Question 8

This question involves the use of simple linear regression on the Auto data set.

library(ISLR2)
## Warning: package 'ISLR2' was built under R version 4.4.2
data(Auto)  # Loads the dataset
  1. Use the lm() function to perform a simple linear regression with mpg as the response and horsepower as the predictor. Use the summary() function to print the results. Comment on the output.
    • There is strong evidence that horsepower negatively affects mpg.
    • Since the R² value is 0.6059, horsepower explains about 60.59% of the variation in mpg, meaning other factors must be considered for better predictions.
    • Since the p-value for horsepower is extremely low**, we confidently conclude that horsepower is a significant predictor of mpg.
# Fit the regression model
model <- lm(mpg ~ horsepower, data = Auto)

# Print the summary of the model
summary(model)
## 
## Call:
## lm(formula = mpg ~ horsepower, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.5710  -3.2592  -0.3435   2.7630  16.9240 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 39.935861   0.717499   55.66   <2e-16 ***
## horsepower  -0.157845   0.006446  -24.49   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.906 on 390 degrees of freedom
## Multiple R-squared:  0.6059, Adjusted R-squared:  0.6049 
## F-statistic: 599.7 on 1 and 390 DF,  p-value: < 2.2e-16
  1. Plot the response and the predictor. Use the abline() function to display the least squares regression line.
# Create the scatter plot
plot(Auto$horsepower, Auto$mpg, 
     main = "MPG vs Horsepower",
     xlab = "Horsepower", 
     ylab = "MPG", 
     col = "blue", 
     pch = 16)

# Add the least squares regression line
abline(model, col = "red", lwd = 2)

  1. Use the plot() function to produce diagnostic plots of the least squares regression ft. Comment on any problems you see with the fit.

The diagnostic plots reveal several issues with the linear regression model predicting mpg from horsepower. The Residuals vs. Fitted plot shows a clear curved pattern, suggesting that the relationship between mpg and horsepower is non-linear, meaning a simple linear model may not be the best fit. Additionally, the Q-Q plot indicates that the residuals deviate significantly from normality, especially in the tails, which suggests the presence of outliers or skewness in the data. The Scale-Location plot further reveals signs of heteroscedasticity, where the variance of residuals increases as fitted values increase, indicating that the model does not explain variability in mpg equally across all levels of horsepower. Finally, the Residuals vs. Leverage plot identifies high-leverage points, such as observations 334, 387, and 17, which may have an outsized influence on the model. These issues suggest that improvements could be made by adding non-linear terms, applying a log transformation to mpg, or using a weighted least squares regression to stabilize variance.

# Generate diagnostic plots
par(mfrow = c(2, 2))  # Arrange plots in a 2x2 grid
plot(model)

par(mfrow = c(1, 1))  # Reset layout

Question 10

This question should be answered using the Carseats data set.

# Load the Carseats dataset
data(Carseats)
  1. Fit a multiple regression model to predict Sales using Price, Urban, and US.
# Fit the multiple regression model
model <- lm(Sales ~ Price + Urban + US, data = Carseats)

# Display the regression summary
summary(model)
## 
## 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

  1. Provide an interpretation of each coeffcient in the model. Be careful—some of the variables in the model are qualitative!

    Price Coefficient (p-value < 2e-16)

    • Since the p-value is extremely small (< 0.001), we reject the null hypothesis H0:β1=0H_0: _1 = 0H0​:β1​=0, meaning that Price significantly affects Sales.

    • Interpretation: For each $1 increase in Price, Sales decrease by 0.0545 units, holding all else constant.

    • The negative sign confirms an inverse relationship between Price and Sales.

    Urban Coefficient (p-value = 0.936)

    • The p-value is very high (0.936), meaning that Urban does not significantly affect Sales.

    • Interpretation: Whether the store is located in an urban or rural area does not have a meaningful impact on Sales.

    • Actionable Insight: Since this variable is not significant, we might consider removing it from the model.

    US Coefficient (p-value = 4.86e-06)

    • The p-value is very small (< 0.001), meaning that we reject the null hypothesis H0:β3=0H_0: _3 = 0H0​:β3​=0, confirming that US location significantly affects Sales.

    • Interpretation: Stores in the US have 1.2 more units of Sales on average compared to non-US stores, holding all else constant.

    • This suggests that US stores generally have higher Sales than international stores.

  2. Write out the model in equation form, being careful to handle the qualitative variables properly.

    Sales=13.04−0.0545×Price−0.0219×UrbanYes+1.2006×USYes

  3. For which of the predictors can you reject the null hypothesis \(H_0: \beta_j = 0\)?

    We can reject the null hypothesis for Price and US, meaning they significantly affect Sales.

  4. 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.

# Fit the reduced multiple regression model (excluding Urban)
reduced_model <- lm(Sales ~ Price + US, data = Carseats)

# Display the summary of the new model
summary(reduced_model)
## 
## 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

  1. How well do the models in (a) and (e) fit the data?

    Both models fit the data equally well, but Model (e) is preferable because RSE is slightly lower, Adjusted \(R^2\) is slightly higher, The F-statistic is higher, and The Urban variable was not significant.

  2. Using the model from (e), obtain 95 % confdence intervals for the coeffcient(s).

# Obtain 95% confidence intervals for the coefficients
confint(reduced_model, level = 0.95)
##                   2.5 %      97.5 %
## (Intercept) 11.79032020 14.27126531
## Price       -0.06475984 -0.04419543
## USYes        0.69151957  1.70776632
  1. Is there evidence of outliers or high leverage observations in the model from (e)?
plot(reduced_model, which = 4)  # Cook's Distance plot

Question 14

This problem focuses on the collinearity problem.

  1. Perform the following commands in R:
set.seed(1)
x1 <- runif(100)
x2 <- 0.5 * x1 + rnorm(100) / 10
y <- 2 + 2 * x1 + 0.3 * x2 + rnorm(100)

The last line corresponds to creating a linear model in which y is a function of x1 and x2. Write out the form of the linear model. What are the regression coeffcients?

# Fit the linear model
model <- lm(y ~ x1 + x2)

# Display the summary to get regression coefficients
summary(model)
## 
## Call:
## lm(formula = y ~ x1 + x2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.8311 -0.7273 -0.0537  0.6338  2.3359 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.1305     0.2319   9.188 7.61e-15 ***
## x1            1.4396     0.7212   1.996   0.0487 *  
## x2            1.0097     1.1337   0.891   0.3754    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.056 on 97 degrees of freedom
## Multiple R-squared:  0.2088, Adjusted R-squared:  0.1925 
## F-statistic:  12.8 on 2 and 97 DF,  p-value: 1.164e-05
  1. What is the correlation between x1 and x2? Create a scatterplot displaying the relationship between the variables.
# Compute correlation
correlation <- cor(x1, x2)
print(paste("Correlation between x1 and x2:", round(correlation, 2)))
## [1] "Correlation between x1 and x2: 0.84"
# Create scatterplot
plot(x1, x2, main = paste("Scatterplot of x1 vs x2 (Correlation:", round(correlation, 2), ")"),
     xlab = "x1", ylab = "x2", pch = 19, col = "blue")
grid()

  1. Using this data, fit a least squares regression to predict \(y\) using \(x_1\) and \(x_2\). Describe the results obtained. What are \(\hat{\beta}_0\), \(\hat{\beta}_1\), and \(\hat{\beta}_2\)? How do these relate to the true \(\beta_0\), \(\beta_1\), and \(\beta_2\)? Can you reject the null hypothesis \(H_0: \beta_1 = 0\)? How about the null hypothesis \(H_0: \beta_2 = 0\)?

    • x1​ is statistically significant (p=0.0487p = 0.0487p=0.0487), though close to the 0.05 threshold.

    • x2x_2x2​ is not significant (p=0.3754p = 0.3754p=0.3754), likely due to multicollinearity with x1x_1x1​.

# Extract p-values
summary(model)$coefficients[, 4]
##  (Intercept)           x1           x2 
## 7.606713e-15 4.872517e-02 3.753565e-01
library(car)
## Loading required package: carData
vif(model)
##       x1       x2 
## 3.304993 3.304993
  1. Now fit a least squares regression to predict \(y\) using only \(x_1\). Comment on your results. Can you reject the null hypothesis \(H_0: \beta_1 = 0\)? Since this is much smaller than 0.05, we reject the null hypothesis
# Fit least squares regression using only x1
model_x1 <- lm(y ~ x1)

# Display summary for model_x1
summary(model_x1)
## 
## Call:
## lm(formula = y ~ x1)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.89495 -0.66874 -0.07785  0.59221  2.45560 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.1124     0.2307   9.155 8.27e-15 ***
## x1            1.9759     0.3963   4.986 2.66e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.055 on 98 degrees of freedom
## Multiple R-squared:  0.2024, Adjusted R-squared:  0.1942 
## F-statistic: 24.86 on 1 and 98 DF,  p-value: 2.661e-06
# Extract p-value for H0: β1 = 0
summary(model_x1)$coefficients["x1", 4]
## [1] 2.660579e-06
  1. Now fit a least squares regression to predict \(y\) using only \(x_2\). Comment on your results. Can you reject the null hypothesis \(H_0: \beta_1 = 0\)? Since this is much smaller than 0.05, we reject the null hypothesis
# Fit least squares regression using only x2
model_x2 <- lm(y ~ x2)

# Display summary for model_x2
summary(model_x2)
## 
## Call:
## lm(formula = y ~ x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.62687 -0.75156 -0.03598  0.72383  2.44890 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.3899     0.1949   12.26  < 2e-16 ***
## x2            2.8996     0.6330    4.58 1.37e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.072 on 98 degrees of freedom
## Multiple R-squared:  0.1763, Adjusted R-squared:  0.1679 
## F-statistic: 20.98 on 1 and 98 DF,  p-value: 1.366e-05
# Extract p-value for H0: β1 = 0 in x2 model
summary(model_x2)$coefficients["x2", 4]
## [1] 1.36643e-05
  1. Do the results obtained in (c)–(e) contradict each other? Explain your answer. The results in (c)–(e) do contradict each other, but this contradiction is explained by multicollinearity. If two variables are highly correlated, their individual impact is difficult to separate, leading to misleading statistical significance results.
  2. Now suppose we obtain one additional observation, which was unfortunately mismeasured.

x1 <- c(x1, 0.1)

x2 <- c(x2, 0.8)

y <- c(y, 6)

Re-ft the linear models from (c) to (e) using this new data. What efect does this new observation have on the each of the models? In each model, is this observation an outlier? A high-leverage point? Both? Explain your answers

# Add the new observation
x1 <- c(x1, 0.1)
x2 <- c(x2, 0.8)
y  <- c(y, 6)

# Fit full model
model_full <- lm(y ~ x1 + x2)

# Display summary
summary(model_full)
## 
## Call:
## lm(formula = y ~ x1 + x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.73348 -0.69318 -0.05263  0.66385  2.30619 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.2267     0.2314   9.624 7.91e-16 ***
## x1            0.5394     0.5922   0.911  0.36458    
## x2            2.5146     0.8977   2.801  0.00614 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.075 on 98 degrees of freedom
## Multiple R-squared:  0.2188, Adjusted R-squared:  0.2029 
## F-statistic: 13.72 on 2 and 98 DF,  p-value: 5.564e-06
# Fit model using only x1
model_x1 <- lm(y ~ x1)

# Display summary
summary(model_x1)
## 
## Call:
## lm(formula = y ~ x1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.8897 -0.6556 -0.0909  0.5682  3.5665 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.2569     0.2390   9.445 1.78e-15 ***
## x1            1.7657     0.4124   4.282 4.29e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.111 on 99 degrees of freedom
## Multiple R-squared:  0.1562, Adjusted R-squared:  0.1477 
## F-statistic: 18.33 on 1 and 99 DF,  p-value: 4.295e-05
# Fit model using only x2
model_x2 <- lm(y ~ x2)

# Display summary
summary(model_x2)
## 
## Call:
## lm(formula = y ~ x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.64729 -0.71021 -0.06899  0.72699  2.38074 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.3451     0.1912  12.264  < 2e-16 ***
## x2            3.1190     0.6040   5.164 1.25e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.074 on 99 degrees of freedom
## Multiple R-squared:  0.2122, Adjusted R-squared:  0.2042 
## F-statistic: 26.66 on 1 and 99 DF,  p-value: 1.253e-06
# Compute leverage values for the full model
leverage_values <- hatvalues(model_full)

# Extract leverage for the new observation (last data point)
new_obs_index <- length(y)
leverage_values[new_obs_index]
##       101 
## 0.4147284
# Residuals vs. Leverage plot
plot(model_full, which = 5)