Why isn't a t test used when comparing two proportions? All the examples I've seen say to use a z test to compare two proportions. For example, n=13, x=0.22 versus n=10, x=0.44. Then all the examples warn that the z test doesn't work with low sample sizes.
So why can't we just use a t test, which provides p values appropriate to the sample size?

Stefan may have already addressed your concerns, but here is the basic rationale for the normal approximation for comparing proportions and why the t-test is not useful:
Although sample proportions are not normally distributed for any finite sample size, they approach a normal distribution as the sample size approaches infinity. In particular, if $p$ is the true proportion in the population, then a ramdom sample of size N will have the number of "successes" will be binomially distributed with success probability p and sample parameter N. Of course, the binomial approaches the normal distribution as N gets large. For paired data where you are comparing differences in proportions, you need BOTH samples large so you can assume that the difference is also normally distributed.
So why not the t-test for small samples? The t-test requires that the sample be from a normal population, which as Stefan pointed out, it is not. 
