Pearson's r

The Pearson product-moment correlation coefficient is a measure of the strength of the linear relationship between two variables. It is referred to as Pearson's correlation or simply as the correlation coefficient. If the relationship between the variables is not linear, then the correlation coefficient does not adequately represent the strength of the relationship between the variables.

The symbol for Pearson's correlation is "$\rho$" when it is measured in the population and "r" when it is measured in a sample. Because we will be dealing almost exclusively with samples, we will use $r$ to represent Pearson's correlation unless otherwise noted.

Pearson's $r$ can range from $-1$ to $1$. An $r$ of $-1$ indicates a perfect negative linear relationship between variables, an $r$ of 0 indicates no linear relationship between variables, and an $r$ of 1 indicates a perfect positive linear relationship between variables. Figure 1 shows a scatter plot for which $r=1$.

With real data, you would not expect to get values of $\mathrm{r}$ of exactly $\mathrm{-1, \, 0,}$ or $\mathrm{1}$. The data for spousal ages shown in Figure 4 and described in the introductory section has an r of 0.97.

The relationship between grip strength and arm strength depicted in Figure 5 (also described in the introductory section) is $\mathrm{0.63}$.

Source: David M. Lane, https://onlinestatbook.com/2/describing_bivariate_data/pearson.html
This work is in the Public Domain.

A basic property of Pearson's $r$ is that its possible range is from $-1$ to $1$.  A correlation of $\mathrm{-1}$ means a perfect negative linear relationship, a correlation of 0 means no linear relationship, and a correlation of $\mathrm{1}$ means a perfect positive linear relationship.

Pearson's correlation is symmetric in the sense that the correlation of $X$ with $Y$ is the same as the correlation of $Y$ with $X$. For example, the correlation of Weight with Height is the same as the correlation of Height with Weight.

A critical property of Pearson's $r$ is that it is unaffected by linear transformations. This means that multiplying a variable by a constant and/or adding a constant does not change the correlation of that variable with other variables. For instance, the correlation of Weight and Height does not depend on whether Height is measured in inches, feet, or even miles. Similarly, adding five points to every student's test score would not change the correlation of the test score with other variables such as GPA.

There are several formulas that can be used to compute Pearson's correlation. Some formulas make more conceptual sense whereas others are easier to actually compute. We are going to begin with a formula that makes more conceptual sense.

We are going to compute the correlation between the variables $X$ and $Y$ shown in Table 1. We begin by computing the mean for $X$ and subtracting this mean from all values of $X$. The new variable is called "$x$". The variable "$y$" is computed similarly. The variables $x$ and $y$ are said to be deviation scores because each score is a deviation from the mean. Notice that the means of $x$ and $y$ are both $0$. Next we create a new column by multiplying $x$ and $y$.

Before proceeding with the calculations, let's consider why the sum of the $x y$ column reveals the relationship between $X$ and $Y$. If there were no relationship between $X$ and $Y$, then positive values of $X$ would be just as likely to be paired with negative values of $y$ as with positive values. This would make negative values of $x y$ as likely as positive values and the sum would be small. On the other hand, consider Table 1 in which high values of $X$ are associated with high values of $Y$ and low values of $X$ are associated with low values of $Y$. You can see that positive values of $x$ are associated with positive values of $y$ and negative values of $x$ are associated with negative values of $y$. In all cases, the product of $x$ and $y$ is positive, resulting in a high total for the $x y$ column. Finally, if there were a negative relationship then positive values of $x$ would be associated with negative values of $y$ and negative values of $x$ would be associated with positive values of $y$. This would lead to negative values for $x y$.

Table 1. Calculation of $r$.

$r=\frac{\sum x y}{\sqrt{\sum x^{2} \sum y^{2}}}$

and therefore

$\mathrm{r}=\frac{30}{\sqrt{(16)(60)}}=\frac{30}{\sqrt{960}}=\frac{30}{30.984}=0.968$

An alternative computational formula that avoids the step of computing deviation scores is:

$\mathrm{r}=\frac{\sum_{\mathrm{XY}-} \frac{\sum \mathrm{x} \sum \mathrm{Y}}{\mathrm{N}}}{\sqrt{\left(\sum \mathrm{X}^{2}-\frac{\left(\sum \mathrm{x}\right)^{2}}{N}\right)} \sqrt{\left(\sum \mathrm{Y}^{2}-\frac{\left(\sum \mathrm{Y}\right)^{2}}{N}\right)}}$