The goal is to estimate the proportion of accounts that are bots. Let that number equal p. The variance of the proportion is p(1-p). The highest that can be is (0.5)0.5. Then, the standard deviation of that is the square root, which is 0.5.
Now, we want to know the standard error for our estimate of the bot proportion. That is sqrt(p(1-p)/n). Suppose 50% of accounts are bots (I assume that would be very high), then our estimate of p would be 0.5 and our standard error with a sample of 100 would be 0.05. Hence, our 95% confidence interval is roughly 0.4–0.6 in the worst case (with a sample of 100).
If the proportion is under 0.1 (let's assume 0.05), then the standard error would be sqrt(0.05(1-0.05)/100) = 0.022. Our 95% confidence interval in this case would be roughly 0.01–0.09.
These seem like large ranges to me. Hence, I would expect them to use a larger sample too.
Now, we want to know the standard error for our estimate of the bot proportion. That is sqrt(p(1-p)/n). Suppose 50% of accounts are bots (I assume that would be very high), then our estimate of p would be 0.5 and our standard error with a sample of 100 would be 0.05. Hence, our 95% confidence interval is roughly 0.4–0.6 in the worst case (with a sample of 100).
If the proportion is under 0.1 (let's assume 0.05), then the standard error would be sqrt(0.05(1-0.05)/100) = 0.022. Our 95% confidence interval in this case would be roughly 0.01–0.09.
These seem like large ranges to me. Hence, I would expect them to use a larger sample too.