The statistic. With \(\hat\rho\) the vector of correlations between \(v\) and the reduced-form errors,
$$W = \hat\rho' \,[\widehat{\mathrm{Var}}(\hat\rho)]^{-1}\, \hat\rho \;\sim\; \chi^2(q),$$
\(q\) the number of endogenous regressors. With a single endogenous regressor it is exactly the squared z-ratio already printed for rho_ in the fit.
Why this is an ordinary chi-square. \(\rho\) lives in the open unit ball, so \(\rho = 0\) is an interior point of the parameter space. That is the opposite of the situation inefficiency_test faces, where \(\sigma_u = 0\) sits on a boundary and the null distribution is a chi-bar-square mixture. No mixture is needed here.
Standard errors for \(\rho\). ivsfm parameterises the correlations as \(\rho = t/\sqrt{1 + t't}\), which keeps them inside the unit ball for any real \(t\). Before version 1.2.0 the standard errors of \(\rho\) were reported as NA on the ground that this map has a non-diagonal Jacobian. It does, but the Jacobian is short --
$$\partial \rho_i/\partial t_j = \delta_{ij}/s - t_i t_j / s^3, \qquad s = \sqrt{1 + t't},$$
that is \(J = (I - \rho\rho')/s\) -- so \(\mathrm{Var}(\hat\rho) = J\,\mathrm{Var}(\hat t)\,J'\) by the delta method. At the null \(J = I\), which is what makes the test well calibrated there. ivsfm now reports these standard errors and carries the full \(\mathrm{Var}(\hat\rho)\) as $vcov_rho, which is what the joint test needs.
Reading a non-rejection. Failing to reject is not evidence for exogeneity. \(\rho\) is identified through the instruments, so weak instruments make \(\hat\rho\) imprecise and the test powerless; the print method says so rather than letting the p-value speak alone.