I was calculating the no. of people I need to survey to represent my fairly large population.
I used an online calculator: https://www.surveymonkey.com/mp/sample-size-calculator/
When I entered the population size to be 10M, 100M and 1B, the required sample size barely changes at 99% confidence level with 1% margin of error.
May I check why this is so? Should it not increase to represent a bigger population?
No. The population size is basically irrelevant to the sample required. The exception is when the sample is very large, relative to the population, in which case you will need a finite population correction.
You don't say exactly what you mean by "represent" - are you estimating a mean? A proportion? Or what?
In your title you say 16600ish. That is a very large sample. It makes me wonder if you plugged the right things into the sample size calculator. But the fact that the required sample doesn't change is not a problem.
What matters is what you want to estimate, how precisely you want to estimate it, what value it is likely to take (for proportions), and, for hypothesis tests, your desired power and acceptable type 1 error rate. Also, the type of sample can matter a lot.
R you could compute that sqrt(16600 / 4) * qnorm(0.995) / 16600 works out to 1%, corroborating the OP's characterization of the solution and revealing specifically what it means.
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The correct answer to your question, "should it not increase to represent a bigger population," is yes -- but there are rapidly diminishing returns and the sample size quickly approaches a limiting value.
It is clear from your description that the app is solving the following problem:
In a survey where a random sample without replacement will be obtained from a population and a binary response will be observed, what is the smallest sample size needed to estimate the mean of that response with at most a 1% margin of error with 99% confidence?
Here are some implicit, basic considerations:
Because such a sample might have to be fairly large, it suffices to compute the margin of error using either a Student $t$ distribution or (for very large samples) a Normal distribution. (When the sample size or the size of the unsampled populations are small -- say, 30 or less -- a more accurate calculation based on Binomial distributions can be used.)
Because nothing is assumed about the response, we have to consider the worst case situation.
The following reasoning leads to a solution.
Let $p$ be the average response in the population, let $N$ be the population size, and let $K$ be any sample size. Probability theory tells us the variance of the sample mean is
$$V(K,p,N) = \frac{N-K}{K(N-1)} p(1-p).$$
The square root of this variance is the standard error (SE) of the mean. The margin of error is a multiple $t(0.99, K)$ of the SE. (99% of the probability of a Student $t$ distribution with $K$ "degrees of freedom" lies between $-t(0.99,K)$ and $t(0.99,K).$)
Thus, the mathematical statement of the sample size question is
What is the smallest value of the positive whole number $K$ for which $t(0.99,K) \sqrt{V(K,p,N)}$ is less than or equal to 1% no matter what value $p$ might have?
Because the only part of this expression that depends on $p$ is proportional to $p(1-p),$ which is largest when $p=1/2,$ the worst case occurs when the population is split 50:50. Given $N$ and assuming this worst case, we are left with a constraint on $K$ that we have to solve.
To understand the constraint, observe these things:
Because $$V(K,p,N) = \frac{Np(1-p)}{N-1}\times \frac{1}{K} - \frac{p(1-p)}{N-1}$$ is inversely proportional to $K,$ the variance (and therefore the SE with it) decreases as $K$ increases.
Because $$V(K,p,N) = p(1-p)\left(\frac{1}{K} - \left(1 - \frac{1}{K}\right)\frac{1}{N-1}\right)$$ is the difference between a fixed value and a quantity inversely proportional to $N-1,$ the variance increases to a limiting value of $p(1-p)/K$ as $N$ grows.
These behaviors conspire to require larger sample sizes as $N$ grows, but the sample size reaches a limiting value given by an "infinite" (arbitrarily large) population.
This plot shows how the requisite sample size changes with population size when $p=1/2$ and 99% confidence in a 1% MoE are required. (Notice the logarithmic scale for the population size.)
The sample sizes you computed all occur in the upper right corner of the plot where it has nearly leveled off: that's why they were all nearly the same.
Here's a simple example that might give some intuition. Suppose you have a coin that you flip $N_1 = 1,000,000$ times, on the basis of which you compute a probability estimate $\hat{p}$ for the coin coming up heads. With this amount of data, $\hat{p}_1$ is very likely an extremely good estimate for the true probability $p$. Now, you decide to flip the coin even more times, to get $N_2 = 10^{100}$ observations. The addition of all these new observations does not suddenly make $\hat{p}_1$ a worse estimate for $p$, and revising the estimate using the new data to obtain $\hat{p}_2$ will give an estimate that is likely correct at more decimal places, but which in absolute value is almost identical to $\hat{p}_1$.
The limiting factor here is that you have a die with $2$ sides, and there is only so much data you need to collect before you can narrow down the probability $p$ with great precision. If you were rolling ten dice, each with a million sides, which may be correlated with each other, you would need much more data to get a good estimate on the probabilities, but ultimately those estimates are independent of how many times you can roll the dice to generate more data, which is in principle infinite.
