Absorption line detected with a significance of $2.2\sigma$ What do we mean by the following statement?

The SV $\lambda$ 786.46 line is detected with a equivalent width of $W = 22.7\pm10.2$ corresponding to a significance of $2.2\sigma$

How does one calculate $\sigma$?
 A: In this statement $\sigma = 10.2$, the uncertainty in the equivalent width. Overall, the statement means that you have a confidence of 2.2$\sigma$ that the absorption line is present (i.e. the equivalent width is 2.2 times its uncertainty bigger than zero, corresponding to a <5% chance that the absorption dip seen is due to measurement uncertainty). This confidence level is only true if the experiment was expressly designed to look for an absorption line at that particular wavelength.
However, how you "calculate" $\sigma$ depends on the data you have, the shape of the line, whether it is in absorption or emission.
A useful expression attributable to Cayrel (1988)  http://adsabs.harvard.edu/abs/1988IAUS..132..345C
is 
$$\Delta W = \sigma \simeq 1.5 \frac{(r p)^{0.5}}{S}\, ,$$
where $r$ is the FWHM of a (assumed) Gaussian absorption line (in wavelength units), $p$ is the pixel size (in the same wavelength units) and $S$ is the signal-to-noise ratio per pixel. $S$ can either be calculated using the number of detected photons and characteristics of the detector (e.g. readout noise), or could be estimated by fitting a featureless part of the same spectrum with a smooth function and taking the residuals to this fit to represent the noise.
This can only be an approximation - it assumes the line is Gaussian and the signal-to-noise ratio is the same for all pixels.
