Linear Regression Calculator

Worked example

Twelve pairs with these sums fit y = 19.0909 + 2.4476x. At x = 10 the line predicts 43.566, with a standard error of the estimate of 2.082 — so a prediction quoted to four decimals is claiming precision the residuals do not support.

How to work it out yourself

1. 1.Gather n, Σx, Σy, Σxy, Σx² and Σy² from the data.
2. 2.Slope is Sxy over Sxx, where Sxy = Σxy − ΣxΣy/n and Sxx = Σx² − (Σx)²/n.
3. 3.Intercept is the mean of y minus the slope times the mean of x, so the line always passes through the point of both means.
4. 4.Predict by substituting x into y = a + bx — but only inside the range the data covered.

The formula

1. Mean of x78 / 12
2. Mean of y420 / 12
3. Σxy − (Σx)(Σy)/n3080 - (78 * 420) / 12
4. Σx² − (Σx)²/n650 - (78 ^ 2) / 12
5. Σy² − (Σy)²/n15600 - (420 ^ 2) / 12
6. Slope350 / 143
7. Intercept, a35 - 2.4475524 * 6.5
8. r², variance explained(350 ^ 2) / (143 * 900)
9. Standard error of the estimatesqrt(43.356643 / (12 - 2))
10. Standard error of the slope2.0822258 / sqrt(143)
11. Predicted y19.090909 + 2.4475524 * 10

Source: NIST/SEMATECH e-Handbook — least squares regression, NIST/SEMATECH e-Handbook of Statistical Methods

Questions people actually ask

How do you find the line of best fit?

By minimising the sum of the squared vertical distances from the points to the line. That criterion has a closed-form answer: slope = Sxy/Sxx and intercept = ȳ − b·x̄. No iteration and no guessing — the arithmetic above is the whole method.

Why squared distances rather than plain ones?

Two reasons, one good and one convenient. Squaring makes the criterion differentiable, so the minimum has a formula. It also weights a point twice as far away four times as heavily, which is a modelling choice rather than a law — least absolute deviations gives a different, more outlier-resistant line.

What does the standard error of the estimate tell me?

The typical size of a residual — how far a real point sits from the line, in the units of y. A prediction is only as good as that figure: reporting 43.5664 when the standard error is 2.08 quotes four decimals of a number that is uncertain in the units column.

Can I predict outside the range of my data?

You can compute it; you cannot rely on it. A line fitted between x = 1 and x = 12 says nothing about x = 100, because nothing in the data shows the relationship stays linear out there. Extrapolation is where regressions produce their most confident wrong answers.

Which variable goes on which axis?

It matters. Regressing y on x minimises vertical distances; regressing x on y minimises horizontal ones, and the two lines are different unless r = 1. Put the thing you want to predict on y — and if neither predicts the other, a correlation is the honest summary, not a line.

Related

• Correlation CalculatorPearson correlation from the sums a class is given, with r², the slope it implies, and the sample size at which it stops being noise.
• Percentage Change CalculatorPercentage increase or decrease between two values, with the absolute difference and the multiplier.
• Slope CalculatorSlope, intercept, angle and the full line equation from two points, with the perpendicular slope beside it.
• Z-Score to P-Value CalculatorExact p-value from a z-score, one-tailed or two-tailed, with the percentile the score sits at.
• Grade CalculatorYour weighted course grade from up to eight categories, by percentage or by points, or the score you need on the final to reach the grade you want.
• Confidence Interval CalculatorThe confidence interval around a sample mean, with the margin of error and what the interval does and does not claim.

Put this calculator on your site

Free, no attribution required beyond the link.

<iframe src="https://rulecalculators.com/embed/linear-regression" width="100%" height="420" style="border:1px solid #e7e4de;border-radius:12px" title="Linear Regression Calculator"></iframe>