Linear regression (lm) in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to fit a linear model predicting mpg from hp.

R Programming

model <- lm(mpg ~ [1], data = mtcars)

Drag options to blanks, or click blank then click option'

Ahp

Bmpg

Ccyl

Dwt

Attempts:

3 left

💡 Hint

Common Mistakes

Using the response variable 'mpg' as predictor.

Using unrelated variables like 'cyl' or 'wt' without instruction.

The formula mpg ~ hp fits a linear model predicting miles per gallon (mpg) using horsepower (hp) as the predictor.

2fill in blank

medium

Complete the code to view the summary of the linear model.

R Programming

summary([1])

Drag options to blanks, or click blank then click option'

Amodel

Bmtcars

Clm

Dsummary

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the dataset 'mtcars' instead of the model.

Passing the function name 'lm' or 'summary' itself.

The summary() function shows details about the fitted linear model object named model.

3fill in blank

hard

Fix the error in the code to extract coefficients from the model.

R Programming

coef <- [1](model)

Drag options to blanks, or click blank then click option'

Acoefficients

Bsummary

Ccoef

Dcoeff

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'summary' which returns a summary object, not coefficients directly.

Using 'coefficients' or 'coeff' which are not functions.

The function coef() extracts the coefficients from a fitted model object.

4fill in blank

hard

Fill both blanks to create a new data frame with predicted mpg values.

R Programming

new_data <- data.frame(hp = c(100, 150, 200))
predictions <- predict([1], newdata = [2])

Drag options to blanks, or click blank then click option'

Amodel

Bnew_data

Cmtcars

Dlm

Attempts:

3 left

💡 Hint

Common Mistakes

Using the dataset 'mtcars' instead of new data for predictions.

Using 'lm' instead of the model object.

Use the fitted model object 'model' and the new data frame 'new_data' to get predictions.

5fill in blank

hard

Fill all three blanks to create a scatter plot with regression line.

R Programming

plot(mtcars$hp, mtcars$mpg, xlab = [1], ylab = [2], main = [3])
abline(model, col = "red")

Drag options to blanks, or click blank then click option'

A"Horsepower"

B"Miles per Gallon"

C"MPG vs Horsepower"

D"Weight"

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up x and y labels.

Using unrelated labels like 'Weight'.

The x-axis label is 'Horsepower', y-axis label is 'Miles per Gallon', and the plot title is 'MPG vs Horsepower'.

Practice

1. What does the lm() function in R do?

easy

A. Fits a linear regression model to predict a variable

B. Creates a plot of data points

C. Calculates the mean of a numeric vector

D. Performs a t-test between two groups

Solution

1. Step 1: Understand the purpose of lm() and compare options

   The lm() function is used to fit linear models, which means it finds a line that best predicts one variable from others. Options A, B, and D describe other functions or analyses, not linear regression modeling.

2. Final Answer:

   Fits a linear regression model to predict a variable -> Option A

3. Quick Check:

   lm() fits linear regression [OK]

Hint: Remember: lm() stands for linear model fitting [OK]

Common Mistakes:

• Confusing lm() with plotting functions
• Thinking lm() calculates summary statistics only
• Mixing lm() with hypothesis testing functions

2. Which of the following is the correct syntax to fit a linear model predicting y from x in R?

easy

A. model <- lm(y ~ x, data = mydata)

B. model <- lm(x ~ y, data = mydata)

C. model <- lm(y, x, data = mydata)

D. model <- lm(data = mydata, y = x)

Solution

1. Step 1: Recall the formula syntax for lm() and check options

   The correct syntax uses lm(response ~ predictor, data = dataset). Here, y is the response and x the predictor. model <- lm(y ~ x, data = mydata) matches. model <- lm(x ~ y, data = mydata) reverses response and predictor. model <- lm(y, x, data = mydata) uses wrong argument order. model <- lm(data = mydata, y = x) uses = instead of ~.

2. Final Answer:

   model <- lm(y ~ x, data = mydata) -> Option A

3. Quick Check:

   lm(response ~ predictor, data) [OK]

Hint: Formula is always response ~ predictor in lm() [OK]

Common Mistakes:

• Swapping response and predictor variables
• Passing variables without formula syntax
• Misordering arguments in lm()

3. What will be the output of this R code?

data <- data.frame(x = 1:5, y = c(2, 4, 6, 8, 10))
model <- lm(y ~ x, data = data)
coef(model)

medium

A. Intercept = 0, Slope = 1

B. Intercept = 2, Slope = 1

C. Intercept = 1, Slope = 2

D. Intercept = 0, Slope = 2

Solution

1. Step 1: Analyze data and model fit

   The data has x = 1 to 5 and y = 2,4,6,8,10 which is exactly y = 2*x. Since y = 2*x perfectly, intercept is 0 and slope is 2.

2. Final Answer:

   Intercept = 0, Slope = 2 -> Option D

3. Quick Check:

   y = 2*x means intercept 0, slope 2 [OK]

Hint: Check if y is exactly a multiple of x for slope [OK]

Common Mistakes:

• Assuming intercept is 1 by default
• Mixing up slope and intercept values
• Not checking data pattern before fitting

4. Identify the error in this R code for linear regression:

data <- data.frame(x = 1:4, y = c(3, 6, 9, 12))
model <- lm(y = x, data = data)
summary(model)

medium

A. Missing library call for lm()

B. Incorrect formula syntax in lm()

C. summary() cannot be used on lm objects

D. Data frame creation is wrong

Solution

1. Step 1: Check lm() call and verify other code

   The lm() function requires a formula like response ~ predictor. Here, y = x is incorrect syntax. Data frame creation is correct. summary() works on lm objects. lm() is base R, no library needed.

2. Final Answer:

   Incorrect formula syntax in lm() -> Option B

3. Quick Check:

   lm() needs formula, not y = x [OK]

Hint: lm() needs formula like y ~ x, not y = x [OK]

Common Mistakes:

• Using = instead of ~ in lm() formula
• Assuming summary() is not for lm objects
• Thinking lm() requires extra libraries

5. You have a dataset with variables height and weight. You want to predict weight from height using linear regression. After fitting model <- lm(weight ~ height, data = df), the summary shows a very low R-squared value. What is the best next step?

hard

A. Remove the intercept term to force a better fit

B. Use the model as is because R-squared is not important

C. Try adding more predictor variables to improve the model

D. Switch to a t-test instead of regression

Solution

1. Step 1: Interpret low R-squared and next steps

   A low R-squared means the model explains little of the variation in weight using height alone. Adding more predictors can help explain more variation. Removing intercept or ignoring R-squared is not good practice. A t-test is unrelated.

2. Final Answer:

   Try adding more predictor variables to improve the model -> Option C

3. Quick Check:

   Low R-squared? Add predictors [OK]

Hint: Low R-squared? Add more predictors to improve [OK]

Common Mistakes:

• Removing intercept to fix low R-squared
• Ignoring model fit quality
• Confusing regression with hypothesis testing