What is a good r-squared value? It depends

Q: Does a higher r-squared always mean a better model?

No. Adding predictors to a model can never lower r², and a single-predictor r² stays high on models with the wrong shape. The square-plot example above has r² = 0.9529 with a residual plot that rules the line out. Judge the model on the residuals first.

Q: Is r-squared the same as r squared?

For a simple linear regression with one predictor, yes: r² is exactly the square of the correlation. The reverse does not work cleanly, because taking the square root of r² loses the sign; you need the slope or the scatterplot to know the direction.

Q: Does a high r-squared mean x causes y?

No. r² measures how well a line describes the data, nothing more. A confounding variable, reverse causation, or coincidence can produce a very high r² with no causal link at all. Only a randomized experiment supports a cause-and-effect conclusion.

By Jude Wallis · Published July 31, 2026

There is no universal threshold. What counts as good depends on the field: 0.30 can be strong for a human outcome and 0.95 poor in a physics lab. R-squared is the proportion of the variation in y explained by the linear relationship with x, and a high value does not prove the model fits.

AP Statistics: Unit 5 (topics 5.4 Residuals, 5.5 Least-Squares Regression). The coefficient of determination appears in Unit 5 topic 5.5 of the Fall 2026 AP Statistics course, where you interpret it from regression output rather than compute it by hand, and topic 5.4 covers the residual plot that decides whether a linear model is appropriate in the first place.

The honest answer: there is no cutoff

Search for a threshold and you will find someone confidently offering one: above 0.7 is strong, above 0.9 is excellent, below 0.3 is useless. Those numbers are conventions from particular fields, repeated until they sounded universal. None of them is a statistical rule, and the AP exam never asks you to classify an $r^2$ as good or bad against a fixed number.

The reason is that $r^2$ compares your model against the variation that happens to be in your data, and the amount of unexplained variation is a fact about the subject, not a grade on your work. Predicting the circumference of a circle from its diameter should give an $r^2$ near 1, because there is no noise to explain. Predicting a person's exam score from how long they slept should not, because dozens of other things move that score.

So the useful question is never "is this a good $r^2$ ". It is "is this $r^2$ high or low compared with what this kind of relationship usually gives, and does the residual plot say the model is even the right shape".

What r-squared actually measures

$r^2$ , the coefficient of determination, is the fraction of the variation in $y$ that the least-squares line on $x$ accounts for. The definition is a ratio of two sums of squares:

r^2 = 1 - \frac{\sum (y_i - \hat{y}_i)^2}{\sum (y_i - \bar{y})^2}

The bottom, often written SST, is the total squared distance of the responses from their own mean $\bar{y}$ : how much $y$ varies when you ignore $x$ entirely. The top, often written SSE, is the total squared residual left over after fitting the line: how much $y$ still varies once $x$ has had its say. Divide, subtract from 1, and you have the share of variation the line removed.

Three consequences follow directly.

$r^2$ runs from 0 to 1, and is usually reported as a percentage. For a simple linear regression it is literally the square of the correlation, so $r = -0.8$ and $r = 0.8$ both give $r^2 = 0.64$ .
$r^2$ has no units. Rescaling either variable leaves it unchanged.
$r^2 = 0.66$ means 66% of the variation in $y$ is explained and 34% is not. It does not mean 66% of the points are on the line, and it does not mean predictions are right 66% of the time.

The interpretation sentence that earns credit names both variables in context: "about 66% of the variation in exam scores is explained by the linear relationship with hours of sleep". The distinction between $r$ and $r^2$ is worked through in r vs r-squared.

Why the answer depends on the field

Every $r^2$ is a signal-to-noise measurement, and the noise floor belongs to the subject.

When the relationship is close to deterministic, as in a physics or chemistry lab, nearly all variation in $y$ is genuine signal. An $r^2$ of 0.95 there is a warning, not a success: the missing 5% probably means a miscalibrated instrument or a curved relationship you fitted with a line.

When the response is a human outcome driven by many causes at once, a single predictor cannot explain most of it even when the effect is completely real. An $r^2$ of 0.30 in that setting can be a substantial, useful finding, and demanding 0.9 would only push you toward models that fit noise.

Two practical consequences:

Compare $r^2$ against models of the same thing, not against a fixed bar. The right benchmark is what other work on the same response variable achieves.
A small $r^2$ does not mean no relationship. A slope can be clearly nonzero while $r^2$ stays low; that combination just says $x$ matters and many other things matter too.

A high r-squared on a model that is wrong

This is the failure mode that matters most, and the one $r^2$ cannot warn you about.

Measure the area of a square plot at side lengths 1 through 8 meters. Area is exactly the square of the side, so the true relationship is a perfect parabola with zero noise. Fit a straight line to those eight points anyway and you get $\hat{y} = -15 + 9x$ with $r^2 = 0.9529$ : by any conventional standard, an excellent fit.

It is not an excellent fit. It is a wrong model with a flattering number. The residuals, in side-length order, are

7,\; 1,\; -3,\; -5,\; -5,\; -3,\; 1,\; 7

Positive at both ends, negative through the middle: a clean U. That pattern is the signature of curvature, and it is decisive evidence the linear model is inappropriate no matter what $r^2$ says. The intercept confirms it independently, since $-15$ predicts a negative area for a square with side 0.

So the rule is: read the residual plot first and $r^2$ second. A high $r^2$ with a patterned residual plot means a bad model that happens to be pointed in roughly the right direction. How to interpret residual plots covers what the shapes mean.

What r-squared does not tell you

Four things it is silent about.

Causation. $r^2 = 0.90$ between two variables says nothing about which one causes the other, or whether a lurking variable causes both. Only a randomized experiment supports a causal claim; see correlation vs causation.
Whether a line is the right model. The square-plot data above is the proof.
Whether outliers or influential points are driving it. One point far out in the $x$ direction can lift or crush $r^2$ on its own. Drag a point around in the influential point interactive and watch it move.
Whether the relationship exists at all outside your range. $r^2$ describes the $x$ values you actually observed, and says nothing about what happens beyond them.

And the mirror-image warning: a low $r^2$ does not license "no relationship". A perfect nonlinear relationship can produce $r^2 = 0$ , which is the case worked out in does a correlation of 0 mean no relationship.

How AP questions actually ask about it

In the Fall 2026 course, $r^2$ appears in Unit 5 alongside residuals (5.4) and least-squares regression (5.5). You will read it off computer output rather than compute it by hand, and the labels vary: on typical regression output it shows as "R-sq" or "R-Sq", sometimes with a second line for "R-sq(adj)" that you do not need.

What the exam asks for is the interpretation, and the scoring is unforgiving about wording. Two habits:

Say variation, not variance or variability of the data in general, and attach it to $y$ : "the proportion of the variation in y explained by the linear relationship with x".
Name the variables in context with units where they help. "96% of the variation is explained" earns less than "about 96% of the variation in fuel use is explained by the linear relationship with vehicle weight".

When a question asks whether a linear model is appropriate, answer from the residual plot, not from $r^2$ . Reading the rest of the output is covered in reading calculator output, and you can generate output for your own data with the regression calculator.

Computing r-squared from the two sums of squares

Eight students report hours of sleep before an exam and their score: $(5, 68)$ , $(6, 71)$ , $(6, 79)$ , $(7, 74)$ , $(7, 83)$ , $(8, 80)$ , $(8, 88)$ , $(9, 85)$ . Find $r^2$ and interpret it. Is it a good value?

The least-squares line is $\hat{y} = 48.1667 + 4.3333x$ , with correlation $r = 0.8117$ .
Find $\bar{y}$ : the scores sum to 628, so $\bar{y} = \frac{628}{8} = 78.5$ .
Compute SST, the total variation in $y$ about its mean: $(-10.5)^2 + (-7.5)^2 + (0.5)^2 + (-4.5)^2 + (4.5)^2 + (1.5)^2 + (9.5)^2 + (6.5)^2 = 342$ .
Compute the residuals as observed minus predicted: $-1.8333$ , $-3.1667$ , $4.8333$ , $-4.5$ , $4.5$ , $-2.8333$ , $5.1667$ , $-2.1667$ .
Compute SSE, the sum of the squared residuals: $116.6667$ .
Apply the definition: $r^2 = 1 - \frac{116.6667}{342} = 1 - 0.3411 = 0.6589$ .
Cross-check by squaring the unrounded correlation, $r = 0.811708$ : $r^2 = 0.6589$ . The two routes agree.
Read it in context: the line removed $342 - 116.6667 = 225.3333$ of the 342 units of squared variation, which is 65.89% of it.

$r^2 = 0.6589$ , so about 65.89% of the variation in exam scores is explained by the linear relationship with hours of sleep, leaving about 34.11% unexplained. Whether that is good has no context-free answer: for a single predictor of a human outcome it is substantial, and for a physical law it would be terrible.

An r-squared of 0.95 on a model that is definitely wrong

The area of a square plot is measured for side lengths 1 through 8 meters, giving $(1, 1)$ , $(2, 4)$ , $(3, 9)$ , $(4, 16)$ , $(5, 25)$ , $(6, 36)$ , $(7, 49)$ , $(8, 64)$ in square meters. Fit a least-squares line, find $r^2$ , and decide whether a linear model is appropriate.

Fit the line: $\hat{y} = -15 + 9x$ , with $r = 0.976187$ .
Square the correlation: $r^2 = (0.976187)^2 = 0.9529$ , about 95.29%. Square the unrounded $r$ here, since squaring a value rounded to four places would give 0.9530 instead.
Confirm it with the sums of squares. The mean area is $\bar{y} = \frac{204}{8} = 25.5$ , giving $\text{SST} = 3570$ .
Predicted areas are $-6, 3, 12, 21, 30, 39, 48, 57$ , so the residuals (observed minus predicted) are $7, 1, -3, -5, -5, -3, 1, 7$ .
Square and add them: $\text{SSE} = 49 + 1 + 9 + 25 + 25 + 9 + 1 + 49 = 168$ .
Then $r^2 = 1 - \frac{168}{3570} = 0.9529$ , matching the square of $r$ .
Now look at the residuals in order: positive, small, negative, negative, negative, negative, small, positive. That U shape is systematic curvature, not random scatter.
Check the intercept as a second red flag: $-15$ square meters is the predicted area of a square with side 0, which is impossible.

$r^2 = 0.9529$ , which looks excellent and is not. The residual plot is a clear U, so the linear model is not appropriate; the true relationship is exactly $y = x^2$ . A high $r^2$ is not evidence that a line fits, which is why the residual plot decides the question.

Frequently asked questions

Is 0.7 a good r-squared value?

It depends on what you are modeling. For a single predictor of a human outcome, 0.7 is high. For a calibration curve in a lab, it is poor enough to suggest an equipment problem. Compare it with other models of the same response, and check the residual plot before calling any value good.

Can r-squared be negative?

Not for a least-squares line with an intercept, which is the only kind on the AP exam. Since $r^2$ is the square of $r$ , it lies between 0 and 1. Software fitting a line through a forced origin can report a negative pseudo r-squared, but that is a different quantity.

Does a higher r-squared always mean a better model?

No. Adding predictors to a model can never lower $r^2$ , and a single-predictor $r^2$ stays high on models with the wrong shape. The square-plot example above has $r^2 = 0.9529$ with a residual plot that rules the line out. Judge the model on the residuals first.

Is r-squared the same as r squared?

For a simple linear regression with one predictor, yes: $r^2$ is exactly the square of the correlation. The reverse does not work cleanly, because taking the square root of $r^2$ loses the sign; you need the slope or the scatterplot to know the direction.

Does a high r-squared mean x causes y?

No. $r^2$ measures how well a line describes the data, nothing more. A confounding variable, reverse causation, or coincidence can produce a very high $r^2$ with no causal link at all. Only a randomized experiment supports a cause-and-effect conclusion.