How to find the correlation coefficient r by hand

The z-score formula for r

The correlation coefficient $r$ measures the strength and direction of the linear association between two quantitative variables. Written with z-scores, the formula is short:

The symbol $\sum$ (capital sigma) means "add up over all the data points", and the two z-scores are

where $\bar{x}$ (x-bar) and $\bar{y}$ (y-bar) are the sample means and $s_x$ and $s_y$ are the sample standard deviations. The recipe is three moves: standardize both variables, multiply the pair of z-scores at each point, then average those products using $n - 1$ as the divisor.

If standardizing is new, work through how to find a z-score first, and standard deviation by hand for the $s_x$ and $s_y$ step.

Why this formula makes r unitless and bounded

Units cancel inside every z-score. If $x$ is measured in volunteer shifts, then $x - \bar{x}$ is in shifts and $s_x$ is in shifts, so the ratio is a pure number. The same happens for $y$. Multiplying two pure numbers gives a pure number, so $r$ has no units at all. That is why switching from inches to centimeters, or from dollars to euros, leaves $r$ completely unchanged, a point spelled out in does changing units change r.

The bound comes from a sum you can check. For any sample, $\sum z_x^2 = n - 1$ exactly, because $s_x^2 = \frac{\sum (x - \bar{x})^2}{n-1}$ rearranges to $\sum (x - \bar{x})^2 = (n-1)s_x^2$, and dividing through by $s_x^2$ leaves $\sum z_x^2 = n - 1$. The same holds for $y$.

Now ask how large $\sum z_x z_y$ could possibly be. The products are largest when $z_y$ matches $z_x$ at every point, which gives $\sum z_x^2 = n - 1$ and therefore $r = \frac{n-1}{n-1} = 1$. The most negative case is $z_y = -z_x$ everywhere, giving $r = -1$. Every other arrangement falls between, so $-1 \le r \le 1$, inclusive.

The AP course states the same two facts about $r$ in Unit 5 topic 5.2: it is unit-free, and it always lies between -1 and 1.

Why points in matching quadrants push r positive

Picture crosshairs drawn on the scatterplot at $(\bar{x}, \bar{y})$, splitting the plot into four quadrants. The sign of each z-score tells you which quadrant a point sits in, and the sign of the product tells you which way that point votes.

• Upper right: $z_x > 0$ and $z_y > 0$, so the product is positive.
• Lower left: $z_x < 0$ and $z_y < 0$, so the product is positive again, since a negative times a negative is positive.
• Upper left and lower right: the two z-scores have opposite signs, so the product is negative.

Each point casts a signed vote, and $r$ is the average of those votes. A cloud that piles up in the lower-left and upper-right quadrants produces mostly positive products and a positive $r$. A cloud in the other two quadrants produces mostly negative products and a negative $r$. Points scattered evenly across all four quadrants produce products that cancel, pushing $r$ toward 0.

Size matters as well as sign. A point far from both means has two large z-scores, so its product dominates the sum. A point sitting near the crosshairs contributes almost nothing either way.

The full calculation in a table

An animal shelter recorded, for five weeks, the number of volunteer shifts filled ($x$) and the number of adoption applications received ($y$). The summary statistics are $\bar{x} = 20$ shifts, $s_x = 5$ shifts, $\bar{y} = 70$ applications, and $s_y = 10$ applications; the worked example below derives them.

Read the last column as five votes. Weeks 1 and 5 sit in matching quadrants (both below the means, then both above), so they contribute $+1.40$ and $+1.96$. Week 3 is below both means too, a small positive $+0.04$.

Week 2 has fewer shifts than average but more applications than average, the one dissenting vote at $-0.16$. Week 4 sits exactly at the mean number of applications, so $z_y = 0$ and it contributes nothing.

Add the products and divide by $n - 1 = 4$:

Two checks confirm the table. Both z-score columns sum to 0, which they must, since deviations from a mean always cancel. And $\sum z_x^2 = 1.00 + 0.64 + 0.04 + 0.36 + 1.96 = 4$ while $\sum z_y^2 = 1.96 + 0.04 + 0.04 + 0.00 + 1.96 = 4$, matching the $n - 1 = 4$ rule from the section above.

The practical routes: technology and the slope

On the AP exam you are not expected to grind through that table. The Fall 2026 course says the correlation coefficient is calculated using technology, and a graphing calculator with statistical capability is expected. Entering the two lists and running the linear regression command reports $r$ along with the slope and intercept; reading calculator output shows where each value appears. The regression calculator does the same thing in a browser.

The second route works when you already have a regression line. The slope of the least-squares line is

so rearranging gives

For the shelter data, $b = 0.81 \times \frac{10}{5} = 0.81 \times 2 = 1.62$ applications per shift. Going back the other way, $r = 1.62 \times \frac{5}{10} = 1.62 \times 0.5 = 0.81$, which returns the value from the table. Notice that $r$ and $b$ always share a sign, since $s_x$ and $s_y$ are never negative. More on that line in least-squares regression line.

A third route starts from $r^2$, the coefficient of determination. Here $r^2 = 0.81^2 = 0.6561$, so about 65.61% of the variation in applications is explained by the linear relationship with shifts. Going backward from $r^2$ to $r$ needs a sign from somewhere else, because both 0.81 and -0.81 square to 0.6561; r vs r-squared explains why the sign is lost.

Why r is not resistant to outliers

Every ingredient in $r$ is nonresistant. The AP course classifies the mean as a nonresistant measure of center and the standard deviation as a nonresistant measure of variability, and $r$ is built from both, so one unusual point can move it a long way.

Suppose week 5's applications were entered as 24 instead of 84, a single typing slip. Rerun the calculation: the mean drops to $\bar{y} = \frac{56 + 72 + 68 + 70 + 24}{5} = \frac{290}{5} = 58$ applications, the deviations become -2, 14, 10, 12, and -34, and $\sum (y - \bar{y})^2 = 4 + 196 + 100 + 144 + 1156 = 1600$, so $s_y = \sqrt{1600/4} = \sqrt{400} = 20$ applications.

The new z-scores for $y$ are -0.1, 0.7, 0.5, 0.6, and -1.7. Multiplying by the unchanged $z_x$ column gives products of 0.10, -0.56, -0.10, 0.36, and -2.38, summing to -2.58, so

One changed value flipped $r$ from $+0.81$ to $-0.645$. The last point carries most of the damage: its product alone is -2.38 out of a total of -2.58. Drag a point around in the influential point interactive and watch $r$ swing in real time.

What r does not tell you

A large $r$ answers one narrow question, and students routinely stretch it into claims it cannot support.

• **$r$ only measures linear association.** A perfect U-shaped pattern can produce an $r$ near 0 even though $x$ predicts $y$ exactly. Always look at the scatterplot first.
• **$r$ close to 1 does not mean a line is the right model.** The AP course says this outright in topic 5.2. A gently curved pattern can still return $r = 0.98$, and only the residual plot exposes it, as shown in how to interpret residual plots.
• **$r$ does not establish causation.** A relationship between two variables does not mean changes in one cause changes in the other, the point developed in correlation vs causation.
• **$r$ is symmetric.** Swapping which variable you call $x$ and which you call $y$ leaves $r$ unchanged, because the formula multiplies the two z-scores and multiplication does not care about order. The slope $b$ does change, which is why the two are not interchangeable.

One arithmetic warning: use $n - 1$, not $n$, as the divisor, and use sample standard deviations in the z-scores. Mixing the population divisor into one step and the sample divisor into another produces a value that is close to right and still wrong. Test yourself on the regression practice set.