Radar signature is measured in dBsm, which is a decibel / logarithmic scale. To convert from decibel square metres to normal square metres, you use the formula:
area in square metre = 10 ^ (dBsm / 10)
Then to convert an area in square metres back into dBsm, you use the formula (where the log is using base 10):
area in dBsm = 10 x log(area in square metres)
Due to the nature of logarithms and powers, the more negative a value is in dBsm, the smaller an area it represents.
So according to that chart (which is going to be speculation or from an incomplete simulation keep in mind), the F-117A has an average radar signature of -14dBsm in radar bands A-D, which roughly equals 0.04m^2, and a radar signature of -24dBsm in radar bands E-M, which roughly equals 0.004m^2.
The F-22 meanwhile apparently has an average radar signature in the A-D bands of -20dBsm or 0.01m^2 (4x smaller than the F-117), and an average radar signature in the E-M bands of -30dBsm or 0.001m^2 (again 4x smaller than the F-117).
Thank you for this outstanding explanation in Layman's terms. I think I have an understanding of this now. I suspected that the lower the negative value, the better its stealthiness. Time to put this new knowledge to test on some missions.