Linear regression (lm) in R Programming - Step-by-Step Execution

Concept Flow - Linear regression (lm)

Start with data

↓

Choose dependent and independent variables

↓

Fit linear model lm(y ~ x)

↓

Calculate coefficients (intercept, slope)

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Use model to predict y from x

↓

Evaluate model fit (summary)

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End

Linear regression fits a straight line to data by finding coefficients that best predict y from x.

Execution Sample

R Programming

x <- c(1, 2, 3, 4, 5)
y <- c(2, 4, 5, 4, 5)
model <- lm(y ~ x)
summary(model)

Fit a linear model predicting y from x and show the summary of the fit.

Execution Table

Step	Action	Evaluation	Result
1	Create vector x	x = c(1,2,3,4,5)	x = 1,2,3,4,5
2	Create vector y	y = c(2,4,5,4,5)	y = 2,4,5,4,5
3	Fit model lm(y ~ x)	Calculate coefficients	model with intercept and slope
4	Extract coefficients	coef(model)	Intercept ≈ 2.2, Slope ≈ 0.6
5	Predict y for x=3	predict(model, newdata=data.frame(x=3))	Predicted y ≈ 3.8
6	Show summary(model)	summary(model)	R-squared ≈ 0.6, p-values, residuals
7	End	Model fit complete	Ready for predictions or diagnostics

💡 All steps complete, model fitted and summary available

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
x	NULL	[1,2,3,4,5]	[1,2,3,4,5]	[1,2,3,4,5]	[1,2,3,4,5]	[1,2,3,4,5]
y	NULL	NULL	[2,4,5,4,5]	[2,4,5,4,5]	[2,4,5,4,5]	[2,4,5,4,5]
model	NULL	NULL	NULL	lm object	lm object	lm object
coefficients	NULL	NULL	NULL	NULL	Intercept=2.2, Slope=0.6	Intercept=2.2, Slope=0.6

Key Moments - 3 Insights

Why does the model have an intercept and slope?

The intercept and slope come from fitting the line y = intercept + slope * x that best matches the data (see execution_table step 4).

What does the summary(model) tell us?

It shows how well the line fits the data, including R-squared and p-values (see execution_table step 6).

Why do we use predict() after fitting the model?

predict() uses the fitted line to estimate y for new x values (see execution_table step 5).

Visual Quiz - 3 Questions

Look at the execution table, what is the approximate slope coefficient after fitting the model?

A 4.0

B 2.2

C 0.6

D 5.0

💡 Hint

Check the 'Extract coefficients' row in the execution_table where coefficients are shown.

At which step do we get the model object created?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look for the step where lm(y ~ x) is called and the model is fitted.

If we add a new x value 6 and predict y, which step would this action be similar to?

A Step 5

B Step 1

C Step 3

D Step 6

💡 Hint

Prediction of y for a new x value is done in the step where predict() is used.

Concept Snapshot

Linear regression fits a line y = intercept + slope * x
Use lm(y ~ x) in R to fit the model
Coefficients show intercept and slope
summary(model) shows fit quality
predict(model, newdata) estimates y for new x

Full Transcript

This visual execution traces fitting a linear regression model in R using lm(). We start by creating numeric vectors x and y. Then we fit the model lm(y ~ x), which calculates the best intercept and slope to predict y from x. The coefficients are extracted and show the intercept around 2.2 and slope around 0.6. Using predict(), we estimate y for a new x value. Finally, summary(model) gives statistics about the fit quality like R-squared. Variables x, y, model, and coefficients change as we progress through these steps.

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