# Correlation Coefficient Calculator with Examples

## Testing Significance of Linear Relationship

A test of significance for a linear relationship between the variables $x$ and $y$ can be performed using the sample correlation coefficient $r_{xy}$.

## Testing Correlation Coefficient rho = 0

Use Correlation Coefficient calculator for testing significance of correlation coefficient.

T test Calculator for testing correlation coefficient
Data 1 : X Data 2 : Y
Enter Data (Separated by comma ,)
Level of Significance ($\alpha$)
Tail Left tailed
Right tailed
Two tailed
Results
Number of Observations (n):
Pearson's Coefficient of Correlation: $r$
Test Statistic: $t$
Degrees of Freedom: $df$
Critical value(s):
P-value :

## How to use correlation coefficient calculater for testing significance?

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

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

Step 3 - Enter the level of significance $\alpha$

Step 4 - Select the type of alternative hypothesis (Two-tailed/right-tailed/left-tailed)

Step 5 - Click on Calculate button to get the results

Step 6 - Gives the output as number of pairs $n$

Step 7 - Gives the output as Pearson's correlation coefficient

Step 8 - Gives the output of the t test statistic

Step 9 - Gives the output of the degrees of freedom,

Step 10 - Calculate critical value for correlation coefficient and p-value

## Example -1 Correlation Coefficient Calculator

For a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears. The Sample correlation coefficient between the chest size and weight of bears is $r=0.744$ for $n=8$ bears. Using $\alpha=0.05$, determine if there is a positive linear correlation between chest size and weight.

#### Solution

Given that $n = 8$ pair of observations, sample correlation coefficient is $r= 0.744$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \rho = 0$ against $H_1 : \rho > 0$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{r}{\sqrt{1-r^2}}\sqrt{n-2} \end{aligned}

which follows $t$ distribution with $n-2$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{right-tailed}$, the critical value of $t$ $\text{is}$ $1.943$.

The rejection region (i.e. critical region) is $\text{t > 1.943}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{r}{\sqrt{1-r^2}}\sqrt{n-2}\\ &= \frac{0.744}{\sqrt{1-0.744^2}}\sqrt{8 -2}\\ &= 2.727 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =2.727$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision (p-value Approach)

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=2.727$) is p-value = $0.0172$.

The p-value is $0.0172$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

#### Interpretation

There is sufficient evidence to conclude that there is a significant positive linear relationship between chest size and weight of bears.

## Example -2 Linear Correlation Coefficient Calculator

Following is the data about the demand and price of a commodity for 8 periods.

Demand 16 20 18 21 13 15 17 22
Price 10 8 12 6 13 9 11 7

It was expected to estimate a linear regression for demand and price of a commodity.

Test whether there is a significant negative relationship between price and demand of a product.

#### Solution

Let $x$ denote the price of a commodity and $y$ denote the demand of a commodity.

The number of pairs $n= 8$.

$x$ $y$ $x^2$ $y^2$ $xy$
1 10 16 100 256 160
2 8 20 64 400 160
3 12 18 144 324 216
4 6 21 36 441 126
5 13 13 169 169 169
6 9 15 81 225 135
7 11 17 121 289 187
8 7 22 49 484 154
Total 76 142 764 2588 1307

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}{8-1}\bigg(764-\frac{(76)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(764-\frac{5776}{8}\bigg)\\ &= \frac{1}{7}\bigg(764-722\bigg)\\ &= \frac{42}{7}\\ &= 6. \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}{8-1}\bigg(2588-\frac{(142)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(2588-\frac{20164}{8}\bigg)\\ &= \frac{1}{7}\bigg(2588-2520.5\bigg)\\ &= \frac{67.5}{7}\\ &= 9.6429. \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}{8-1}\bigg(1307-\frac{(76)(142)}{8}\bigg)\\ &= \frac{1}{7}\bigg(1307-\frac{10792}{8}\bigg)\\ &= \frac{1}{7}\bigg(1307-1349\bigg)\\ &= \frac{-42}{7}\\ &= -6. \end{aligned}

The Karl Pearson's sample correlation coefficient between price of a commodity and demand of a commodity is

 \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{-6}{\sqrt{6\times 9.6429}}\\ &=\frac{-6}{\sqrt{57.8574}}\\ &=-0.789. \end{aligned}

The correlation coefficient between price of a commodity and demand of a commodity is $-0.789$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \rho = 0$ against $H_1 : \rho < 0$ ($\text{left-tailed}$)

#### Step 2 Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{r}{\sqrt{1-r^2}}\sqrt{n-2} \end{aligned}
which follows t distribution with $n-2$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value for Correlation Coefficient

As the alternative hypothesis is $\text{left-tailed}$, the critical value of $t$ $\text{is}$ $-1.943$.

The rejection region (i.e. critical region) is $\text{t < -1.943}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{r}{\sqrt{1-r^2}}\sqrt{n-2}\\ &= \frac{-0.789}{\sqrt{1--0.789^2}}\sqrt{8 -2}\\ &= -3.146 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =-3.146$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

#### Step 6 Decision (p-value Approach)

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=-3.146$) is p-value = $0.01$.

The p-value is $0.01$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

#### Interpretation

There is sufficient evidence to conclude that there is a significant negative linear relationship between demand and price of a commodity.

## Example -3 Correlation Coefficient Significance Calculator

Following is the data about the exam scores of 10 randomly selected students and the number of hours they studied for the exam.

Hours studied 4 5 6 9 10 8 10 7 3 8 5
Exam score 68 65 85 84 82 86 83 76 67 74 69

Test whether there is a significant correlation between hours studied and examination score. Use $\alpha=0.01$.

#### Solution

Let $x$ denote the hours studied and $y$ denote the exam score.

The number of pairs $n= 11$.

$x$ $y$ $x^2$ $y^2$ $xy$
1 4 68 16 4624 272
2 5 65 25 4225 325
3 6 85 36 7225 510
4 9 84 81 7056 756
5 10 62 100 3844 620
6 8 86 64 7396 688
7 10 83 100 6889 830
8 7 76 49 5776 532
9 3 67 9 4489 201
10 8 74 64 5476 592
11 5 69 25 4761 345
Total 75 819 569 61761 5671

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}{11-1}\bigg(569-\frac{(75)^2}{11}\bigg)\\ &= \frac{1}{10}\bigg(569-\frac{5625}{11}\bigg)\\ &= \frac{1}{10}\bigg(569-511.3636\bigg)\\ &= \frac{57.6364}{10}\\ &= 5.7636. \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}{11-1}\bigg(61761-\frac{(819)^2}{11}\bigg)\\ &= \frac{1}{10}\bigg(61761-\frac{670761}{11}\bigg)\\ &= \frac{1}{10}\bigg(61761-60978.2727\bigg)\\ &= \frac{782.7273}{10}\\ &= 78.2727. \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}{11-1}\bigg(5671-\frac{(75)(819)}{11}\bigg)\\ &= \frac{1}{10}\bigg(5671-\frac{61425}{11}\bigg)\\ &= \frac{1}{10}\bigg(5671-5584.0909\bigg)\\ &= \frac{86.9091}{10}\\ &= 8.6909. \end{aligned}
The Karl Pearson's sample correlation coefficient between hours studied and exam score is

 \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{8.6909}{\sqrt{5.7636\times 78.2727}}\\ &=\frac{8.6909}{\sqrt{451.1325}}\\ &=0.409. \end{aligned}
The sample correlation coefficient between hours studied and exam score is $0.409$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \rho = 0$ against $H_1 : \rho \neq 0$ ($\text{two-tailed}$)

#### Step 2 Test Statistic

The test statistic is
 \begin{aligned} t& =\frac{r}{\sqrt{1-r^2}}\sqrt{n-2} \end{aligned}
which follows t distribution with $n-2$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.01$.

#### Step 4 Critical Value for Correlation Coefficient Calculation

As the alternative hypothesis is $\text{two-tailed}$, the critical value of $t$ $\text{are}$ $-3.25 and 3.25$.

The rejection region (i.e. critical region) is $\text{t < -3.25 or t > 3.25}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{r}{\sqrt{1-r^2}}\sqrt{n-2}\\ &= \frac{0.409}{\sqrt{1-0.409^2}}\sqrt{11 -2}\\ &= 1.345 \end{aligned}

#### Step 6 Decision (Traditional Approach)

The test statistic is $t =1.345$ which falls $\text{outside}$ the critical region, we $\text{fail to reject}$ the null hypothesis.

OR

#### Step 6 Decision (p-value Approach)

This is a $\text{two-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=1.345$) is p-value = $0.2117$.

The p-value is $0.2117$ which is $\text{greater than}$ the significance level of $\alpha = 0.01$, we $\text{fail to reject}$ the null hypothesis.

#### Interpretation

There is insufficient evidence to conclude that there is a significant linear relationship between hours studied and examination score because the correlation coefficient between $x$ and $y$ is not significantly different from zero.

## Conclusion

In this tutorial, you learned about the step by step procedure for testing significance of correlation coefficient. You also learned about how to solve numerical problems for testing significance of correlation coefficient.