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.

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.

To learn more about other correlation and regression, please refer to the following tutorials:

Descriptive Statistics

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