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At this website:

The passage at the bottom states that a V-band magnitude of 17.62, with an error $\pm$0.02 is a 49.4-$\sigma$ detection significance.

How is this value calculated? Could you provide the working? I have a similar problem, with a set of magnitudes and errors, and would like to know at what magnitude limit I can claim a statistically significant (6-sigma) detection.


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I don't know the answer, but I'm pretty sure that it's not derivable from just the above information. At a quick glance through that Web page I don't think that the required information to derive that significance is there either. In general, to assess the significance of a detection like this, I think you'd need to know something like the background sky brightness. – Ted Bunn Feb 19 '11 at 20:32

I believe those are two different measures. The detection significance is the likelihood that a signal is real; the error is a statement about how precise your measurement is. It's the old accuracy-versus-precision distinction. The detection significance is measured by comparing how strong your signal is to the background noise of the image.

If you want the detection significance, it is calculated as $\frac {N_{p} - N_{b}}{\sigma_{b}}$ where

  • $N_{p}$ is the number of raw counts at the peak of the star's point spread function,
  • $N_{b}$ is the average number of raw counts in the background near the star, and
  • $\sigma_{b}$ is the standard deviation in the raw counts of the background near the star.

You'll need the actual raw image to do this calculation.

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This is correct in spirit, but note that most astronomical imaging systems are oversampled, so that an object (even a point source like a star) is spread over several image pixels. The overall significance of the detection is greater than that implied by the peak counts alone. – coneslayer Mar 4 '11 at 16:42
@Coneslayer yes - but if you wanted to do this right, it's too hard to do by hand. Detection significance is typically spit out by IRAF automatically in my experience. As long as your PSF is relatively tight, this is a good approximation by hand, and it's only an underestimate of significance, which is good. – spencer nelson Mar 4 '11 at 17:25

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