here's the thing I've trying to wrap my head around. Please point out any error in reasoning.

Imagine you want to estimate the number of people in a field of known total area ($A_{T}$). People are walking around randomly and the distribution of people in the field is uniform. So, your method consists of taking a snapshot of a smaller area ($A_{S}$), counting the number of people in this smaller square ($S$), then extrapolating for the whole area. Your estimate for the total number of people ($T$) would be: $$ T = S . \frac{A_{T}}{A_{S}} $$

Now, say the counting method isn't perfect. Observers might differ, since people are moving, the uniformity of the distribution of people isn't perfect either, the same observer counting two different samples from the same field will get different results, even the same sample counted twice might give different results. If that's the case, there will be a random error upon the counts of the samples.

Say the total random error in the counts is $\varepsilon$. That translates to an estimate error of about $T = \varepsilon . \frac{A_{T}}{A_{S}}$, right?

So, for the concrete example, if I have the 100 x 100 field, then count $10 \pm 3 $ people in a random 10 x 10 sample square (1/100 of the area), I get an estimate of $T = 10 \pm 3 . \frac{100}{1} = 1000 \pm 300$.

The thing is, the error happens in the counts of the sample. The error of the actual estimate will depend on the size of my sample (that makes sense, larger samples mean better estimates) but will also depend on the total area. So, reducing total area would improve my estimates? That is, in the sense that the error in the counts wouldn't reflect so much on the final estimate error.

I'm trying to get the intuition behind why this happens. Is it simply because in smaller fields a smaller sample is more representative than in larger ones? Because if it's true, it means "diluting" the people over a larger field (or concentrating them on a smaller one) before taking the sample would make my estimate less precise (or more precise). In my example, if we asked all people to move to one half of the field before taking the samples (of same area) we'd get an estimate of about $T = 20 \pm 3 . \frac{50}{1} = 1000 \pm 150$ (assuming the errors in counting are independent of the number of people counted in a sample, which might not be a good assumption). Does that make sense?

Hopefully someone can help or point to where I'd find a formal statistical treatment to problems of this kind. Thank you!


1 Answer 1


First of all, I believe you are denoting both the total population and the total error using the symbol T. You should use a different symbol, possibly $ε_T$, to denote the total error.

As for your actual question, I'll add some terminology that I think is appropriate. Based on numerous hypothetical counts, you are modeling the total population of a 10x10 sample square as a normal random variable with mean = 10 and (I think you mean) standard deviation = 3. Then, you say that you can model the population of the entire space as a sum of this and 99 other identical normal random variables. This sum is also a normal random variable, with mean equal to 1000, as you say. It is not true, however, that the standard deviation of the total population is equal to 100 times the standard deviation of the sample. Actually, it is the variance, or standard deviation squared, that behaves this way. If the standard deviation of your sample is 3, the variance is $3^2=9$, meaning the variance of the total is $9*100=900$ and the standard deviation of the total is $\sqrt(900)$, or 30, much smaller than 300!.

As for your other question, I actually think it's not unreasonable to say that a counter will make a similar error regardless of the count. Let's assume this is the case and say you double your sampling area and get the estimate you mentioned of mean=20 and stddev=3. In this case the stddev of the total will be equal to $\sqrt(50*3^2) \approx 21$. So as you assumed, by increasing the sample size we are more confident of the total, but there is a diminishing return - we have done twice as much work counting, but our confidence interval is not twice as skinny.

I think this answer gives a more thorough explanation for a related question:

How to 'sum' a standard deviation?


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