An approximate \((1-\alpha)100\%\) confidence interval for \(\rho\) is \(L\leq \rho \leq U\) where:

\(L=\dfrac{1+R-(1-R)\text{exp}(2z_{\alpha/2}/\sqrt{n-3})}{1+R+(1-R)\text{exp}(2z_{\alpha/2}/\sqrt{n-3})}\)

and

\(U=\dfrac{1+R-(1-R)\text{exp}(-2z_{\alpha/2}/\sqrt{n-3})}{1+R+(1-R)\text{exp}(-2z_{\alpha/2}/\sqrt{n-3})}\)

Proof

We previously learned that:

\(Z=\dfrac{\dfrac{1}{2}ln\dfrac{1+R}{1-R}-\dfrac{1}{2}ln\dfrac{1+\rho}{1-\rho}}{\sqrt{\dfrac{1}{n-3}}}\)

follows at least approximately a standard normal \(N(0,1)\) distribution. So, we can do our usual trick of starting with a probability statement:

\(P\left(-z_{\alpha/2} \leq \dfrac{\dfrac{1}{2}ln\dfrac{1+R}{1-R}-\dfrac{1}{2}ln\dfrac{1+\rho}{1-\rho}}{\sqrt{\dfrac{1}{n-3}}} \leq z_{\alpha/2} \right)\approx 1-\alpha\)

and manipulating the quantity inside the parentheses:

\(-z_{\alpha/2} \leq \dfrac{\dfrac{1}{2}ln\dfrac{1+R}{1-R}-\dfrac{1}{2}ln\dfrac{1+\rho}{1-\rho}}{\sqrt{\dfrac{1}{n-3}}} \leq z_{\alpha/2}\)

to get ..... can you fill in the details?! ..... the formula for a \((1-\alpha)100\%\) confidence interval for \(\rho\):

\(L\leq \rho \leq U\)

where:

\(L=\dfrac{1+R-(1-R)\text{exp}(2z_{\alpha/2}/\sqrt{n-3})}{1+R+(1-R)\text{exp}(2z_{\alpha/2}/\sqrt{n-3})}\) and \(U=\dfrac{1+R-(1-R)\text{exp}(-2z_{\alpha/2}/\sqrt{n-3})}{1+R+(1-R)\text{exp}(-2z_{\alpha/2}/\sqrt{n-3})}\)

as was to be proved!