What is statistically more accurate - the average of a dataset of calculated concentrations or the sum of the total mass divided by the total volume?

Question

Hopefully an easy question to answer.

But if I have a series of five water samples, each sample is a different volume with a different number of plastic particles contained within. I can calculate the concentration of each sample by dividing the mass of particles by the volume of the sample (e.g. 500 mg of particles and 1 litre of water = 500 mg/L).

If I want to understand the true average of the dataset, would it be more appropriate to take the average of the concentrations of all the samples OR take the total mass of all the samples combined and divide that by the total volume to get an average concentration? See below for example:

Total	Sample volume (L)	Sample mass (mg)	Sample concentration (mg/L)
	4	253	63.3
	6	439	73.2
	5	205	41.0
	9	226	25.1
	4	30	7.5
Total	28	1153

Would the correct average in this case be 42.0 mg/L (the average of the calculated concentrations) OR the calculated average that divides the total sum of the sample mass (i.e. 1153 mg) by the total sample volume (28 L), i.e. 41.2 mg/L?

I know in this case the discrepancy between the values is small, however if you expand the dataset this discrepancy will increase.

I feel the answer to this question is a simple mathematical response, but I just can't seem to come to it myself.

I appreciate the help!

Answer 1

Generally speaking, the total mass divided by the total volume is the more accurate estimate of the mean particulate concentration, under the assumption that samples are independent and taken at random from a population that doesn't change from sample to sample.

To understand why, consider the following extreme scenario: you take two samples, one that is $1000$ liters, and one that is $0.00001$ liters (i.e., $0.01$ mL). The first sample has $42$ grams of particulates. The second sample has $0$ mg, because the sample volume is so small that your equipment cannot measure any detectable particulates.

Yet if you take the naive average of these two, you would obtain an estimate of $21$ mg per liter. This is plainly contradictory to one's intuition. Although this is just a thought experiment, it shows that we must take into consideration the volume of each sample as a weighting factor for how much influence its concentration contributes to the average, since a large volume that was sampled should be in a sense be "more representative" of the true mean particulate concentration than a small sample.

That said, if all volumes sampled were identical, the two methods of computing the sample mean would be equivalent: if you captured, say, exactly $5$ L every time, it would make no difference which method you use. But it is precisely because some volumes were substantially larger, and others smaller, that a weighted average becomes necessary.

Algebraically, if we have $n$ samples and the $i^{\rm th}$ sample has volume $w_i$ and particulate mass $x_i$, then the sample mean concentration should be calculated as $$\bar x = \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n w_i}.$$

Answer 2

Hopefully an easy question to answer

Hopefully my answer is easy to grasp.
First: your table is missing three columns, the error of every meassured item and the error of the calculated result. When you note for the first record a volume of 4 l, how precisely is this data? Are this 4.0 litres or 4.00 litres or even more precisely? Take a closer look to your instruments how exact you may meassure the volume. This will be the absolute error of every volume determination. If this absolute error is the same for 4 litres as for 9 litres, the relative error for the 9 litres is lower. Ok?
A similar error estimation is needed for the weight of each sample -- how exact is your balance, is the used balance apt to give exact results for the 30 mg sample? Take a look to the specification plate or see its manual. This will result in an error estimation for every weight measure -- also an absolute error. The relation to the corresponding weight is the relative error of this meassurement.
Now the third column of your table is a calculated value (and belongs to the evaluation, not to the log of the meassurements). For addition and substraction the absolute errors sum up, while for multipication and division the relative errors sum up.

If $e$ stands for error, $m$ for mass and $v$ for volume, and $k$ for the number of samples compare the error of totaling samples $$\frac{\sum_k{e_{mk}}}{\sum_k{m_k}}\cdot\frac{\sum{e_{vk}}}{\sum_k{v_k}}$$ against the error of average results $$\sum_k{\frac{e_{mk}}{m_k}}\cdot\sum_k{\frac{e_{vk}}{v_k}}$$ HTH

Answer 3

In order to answer this question you would have to know more about what is causing the variation in your samples.

In general, say you are trying to estimate some true value $M$, and you have several independent estimates $m_1, m_2,...,m_n$. If each of the estimates $m_i$ is "unbiased," meaning $E[m_i] = M$ (where $E$ denotes the expected value from statistics), then any weighted average of the $m_i$ will also be unbiased. But the optimal weighting will depend on the accuracy (or "variance") of the individual estimates $m_i$.

Intuitively you will want to give more weight to the more accurate estimates, and less weight to the less accurate estimates. In particular, if you want to minimize mean square error of your final estimate, you would want to weight each $m_i$ with a weight inversely proportional to the variance of that estimate. https://en.wikipedia.org/wiki/Inverse-variance_weighting

In many contexts larger samples will be more accurate. For example, if we are sampling balls from an urn by taking handfulls of balls and counting how many are red, then when we take larger handfuls of balls we would expect to get a less noisy estimate of the true percentage of red ballls. If this is the ONLY reason that makes larger handfulls differently accurate than smaller handfulls, then we would want to weight precisely by the size of the handful. This is mathematecally equivelant to just adding up all the red balls in all the samples and dividing by the total number of balls in all the samples.

This works because each ball can be considered independent of the other balls within each handful, and there is no other source of variation in our samples.

In other cases, this may not be optimal. For example, if you are sampling to find plastic concentration within a lake, and the lake is not well "stirred up" such that the plastic concentration is much higher in some parts of the lake than others, then the primary source of variation is not based on how many liters are in the sample but from where in the lake you drew it out. In this case it woud probably make sense to weight each sample equally. Another source of variation might be the measurement error for each sample based on whatever procedure you are using to count the particles. If this procedure is more accurate on larger samples, then it makes more sense to weight the larger samples more highly.

One last thing, if you have lots of samples it is possible empirically examine the pattern of variation in your samples to tell what is going on. Eyeballing your table it seems unlikely to me that your sampling variation is like the well-stirred balls in the urn. The variation in concentration appears too large. I would guess the variation is caused more by where/when you took the sample, or by something about how you are counting the particles. In this case equally weighting the samples may well be more accurate than over-weighting the large samples (which, again, is what happens if you use total mass divided by total volume).