What is the difference between R^2 and the correlation coefficient r?

The correlation coefficient r measures both strength and direction of a linear relationship and ranges from -1 to 1. R^2 is simply r squared, ranges from 0 to 1, and shows only the proportion of variation explained; the sign (direction) is lost when you square.

Can R^2 be greater than 1 or negative?

No, R^2 must always be between 0 and 1 (or 0% and 100%) when calculated correctly. If you get a value outside this range, you have made a calculation error, such as using the wrong formula or accidentally reversing residual and total sums of squares.

Why should I care about R^2 if I already have a regression equation?

R^2 tells you how much to trust your equation's predictions. A high R^2 (near 1) means the equation fits well and your predictions will be reliable; a low R^2 (near 0) means other factors strongly influence the outcome, so predictions based on the equation alone will be unreliable.

Does a high R^2 guarantee my model is useful?

High R^2 is necessary but not sufficient for a good model. You must also verify that the relationship is truly linear, that residuals are scattered randomly with no pattern, that there are no influential outliers, and that the model makes practical sense in your specific context.

Coefficient of Determination

Coefficient of Determination (R²) indicates how much of the variation in your outcome variable is explained by a regression model's predictors.

R^2 = \text{proportion of variation explained}

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What each symbol means

What Coefficient of Determination takes

Symbol	Meaning
$R$	Represents the correlation coefficient from which R² is derived; it ranges from -1 to 1 and measures the strength and direction of a linear relationship. Confusing R with R² is common—squaring the correlation gives R², which is always between 0 and 1 and represents only the proportion of variation explained, regardless of direction.

When to use it

Use R² when you fit a regression model to measure what proportion of the variation in your outcome is captured by the predictors.

Level

Usually taught in: Statistics

Worked examples

1. Calculating R² for Study Hours vs Test Scores

Problem

A teacher fits the regression line

\hat{y} = 50 + 5x

to model test scores (

y

) based on study hours (

x

) for 5 students. The observed test scores were 61, 68, 82, 84, and 92 when study hours were 2, 4, 6, 7, and 8 respectively. Calculate

R^2

$\hat{y}_1 = 60, \quad \hat{y}_2 = 70, \quad \hat{y}_3 = 80, \quad \hat{y}_4 = 85, \quad \hat{y}_5 = 90$
Substitute each study hour into $\hat{y} = 50 + 5x$ to find the predicted test score for each student.
$e_1 = 61 - 60 = 1, \quad e_2 = 68 - 70 = -2, \quad e_3 = 82 - 80 = 2, \quad e_4 = 84 - 85 = -1, \quad e_5 = 92 - 90 = 2$
Calculate residuals as the difference between observed and predicted scores for each student.
$\text{RSS} = 1^2 + (-2)^2 + 2^2 + (-1)^2 + 2^2 = 1 + 4 + 4 + 1 + 4 = 14$
Square each residual and sum them to find the residual sum of squares.
$\bar{y} = \frac{61 + 68 + 82 + 84 + 92}{5} = \frac{387}{5} = 77.4$
Find the mean of all observed test scores.
$\text{TSS} = (61 - 77.4)^2 + (68 - 77.4)^2 + (82 - 77.4)^2 + (84 - 77.4)^2 + (92 - 77.4)^2 = 268.96 + 88.36 + 21.16 + 43.56 + 213.16 = 635.2$
Calculate the total sum of squares by summing the squared deviations of each score from the mean.
$R^2 = 1 - \frac{14}{635.2} = 1 - 0.0220 = 0.978$
Apply the formula $R^2 = 1 - \frac{\text{RSS}}{\text{TSS}}$ to find the coefficient of determination.

Answer: $R^2 \approx 0.978 \text{ or } 97.8\%$

This high $R^2$ value means that approximately 97.8% of the variation in test scores is explained by study hours. The regression model fits the data very well, showing that study hours are a strong predictor of test performance.

2. Calculating R² for Altitude vs Temperature

Problem

A scientist measures air temperature at different altitudes. The regression line is

\hat{y} = 20 - 6x

where

x

is altitude in km and

y

is temperature in °C. For altitudes 1, 2, 3, and 4 km, the observed temperatures were 12.5, 9.8, 3.2, and -3.5 °C. Calculate

R^2

$\hat{y}_1 = 20 - 6(1) = 14, \quad \hat{y}_2 = 8, \quad \hat{y}_3 = 2, \quad \hat{y}_4 = -4$
Substitute each altitude into the regression equation $\hat{y} = 20 - 6x$ to get the predicted temperature.
$e_1 = 12.5 - 14 = -1.5, \quad e_2 = 9.8 - 8 = 1.8, \quad e_3 = 3.2 - 2 = 1.2, \quad e_4 = -3.5 - (-4) = 0.5$
Calculate residuals as the difference between observed and predicted temperatures at each altitude.
$\text{RSS} = (-1.5)^2 + (1.8)^2 + (1.2)^2 + (0.5)^2 = 2.25 + 3.24 + 1.44 + 0.25 = 7.18$
Square each residual and sum them to get the residual sum of squares.
$\bar{y} = \frac{12.5 + 9.8 + 3.2 + (-3.5)}{4} = \frac{22}{4} = 5.5$
Find the mean of the observed temperatures.
$\text{TSS} = (12.5 - 5.5)^2 + (9.8 - 5.5)^2 + (3.2 - 5.5)^2 + (-3.5 - 5.5)^2 = 49 + 18.49 + 5.29 + 81 = 153.78$
Calculate the total sum of squares as the sum of squared deviations from the mean.
$R^2 = 1 - \frac{7.18}{153.78} = 1 - 0.0467 = 0.953$
Use the formula $R^2 = 1 - \frac{\text{RSS}}{\text{TSS}}$ to find the coefficient of determination.

Answer: $R^2 \approx 0.953 \text{ or } 95.3\%$

An $R^2$ of 0.953 indicates that 95.3% of temperature variation is explained by altitude alone, showing an excellent fit. The remaining 4.7% of variation might come from seasonal effects, weather patterns, or measurement error.

3. Calculating R² for Video Game Level vs Points

Problem

In a video game, a player's total points should grow with their level. The regression model is

\hat{y} = 100 + 50x

where

x

is the level and

y

is total points earned. For levels 1 through 5, the actual points were 145, 195, 305, 395, and 510. Calculate

R^2

$\hat{y}_1 = 150, \quad \hat{y}_2 = 200, \quad \hat{y}_3 = 250, \quad \hat{y}_4 = 300, \quad \hat{y}_5 = 350$
Calculate expected points for each level using $\hat{y} = 100 + 50x$ .
$e_1 = 145 - 150 = -5, \quad e_2 = 195 - 200 = -5, \quad e_3 = 305 - 250 = 55, \quad e_4 = 395 - 300 = 95, \quad e_5 = 510 - 350 = 160$
Find residuals showing how actual points differ from the model's predicted values.
$\text{RSS} = (-5)^2 + (-5)^2 + 55^2 + 95^2 + 160^2 = 25 + 25 + 3025 + 9025 + 25600 = 37700$
Square each residual and sum them.
$\bar{y} = \frac{145 + 195 + 305 + 395 + 510}{5} = \frac{1550}{5} = 310$
Calculate the mean of all observed point values.
$\text{TSS} = (145 - 310)^2 + (195 - 310)^2 + (305 - 310)^2 + (395 - 310)^2 + (510 - 310)^2 = 87700$
Calculate the total sum of squares from all deviations from the mean.
$R^2 = 1 - \frac{37700}{87700} = \frac{50000}{87700} \approx 0.571$
Apply the formula to find $R^2$ .

Answer: $R^2 \approx 0.571 \text{ or } 57.1\%$

An $R^2$ of approximately 0.571 means only about 57% of points variation is explained by level alone. The remaining 43% comes from other factors like bonus multipliers, combo streaks, or difficulty mode, showing the limitations of using level alone to predict points.

Common mistakes

Where Coefficient of Determination usually goes wrong

The mistake	Why it is wrong	The fix
I calculated the correlation coefficient $r = -0.8$ , so $R^2 = -0.8$ .	The correlation coefficient $r$ can be negative because it shows direction, but $R^2$ is always non-negative because it is $r$ squared, and squaring a negative number gives a positive result.	Square the correlation: $R^2 = (-0.8)^2 = 0.64$ , meaning 64% of the variation is explained by the relationship.
$R^2 = 0.85$ means 85% of my predictions were correct.	$R^2$ measures the proportion of variation explained by the model, not the percentage of correct predictions; it describes how well the line fits the data, not how often predictions hit the mark.	Interpret correctly: $R^2 = 0.85$ means the regression line explains 85% of the observed variation in the response variable, but says nothing about prediction accuracy.
My $R^2$ is 0.5, so the model is bad and I should choose a different method.	Whether $R^2 = 0.5$ is good depends entirely on the context and field; in social sciences and psychology, 0.5 is often considered very good, while in physics or engineering, it might indicate a poor fit.	Always interpret $R^2$ in the context of your field: compare to similar published studies, consider the practical importance of the unexplained variation, and consult domain experts.

Tips and when to use something else

$R^2$ always falls between 0 and 1 (or 0% and 100%); if you calculate a value outside this range, you made an arithmetic error, such as swapping RSS and TSS or using the wrong formula.
Never rely on $R^2$ alone—always plot your data and residuals visually, because $R^2$ can be high even when the relationship is actually nonlinear or when a few outliers distort the fit.
When comparing two regression models with different numbers of predictors, use adjusted $R^2$ instead of regular $R^2$ , because adjusted $R^2$ penalizes adding unnecessary predictors and prevents overfitting.
For a simple two-variable relationship, $R^2$ equals the square of Pearson's correlation coefficient: $R^2 = r^2$ , so you can convert between them quickly.

Related formulas

Correlation CoefficientCorrelation coefficient measures the strength and direction of a linear relationship between two quantitative variables, ranging from -1 to 1.Least Squares Regression LineThe line of best fit that minimizes squared prediction errors, used to predict one variable from another in a linear relationship.Sample VarianceSample variance measures how spread out data points are around the mean, using n-1 instead of n to account for sample uncertainty.Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.

Frequently asked questions

What is the difference between $R^2$ and the correlation coefficient $r$ ?: The correlation coefficient $r$ measures both strength and direction of a linear relationship and ranges from -1 to 1. $R^2$ is simply $r$ squared, ranges from 0 to 1, and shows only the proportion of variation explained; the sign (direction) is lost when you square.
Can $R^2$ be greater than 1 or negative?: No, $R^2$ must always be between 0 and 1 (or 0% and 100%) when calculated correctly. If you get a value outside this range, you have made a calculation error, such as using the wrong formula or accidentally reversing residual and total sums of squares.
Why should I care about $R^2$ if I already have a regression equation?: $R^2$ tells you how much to trust your equation's predictions. A high $R^2$ (near 1) means the equation fits well and your predictions will be reliable; a low $R^2$ (near 0) means other factors strongly influence the outcome, so predictions based on the equation alone will be unreliable.
Does a high $R^2$ guarantee my model is useful?: High $R^2$ is necessary but not sufficient for a good model. You must also verify that the relationship is truly linear, that residuals are scattered randomly with no pattern, that there are no influential outliers, and that the model makes practical sense in your specific context.

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Reviewed 2026-09-18