Ratio that accounts for different sample sizes I'm trying to get a measure for a set of data that indicates someone's "spam score". Essentially, the higher the spam score, the more likely they are to be spammers.
Right now, I'm measuring a person's spam score as a ratio: bad posts/total posts. If this ratio is high, then they're likely to be spammers.
The difficulty I'm having is trying to compare different people - for example, a person with 6/8 bad posts is not AS likely to be as spammy as someone who has 600/800 bad posts (the person who's made 600 bad posts is clearly a spammer, but the one who's made 6 has not proven themselves to be a spammer to that extent). 
However, right now they are being assigned the same spam score of 6/8 = 0.75. Is there a way I can account for the size of the sample in my spam score?
 A: A simple way to do this is to use a test that the proportion is greater than some threshold--for example, that your proportion, $p$, is greater than 0.5.  The most basic one sample proportion test is $$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$
The $z$ score can then be tested in the usual way to get a $p$-value.
For instance, in your example if you had 600 out of 800 posts and you wanted to compare to 0.5, then you would calculate $$\frac{0.75-0.5}{\sqrt{\frac{0.5^2}{800}}}=14.14$$
If we want to compare 600 out of 800 to 6 out of 8, note that the only difference will be in the $n$ in the denominator, so we'll just cheat and calculate 14.14/10 = 1.414.
If we plug these two numbers into our handy normal probability calculators, we'll get $p$-values of basically 0 for the first one and 0.0787 for the second.
This test is not perfect, but it will probably work just fine for your problem.  All that's left would be calibrating it with the values of $p_0$ and $\alpha$ (what you'll compare your $p$-value to) that will give you satisfactory results.
