1)
tbl_regression(model, intercept = TRUE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| (Intercept) | -17 | -26, -8.1 | <0.001 |
| cylinders | -0.49 | -1.1, 0.14 | 0.13 |
| displacement | 0.02 | 0.01, 0.03 | 0.008 |
| horsepower | -0.02 | -0.04, 0.01 | 0.2 |
| weight | -0.01 | -0.01, -0.01 | <0.001 |
| acceleration | 0.08 | -0.11, 0.27 | 0.4 |
| year | 0.75 | 0.65, 0.85 | <0.001 |
| origin | 1.4 | 0.88, 2.0 | <0.001 |
| Abbreviation: CI = Confidence Interval | |||
Answer: Yes, there is a relationship between the predictors and the response variables. The statistically significant variables are to MPG are displacement, weight, year, and origin. The year variables coefficient suggests that for every 1-unit increase in MPG, the car tends to be .75 years newer.
(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(model)
Answer: The residual vs fitted plot indicates that the
data does not move in a straight line and that it has a heavy lift-off
on the right side of the plot. The scale-location plot shows this
non-linear patern in the data further. The residual vs leverage plot
indicates that 327,394, and 141 are outliers that may affect the
model.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
model1 = lm(mpg ~ displacement * weight + year:origin, data = auto1)
tbl_regression(model1, intercept = TRUE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| (Intercept) | 50 | 45, 55 | <0.001 |
| displacement | -0.06 | -0.09, -0.04 | <0.001 |
| weight | -0.01 | -0.01, -0.01 | <0.001 |
| displacement * weight | 0.00 | 0.00, 0.00 | <0.001 |
| year * origin | 0.01 | 0.00, 0.02 | 0.009 |
| Abbreviation: CI = Confidence Interval | |||
Answer: The interaction terms displacement + weight and year + origin appear to be statistically significant. All terms appear significant.
(f) Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings
model2 = lm(mpg ~ sqrt(displacement) + log(weight) + year + I(origin^2), data = auto1)
tbl_regression(model2, intercept = TRUE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| (Intercept) | 123 | 101, 146 | <0.001 |
| sqrt(displacement) | 0.07 | -0.18, 0.33 | 0.6 |
| log(weight) | -20 | -23, -17 | <0.001 |
| year | 0.78 | 0.69, 0.88 | <0.001 |
| I(origin^2) | 0.18 | 0.05, 0.30 | 0.006 |
| Abbreviation: CI = Confidence Interval | |||
Answer: After using some transformations on the variables all variables are steadily significant. The R squared at 84% is an indicator that this is a good model. The R squared shows that 84% of change in MPG can be explained by displacement, weight, year, and orgin in this regression model.
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.
m = lm(Sales ~ Price + Urban + US, data = Carseats)
table = tbl_regression(m, intercept = TRUE)
as_gt(table) |> gt::tab_header("Linear Regression Model: Sales")| Linear Regression Model: Sales | |||
| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| (Intercept) | 13 | 12, 14 | <0.001 |
| Price | -0.05 | -0.06, -0.04 | <0.001 |
| Urban | |||
| No | — | — | |
| Yes | -0.02 | -0.56, 0.51 | >0.9 |
| US | |||
| No | — | — | |
| Yes | 1.2 | 0.69, 1.7 | <0.001 |
| Abbreviation: CI = Confidence Interval | |||
(b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
Answer: Per the statistically significant variables in this model for every .05 cent decrease in price there is a one unit increase in Sales. Also, the US made products tend to be larger sales drivers. Note that the R squared on this model is only 23%.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
Answer: \(Sales = \beta_0 + \beta_1(Price) + \beta_2(Urban) + \beta_3(US) + \varepsilon\)
(d) For which of the predictors can you reject the null hypothesis H0 : βj =0?
Answer: For predictors Price and USyes we reject the NULL hypothesis which states there is no relationship between variables
(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.
m1 = lm(Sales ~ Price + US, data = Carseats)
table1 = tbl_regression(m1, intercept = TRUE)
as_gt(table1) |> gt::tab_header("Linear Regression Model: Significant Predictors")| Linear Regression Model: Significant Predictors | |||
| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| (Intercept) | 13 | 12, 14 | <0.001 |
| Price | -0.05 | -0.06, -0.04 | <0.001 |
| US | |||
| No | — | — | |
| Yes | 1.2 | 0.69, 1.7 | <0.001 |
| Abbreviation: CI = Confidence Interval | |||
(f) How well do the models in (a) and (e) fit the data?
Answer: Both these models do not fit the data very well shown in the R^2 of 24%. Although Price and USyes are statistically significant predictors the there is a still 76% of the variance of Sales that cannot be explained by this model.
(g) Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(m1)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632Answer: For every unit of increase in Sales 1,200 of them are made in the US and Price decreases by .5 cents. The R squared of 24% stayed the same for both models proving that removing statistically insignificant variables did not help.
(h) Is there evidence of outliers or high leverage observations in the model from (e)?
Carseats |>
ggplot(aes(Price)) +
geom_histogram(binwidth =10,
fill="darkgreen",
color="white") +
labs(title= "Price Outlier Graph")
Carseats |>
ggplot(aes(Sales)) +
geom_histogram(binwidth =1,
fill="darkgreen",
color="white") +
labs(title= "Sales Outlier Graph")
Answer: In the Sales section the range of sales sits between 0 - 20 without any real evidence of outliers. With the price variable there seems to be some outliers with a few price points in the 170 -180 range.
Question 12
This problem involves simple linear regression without an intercept.
(a) 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?
Answer: When X and Y can both be used to predict each other the coefficient estimate stays the same. For example, if sales can predict price and price can predict sales the coefficient estimate would be 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.
market = Smarket |> head(100)
m3 = lm(Lag1 ~ Volume, data = market)
tbl_regression(m3, intercept = FALSE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| Volume | 1.3 | -0.40, 3.0 | 0.13 |
| Abbreviation: CI = Confidence Interval | |||
m4 = lm(Volume ~ Lag1, data = market)
tbl_regression(m4, intercept = FALSE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| Lag1 | 0.02 | -0.01, 0.04 | 0.13 |
| Abbreviation: CI = Confidence Interval | |||
Answer: Using the Volume and Lag5 variables here we can see that these two variables work in different ways. The coefficient value for Volume is -0.7032 and the coefficient value for Lag1 is -0.00835 are completely different
(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.
market = Smarket |> head(100)
m5 = lm(Lag1 ~ Lag2, data = market)
tbl_regression(m5, intercept = FALSE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| Lag2 | -0.01 | -0.21, 0.19 | >0.9 |
| Abbreviation: CI = Confidence Interval | |||
m6 = lm(Lag2 ~ Lag1, data = market)
tbl_regression(m6, intercept = FALSE)| Characteristic | Beta | 95% CI | p-value |
|---|---|---|---|
| Lag1 | -0.01 | -0.21, 0.19 | >0.9 |
| Abbreviation: CI = Confidence Interval | |||
Answer: Using 100 observations from the Smarket dataset, Lag1 and Lag2 can be used as an example in which the coefficient estimates are the same for both scenarios of flipping the variables. The 0.000156 R squared and the .12 show that the two models are almost identical.