Testing homogeneity of two correlation coefficient calculator
Use this calculator to test the homogeneity of two correlation coefficients using $Z$transformation.
Homogeneity of two corr coeff  

Sample 1  Sample 2  
Number of observations  
Correlation Coefficient  
Level of Significance ($\alpha$)  
Tail  Left tailed Right tailed Two tailed 

Results  
Z1:  
Z2:  
Zstatistic:  
ZCritical value(s):  
pvalue:  
How to test homogeneity of two correlation coefficient?
Step 1  Enter the number of pairs $n_1$ and $n_2$
Step 2  Enter the correlation coefficients $r_1$ and $r_2$
Step 3  Enter the level of significance $\alpha$
Step 4  Select the type of alternative hypothesis
Step 5  Click calculate button to calculate test statistic
Step 6  Gives the $z$ transformation values
Step 7  Gives the $Z$ test statistic
Step 8  Gives the $Z$ critical value(s)
Step 9  Gives the $p$ value of the homogeneity test
Testing homogeneity of correlation coefficient Example 1
The correlation between the mathematics scores and statistics scores of group of 50 male students is 0.358 and the correlation between the mathematics scores and statistics scores of groups of 40 female students is 0.569.
Is the relationship between mathematics scores and statistics scores for male students is weaker than for female students?
Solution
Let $\rho_1$ be the population correlation coefficient between Mathematics score and Statistics score for males and $\rho_2$ be the population correlation coefficient between Mathematics score and Statistics score for females.
Let $X$ denote the Mathematics score and $Y$ denote the Statistics score. Given that for sample of $n_1 =50$ (males) the sample correlation between $X$ and $Y$ is $0.358$ and for sample of $n_2 =40$ (female) the sample correlation between $X$ and $Y$ is $0.669$.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \rho_1 = \rho_2$ against $H_1 : \rho_1 < \rho_2$ ($\text{lefttailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} Z&=\dfrac{U_1U_2}{\sqrt{\frac{1}{n_13}+\frac{1}{n_23}}} \end{aligned} $$
where
$$ \begin{aligned} U_1&=\frac{1}{2}\log_e \bigg(\frac{1+r_1}{1r_1}\bigg) \end{aligned} $$
and
$$ \begin{aligned} U_2 & =\frac{1}{2}\log_e \bigg(\frac{1+r_2}{1r_2}\bigg) \end{aligned} $$
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
$$ \begin{aligned} U_1&=\frac{1}{2}\log_e \bigg(\frac{1+r_1}{1r_1}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.358}{10.358}\bigg)\\ &=0.5\times \log_e\big(2.1153\big)\\ &=0.5\times 0.7492\\ &= 0.3746 \end{aligned} $$
and
$$ \begin{aligned} U_2&=\frac{1}{2}\log_e \bigg(\frac{1+r_2}{1r_2}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.669}{10.669}\bigg)\\ &=0.5\times \log_e\big(5.0423\big)\\ &=0.5\times 1.6179\\ &= 0.8089 \end{aligned} $$
The test statistic under the null hypothesis is
$$ \begin{aligned} Z&=\dfrac{U_1U_2}{\sqrt{\frac{1}{n_13}+\frac{1}{n_23}}}\\ &=\dfrac{0.37460.8089}{\sqrt{\frac{1}{503}+\frac{1}{403}}}\\ &=\dfrac{0.4343}{\sqrt{\frac{1}{47}+\frac{1}{37}}}\\ &=\dfrac{0.4343}{0.2198}\\ &=1.9762 \end{aligned} $$
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =1.9762$ 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 $\text{left}$ of the test statistic ($Z_{obs}=1.9762$) is pvalue = $0.0241$.
The pvalue is $0.0241$ 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
The data provide enough evidence to conclude that the relationship between mathematics scores and statistics scores in male students is weaker than that of female students.
Testing homogeneity of correlation coefficient Example 2
The correlation coefficient between weight and heart rate of 100 males is 0.3 and the correlation coefficient between weight and heart rate of 100 females is 0.5. Test whether there is significant difference between the two correlation coefficients at $\alpha=0.05$.
Solution
Let $\rho_1$ be the population correlation coefficient between weight and heart rate for males and $\rho_2$ be the population correlation coefficient between weight and heart rate for females.
Let $X$ denote the weight and $Y$ denote the heart rate. Given that for sample of $n_1 =50$ (males) the sample correlation between $X$ and $Y$ is $0.3$ and for sample of $n_2 =60$ (female) the sample correlation between $X$ and $Y$ is $0.5$.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \rho_1 = \rho_2$ against $H_1 : \rho_1 \neq \rho_2$ ($\text{twotailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} Z&=\dfrac{U_1U_2}{\sqrt{\frac{1}{n_13}+\frac{1}{n_23}}} \end{aligned} $$
where
$$ \begin{aligned} U_1&=\frac{1}{2}\log_e \bigg(\frac{1+r_1}{1r_1}\bigg) \end{aligned} $$
and
$$ \begin{aligned} U_2 & =\frac{1}{2}\log_e \bigg(\frac{1+r_2}{1r_2}\bigg) \end{aligned} $$
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 $\text{twotailed}$, the critical value of $Z$ $\text{are}$ $1.96 and 1.96$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < 1.96 or Z > 1.96}$.
Step 5 Computation
$$ \begin{aligned} U_1&=\frac{1}{2}\log_e \bigg(\frac{1+r_1}{1r_1}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.3}{10.3}\bigg)\\ &=0.5\times \log_e\big(1.8571\big)\\ &=0.5\times 0.619\\ &= 0.3095 \end{aligned} $$
and
$$ \begin{aligned} U_2&=\frac{1}{2}\log_e \bigg(\frac{1+r_2}{1r_2}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.5}{10.5}\bigg)\\ &=0.5\times \log_e\big(3\big)\\ &=0.5\times 1.0986\\ &= 0.5493 \end{aligned} $$
The test statistic under the null hypothesis is
$$ \begin{aligned} Z&=\dfrac{U_1U_2}{\sqrt{\frac{1}{n_13}+\frac{1}{n_23}}}\\ &=\dfrac{0.30950.5493}{\sqrt{\frac{1}{503}+\frac{1}{603}}}\\ &=\dfrac{0.2398}{\sqrt{\frac{1}{47}+\frac{1}{57}}}\\ &=\dfrac{0.2398}{0.197}\\ &=1.217 \end{aligned} $$
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =1.217$ 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 $\text{twotailed}$ test, so the pvalue is the area to the $\text{extreme}$ of the test statistic ($Z_{obs}=1.217$) is pvalue = $0.2236$.
The pvalue is $0.2236$ 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
The data do not provide enough evidence to conclude that the relationship between weight and heart rate for males is different than that of females. That is the difference between the correlations more likely to be due to chance.
Conclusion
In this tutorial, you learned about the step by step procedure for testing homogeneity of two correlation coefficients. You also learned about how to interpret the results of homogeneity of correlation coefficients.
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 homogeneity of two correlation coefficients and your thought on this article.