p-values from t.test and prop.test differ considerably Testing a difference in hits (ones versus zeros) between two independent groups $X$ and $Y$ should be possible with a t-test, according to the following considerations:

*

*$x_i\in\{0,1\}$ is the measurement for the $i$-th item im group $X$, and $y_i\in\{0,1\}$ the same for group $Y$

*the proportion in each group is the mean value of the measurements, i.e. $\mu_X=\sum_i x_i/n$ and $\mu_Y=\sum_i y_i/n$

*the difference of mean values $\mu_X$ and $\mu_Y$ in two groups can be tested with a t-test

In this particular case, a proportion test (function prop.test in R) is an alternative test option. Interestingly, the results are quite different:
> x <- c(rep(1, 10), rep(0, 90))
> y <- c(rep(1, 20), rep(0, 80))
> t.test(x,y,paired=FALSE)
t = -1.99, df = 183.61, p-value = 0.04808
> prop.test(c(10,20), c(100,100))
X-squared = 3.1765, df = 1, p-value = 0.07471

Note the higher p-value of the prop.test. Does this mean that a t-test has a higher power, i.e., can distinguish between $H_0$ and its alternative already for smaller $n$? Is there a reason why a t-test should not be used in this case?
Addition (Edit: resolved in a comment under the answer by Thomas Lumley below): The result of the t-test is even more surprising in the light of the observation that even the asymptotic ("Wald") 95% confidence intervals of both measurements overlap (0.1587989 > 0.1216014):
> library(binom)
> binom.confint(10, 100, method="asymptotic")
      method  x   n mean      lower     upper
1 asymptotic 10 100  0.1 0.04120108 0.1587989
> binom.confint(20, 100, method="asymptotic")
      method  x   n mean     lower     upper
1 asymptotic 20 100  0.2 0.1216014 0.2783986

As confidence intervals based on the t-distribution should be even wider than those based on the normal distribution (i.e. $z_{1-\alpha/2}$), I do not understand why the t-test reports a significant difference at the 5% level.

The t-test can be quite robust to deviations from the normality assumption, particularly when sample sizes are large, so I understand why one might want to use a t-test for this task.
However, you know the parametric family; since the outcome is either $0$ or $1$, the distribution is completely characterized by the relative proportion, thus Bernoulli. Consequently, you can rely on a parametric test designed for a Bernoulli variable, which the t-test is not.
Methods that are robust to deviations from parametric assumptions are wonderful, since we typically do not know the type of population distribution. (If we did, why did we not determine the population parameters when we had the chance!?) However, the case of a binary variable is unique in how it is completely defined by the relative proportion and must be Bernoulli (or easy to represent as Bernoulli, such as calling “heads” and “tails” of a coin $0$ and $1$, respectively).
