Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.
However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).
But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.
So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.
EDIT: re: updated question.
(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.
(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.
So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.
Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.
These results are the best you can do, there is no way round them.
When talking about abstract quantities, we generally stop talking about "noise" and instead start talking about probability distribution functions (PDFs) and confidence levels.
You might be able to say, for example, that the answer (position) is correct to within 0.1 meters 95% of the time. You need to establish what your correctness criteria are for your application, and analyze how the PDF of the raw data gets transformed into the PDF of the result.
For a more detailed description, do some research into how the results are described for GPS receivers.
Best Answer
-3dB is the point of 50% power. The point when the tipping point changes from a majority of the power to a minority of the power gets through.
The native form of the equation is: $$ SNR=10\log_{10}\left(\frac{P_{SIGNAL}}{P_{NOISE}}\right) $$
Power is proportional the square amplitude (think \$P= I^{2}R\$ or \$P = \frac{V^{2}}{R}\$). Which is why if you use dB to calculate amplitude, you get:
$$ SNR=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right) $$
Changing the coefficient from "10" to "20" applies an extra "2" which works it's way into the exponent through the log function which effectively squares the amplitude to end up with the dB which is supposed to represent power.
So the thing to understand is that dB ALWAYS talks about power. ALWAYS. There is no such thing like dB for amplitude. It's always dB for power. There's no such thing as "it increased in amplitude by 3dB". It's always "it increased in power by 3dB", but you can use that to calculate the equivalent increase in amplitude.
So now, to answer your question:
$$ SNR=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right)-3dB$$
is wrong as far as I can tell. Subtracting 3dB just makes the SNR half the power which makes no sense. If you provided more context maybe we can be more clear. I am also really curious where you saw this if you saw it more than once (especially if you saw it in more than one paper).
EDIT: I think I know what's going on. It's still wrong though.
I think whoever wrote that was thought that the 10log() equation was dB for power, the 20log() equation was dB for amplitude (something that doesn't exist), but knew that SNR in dB was always in terms of power.
And then they tried to come up with a quick shortcut between the dB equations for power and amplitude. Their minus 3dB was supposed to be a shortcut way to undo the 20log() and turn it back into a 10log() without replacing the amplitude ratios with power ratios inside the log function.
However, they messed up because the multiplication by two which turns the 10log() into a 20log() actually represents a power of two since the log() function places it in the exponent. It appears in the equation as a multiplication of two but isn't actually a multiplication of two on a linear scale.
Meanwhile, subtracting 3dB equates to a division of two on a linear scale, which was supposed to undo the previous multiplication by two...except as we established, the previous multiplication by two wasn't actually a multiplication of two on a linear scale so it doesn't actually undo the squaring that took place.
Of course, none of that was necessary since the 10log() and 20log() equations both represent the same thing: dB which is a measurement of power, not amplitude.