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My boss asked a senior scientist in his network for recommendations about testing whether two Poisson random variables have the same mean. The context is differences in measurements from two radioactive samples. We need to decide whether the samples come from the same source and measurement differences are due to the randomness of the process, or whether they might come from different sources/have been handled differently.

My boss trusts the senior scientist and would like to apply the guidelines that person gave us. But we're having a hard time parsing these guidelines (and it's hard to get back in touch with that person to ask for clarifications). I also want to believe that person though I can't help but wonder whether they might have been in a rush and sent an ad hoc somewhat improvised rule of thumb.

The question was "how close must the count-per-minute be to confidently reject the hypothesis that the processes have the same mean?". To which the senior scientist responded:

"The samples should be within 100%(2sqrt(N)/N+0.05) of each other. This allows for statistical variation (two sigma) and sample pipetting error (5%). Note: N is count-per-minute*count-duration-in-minutes."

There are a number of things we don't understand or are unsure about:

  • The meaning of $N$: Is $N$ supposed to be the sum of all detections across the two samples? The average number of detections across the two samples?
  • Order of operations: is it $[2*\sqrt(N)/(N+0.05)]$ or $([2*\sqrt(N)/N]+0.05)$?
  • Why the extra-term for pipetting error if such errors can be viewed as differentially affecting the count-per-minute which should itself be detected by a "mere" statistical test of mean equality?
  • If the order of operations is $([2*\sqrt(N)/N]+0.05)$, why not report it as $([2/\sqrt(N)]+0.05)$? Could that suggest there's a typo and the $N$ at the bottom of the fraction should be something else? -...

So my question is: Can anyone help us make sense of the above formula? Does it resemble any valid test for equality of the mean of two Poissons you're aware of? Or perhaps can you confirm that it's a kind of rule of thumb?

I found a lot of useful information on testing the equality of the mean of two Poissons, both here (Checking if two Poisson samples have the same mean, Differences between poisson.test and E-test when testing Poisson parameters) and elsewhere (https://www.google.com/books/edition/Sample_Size_Calculations/3uqZ_81If4UC?hl=en&gbpv=1&bsq=Poisson 5.1). But the closest I've been to matching the above formula with a proper test is

  • Trying to match the normal approximate test from 5.2.1.1 in Practical Methods for Engineers and Scientists by Paul Mathews (2010),
  • Putting aside the "pipetting" term,
  • Assuming that $N =$ sum of all detections across the two samples,
  • Assuming there is a typo and the formula should read [2*\sqrt(N)/count-duration-in-minutes], and
  • Assuming that 2 is an approximation of $z_{0.025} = 1.96$ (I guess you could also say that $2$ is just $z_\alpha$ for some other value of $\alpha < 0.025$).

If anyone has a better guess, I'd love to hear it.

FZS
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    You write, "We need to decide whether the samples come from the same source . . . or whether they might come from different sources/have been handled differently." I just like to clarify that this is $-$ in fact $-$ the only question you mean to answer, and that the "same mean" is just some off-hand way in which someone thought that (maybe) you could test that hypothesis. "Same mean" is one question; "same source" is another. Just checking, because if "same source" is the question, I suggest we relegate "same mean" to necessary-but-not-sufficient status. – Peter Leopold Sep 24 '22 at 00:36
  • @PeterLeopold Thanks for asking. Sorry if I've been a little loose in introducing the context of the question. We do understand that "same mean" is necessary-but-not-sufficient to conclude that the two samples actually come from the same source and have been handled identically throughout the measurement process. So it's just one of the checks we'd like to run in the hope of at least ruling out some bad --- or badly processed --- batches. – FZS Sep 24 '22 at 00:46
  • Also, what are the half-lives of your radioactive components? All same type of decay event? Or mixture? Are the event time distributions better fit with a negative binomial distribution than a Poisson? Most Poisson process are far from pure, so most datasets are over-disperse and a negative binomial might be a better fit. The only area where purity might be the norm is, in fact, in radioactive decay. So I have to ask . . . – Peter Leopold Sep 24 '22 at 00:46
  • The machine use measures radioactive decays in blood samples from patients who have been the subject of radiation therapy. I am unfortunately not the radiation expert of the team, so I don't know the full details about the nature of the decays. It's a shame, but I don't even know the half-lives of the components we're observing. But I know that we do not observe a significant drop in the rate of detected decays throughout our observation windows, which suggests half-lifes much larger than said windows. – FZS Sep 24 '22 at 00:56
  • Also, my guess is the scientist who sent the guidelines had a Poisson model in mind. Though if assuming a negative binomial helps understands their suggestion, I'd be very interested in hearing about that. More generally, if you're able to share a reference with more details about the way a negative binomial could be applied here, I'd very much appreciate it (I looked up https://en.wikipedia.org/wiki/Negative_binomial_distribution and I am having a hard time seeing how to match the parameters $r$ and $p$ with the situation at stake). – FZS Sep 24 '22 at 01:02
  • If you look up Poisson distribution in wikipedia and scroll down to posterior predictive distribution for assumptions using a gamma function prior, the negative binomial (NB) appears. That is, the data you collect would seem NB (not Poisson) for a Poisson process with a gamma-over-disperse distribution of the Poisson parameter $\lambda$. This is what happens with normal mixtures. But again, we're talking about radioactivity, so over-dispersion may not be a problem with single-type radio sources. In any event, I'll write it up below . . . – Peter Leopold Sep 24 '22 at 01:20
  • @PeterLeopold Do you mean for the pooled/marginal distribution of all data points rather than the conditional distribution? – Dave Sep 24 '22 at 03:04
  • @Dave, yes. The posterior predictive distribution (a.k.a., pooled distribution) of a Poisson likelihood with a gamma prior is negative binomial. Small world, huh? I'll cop to being gob-smacked when I learned that. In some quarters, so I hear, the negative binomial is known as the Poisson-gamma distribution. – Peter Leopold Sep 24 '22 at 03:09

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