Does a correlation of 0 mean no relationship?

Q: Does r = 0 mean the two variables are independent?

No. Independence is much stronger. The ball-toss data have r = 0 while height is a deterministic function of time, so knowing the time tells you the height exactly. Independence implies zero correlation, but zero correlation does not imply independence.

Q: If r = 0, should I still fit a least-squares line?

You can, but it tells you nothing useful. The slope is r · s_y / s_x, which is 0 when r is 0, so the line is flat at y-bar and predicts the same value for every x. Describe the curved pattern instead.

Q: What does r = 0 look like on a residual plot?

With a flat fitted line, the residuals are just y - y-bar, so the residual plot has the same shape as the scatterplot. For the ball data that is an arch running -25, 0, 15, 20, 15, 0, -25, which is exactly the curvature that says a linear model is inappropriate.

By Jude Wallis · Published July 31, 2026

No. A correlation of 0 means no linear relationship, not no relationship. A data set following a perfect symmetric curve can have r exactly 0 while y is completely determined by x. Always look at the scatterplot before concluding that two variables are unrelated.

AP Statistics: Unit 5 (topics 5.1 Graphical Representations Between Two Quantitative Variables, 5.2 Correlation). Making and describing a scatterplot is Unit 5 topic 5.1 of the Fall 2026 AP Statistics course and correlation is topic 5.2, which asks you to interpret r as a measure of linear association only. Unit 5 contains no topic on transforming data to achieve linearity, so a curved pattern is described rather than re-expressed.

r measures one specific thing

The correlation coefficient $r$ measures the direction and strength of the linear association between two quantitative variables. That word is doing all the work, and it is the word people drop when they summarize.

The formula makes the restriction visible:

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

Every term is a product of two standardized deviations. A point above the mean in both variables contributes a positive term, a point above in $x$ but below in $y$ contributes a negative term, and $r$ is essentially the average of those signed products. So $r$ near 0 means the positive and negative contributions cancel out. Cancellation can happen because there is no pattern at all, or because a strong pattern is arranged so that its two halves cancel exactly. The number cannot tell those two situations apart. Your eyes can.

A data set with r = 0 and a perfect relationship

A ball is launched straight upward at 30 meters per second. Idealizing gravity at 10 meters per second squared, its height above the launch point each second is:

Time $t$ (s)	0	1	2	3	4	5	6
Height $h$ (m)	0	25	40	45	40	25	0

Height is exactly determined by time here: every value satisfies $h = 30t - 5t^2$ , equivalently $h = 45 - 5(t - 3)^2$ , with no scatter and no measurement error. If you know the time, you know the height to the meter. That is as strong as a relationship gets.

Run the correlation on these seven points and you get $r = 0$ . Not 0.02, not 0.1. Exactly 0, and therefore $r^2 = 0$ as well. The least-squares line is $\hat{h} = 25$ , a flat line through the mean height, with slope 0.

So here is a data set where $y$ is a perfect function of $x$ and where $r$ reports nothing. Anyone who reads " $r = 0$ , so time and height are unrelated" gets the physics exactly backwards.

Why the arithmetic collapses to zero

The cancellation is easy to see term by term. For the ball data, $\bar{t} = 3$ and $\bar{h} = 25$ . The products $(t_i - \bar{t})(h_i - \bar{h})$ that drive the correlation are:

$t$	$t - 3$	$h - 25$	Product
0	$-3$	$-25$	75
1	$-2$	0	0
2	$-1$	15	$-15$
3	0	20	0
4	1	15	15
5	2	0	0
6	3	$-25$	$-75$

The products come in matched pairs with opposite signs: 75 with $-75$ , and $-15$ with 15. They sum to 0, so the numerator of $r$ is 0 and $r = 0$ regardless of the standard deviations underneath.

That is the general mechanism, and it takes two conditions at once: the $x$ values placed symmetrically about $\bar{x}$ , so every point has a partner the same distance away on the other side, and a $y$ that depends only on that distance. When both hold, the rising half contributes positive products, the falling half contributes equal negative products, and the sum is 0. The ball data satisfy both, which is why the products above pair off so neatly.

A symmetric curve on its own is not enough. Take $x = 0, 1, 2, 3, 6$ , so $\bar{x} = 2.4$ , with $y = -(x - 2.4)^2$ : the parabola's vertex sits exactly at $\bar{x}$ , yet $r = -0.6118$ , because those five $x$ values are not placed symmetrically about 2.4, so no point has a partner an equal distance away on the other side. Respace them symmetrically, as $0, 1.2, 2.4, 3.6, 4.8$ , and $r$ returns to exactly 0.

The relationship is still there, and r can find part of it

Cut the same data at the peak and the correlation reappears immediately.

Only the way up, $t = 0$ to $t = 3$ : $r = 0.9583$ , a strong positive linear association.
Only the way down, $t = 3$ to $t = 6$ : $r = -0.9583$ , equally strong and negative.

Same ball, same seven points. Restricting the range flips $r$ from 0 to nearly $\pm 1$ . This is worth remembering as a second lesson: $r$ describes the data over the range you fed it, and it can change dramatically when that range changes.

It also shows what $r = 0$ over the full range actually reported. It did not say "no association". It said "no net linear trend", which is true: a straight line fitted to all seven points really is flat, because the rise and the fall balance.

The fix: plot first, always

There is no numerical repair for this within the tools of this course. No statistic you will compute on the AP exam distinguishes a formless cloud from a perfect parabola if both have $r = 0$ ; that is why the course puts the scatterplot before the correlation, not after.

Two habits handle it:

Make the scatterplot before you compute anything. If the points trace a curve, $r$ is the wrong summary and you say so instead of quoting it.
Check the residual plot after fitting a line. A clear curve or U shape in the residuals is evidence the linear model is wrong even when the numbers look acceptable. For the ball data, the residuals from the flat line are $-25, 0, 15, 20, 15, 0, -25$ , an unmistakable arch. Reading those plots is covered in how to interpret residual plots.

The same warning runs the other way too: $r$ near $\pm 1$ does not guarantee a line is appropriate, and a high $r^2$ can sit on top of a badly curved fit. That case is worked out in what is a good r-squared value.

What to write on the AP exam

Unit 5 of the Fall 2026 course covers scatterplots (5.1), correlation (5.2), linear regression models (5.3), residuals (5.4), and least-squares regression (5.5). The course has no topic on transforming data to achieve linearity, so when a pattern is curved the expected response is to describe it and to say that a linear model is not appropriate, not to re-express the variables.

Phrasing that earns credit:

"There is no linear association between $x$ and $y$ ( $r = 0$ ), but the scatterplot shows a strong curved association, so $r$ is not an appropriate measure here."

Phrasing that loses credit:

" $r = 0$ , so there is no association between $x$ and $y$ ."

Also keep the independence claim out of your answer. Zero correlation does not mean the variables are independent; the ball data are the counterexample, since knowing the time pins the height exactly. Independence is a probability idea, checked with the rules for independent events, not with $r$ .

Computing r for the ball data and confirming it is zero

Using the seven ball-toss points $(0, 0)$ , $(1, 25)$ , $(2, 40)$ , $(3, 45)$ , $(4, 40)$ , $(5, 25)$ , $(6, 0)$ , compute $r$ and $r^2$ , find the least-squares line, and state what they do and do not tell you.

Find the means: $\bar{t} = \frac{0 + 1 + 2 + 3 + 4 + 5 + 6}{7} = \frac{21}{7} = 3$ and $\bar{h} = \frac{0 + 25 + 40 + 45 + 40 + 25 + 0}{7} = \frac{175}{7} = 25$ .
Find the sample standard deviations: $s_t = 2.1602$ seconds and $s_h = 18.7083$ meters. Both are positive, so neither variable is constant and $r$ is defined.
Sum the cross products $(t_i - 3)(h_i - 25)$ : $75 + 0 + (-15) + 0 + 15 + 0 + (-75) = 0$ .
Since $r$ is that sum divided by $(n - 1)s_t s_h$ , and the sum is 0 while the denominator is positive, $r = 0$ exactly. Then $r^2 = 0$ as well.
Find the slope: $b = r \cdot \frac{s_h}{s_t} = 0 \times \frac{18.7083}{2.1602} = 0$ .
Find the intercept: $a = \bar{h} - b\bar{t} = 25 - 0(3) = 25$ , so the least-squares line is $\hat{h} = 25$ .
Compute the residuals as observed minus predicted: $0 - 25 = -25$ , $25 - 25 = 0$ , $40 - 25 = 15$ , $45 - 25 = 20$ , $40 - 25 = 15$ , $25 - 25 = 0$ , $0 - 25 = -25$ .
Those residuals rise then fall in a clean arch, which is the graphical confirmation that a line is the wrong model, not that the variables are unrelated.

$r = 0$ , $r^2 = 0$ , and the least-squares line is the flat $\hat{h} = 25$ . The correct conclusion is that there is no linear relationship between time and height. The relationship itself is perfect: every height satisfies $h = 30t - 5t^2$ exactly, and the arched residual pattern says so.

Splitting the same data restores a strong correlation

Take only the first four ball-toss points, $(0, 0)$ , $(1, 25)$ , $(2, 40)$ , $(3, 45)$ , then only the last four, $(3, 45)$ , $(4, 40)$ , $(5, 25)$ , $(6, 0)$ . Compute $r$ for each half and explain the contrast with $r = 0$ for all seven.

First half means: $\bar{t} = \frac{0 + 1 + 2 + 3}{4} = 1.5$ and $\bar{h} = \frac{0 + 25 + 40 + 45}{4} = 27.5$ .
First half correlation: $r = 0.9583$ , so $r^2 = 0.9184$ . A line explains about 91.84% of the variation in height over that range.
Second half means: $\bar{t} = \frac{3 + 4 + 5 + 6}{4} = 4.5$ and the same $\bar{h} = \frac{45 + 40 + 25 + 0}{4} = 27.5$ .
Second half correlation: $r = -0.9583$ , with the same $r^2 = 0.9184$ . The strength matches; only the direction flips.
The fitted slopes are $+15$ meters per second going up and $-15$ meters per second coming down, mirror images about the vertex at $t = 3$ .
Over the full flight the rising and falling halves contribute cross products that cancel term for term, which is why all seven points together give $r = 0$ .

$r = 0.9583$ on the way up and $r = -0.9583$ on the way down, against $r = 0$ for the whole flight. The association never disappeared; the full-range correlation is 0 only because a rising half and a falling half of equal strength cancel.

Frequently asked questions

Does r = 0 mean the two variables are independent?

No. Independence is much stronger. The ball-toss data have $r = 0$ while height is a deterministic function of time, so knowing the time tells you the height exactly. Independence implies zero correlation, but zero correlation does not imply independence.

Can a curved relationship really give r exactly 0, not just close to 0?

Yes, when the scatterplot is symmetric about the vertical line $x = \bar{x}$ , meaning the $x$ values sit symmetrically about $\bar{x}$ and $y$ depends only on the distance from it. Then the cross products cancel in matched pairs and $r$ is exactly 0. The ball data are one example. Real data are rarely that tidy, so in practice you see $r$ near 0 with an obvious curve in the plot.

If r = 0, should I still fit a least-squares line?

You can, but it tells you nothing useful. The slope is $r \cdot s_y / s_x$ , which is 0 when $r$ is 0, so the line is flat at $\bar{y}$ and predicts the same value for every $x$ . Describe the curved pattern instead.

What does r = 0 look like on a residual plot?

With a flat fitted line, the residuals are just $y - \bar{y}$ , so the residual plot has the same shape as the scatterplot. For the ball data that is an arch running $-25, 0, 15, 20, 15, 0, -25$ , which is exactly the curvature that says a linear model is inappropriate.

How do I tell a genuine no-relationship r = 0 from a curved one?

Only by looking. A formless cloud of points and a perfect parabola can both return $r = 0$ , and no other statistic in this course separates them. Make the scatterplot first; if it shows a pattern, report the pattern rather than the correlation.