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# Pearson Correlation Coefficient Calculator Examples

## Karl Pearson’s Correlation Coefficient

Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations then the Karl Pearson’s coefficient of correlation between two variables $X$ and $Y$ is denoted by $r_{xy}$ or $r$ and is given by

### $r = \dfrac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}}$

where

\begin{aligned} Cov(x,y) =s_{xy}&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})(y_i-\overline{y})\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_iy_i - \frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}\bigg) \end{aligned}

• the sample variance of $x$ is

\begin{aligned} V(x) =s_{x}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_i^2 - \frac{(\sum_{i=1}^n x_i)^2}{n}\bigg) \end{aligned}

• the sample variance of $y$ is

\begin{aligned} V(y) =s_{y}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(y_i -\overline{y})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n y_i^2 - \frac{(\sum_{i=1}^n y_i)^2}{n}\bigg) \end{aligned}

• the sample mean of $x$ is

\begin{aligned} \overline{x}&=\frac{1}{n}\sum_{i=1}^n x_i \end{aligned}

• the sample mean of $y$ is

\begin{aligned} \overline{y}&=\frac{1}{n}\sum_{i=1}^n y_i \end{aligned}

Thus,
\begin{aligned} r_{xy}&=\dfrac{s_{xy}}{s_x\cdot s_y} \end{aligned}

The correlation coefficient $r$ can not exceed unity numerically. i.e. $|r|\leq 1 \implies -1 \leq r \leq +1$.

Two independent variables are uncorrelated. But the converse is not necessarily true.

## Interpretation

• If $r = 0$, then there is no correlation between the ranks.
• If $r > 0$, then there is a positive correlation between the ranks.
• If $r = 1$, then there is a perfect positive correlation between the ranks.
• If $0 < r < 1$, then there is a partially positive correlation between the ranks.
• If $r < 0$, then there is a negative correlation between the ranks.
• If $r = -1$, then there is a perfect negative correlation between the ranks.
• If $-1 < r < 0$, then there is a partially negative correlation between the ranks.

## Karl Pearson’s Correlation Coefficient Calculator

Use this calculator to calculate the Karl Pearson’s correlation coefficient.

Pearson’s Correlation Coefficient Calculator
Data 1 : X Data 2 : Y
Enter Data (Separated by comma ,)
Results
Number of Observations (n):
Variance of X:
Variance of Y:
Covariance between X and Y:
Pearson’s Coefficient of Correlation: $r$
Coefficient of Determination: $r^2$

## How to calculate Pearson’s Correlation Coefficient?

Step 1 – Enter the $X$ values separated by commas

Step 2 – Enter the $Y$ values separated by commas

Step 3 – Click calculate button to calculate correlation coefficient

Step 4 – Gives the number of pairs of observations

Step 5 – Gives the sample variance of $X$

Step 6 – Gives the sample variance of $Y$

Step 7 – Gives the sample covariance between $X$ and $Y$

Step 8 – Gives the sample Pearson’s correlation coefficient and coefficient of determination.

## Pearson’s Correlation Coefficient Example 1

A study was conducted to analyze the relationship between advertising expenditure and sales. The following data were recorded:

X Advertising (in \$) 20 24 30 32 35 Y Sales (in \$) 310 340 400 420 490

Compute the correlation coefficient between advertising expenditure and sales.

#### Solution

Let $x$ denote the advertising expenditure and $y$ denote the sales.

$x$ $y$ $x^2$ $y^2$ $xy$
1 20 310 400 96100 6200
2 24 340 576 115600 8160
3 30 400 900 160000 12000
4 32 420 1024 176400 13440
5 35 490 1225 240100 17150
Total 141 1960 4125 788200 56950

The sample variance of $x$ is

 \begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(4125-\frac{(141)^2}{5}\bigg)\\ &= \frac{1}{4}\bigg(4125-\frac{19881}{5}\bigg)\\ &= \frac{1}{4}\bigg(4125-3976.2\bigg)\\ &= \frac{148.8}{4}\\ &= 37.2. \end{aligned}

The sample variance of $x$ is

 \begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(788200-\frac{(1960)^2}{5}\bigg)\\ &= \frac{1}{4}\bigg(788200-\frac{3841600}{5}\bigg)\\ &= \frac{1}{4}\bigg(788200-768320\bigg)\\ &= \frac{19880}{4}\\ &= 4970. \end{aligned}

The sample covariance between $x$ and $y$ is

 \begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{5-1}\bigg(56950-\frac{(141)(1960)}{5}\bigg)\\ &= \frac{1}{4}\bigg(56950-\frac{276360}{5}\bigg)\\ &= \frac{1}{4}\bigg(56950-55272\bigg)\\ &= \frac{1678}{4}\\ &= 419.5. \end{aligned}

The Karl Pearson’s sample correlation coefficient between advertising expenditure and sales is

 \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{419.5}{\sqrt{37.2\times 4970}}\\ &=\frac{419.5}{\sqrt{184884}}\\ &=0.9756. \end{aligned}

The correlation coefficient between advertising expenditure and sales is $0.9756$. Since the value of correlation coefficient is positive, there is a strong positive relationship between advertising expenditure and sales.

## Pearson’s Correlation Coefficient Example 2

A study of the amount of rainfall and the quantity of air pollution removed produced the following data:

Daily Rainfall (0.01cm) 4.3 4.5 5.9 5.6 6.1 5.2 3.8 2.1 7.5
Particulate Removed ($\mu g/m^3$) 126 121 116 118 114 118 132 141 108

Calculate correlation coefficient between daily rainfall and particulate removed,

#### Solution

Let $x$ denote the daily rainfall (0.01 cm) and $y$ denote the particulate removed ($\mu g/m^3$).

Let $x$ denote the daily rainfall and $y$ denote the particulate removed.

$x$ $y$ $x^2$ $y^2$ $xy$
1 4.3 126 18.49 15876 541.8
2 4.5 121 20.25 14641 544.5
3 5.9 116 34.81 13456 684.4
4 5.6 118 31.36 13924 660.8
5 6.1 114 37.21 12996 695.4
6 5.2 118 27.04 13924 613.6
7 3.8 132 14.44 17424 501.6
8 2.1 141 4.41 19881 296.1
9 7.5 108 56.25 11664 810.0
Total 45.0 1094 244.26 133786 5348.2

The sample variance of $x$ is

 \begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(244.26-\frac{(45)^2}{9}\bigg)\\ &= \frac{1}{8}\bigg(244.26-\frac{2025}{9}\bigg)\\ &= \frac{1}{8}\bigg(244.26-225\bigg)\\ &= \frac{19.26}{8}\\ &= 2.4075. \end{aligned}

The sample variance of $x$ is

 \begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(133786-\frac{(1094)^2}{9}\bigg)\\ &= \frac{1}{8}\bigg(133786-\frac{1196836}{9}\bigg)\\ &= \frac{1}{8}\bigg(133786-132981.7778\bigg)\\ &= \frac{804.2222}{8}\\ &= 100.5278. \end{aligned}

The sample covariance between $x$ and $y$ is

 \begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{9-1}\bigg(5348.2-\frac{(45)(1094)}{9}\bigg)\\ &= \frac{1}{8}\bigg(5348.2-\frac{49230}{9}\bigg)\\ &= \frac{1}{8}\bigg(5348.2-5470\bigg)\\ &= \frac{-121.8}{8}\\ &= -15.225. \end{aligned}

The Karl Pearson’s sample correlation coefficient between daily rainfall and particulate removed is

 \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{-15.225}{\sqrt{2.4075\times 100.5278}}\\ &=\frac{-15.225}{\sqrt{242.0207}}\\ &=-0.9787. \end{aligned}

The correlation coefficient between daily rainfall and particulate removed is $-0.9787$. Since the value of correlation coefficient is negative, there is a strong negative relationship between daily rainfall and particulate removed.

## Pearson’s Correlation Coefficient Example 3

The number of hours 14 students spent studying for a test and their test on that scores are recorded as follows:

Hours spent ($x$) Test Scores ($Y$)
1 41
0 40
1 39
2 48
2 52
3 47
3 49
5 53
6 65
6 70
5 63
7 80
7 87
8 94

Calculate correlation coefficient between hours spent and test scores.

#### Solution

Let $x$ denote the number of hours hours spent studying and $y$ denote the test scores.

Let $x$ denote the no. of hours spent studying for a test and $y$ denote the test scores.

$x$ $y$ $x^2$ $y^2$ $xy$
1 1 41 1 1681 41
2 0 40 0 1600 0
3 1 39 1 1521 39
4 2 48 4 2304 96
5 2 52 4 2704 104
6 3 47 9 2209 141
7 3 49 9 2401 147
8 5 53 25 2809 265
9 6 65 36 4225 390
10 6 70 36 4900 420
11 5 63 25 3969 315
12 7 80 49 6400 560
13 7 87 49 7569 609
14 8 94 64 8836 752
Total 56 828 312 53128 3879

The sample variance of $x$ is

 \begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{14-1}\bigg(312-\frac{(56)^2}{14}\bigg)\\ &= \frac{1}{13}\bigg(312-\frac{3136}{14}\bigg)\\ &= \frac{1}{13}\bigg(312-224\bigg)\\ &= \frac{88}{13}\\ &= 6.7692. \end{aligned}

The sample variance of $x$ is

 \begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{14-1}\bigg(53128-\frac{(828)^2}{14}\bigg)\\ &= \frac{1}{13}\bigg(53128-\frac{685584}{14}\bigg)\\ &= \frac{1}{13}\bigg(53128-48970.2857\bigg)\\ &= \frac{4157.7143}{13}\\ &= 319.8242. \end{aligned}

The sample covariance between $x$ and $y$ is

 \begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{14-1}\bigg(3879-\frac{(56)(828)}{14}\bigg)\\ &= \frac{1}{13}\bigg(3879-\frac{46368}{14}\bigg)\\ &= \frac{1}{13}\bigg(3879-3312\bigg)\\ &= \frac{567}{13}\\ &= 43.6154. \end{aligned}

The Karl Pearson’s sample correlation coefficient between no. of hours spent studying for a test and test scores is

 \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{43.6154}{\sqrt{6.7692\times 319.8242}}\\ &=\frac{43.6154}{\sqrt{2164.954}}\\ &=0.9374. \end{aligned}

The correlation coefficient between no. of hours spent studying for a test and test scores is $0.9374$. Since the value of correlation coefficient is positive, there is a strong positive relationship between no. of hours spent studying for a test and test scores.

## Conclusion

In this tutorial, you learned about the step by step procedure for calculating Pearson’s correlation coefficient. You also learned about how to interpret the correlation coefficient.