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I'm trying to quantify the impact of sample volume and variability on the propensity for false negatives, but I'm having trouble making a connection between the idea and the math to quantify it.

An example of the type of problem: A solution may (or may not) have a very low concentration of particles in it. I want to check a small sample of the solution for the presence of particles. Because of the low concentration, sometimes I will obtain a sample with no particles, even if there are particles in the solution (the particles are elsewhere in the solution). The smaller my sample volume, and the smaller the concentration, the greater the chance of this false negative occuring.

I think it's the fact that my sample size is a volume that is tripping me up. How would you express the impact to false negatives of the sample size and concentration, noting that it's a binary result (pass/fail) and bounded at 0?

Thanks

Kern Hast
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1 Answers1

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Let's assume you have done your best to mix the solution and to keep it mixed while withdrawing a sample, so that the absence of particles in the sample of the solution is random.

We may model a "mixed" solution as one where each particle has an equal chance of being located anywhere and the chances are independent. Independence obviously isn't quite correct, because two particles usually cannot occupy a common location. But provided the particle sizes are small compared to the dimension of the container holding the solution, assuming the particles do not interact (they have no repulsive charges and they don't form clumps), and assuming the sampling procedure does not preferentially recover (or avoid) particles, this can be an extremely accurate model.

In this circumstance, the chance of not including any particles in your sample of the solution is the chance they all are not in the sample. The independence assumption implies this chance can be found by multiplying the chances for each particle.

Thus, when the ratio of the sample volume to the solution is $\rho \le 1$ and there are $n$ particles in the solution, your chance of not having any of them in the sample is

$$(1-\rho)^n.$$

Because you don't know $n,$ you may need to be content with studying this as a function of $\rho$ (which you do know or can easily find out). One approach is to back out a maximum value of $n$ that is plausible given you observed no particles: that is, an upper confidence limit for $n.$ You have to specify the degree of plausibility. For instance, if you're willing to take a 1% chance of being wrong, you would set

$$1/100 \ge (1-\rho)^n$$

whose solution is

$$n \le \frac{\log(1/100)}{\log(1-\rho)}.$$

Generally, you can replace the value $1/100$ by any level of risk you are willing to incur, often called $\alpha,$ and this is called a $1 - \alpha$ upper confidence limit for the number of particles.

For instance, if you sample 5 ml out of a 1 L solution, $\rho = 0.005$. The 99% confidence limit for $n$ works out to be $918.$ Interpretation: if there were $919$ or more particles in the solution, your sample would have had at least a 99% chance of containing one or more of them.

whuber
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