Statistic that combines proportion and 'success' size I have proportion data where I am interested in the proportion of successes, but also in the number of successes.  Is there some statistic I can calculate to summarize this info into one number? 
Sample data:
X   N    P
2   4   .5
40 200  .2
3   10  .3
50 100  .5

Here the first and last data have the same proportion, but I would like the statistic to acknowledge the fact there is a lot more confidence in the last line (maybe incorporate the margin of error some how, but I would like it to take into account X not N).
 A: I'd compute the proportion (X/N=P) along with a 95% confidence interval for that proportion. The width of that confidence interval shows you the margin of error you want. Use this free online calculator to do the math.
A: You might consider shrinkage estimation of $P$ given $X$. One straightforward way to do this is, for each row $i$,
$P_{i}^* = m\frac{c}{c+x_i} + \frac{x_i}{n_i}\frac{x_i}{x_i+c}$, where


*

*$x/n = P$, the proportion of successes

*$x =$ the number of successes

*$m =$ the mean proportion across the whole data set (which, based on the data above, is 0.303) 

*$c =$ a scaling constant (the higher this is, the more shrinkage there is) 


The intuition is that you are creating a weighted average $P^*$ which shrinks each proportion to the mean on the basis of how many successes it has. The left term establishes the base value, which will be more or less of a factor based on the degree to which $x_i$ differs from $c$. The right term pushes the value from $P_i^*$ away from that average, once again based on how $x_i$ differs from $c$. 
The downside if that you have to choose an arbitrary scaling constant $c$. But you can justify this based on how many successes values $x_i$ it would take you to "trust" the proportion estimate $p_i$.
A canonical example of this in action is the imdb top 250, for which movie ratings are scaled to be higher (or lower, for the bottom 100) on the basis of how many votes they have. 
(By the way, it's not clear to me why you trust entries with high $X$ more than high $N$, but you clearly have more domain-specific knowledge.) 
A: You can't summarize this in one statistic,  But p and N the two binomial parameters tell the story.
