# Linear regression calculator (line of best fit)

> Linear regression calculator: least-squares line of best fit with R², standard errors, p-values and predictions, plus polynomial and exponential fits.

Versión interactiva: https://www.calcopenly.com/es/statistics/regression-calculator
Tema: Calculadoras de estadística y probabilidad

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. Each coefficient gets a standard error, a t statistic and a p-value, and R² gives the share of the variation in y that the fit accounts for.

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.

The standard errors and p-values assume independent residuals with constant spread. 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.

## Datos

- **x values**: Weekly study hours (default example).
- **y values**: Exam scores, in the same order as x.
- **Model** (opciones: Linear, Polynomial, Exponential)
- **Degree** (opciones: 2 (quadratic), 3 (cubic), 4 (quartic))
- **Predict y at x**

## Resultados

- Fitted equation — resultado principal
- Intercept b₀
- Slope / x coefficient b₁
- x² coefficient b₂
- x³ coefficient b₃
- x⁴ coefficient b₄
- a in y = a·e^(bx)
- b in y = a·e^(bx)
- Correlation r
- R²
- Adjusted R²
- Residual standard error
- Standard error of b₀
- Standard error of b₁
- Predicted y
- Points n

## Fórmula

$$
b_1 = \frac{\sum (x_i-\bar x)(y_i-\bar y)}{\sum (x_i-\bar x)^2},\quad b_0 = \bar y - b_1\bar x,\quad R^2 = 1 - \frac{\sum (y_i-\hat y_i)^2}{\sum (y_i-\bar y)^2}
$$

## Ejemplos resueltos

### Exact straight line y = 1 + 2x

- x values: 1, 2, 3
- y values: 3, 5, 7
- Model: Linear
- Predict y at x: 4
- **Fitted equation: y = 1 + 2x**
- **Intercept b₀: 1**
- **Slope / x coefficient b₁: 2**
- **R²: 1**
- **Predicted y: 9**
- Fuente de comprobación: Constructed points (1,3), (2,5), (3,7) lie exactly on y = 1 + 2x; at x = 4, y = 9.

### Study hours vs score (linear)

- x values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
- y values: 55, 58, 62, 64, 68, 71, 72, 78, 80, 83
- Model: Linear
- Predict y at x: 7.5
- **Slope / x coefficient b₁: 3.109091**
- **Intercept b₀: 48.890909**
- **R²: 0.993252**
- **Adjusted R²: 0.992408**
- **Residual standard error: 0.822966**
- **Standard error of b₁: 0.090606**
- **Standard error of b₀: 0.643873**
- **Predicted y: 72.209091**
- **Correlation r: 0.99662**
- Fuente de comprobación: Python fractions: b₁ = Sxy/Sxx, b₀ = ȳ − b₁x̄, se = √(SSE/(n − 2)) with decimal square roots

### Quadratic trend

- x values: 1, 2, 3, 4, 5, 6, 7, 8
- y values: 2.1, 3.9, 8.2, 14.8, 24.1, 35.9, 50.2, 66.8
- Model: Polynomial
- Degree: 2 (quadratic)
- **Intercept b₀: 2.85**
- **Slope / x coefficient b₁: -1.980952**
- **x² coefficient b₂: 1.247619**
- **R²: 0.999994**
- **Residual standard error: 0.069007**
- Fuente de comprobación: Python fractions solving the 3×3 normal equations exactly (b₀ = 57/20, b₁ = −208/105, b₂ = 131/105); agrees with numpy.polyfit(x, y, 2)

### Exact quartic y = 1 + x + x² + x³ + x⁴

- x values: 0 1 2 3 4 5 6 7 8 9 10
- y values: 1 5 31 121 341 781 1555 2801 4681 7381 11111
- Model: Polynomial
- Degree: 4 (quartic)
- Predict y at x: 11
- **Intercept b₀: 1**
- **Slope / x coefficient b₁: 1**
- **x² coefficient b₂: 1**
- **x³ coefficient b₃: 1**
- **x⁴ coefficient b₄: 1**
- **R²: 1**
- **Predicted y: 16,105**
- Fuente de comprobación: Constructed data: y = 1 + x + x² + x³ + x⁴ exactly, so every coefficient is 1 and R² = 1; 1 + 11 + 121 + 1331 + 14641 = 16105

### Exponential growth

- x values: 0, 1, 2, 3, 4, 5
- y values: 3.0, 4.4, 6.7, 9.8, 14.9, 22.1
- Model: Exponential
- Predict y at x: 6
- **a in y = a·e^(bx): 2.979832**
- **b in y = a·e^(bx): 0.400696**
- **R²: 0.99985**
- **Predicted y: 32.98473**
- Fuente de comprobación: Python decimal ln(y) at 50 digits, least squares with fractions, a = exp(intercept); numpy.polyfit(x, log(y), 1) gives b = 0.40069633

## Preguntas

### How do you calculate the line of best fit?

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).

### What does R² tell you in regression?

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.

### What is the difference between R² and adjusted R²?

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.

### What does the p-value of the slope mean?

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 should I use an exponential fit instead of a straight line?

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.

### ¿Qué precisión tiene «Linear regression calculator (line of best fit)»?

La precisión depende de tus datos y de los supuestos del método. El cálculo decimal usa 50 cifras significativas, pero las estimaciones, los métodos numéricos y los datos de origen pueden ser menos precisos; el redondeo mostrado no elimina esos límites. Ejemplos resueltos comprobados con fuentes independientes: 5. Por ejemplo, «Exact straight line y = 1 + 2x» se comprueba con Constructed points (1,3), (2,5), (3,7) lie exactly on y = 1 + 2x; at x = 4, y = 9.

### ¿De dónde procede el método?

NIST/SEMATECH e-Handbook of Statistical Methods, §4.1.4.1 Linear least squares regression; NIST Statistical Reference Datasets — linear regression; Draper & Smith, Applied Regression Analysis, 3rd ed. (Wiley, 1998), ch. 1 and 12.

## Fuentes

- [NIST/SEMATECH e-Handbook of Statistical Methods, §4.1.4.1 Linear least squares regression](https://www.itl.nist.gov/div898/handbook/pmd/section1/pmd141.htm)
- [NIST Statistical Reference Datasets — linear regression](https://www.itl.nist.gov/div898/strd/lls/lls.shtml)
- Draper & Smith, Applied Regression Analysis, 3rd ed. (Wiley, 1998), ch. 1 and 12