[btilly]

The total is indeed nearly normally distributed, but the rate of convergence (particularly in the tails) is not fast enough to avoid having those very sensitive tests give wrong results.

Were it otherwise there would have been no need to develop the chi-square test. It would have been entirely redundant. (It actually is redundant because we have the g-test. But evaluating the chi-square test just involves taking squares, while the g-test involves taking natural logarithms. This made the less accurate chi-square test much easier to do when people didn't have computers to calculate it on. Today we should use the g-test, but few people have heard of it.)

[sesqu]

Ah, right. I spent a while drawing up a proper plot of the likelihood of the difference and the normal approximation of the difference, and saw that the normal had too small a variance. The effect is still pretty credible in the OP, though.

[sesqu]

Noprocrast caught out my attempt to edit. The normal variance is too large.

Here's the plot. Black for discretized(n=1000) binomial likelihood, red for normal approximation. The effect is clear, but a t-test won't show it. I'm not familiar with the theory behind the g-test, but there's clearly a lot of room for improvement at these sample sizes.

http://img693.imageshack.us/img693/4880/bindiff.png

[btilly]

It looks like you forgot to rescale the binomial distribution.

If X_i is a series of independent, identically distributed random variables with mean m and variance v, then X_1 + X_2 + ... + X_n is approximately a normal variable with mean nm and variance vn. Therefore

(X_1 + X_2 + ... + X_n - nm)/sqrt(vn)

is approximately a standard normal.

If you draw that graph, visually the two lines should lie right on top of each other. To see the problem you need to zoom in on the tail and blow it up, and only then will you see the issues with the convergence.

[sesqu]

You're right, I did have the wrong scale, and should have realized. I believe this plot to be correct. Black is binomial, red is normal, both for the difference in conversion rates.

http://img267.imageshack.us/img267/6691/binnormdifflikelihoo...

The normal approximation seems reasonable, and so a t-test shouldn't cause problems. I made a plot of the relative error of the normal approximation; the tails are indeed too fat. In particular, there's a bump to the left of 0, so a t-test would slightly overestimate p.

http://img205.imageshack.us/img205/6691/binnormdifflikelihoo...

[btilly]

The problem is that when you're near confidence you're in the tail, and the student t-test is extremely sensitive to the shape of said tail. If n is large enough this difference will be washed away as everything converges to normals. But with smaller sample sizes the difference can be quite significant.