Linear regression calculator: least-squares line of best fit with R², standard errors, p-values and predictions, plus polynomial and exponential fits.
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Each extra unit of x corresponds to a fitted change of 3.1091 in y. The fit accounts for 99.33% of the observed variation in y. At x = 7.5 the fitted value is 72.2091.
| Term | Coefficient | Standard error | t | p-value |
|---|---|---|---|---|
| Intercept | 48.8909 | 0.6439 | 75.9325 | 1.008 × 10⁻¹² |
| x | 3.1091 | 0.0906 | 34.3146 | 5.686 × 10⁻¹⁰ |
Least squares picks the line or curve that makes the sum of squared vertical distances from the points as small as possible. For a straight line the slope is b₁ = Sxy/Sxx and the intercept b₀ = ȳ − b₁x̄; polynomial fits of degree 2 to 4 solve the normal equations, and the exponential model fits a straight line to ln y. When residual degrees of freedom and standard errors permit, coefficients include t statistics and p-values. R² describes variation in y for linear and polynomial fits, and in ln y for exponential fits.
The default relates weekly study hours to exam scores for 10 students. The fitted line is y = 48.89 + 3.109x, so each extra hour goes with about 3.1 more points, R² is 0.9933, and at 7.5 hours the line predicts 72.2.
Standard errors assume independent errors with constant variance and a correct model; exact small-sample t inference also assumes normally distributed errors on the fitted scale. A curved or funnel-shaped residual plot means the model is missing something, and predictions outside the range of the x data are flagged as extrapolations.
The least-squares line y = b₀ + b₁x has slope b₁ = Σ(x − x̄)(y − ȳ) ÷ Σ(x − x̄)² and intercept b₀ = ȳ − b₁x̄. For 10 students' study hours and exam scores, the two sums are 256.5 and 82.5, so b₁ = 3.109, b₀ = 48.89 and the line predicts 72.2 at 7.5 hours.
The data pair hours 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 with scores 55, 58, 62, 64, 68, 71, 72, 78, 80 and 83. Worked by hand:
Least squares picks the only line that makes the sum of squared vertical distances from the points as small as possible (NIST).
The slope is the fitted change in y per unit of x: each extra study hour goes with about 3.1 more points. That is an association in these 10 students, not proof that one more hour causes 3.1 points.
The intercept is the fitted y at x = 0, here 48.89 points for zero hours. No student studied fewer than 2 hours, so that value is an extrapolation that holds the line in place rather than a finding.
R² = 1 − SSE ÷ SST is the share of the variation in y the line accounts for. Here SSE = 5.418 and SST = 802.9, so R² = 0.99325 and hours account for 99.3% of the spread in scores.
For a straight line, R² is the square of the correlation, r = 0.99662. Adjusted R² = 1 − (1 − R²)(n − 1) ÷ (n − k) charges for each coefficient, with k counting the intercept; for n = 10 and k = 2 it is 0.99241.
Plain R² never falls when a term is added, but adjusted R² can. An x² term added to these data gets a coefficient of exactly 0, so R² stays 0.99325 while adjusted R² drops to 0.99132.
The test of a zero slope uses t = b₁ ÷ SE(b₁) on n − 2 degrees of freedom, where SE(b₁) = s ÷ √Sxx and s = √(SSE ÷ (n − 2)) is the residual standard error.
Here s = 0.8230, SE(b₁) = 0.8230 ÷ √82.5 = 0.0906 and t = 3.109 ÷ 0.0906 = 34.3 on 8 df. The two-sided p-value is 5.7 × 10⁻¹⁰, so a slope this steep would be very unlikely if hours and scores had no linear relation.
These standard errors assume independent errors with constant variance and a correctly shaped model. Exact small-sample t inference also assumes normally distributed errors.
A prediction inside the observed x range, like 72.2 at 7.5 hours, is a fitted average for that x, not a guaranteed score. Outside the range the line keeps going: at 20 hours it predicts 111.1, impossible on a 100-point exam.
A polynomial of degree 2 to 4 adds x², x³ and x⁴ terms, and a degree-d curve needs at least d + 1 distinct x values. With exactly d + 1 points it passes through every one and leaves no residual degrees of freedom, which is no evidence of good prediction.
An exponential fit, y = a·e^(bx), fits a straight line to ln y, so every y must be positive. For x = 0 to 5 and y = 3.0, 4.4, 6.7, 9.8, 14.9 and 22.1, the fit is y = 2.98·e^(0.4007x). Each unit of x multiplies y by e^0.4007 = 1.493, a 49.3% increase.
The exponential R², 0.99985, describes ln y. A straight line on the same data has R² = 0.924 on the original scale, so the two numbers measure different things.
| Scenario | Intercept | Slope | Prediction |
|---|---|---|---|
| Default | 48.890909 | 3.109091 | 72.209091 |
| Exact line | 1.000000 | 2.000000 | 9.000000 |
| Negative slope | 11.800000 | -1.800000 | 11.800000 |
| Repeated x | 0.500000 | 2.000000 | 4.500000 |
"Default" is the study-hours line at x = 7.5. "Exact line" uses (1, 3), (2, 5) and (3, 7), predicting at 4.
"Negative slope" uses x = 1 to 5 with y = 10, 8, 7, 4 and 3, predicting at 0. "Repeated x" uses x = 1, 1, 2, 2, 3, 3 with y = 2 to 7, predicting at 2. Repeated x values are fine as long as at least two distinct values remain.
To measure the strength of the relationship directly, or with ranks through Spearman's coefficient, use the correlation coefficient calculator.
Checked against: Constructed points (1,3), (2,5), (3,7) lie exactly on y = 1 + 2x; at x = 4, y = 9.
Checked against: b₁ = Sxy/Sxx, b₀ = ȳ − b₁x̄, se = √(SSE/(n − 2)) with decimal square roots
Checked against: Solving the 3×3 normal equations exactly (b₀ = 57/20, b₁ = −208/105, b₂ = 131/105)
Checked against: Constructed data: y = 1 + x + x² + x³ + x⁴ exactly, so every coefficient is 1 and R² = 1; 1 + 11 + 121 + 1331 + 14641 = 16105
Find the means x̄ and ȳ, then the slope b₁ = Σ(x − x̄)(y − ȳ) / Σ(x − x̄)² and the intercept b₀ = ȳ − b₁x̄. For the default study-hours data, Sxy = 256.5 and Sxx = 82.5, so b₁ = 3.109 and b₀ = 48.89. This least-squares line minimizes the sum of squared vertical distances, as set out in the NIST/SEMATECH e-Handbook (§4.1.4.1).
R² is the fraction of the variation in y that the model accounts for: R² = 1 − SSE/SST. The default fit has R² = 0.9933, so study hours account for 99.3% of the spread in exam scores. For a straight line R² equals the square of the correlation r. A high R² does not show that x causes y, that the model has the right shape, or that predictions beyond the data will hold.
Adjusted R² charges for each extra coefficient: 1 − (1 − R²)(n − 1)/(n − k), where k counts the coefficients including the intercept. R² never falls when a term is added, so a quartic always fits at least as well as a quadratic; adjusted R² falls when the new term explains less than it costs. For the default data, n = 10 and k = 2 turn R² = 0.99325 into 0.99241.
It tests the null hypothesis that the true slope is 0, using t = b₁/SE(b₁) with n − 2 degrees of freedom. For the default data t = 3.109/0.0906 = 34.3 with 8 df, giving p = 5.7 × 10⁻¹⁰: a slope this steep would be very unlikely if hours and scores had no linear relation. The p-value says nothing about how large or important the effect is.
When y grows or shrinks by a roughly constant percentage for each unit of x, so that ln y against x is straight. The exponential example, with y rising from 3.0 to 22.1 as x goes from 0 to 5, fits y = 2.98·e^(0.4007x), an increase of 49.3% per unit of x. The fit is made on ln y, as Excel's LOGEST function does, so R² describes ln y rather than y.
This calculator includes 5 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.