Estimating error from a 1% sample Say we decide to take a 1% random sample (without replacement) of a population to estimate how many individuals have some condition. 
We then observe that X individuals in the sample have this condition (e.g. 800).
That's all we know. We don't know the size of the sample, the population, nor anything else. 
What can we say about:


*

*The number of individuals in the population that have this condition?

*The error in the estimation?

 A: Clearly a good (unbiased) estimate of the number of people in the population with the condition is $X/(1\%)=100X$.  
$X$ has a Binomial distribution--but we don't know its parameters, because we lack information on the population size (except to know it is at least $800/(1\%)=800\times100=80000$). If we assume the proportion of people in the population with the condition is small, then to an excellent approximation $X$ has a Poisson distribution.  A good estimate of the sampling standard deviation of $X$ is its square root.  If that unknown proportion is large, then the sampling standard deviation of $X$ will be smaller than its square root: so let's conservatively use that square root to make sure we produce a confidence interval that isn't overly narrow.
Again because $X$ is large, its sampling distribution will also be approximately Normal.  Thus, to find a two-sided confidence interval of confidence $100-100\alpha\%$, find the upper $100-100\alpha\%$ percentile of a standard normal distribution as $Z_{1-\alpha/2}$ and form the interval
$$CI = [100(X -Z_{1-\alpha/2}\sqrt{X}), 100(X + Z_{1-\alpha/2}\sqrt{X})].$$
This has at least a $100-100\alpha\%$ chance of covering the true value.
With $X=800$ and $\alpha=0.05$ (for a $95\%$ confidence interval), $Z_{1-\alpha/2} = 1.96$ and the interval is
$$CI = [74456, 85544].$$
For a little more insight into this result, we might ask the computer to plot the limits of the CI (using a Normal approximation to a Binomial distribution) as a function of the population size.  The smallest possible size is $80000$, so let's indicate any potential population size as a multiple of this minimum.

The conservative 95% limits are plotted as horizontal gray lines while the Binomial limits are plotted as red curves.  You can see that the Binomial intervals approach the limits quickly: unless most people in the population have the condition, the conservative interval will not be too wide.
A similar plot could be produced for sampling without replacement: the intervals would be narrower at the left, but by the time the multiple reached $10$ or larger, there would be little difference between the two plots.
A: Suppose we have a population of $N$ Bernoulli trials, but $N$ is unknown. Suppose $α ∈ [0, 1]$ (.01 in the example) is known and we draw a simple random sample of size $αN$ (unknown). We observe $X$ successes (known) in the $αN$ trials. We want to estimate $K$, the number of successes among the $N$ trials in the population.
Since the sampling is without replacement, $X$ follows a hypergeometric distribution with population size $N$, sample size $αN$, and number of successes $K$. You could estimate $K$ in a Bayesian fashion, putting priors on $N$ and $K$, or you could find the maximum likelihood estimate of $K$ (which will probably be $X/α$, as you'd expect).
Of course, there's no way to compute the error between the estimate and the true value without knowing the true value. You could get a sense of the uncertainity in the estimate using a credible interval (in the Bayesian case) or a bootstrapped confidence interval (in the MLE case).
