Testing correlation coefficient Exmples 2
Calculator for Testing Correlation Coefficient $\rho = \rho_0$
Use this calculator for testing whether the correlation coefficient equals the specified value, i.e., $H_0:\rho=\rho_0$.
T test Calculator for testing correlation coefficient $\rho=\rho_0$  

Data 1 : X  Data 2 : Y  
Enter Data (Separated by comma ,)  
Value of $\rho_0$)  
Level of Significance ($\alpha$)  
Tail  Left tailed Right tailed Two tailed 

Results  
Number of Observations (n):  
Pearson’s Coefficient of Correlation: $r$  
Test Statistic: $Z$  
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 value of correlation coefficient under null hypothesis $\rho_0$
Step 4 – Enter the level of significance $\alpha$
Step 5 – Select the type of alternative hypothesis (Twotailed/righttailed/lefttailed)
Step 6 – Click on Calculate button to get the results
Step 7 – Gives the output as number of pairs $n$
Step 8 – Gives the output as Pearson’s correlation coefficient
Step 9 – Gives the output of the $t$ test statistic
Step 10 – Gives the output of the degrees of freedom, critical value(s) and $p$value
Testing Correlation Coefficient Example 1
The median records shows that the correlation between the age of the mother and the birth weight of their first child is less than 0.34. A random sample of 8 mother’s age and the birth weight of their first child are as follows:
Age of mother  35  24  28  29  26  30  34  32 

Birth weight of child  2.85  3.50  3.25  3.00  3.25  2.75  2.90  3.00 
Test whether the medical records provide the true information at 5% level of significance.
Solution
Let $x$ denote the age of mother and $y$ denote the birth weight of first child.
The number of pairs $n= 8$.
$x$  $y$  $x^2$  $y^2$  $xy$  

1  35  2.85  1225  8.123  99.75 
2  24  3.50  576  12.250  84.00 
3  28  3.25  784  10.562  91.00 
4  29  3.00  841  9.000  87.00 
5  26  3.25  676  10.562  84.50 
6  30  2.75  900  7.562  82.50 
7  34  2.90  1156  8.410  98.60 
8  32  3.00  1024  9.000  96.00 
Total  238  24.50  7182  75.470  723.35 
The sample variance of $x$ is
$$ \begin{aligned} s_{x}^2 & = \frac{1}{n1}\bigg(\sum x^2  \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{81}\bigg(7182\frac{(238)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(7182\frac{56644}{8}\bigg)\\ &= \frac{1}{7}\bigg(71827080.5\bigg)\\ &= \frac{101.5}{7}\\ &= 14.5. \end{aligned} $$
The sample variance of $x$ is
$$ \begin{aligned} s_{y}^2 & = \frac{1}{n1}\bigg(\sum y^2  \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{81}\bigg(75.47\frac{(24.5)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(75.47\frac{600.25}{8}\bigg)\\ &= \frac{1}{7}\bigg(75.4775.0312\bigg)\\ &= \frac{0.4387}{7}\\ &= 0.0627. \end{aligned} $$
The sample covariance between $x$ and $y$ is
$$ \begin{aligned} s_{xy} & = \frac{1}{n1}\bigg(\sum xy  \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{81}\bigg(723.35\frac{(238)(24.5)}{8}\bigg)\\ &= \frac{1}{7}\bigg(723.35\frac{5831}{8}\bigg)\\ &= \frac{1}{7}\bigg(723.35728.875\bigg)\\ &= \frac{5.525}{7}\\ &= 0.7893. \end{aligned} $$
The Karl Pearson’s sample correlation coefficient between age of mother and birth weight of first child is
$$ \begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{0.7893}{\sqrt{14.5\times 0.0627}}\\ &=\frac{0.7893}{\sqrt{0.9092}}\\ &=0.828. \end{aligned} $$
The correlation coefficient between age of mother and birth weight of first child is $0.828$.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \rho = 0.34$ against $H_1 : \rho < 0.34$ ($\text{lefttailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} Z&=\dfrac{U\xi}{\sqrt{\frac{1}{n3}}} \end{aligned} $$
where
$$ \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1r}\bigg) \end{aligned} $$
and
$$ \begin{aligned} \xi & =\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1\rho_0}\bigg) \end{aligned} $$
Under the null hypothesis the test statistic $Z$ follows $N(0,1)$ distribution.
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 $Z$ $\text{is}$ $1.64$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < 1.64}$.
Step 5 Computation
$$ \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1r}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+(0.828)}{1(0.828)}\bigg)\\ &=0.5\times \log_e\big(0.0941\big)\\ &=0.5\times 2.3635\\ &= 1.1817 \end{aligned} $$
and
$$ \begin{aligned} \xi&=\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1\rho_0}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+(0.34)}{1(0.34)}\bigg)\\ &=0.5\times \log_e\big(0.4925\big)\\ &=0.5\times 0.7082\\ &= 0.3541 \end{aligned} $$
The test statistic under the null hypothesis is
$$ \begin{aligned} Z&=\dfrac{U\xi}{\sqrt{\frac{1}{n3}}}\\ &=\dfrac{1.1817(0.3541)}{\sqrt{\frac{1}{83}}}\\ &=\dfrac{0.8276}{\sqrt{\frac{1}{5}}}\\ &=1.8507 \end{aligned} $$
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =1.851$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.
OR
Step 6 Decision ($p$value Approach)
This is a $\text{lefttailed}$ test, so the pvalue is the area to the $\text{negative}$ of the test statistic ($Z_{obs}=1.851$) is pvalue = $0.0321$.
The pvalue is $0.0321$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis at $\alpha =0.05$ level of significance.
Interpretation
There is enough evidence to conclude that the medical records provide true information at $0.05$ level of significance.
Testing Correlation Coefficient Example 2
The correlation between scores on a traditional aptitude test and scores on a final test is known to be approximately 0.6. A new aptitude test has been developed and is tried on a random sample of 100 students, resulting in a correlation of 0.65. Does this result imply that the new test is better?
Solution
Given that the sample correlation between $X$ and $Y$ is $0.65$ for a sample of $100$ pair of observations.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \rho = 0.6$ against $H_1 : \rho > 0.6$ ($\text{righttailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} Z&=\dfrac{U\xi}{\sqrt{\frac{1}{n3}}} \end{aligned} $$
where
$$ \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1r}\bigg) \end{aligned} $$
and
$$ \begin{aligned} \xi & =\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1\rho_0}\bigg) \end{aligned} $$
Under the null hypothesis the test statistic $Z$ follows $N(0,1)$ distribution.
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 $Z$ $\text{is}$ $1.64$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z > 1.64}$.
Step 5 Computation
$$ \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1r}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.65}{10.65}\bigg)\\ &=0.5\times \log_e\big(4.7143\big)\\ &=0.5\times 1.5506\\ &= 0.7753 \end{aligned} $$
and
$$ \begin{aligned} \xi&=\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1\rho_0}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.6}{10.6}\bigg)\\ &=0.5\times \log_e\big(4\big)\\ &=0.5\times 1.3863\\ &= 0.6931 \end{aligned} $$
The test statistic under the null hypothesis is
$$ \begin{aligned} Z&=\dfrac{U\xi}{\sqrt{\frac{1}{n3}}}\\ &=\dfrac{0.77530.6931}{\sqrt{\frac{1}{1003}}}\\ &=\dfrac{0.0822}{\sqrt{\frac{1}{97}}}\\ &=0.8091 \end{aligned} $$
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =0.809$ which falls $\text{outside}$ the critical region, we $\text{fail to reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.
OR
Step 6 Decision ($p$value Approach)
This is a $\text{righttailed}$ test, so the pvalue is the area to the $\text{right}$ of the test statistic ($Z_{obs}=0.809$) is pvalue = $0.2092$.
The pvalue is $0.2092$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis at $\alpha =0.05$ level of significance.
Interpretation
There is insufficient evidence to conclude that the new test is better.
Conclusion
In this tutorial, you learned about the step by step procedure for testing the correlation coefficients, i.e., $H_0:\rho=\rho_0$. You also learned about how to interpret the results of testing 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 Testing the correlation coefficient using ztransformation and your thought on this article.