"The test statistic is reported as a z..." I was reading a neuroimaging paper in which the authors compare the effect of a chemical compound on a measure of brain connectivity. They measure the value of connectivity for each brain voxel under two conditions: with and without the chemical compound being introduced in the body. The authors continue:

We then computed a one-sample t-test on all the brain voxels. Because
  of the very large sample (> 220,000 voxels), the test statistic is
  reported as a z and effect sizes are included.

I don't really understand what is meant by the phrase "the test statistic is reported as a z". Does it mean that with a large number of t-tests, we can interpret the t-statistic values as z-scores? What is the justification of this process?
Is it possible that they do: t-statistic values -> p values -> z-scores? Is this common?
 A: The with large degrees of freedom, $\nu$, the distribution of a $t_\nu$ is very close to a standard normal.
The text says "a one sample t-test on all the brain voxels", so I assume it's a single t-test across all voxels, with 220000+ df (if it was per-voxel, I'd expect to see "each of" rather than "all"). Even so, it's ambiguous -- if it is one t-test per-voxel, a similar reasoning would apply on the assumption that it's 50 observations per voxel, one per image; many people happily approximate the $t$ by the normal for 49 df.
So while it's still a t-test because the test statistic is a t-statistic, with a t-distribution*, rather than look up a $t_{\nu}$ distribution (for large $\nu$) to compute critical values or p-values, they use z-tables ... which, aside from the extreme tail, will be very close from the exact distribution. 

* as long as the assumptions hold. It's not clear to me why independence would be expected to hold up (under either interpretation above), for example.
I don't think they're converting p-values to z-scores.
A: It is always convinient to use z-scores instead of t-values. A nice transform is goes like this.
$z=\phi^{-1}(F_{sample size}(t))$. Where $\phi$ and $F_{samplesize}$ are the cummulative distribution of the normal and $t_{samplesize}$.
A: It appears from the wording that they probably used Z-test rather than t-test.
