How to Interpret WMAP's statement they have "nailed down the curvature of space to within 0.4% of "flat" Euclidean"” I find myself slightly frustrated when I'm reading a NASA page and they claim they have 

“nailed down the curvature of space to within 0.4% of "flat"
  Euclidean”

0.4% of what? Mathematically this statement is un-useful as you can extract no information out of it.
This is equivalent to saying the neutrino mass is measured within 0.4% to be zero, an equally meaningless statement. 
What researchers DO report are upper and lower bounds on Neutrino (and other particle) masses such as the title of this paper: 

“On the improvement of cosmological neutrino mass bounds”

which actually tells you something you can quickly jump to a table and absorb the information from it.
In the same spirit the spatial curvature of our universe should be described by a statment like (for a made up example):
“

Our analysis concludes that the spatial radius of curvature of our
  universe $R$ is bounded by:$$\frac{1}{\mid R\mid}\leq9\cdot10^{-46}m^{-1}\pm error$$

How can I get this data from their study? 
(If I remember right R here should correspond to spatial components of the Ricci tensor (at cosmological scales anyway)). 
 A: The statement is not meaningless. The curvature parameter, $\Omega_k$, used in cosmology is a dimensionless parameter. So the statement that it is flat to within 0.4% means that the parameter has been measured to be between -0.004 and +0.004. It is a dimensionless number so the number has meaning on its own and does not need an “of what” qualifier. 
The writers of the press release just (reasonably) decided that a message to the public saying “within 0.4% of flat” would be more generally understandable than “$\Omega_k = -0.0027 ^{+0.0039}_{-0.0038}$”
See here for the actual paper:
https://arxiv.org/abs/1212.5226
A: Since Dale Got me going on the right track I'm going to go ahead and Accept his answer I just wanted to Add something to it and figured another answer would suffice for this.
 According to This Caltech Webpage  We can write the scale factor R for a standard FLRW universe as:
$$R_{0}=\frac{c}{H_{0}}[(\Omega-1)/k]^{-1/2}$$
Where $c/H_0$ is the hubble length, Omega is as Dale described above and k is either plus or minus 1 for a closed or open universe respectively. This is really just a rearrangement of the Friedmann equation. 
In the case of $\Omega=1\pm0.004$ and $H_{0}\approx1.3\cdot10^{26}m$ (or about 14 billion light years) One obtains:
$$\mid R_{0}\mid\geq2\cdot10^{27}m$$
Corresponding to a radius of curvature of more than 200 Billion Light Years!! 
Rather than being being flat the universe could just be Really Really big. There's no reason to suppose that it isn't since It is after all, the universe we are talking about. 
I feel I should note that for positive curvature we have a finite volume for the universe, whereas for negative or zero curvature one gets infinite universes (or some boundary? I'm not going to touch that!) 
I get that observations of accelerating expansion all but rule out a closed universe, but that's outside the scope of this question.
