Calculator for Testing Correlation Coefficient Examples
 1 Testing Significance of Linear Relationship
 2 Testing Correlation Coefficient rho = 0
 3 How to use calculater for testing significance of correlation coefficient?
 4 Testing Correlation Coefficient Example 1
 5 Testing Correlation Coefficient Example 2
 6 Testing Correlation Coefficient Example 3
 7 Conclusion
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 this 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):  
Pvalue :  
How to use calculater for testing significance of correlation coefficient?
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 (Twotailed/righttailed/lefttailed)
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, critical value(s) and $p$value
Testing Correlation Coefficient Example 1
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 $H_1 : \rho > 0$ ($\text{righttailed}$)
Step 2 Test Statistic
The test statistic is
$$ \begin{aligned} t& =\frac{r}{\sqrt{1r^2}}\sqrt{n2} \end{aligned} $$
which follows $t$ distribution with $n2$ degrees of freedom.
Step 3 Significance Level
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\text{righttailed}$, 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
$$ \begin{aligned} t&=\frac{r}{\sqrt{1r^2}}\sqrt{n2}\\ &= \frac{0.744}{\sqrt{10.744^2}}\sqrt{8 2}\\ &= 2.727 \end{aligned} $$
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 $\text{righttailed}$ test, so the pvalue is the area to the left of the test statistic ($t=2.727$) is pvalue = $0.0172$.
The pvalue 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.
Testing Correlation Coefficient Example 2
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
$$ \begin{aligned} s_{x}^2 & = \frac{1}{n1}\bigg(\sum x^2  \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{81}\bigg(764\frac{(76)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(764\frac{5776}{8}\bigg)\\ &= \frac{1}{7}\bigg(764722\bigg)\\ &= \frac{42}{7}\\ &= 6. \end{aligned} $$
The sample variance of $x$ is
$$ \begin{aligned} s_{y}^2 & = \frac{1}{n1}\bigg(\sum y^2  \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{81}\bigg(2588\frac{(142)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(2588\frac{20164}{8}\bigg)\\ &= \frac{1}{7}\bigg(25882520.5\bigg)\\ &= \frac{67.5}{7}\\ &= 9.6429. \end{aligned} $$
The sample covariance between $x$ and $y$ is
$$ \begin{aligned} s_{xy} & = \frac{1}{n1}\bigg(\sum xy  \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{81}\bigg(1307\frac{(76)(142)}{8}\bigg)\\ &= \frac{1}{7}\bigg(1307\frac{10792}{8}\bigg)\\ &= \frac{1}{7}\bigg(13071349\bigg)\\ &= \frac{42}{7}\\ &= 6. \end{aligned} $$
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 $H_1 : \rho < 0$ ($\text{lefttailed}$)
Step 2 Test Statistic
The test statistic is
$$ \begin{aligned} t& =\frac{r}{\sqrt{1r^2}}\sqrt{n2} \end{aligned} $$
which follows $t$ distribution with $n2$ degrees of freedom.
Step 3 Significance Level
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\text{lefttailed}$, 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
$$ \begin{aligned} t&=\frac{r}{\sqrt{1r^2}}\sqrt{n2}\\ &= \frac{0.789}{\sqrt{10.789^2}}\sqrt{8 2}\\ &= 3.146 \end{aligned} $$
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 $\text{lefttailed}$ test, so the pvalue is the area to the left of the test statistic ($t=3.146$) is pvalue = $0.01$.
The pvalue 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.
Testing Correlation Coefficient Example 3
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  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
$$ \begin{aligned} s_{x}^2 & = \frac{1}{n1}\bigg(\sum x^2  \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{111}\bigg(569\frac{(75)^2}{11}\bigg)\\ &= \frac{1}{10}\bigg(569\frac{5625}{11}\bigg)\\ &= \frac{1}{10}\bigg(569511.3636\bigg)\\ &= \frac{57.6364}{10}\\ &= 5.7636. \end{aligned} $$
The sample variance of $x$ is
$$ \begin{aligned} s_{y}^2 & = \frac{1}{n1}\bigg(\sum y^2  \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{111}\bigg(61761\frac{(819)^2}{11}\bigg)\\ &= \frac{1}{10}\bigg(61761\frac{670761}{11}\bigg)\\ &= \frac{1}{10}\bigg(6176160978.2727\bigg)\\ &= \frac{782.7273}{10}\\ &= 78.2727. \end{aligned} $$
The sample covariance between $x$ and $y$ is
$$ \begin{aligned} s_{xy} & = \frac{1}{n1}\bigg(\sum xy  \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{111}\bigg(5671\frac{(75)(819)}{11}\bigg)\\ &= \frac{1}{10}\bigg(5671\frac{61425}{11}\bigg)\\ &= \frac{1}{10}\bigg(56715584.0909\bigg)\\ &= \frac{86.9091}{10}\\ &= 8.6909. \end{aligned} $$
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 $H_1 : \rho \neq 0$ ($\text{twotailed}$)
Step 2 Test Statistic
The test statistic is
$$ \begin{aligned} t& =\frac{r}{\sqrt{1r^2}}\sqrt{n2} \end{aligned} $$
which follows $t$ distribution with $n2$ degrees of freedom.
Step 3 Significance Level
The significance level is $\alpha = 0.01$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\text{twotailed}$, 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
$$ \begin{aligned} t&=\frac{r}{\sqrt{1r^2}}\sqrt{n2}\\ &= \frac{0.409}{\sqrt{10.409^2}}\sqrt{11 2}\\ &= 1.345 \end{aligned} $$
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 $\text{twotailed}$ test, so the pvalue is the area to the left of the test statistic ($t=1.345$) is pvalue = $0.2117$.
The pvalue 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:
Let me know in the comments if you have any questions on calculator for testing significance of correlation coefficient with examples and your thought on this article.