Linear regression worked example

The data

Five students report study hours (X) and exam scores (Y):

Formulas

For a straight line ŷ = a + bx fitted by ordinary least squares:

• b = Σ((x − x̄)(y − ȳ)) / Σ((x − x̄)²)
• a = ȳ − bx̄

Equivalently, using centered sums:

• S_xx = Σ(x − x̄)²
• S_xy = Σ(x − x̄)(y − ȳ)
• b = S_xy / S_xx

Step-by-step calculation

First compute the means:

• x̄ = (1 + 2 + 3 + 4 + 5) / 5 = 3
• ȳ = (52 + 57 + 65 + 68 + 76) / 5 = 63.6

Then the centered sums for each row:

Summing the last two columns:

• S_xy = 23.2 + 6.6 + 0 + 4.4 + 24.8 = 59.0
• S_xx = 4 + 1 + 0 + 1 + 4 = 10.0

Therefore:

• b = 59.0 / 10.0 = 5.9
• a = 63.6 − 5.9 × 3 = 63.6 − 17.7 = 45.9

The fitted line is:

ŷ = 45.9 + 5.9x

For one extra study hour, the model predicts an exam score about 5.9 points higher on this sample.

Check with the calculator

Paste the data into the linear regression calculator and click Calculate. It returns the same slope and intercept and adds R², RMSE and a residual table.

Study hours,Exam score
1,52
2,57
3,65
4,68
5,76

R² and RMSE for this fit

Using the fitted values ŷ and residuals y − ŷ:

• SSE = 0.04 + 0.49 + 1.96 + 2.25 + 0.36 = 5.10
• SST = Σ(y − ȳ)² = (−11.6)² + (−6.6)² + (1.4)² + (4.4)² + (12.4)² = 365.20
• R² = 1 − 5.10 / 365.20 ≈ 0.986
• RMSE = √(5.10 / 5) ≈ 1.01

R² is high because the points lie close to a straight line in this sample.

Limitations

• This line describes the five observations. It does not prove that study hours cause higher scores.
• Predictions outside the observed range (1–5 hours) assume the same straight-line relationship continues.
• Five observations are enough for a worked example but not enough to assess real-world reliability.
• Ordinary least squares minimises squared vertical errors; other loss functions would give different fits.