Linear regression (lm) in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of simple linear regression coefficients

What is the output of the following R code that fits a linear model and extracts coefficients?

R Programming

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

A Intercept = 1, x = 2

B Intercept = 2, x = 0

C Intercept = 0, x = 2

D Intercept = 0, x = 1

💡 Hint

Think about the relationship between y and x in the data.

The data y is exactly 2 times x, so the intercept is 0 and slope is 2.

❓ data_output

intermediate

Number of residuals after fitting lm

After fitting a linear model with 10 observations, how many residuals does the model produce?

R Programming

df <- data.frame(x = 1:10, y = rnorm(10))
model <- lm(y ~ x, data = df)
length(residuals(model))

A 10

B 9

C 11

D 1

💡 Hint

Residuals correspond to each observation.

There is one residual per observation, so 10 residuals for 10 data points.

🔧 Debug

advanced

Identify the error in lm formula usage

What error does this R code produce when fitting a linear model?

R Programming

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

A No error, model fits successfully

B Error: object 'y' not found

C Error: argument is missing, with no default

D Error: invalid formula

💡 Hint

Check the formula argument syntax in lm.

lm expects a formula like y ~ x, not y = x.

🚀 Application

advanced

Interpreting adjusted R-squared

You fit a linear model and get an adjusted R-squared of 0.85. What does this tell you about the model?

A 85% of the variance in the response is explained by the predictors, adjusted for number of predictors

B The model predictions are 85% accurate

C The model has 85% chance to predict new data correctly

D The residuals sum to 0.85

💡 Hint

Adjusted R-squared measures explained variance accounting for predictors.

Adjusted R-squared shows how well the model explains data variance, adjusting for predictors.

🧠 Conceptual

expert

Effect of multicollinearity on lm coefficients

What is the main effect of strong multicollinearity among predictors in a linear regression model?

A Model residuals become zero

B Coefficients become unstable and standard errors increase

C Adjusted R-squared becomes negative

D The model cannot be fit at all

💡 Hint

Think about how correlated predictors affect coefficient estimates.

Multicollinearity inflates standard errors and makes coefficients unreliable.

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