How to calculate the statistically significant level for a guassian line profile? In some papers, they fit a line profile with gaussian model. For example, in
this paper, fig 1 draw three gauss line profile, O VII, O VIII, and Ne IX.
But the data seems noisy and the gaussian height (from bottom to the top of the gauss line, especially O VIII, and Ne IX) do not seems above 3*(the height of 1-sigma errorbar) of the residual data. But in the text they say (on the left bottom of page 3): &quot;The centroid wavelengths of three lines are consistent with O VII, O VIII, and Ne IX at zero redshift,respectively. Those lines are statistically significant at the 6.5, 3.1, and 4.5 sigma levels.&quot;. So i'm puzzled with the calculation algorithom of the significance. In my memory that the sigma=sqrt(variance of the residual data) and the height of gauss Intensity=exp(a*(x-b)/c^2), the significance = a/sigma. Then any wrong with my calculation?
It is difficult to say, there isn't a description in the paper as to how they arrive at those numbers. Given that the equivalent widths of the lines have signal to noise levels that mean they are only greater than zero at significance levels of 2-3, then it isn't clear these are meaningful numbers at all.
However, here is how it might have been done.
You fit a continuum model to the spectrum, get a minimum chi-squared $\chi^2_1$ (or maximum [log] likelihood value); you then fit a continuum plus absorption line model and get a new, lower value, of chi-squared $\chi^2_2$ (or higher value of max log likelihood)
You then get the difference in chi-squared $\Delta \chi^2 = \chi^2_1 - \chi^2_2$ (or form the likelihood ratio) of the two models, which should be distributed as chi-squared with the number of degrees of freedom being equal to the number of new parameters in the more advanced model (i.e. 3 for a line centre, width and depth). You then check a chi-squared table for the corresponding p-value and significance.
In other words, if $\Delta \chi^2 = 6.25$, that corresponds to p=0.1, or if $\Delta \chi^2 = 11.35$ for p=0.01 etc.
In your comments you say:

First i use powerlaw to fit the continue get reduced chi-square1 = 1.00277 / dof=179. Then i fit with powerlaw+gauss, the reduce chi-square2 = 0.88091 / dof=176. The Δχ2 = 24.455. Is that right?

Yes. A chisquared test for 3 d.o.f. yields p=0.00002012
To convert to a number of Gaussian sigma, it is the stuff of standard statistics that
$$ n\sigma = \sqrt{2} {\rm erf}^{-1} (1 - p)$$
which in the case above suggests a significance of $4.26\sigma$.