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.