Can the correlation coefficient be greater than 1 or less than -1?

No. The correlation coefficient always ranges from -1 to 1, inclusive. If you calculate a value outside this range, you have made a computational error. r = 1 is perfect positive correlation, r = -1 is perfect negative correlation, and r = 0 indicates no linear relationship.

Why does the correlation formula divide by (n-1) instead of n?

The correlation coefficient is a sample statistic. Dividing by (n-1) instead of n (Bessel's correction) provides an unbiased estimate of the true population correlation. Using n would underestimate the population value, particularly in small samples.

If I find a high correlation, does that prove one variable causes the other?

No. Correlation shows association (two variables move together) but not causation. High correlation can result from direct causation, reverse causation, a confounding third variable, or even coincidence. Causation requires controlled experimentation, temporal order, and elimination of alternative explanations.

What is the difference between r and r^2?

r (the correlation coefficient) measures the strength and direction of a linear relationship, ranging from -1 to 1. r^2 (the coefficient of determination) is the square of r and represents the fraction (or percentage) of variance in one variable explained by the linear relationship with the other. For example, r = 0.8 means r^2 = 0.64, so the linear model explains 64% of the variation in the response variable.

Correlation Coefficient

Correlation coefficient measures the strength and direction of a linear relationship between two quantitative variables, ranging from -1 to 1.

r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_x s_y}

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What each symbol means

What Correlation Coefficient takes

Symbol	Meaning
$r$	The correlation coefficient itself, always ranging from -1 (perfect negative correlation) to 1 (perfect positive correlation); values near 0 indicate weak or no linear relationship.
$x_i$	The $i$ -th individual observation of the first variable; the subscript $i$ identifies which data point (for example, $x_3$ is the third value of $x$ ).
$y_i$	The $i$ -th individual observation of the second variable, paired with $x_i$ from the same subject or time period; both observations must align positionally.
$n$	The total number of paired observations in the dataset; appears as $(n-1)$ in the denominator to apply Bessel's correction for unbiased sample estimation.

When to use it

Use the correlation coefficient when you have two quantitative variables and want to measure how strongly they move together linearly.

Level

Usually taught in: Statistics · Appears on: AP Statistics

Worked examples

1. Correlation coefficient with small integer pairs

Problem

Calculate the correlation coefficient for the dataset:

x = 1, 2, 3

and

y = 2, 3, 4

$\bar{x} = \frac{1+2+3}{3} = 2, \quad \bar{y} = \frac{2+3+4}{3} = 3$
Calculate the mean of each variable by summing all values and dividing by $n = 3$ .
$(x_i - \bar{x}) = -1, 0, 1 \text{ and } (y_i - \bar{y}) = -1, 0, 1$
Subtract the mean from each observation to find deviations from the center.
$(x_i - \bar{x})(y_i - \bar{y}) = 1, 0, 1 \implies \sum = 2$
Multiply corresponding deviations pair by pair: $(-1) \times (-1) = 1$ , $0 \times 0 = 0$ , $1 \times 1 = 1$ ; then sum these products.
$s_x^2 = \frac{(-1)^2 + 0^2 + 1^2}{3-1} = \frac{2}{2} = 1, \quad s_x = 1$
Calculate the sample variance of $x$ by summing squared deviations and dividing by $n-1 = 2$ , then take the square root.
$s_y^2 = \frac{(-1)^2 + 0^2 + 1^2}{3-1} = \frac{2}{2} = 1, \quad s_y = 1$
Calculate the sample variance of $y$ the same way.
$r = \frac{2}{(3-1) \cdot 1 \cdot 1} = \frac{2}{2} = 1$
Plug all values into the correlation formula; $r = 1$ indicates perfect positive linear association.

Answer: $r = 1$

This example shows a perfect linear relationship: as $x$ increases by 1 each time, $y$ also increases by exactly 1. The correlation coefficient captures this perfect association with $r = 1$ , the maximum possible value.

2. Correlation coefficient with negative relationship and decimals

Problem

Find the correlation coefficient for

x = 0.5, 1.5, 2.5

and

y = 3, 1, 2

$\bar{x} = \frac{0.5 + 1.5 + 2.5}{3} = \frac{4.5}{3} = 1.5, \quad \bar{y} = \frac{3 + 1 + 2}{3} = 2$
Calculate the mean of each variable.
$(x_i - \bar{x}) = -1, 0, 1 \text{ and } (y_i - \bar{y}) = 1, -1, 0$
Find deviations: subtract each mean from the corresponding values.
$(x_i - \bar{x})(y_i - \bar{y}) = -1, 0, 0 \implies \sum = -1$
Multiply deviations: $(-1)(1) = -1$ , $(0)(-1) = 0$ , $(1)(0) = 0$ ; the negative product indicates the negative trend.
$s_x^2 = \frac{1 + 0 + 1}{2} = 1, \quad s_x = 1$
Calculate sample variance of $x$ : squared deviations are $1, 0, 1$ ; sum is 2.
$s_y^2 = \frac{1 + 1 + 0}{2} = 1, \quad s_y = 1$
Calculate sample variance of $y$ : squared deviations are $1, 1, 0$ ; sum is 2.
$r = \frac{-1}{(3-1) \cdot 1 \cdot 1} = -\frac{1}{2}$
The negative numerator produces negative $r$ ; this weak negative value is closer to 0 than to $-1$ , showing weak association.

Answer: $r = -\frac{1}{2}$

Here the relationship is weakly negative: as $x$ increases, $y$ tends to decrease, but not consistently. The value $r = -0.5$ reflects this weak trend and lies between 0 (no relationship) and $-1$ (perfect negative relationship).

3. Correlation coefficient for rectangular garden plot fencing and yield

Problem

A farmer plants three rectangular garden plots and records the perimeter of fencing (in meters) and vegetable yield (in kg). Perimeter: 12 m, 20 m, 28 m; Yield: 30 kg, 50 kg, 70 kg. Calculate the correlation coefficient.

$\bar{P} = \frac{12 + 20 + 28}{3} = \frac{60}{3} = 20, \quad \bar{Y} = \frac{30 + 50 + 70}{3} = \frac{150}{3} = 50$
Calculate the mean perimeter and mean yield.
$(P_i - \bar{P}) = -8, 0, 8 \text{ and } (Y_i - \bar{Y}) = -20, 0, 20$
Find deviations for each plot by subtracting the mean from each measurement.
$(P_i - \bar{P})(Y_i - \bar{Y}) = 160, 0, 160 \implies \sum = 320$
Multiply deviations: $(-8)(-20) = 160$ , $(0)(0) = 0$ , $(8)(20) = 160$ ; all non-zero products are positive, strengthening the positive correlation.
$s_P^2 = \frac{64 + 0 + 64}{2} = \frac{128}{2} = 64, \quad s_P = 8$
Calculate sample variance of perimeter: squared deviations $(-8)^2, 0^2, 8^2$ sum to 128; variance is $128 \div 2 = 64$ .
$s_Y^2 = \frac{400 + 0 + 400}{2} = \frac{800}{2} = 400, \quad s_Y = 20$
Calculate sample variance of yield: squared deviations $(-20)^2, 0^2, 20^2$ sum to 800; variance is $800 \div 2 = 400$ .
$r = \frac{320}{(3-1) \cdot 8 \cdot 20} = \frac{320}{2 \cdot 160} = \frac{320}{320} = 1$
Substitute into the formula: numerator is 320, denominator is $(n-1) \cdot s_P \cdot s_Y = 2 \times 8 \times 20 = 320$ .

Answer: $r = 1$

This word problem illustrates a perfect positive linear relationship in a real context: each 8-meter increase in fencing corresponds to exactly 20 kg more yield. Perfect correlations ( $r = 1$ or $r = -1$ ) are rare in real data, but this example shows how the formula captures when one variable perfectly predicts another.

Common mistakes

Where Correlation Coefficient usually goes wrong

The mistake	Why it is wrong	The fix
Calculate the numerator as $\sum x_i y_i$ (raw products of the original data) instead of $\sum (x_i - \bar{x})(y_i - \bar{y})$ (products of deviations from the means).	The formula requires deviations from the means; using raw products $x_i y_i$ ignores where the data center is and produces a value that depends on the scale of the numbers rather than the strength of the linear association.	Always subtract the mean first: compute $(x_i - \bar{x})$ and $(y_i - \bar{y})$ for each pair, then multiply these deviations together and sum the products.
Calculate standard deviations using the population formula $s_x = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}}$ instead of the sample formula $s_x = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$ with $(n-1)$ in the denominator.	The correlation coefficient formula requires the sample standard deviation (dividing by $n-1$ ); using the population formula (dividing by $n$ ) produces an incorrect $r$ value, especially noticeable with small sample sizes.	Always use $(n-1)$ in the denominator when calculating $s_x$ and $s_y$ for correlation: $s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$ , because the correlation coefficient is a sample statistic.
Interpret $r = 0.8$ to mean that '80% of the relationship is explained' or that a value of $r \geq 0.7$ automatically indicates a 'strong' or 'useful' relationship.	The correlation coefficient $r$ ranges from $-1$ to $1$ and does not represent a percentage; the proportion of variance explained is $r^2$ (the coefficient of determination), and whether a correlation is 'strong' depends on the field, context, and stakes of the problem.	Remember that $r$ is a measure on the $-1$ to $1$ scale, not a percentage. To find the proportion of variance explained, square $r$ to get $r^2$ (e.g., $r = 0.8 \implies r^2 = 0.64 \approx 64%$ ). Always evaluate the practical significance of a correlation within your specific context rather than using a universal threshold.

Tips and when to use something else

Always create a scatter plot before interpreting $r$ ; the correlation coefficient can hide non-linear relationships, outliers, or clusters that would jump out visually.
If $r \approx 0$ , do not assume the variables are unrelated—they may share a strong curved or non-linear relationship; consider using Spearman's rank correlation for ordinal data or fitting a non-linear regression model.
The correlation coefficient is symmetric: $r(x, y) = r(y, x)$ , so swapping which variable is called $x$ and which is called $y$ does not change the result.
A practical rule (context-dependent): $|r| > 0.7$ typically suggests strong linear association, $0.3 < |r| < 0.7$ suggests moderate association, and $|r| < 0.3$ suggests weak linear association—but always check the scatter plot and consider your field's standards.

Related formulas

Coefficient of DeterminationCoefficient of Determination (R²) indicates how much of the variation in your outcome variable is explained by a regression model's predictors.Least Squares Regression LineThe line of best fit that minimizes squared prediction errors, used to predict one variable from another in a linear relationship.Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.Arithmetic MeanArithmetic Mean is the average value found by adding all numbers in a dataset and dividing by how many numbers there are.

Frequently asked questions

Can the correlation coefficient be greater than 1 or less than -1?: No. The correlation coefficient always ranges from $-1$ to $1$ , inclusive. If you calculate a value outside this range, you have made a computational error. $r = 1$ is perfect positive correlation, $r = -1$ is perfect negative correlation, and $r = 0$ indicates no linear relationship.
Why does the correlation formula divide by $(n-1)$ instead of $n$ ?: The correlation coefficient is a sample statistic. Dividing by $(n-1)$ instead of $n$ (Bessel's correction) provides an unbiased estimate of the true population correlation. Using $n$ would underestimate the population value, particularly in small samples.
If I find a high correlation, does that prove one variable causes the other?: No. Correlation shows association (two variables move together) but not causation. High correlation can result from direct causation, reverse causation, a confounding third variable, or even coincidence. Causation requires controlled experimentation, temporal order, and elimination of alternative explanations.
What is the difference between $r$ and $r^2$ ?: $r$ (the correlation coefficient) measures the strength and direction of a linear relationship, ranging from $-1$ to $1$ . $r^2$ (the coefficient of determination) is the square of $r$ and represents the fraction (or percentage) of variance in one variable explained by the linear relationship with the other. For example, $r = 0.8$ means $r^2 = 0.64$ , so the linear model explains 64% of the variation in the response variable.

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Reviewed 2026-09-18