6.3  Testing for Partial Correlation
6.3  Testing for Partial CorrelationWhen discussing ordinary correlations we looked at tests for the null hypothesis that the ordinary correlation is equal to zero, against the alternative that it is not equal to zero. If that null hypothesis is rejected, then we look at confidence intervals for the ordinary correlation. Similar objectives can be considered for the partial correlation.
First, consider testing the null hypothesis that a partial correlation is equal to zero against the alternative that it is not equal to zero. This is expressed below:
\(H_0\colon \rho_{jk\textbf{.x}}=0\) against \(H_a\colon \rho_{jk\textbf{.x}}\ne 0\)
Here we will use a test statistic that is similar to the one we used for an ordinary correlation. This test statistic is shown below:
\(t = r_{jk\textbf{.x}}\sqrt{\frac{n2k}{1r^2_{jk\textbf{.x}}}}\) \(\dot{\sim}\) \(t_{n2k}\)
The only difference between this and the previous one is what appears in the numerator of the radical. Before we just took n  2. Here we take n  2  k, where k is the number of variables upon which we are conditioning. In our Adult Intelligence data, we conditioned on two variables so k would be equal to 2 in this case.
Under the null hypothesis, this test statistic will be approximately tdistributed, also with n  2  k degrees of freedom.
We would reject \(H_{o}\colon\) if the absolute value of the test statistic exceeded the critical value from the ttable evaluated at \(α\) over 2:
\(t > t_{n2k, \alpha/2}\)
Example 63: Wechsler Adult Intelligence Data
For the Wechsler Adult Intelligence Data we found a partial correlation of 0.711879, which we enter into the expression for the test statistic as shown below:
\(t = 0.711879 \sqrt{\dfrac{3722}{10.711879^2}}=5.82\)
The sample size is 37, along with the 2 variables upon which we are conditioning is also substituted in. Carry out the math and we get a test statistic of 5.82 as shown above.
Here we want to compare this value to a tdistribution with 33 degrees of freedom for an α = 0.01 level test. Therefore, we are going to look at the critical value for 0.005 in the table (because 33 does not appear to use the closest df that does not exceed 33 which is 30). In this case it is 2.75, meaning that \(t _ { ( d f , 1  \alpha / 2 ) } = t _ { ( 33,0.995 ) } \) is 2.75.
Because \(5.82 > 2.75 = t _ { ( 33,0.995 ) }\), we can reject the null hypothesis, \(H_{o}\) at the \(\alpha = 0.01\) level and conclude that there is a significant partial correlation between these two variables. In particular, we would include that this partial correlation is positive indicating that even after taking into account Arithmetic and Picture Completion, there is a positive association between Information and Similarities.
Confidence Interval for the partial correlation, \(\rho_{jk\textbf{.x}}\)
The procedure here is very similar to the procedure we used for ordinary correlation.
Steps

Compute the Fisher's transformation of the partial correlation using the same formula as before.
\(z_{jk} = \dfrac{1}{2}\log \left( \dfrac{1+r_{jk\textbf{.X}}}{1r_{jk\textbf{.X}}}\right) \)
In this case, for a large n, this Fisher transform variable will be possibly normally distributed. The mean is equal to the Fisher transform for the population value for this partial correlation, and variance equal to 1 over n3k.
\(z_{jk}\) \(\dot{\sim}\) \(N \left( \dfrac{1}{2}\log \dfrac{1+\rho_{jk\textbf{.X}}}{1\rho_{jk\textbf{.X}}}, \dfrac{1}{n3k}\right)\)

Compute a \((1  α) × 100\%\) confidence interval for the Fisher transform correlation. This expression is shown below:
\( \dfrac{1}{2}\log \dfrac{1+\rho_{jk\textbf{.X}}}{1\rho_{jk\textbf{.X}}}\)
This yields the bounds \(Z_{l}\) and \(Z_{u}\) as before.
\(\left(\underset{Z_l}{\underbrace{Z_{jk}\dfrac{Z_{\alpha/2}}{\sqrt{n3k}}}}, \underset{Z_U}{\underbrace{Z_{jk}+\dfrac{Z_{\alpha/2}}{\sqrt{n3k}}}}\right)\)

Back transform to obtain the desired confidence interval for the partial correlation  \(\rho_{jk\textbf{.X}}\)
\(\left(\dfrac{e^{2Z_l}1}{e^{2Z_l}+1}, \dfrac{e^{2Z_U}1}{e^{2Z_U}+1}\right)\)
Example 63: Wechsler Adult Intelligence Data (Steps Shown)
The confidence interval is calculated substituting in the results from the Wechsler Adult Intelligence Data into the appropriate steps below:
Step 1: Compute the Fisher transform:
\begin{align} Z_{12} &= \dfrac{1}{2}\log \frac{1+r_{12.34}}{1r_{12.34}}\\[5pt] &= \dfrac{1}{2} \log \frac{1+0.711879}{10.711879}\\[5pt] &= 0.89098 \end{align}
Step 2: Compute the 95% confidence interval for \( \frac{1}{2}\log \frac{1+\rho_{12.34}}{1\rho_{12.34}}\) :
\begin{align} Z_l &= Z_{12}Z_{0.025}/\sqrt{n3k}\\[5pt] & = 0.89098  \dfrac{1.96}{\sqrt{3732}}\\[5pt] &= 0.5445 \end{align}
\begin{align} Z_U &= Z_{12}+Z_{0.025}/\sqrt{n3k}\\[5pt] &= 0.89098 + \dfrac{1.96}{\sqrt{3732}} \\[5pt] &= 1.2375 \end{align}
Step 3: Backtransform to obtain the 95% confidence interval for \(\rho_{12.34}\) :
\(\left(\dfrac{\exp\{2Z_l\}1}{\exp\{2Z_l\}+1}, \dfrac{\exp\{2Z_U\}1}{\exp\{2Z_U\}+1}\right)\)
\(\left(\dfrac{\exp\{2\times 0.5445\}1}{\exp\{2\times 0.5445\}+1}, \dfrac{\exp\{2\times 1.2375\}1}{\exp\{2\times 1.2375\}+1}\right)\)
\((0.4964, 0.8447)\)
Based on this result, we can conclude that we are 95% confident that the interval (0.4964, 0.8447) contains the partial correlation between Information and Similarities scores given scores on Arithmetic and Picture Completion.