Frequencies/percentages different from zero - which test? I conducted an experiment and had participants indicate binary judgments by either pressing "1" or "0" when seeing different types of stimuli. Now I want to test whether the proportions of number presses differ from chance likelihood of 50% for the different stimulus conditions. My supervisor always used to run a one-sample t-test against 0.5 on the subject-level aggregated data (with the resulting mean representing the percentage of 1-presses); however, I am wondering this is the best way to do it. Since I am dealing with a binary variable (1/0), wouldn't it be more appropriate to run a chi-square goodness of fit test on the unaggregated data?
Any input is highly appreciated!
[Update] This is a fictive example for my data structure:
Subj    type    response
1   target_top      1
1   target_bottom   0
1   target_bottom   0
1   target_top      1
1   target_bottom   0
1   target_top      1
2   target_bottom   0
2   target_top      0
2   target_top      0
2   target_bottom   1
2   target_bottom   1
2   target_top      0

As can be seen, it is a repeated measures design. Stimuli differ in their vertical position and we test the hypothesis that spatial position influences judgments (i.e., for example, 1-presses when presented on top position significantly differ from chance likelihood). 
 A: The presumptive appropriate approach would be mixed effects logistic regression, where the dependent variable is Response, the independent variable is Type, and Subject is treated as the random subject. This is the approach described in the answer by @Noah. 
Using a simple chi-square goodness-of-fit test or binomial test ignores the fact that the same subject is responding several times. 
A t-test is probably not a good solution considering that the dependent variable is binomial and not continuous, as well as the fact that a t-test ignores the repeated responses by each subject.
A: My inclination is to use multilevel logistic regression. Your outcome is whether the observation was 0 or 1. Your level-1 predictor is the target location. Your clustering variable is participant ID. Alternatively, logistic regression with cluster-robust standard errors or fixed effects for participant ID would work as well.
Your output would be a coefficient on target location, which indicates the difference in the log odds of selecting 1 vs. 0 between the target locations, and a constant, which would represent the probability of selecting 1 in the baseline target location. With these, you could transform the log odds ratio into a difference in probabilities, but the significance test on the coefficient would be the test of whether target location affects the probability of selecting 0 or 1.
A: If I am understanding you correctly, you wish to see if the proportion of 1's in your data is significantly different from .5. You can calculate the sample proportion, say, $\hat{p}=\frac{\# \ of \ 1}{N}$. You can then calculate a $z$-statistic: $z=\frac{\hat{p}-p_0}{\frac{p_0(1-p_0)}{N}}$ where $p_0=.5$ and $N$ is your total sample size. Then you can calculate a $p$ value given the $z$ statistics and conclude statistical significance or lack thereof.
This can be done for each type of stimuli. If the stimulus are independent of each other and the order is randomized, a Bonferroni correction can be applied: you should compare your $p$ value to $\frac{\alpha}{ \# \ of \ stimulus}$, e.g., you have 4 different types and your overall significance level is .05, then you should compare your $p$ values to .0125.
A: The probably most obvious choice to test, whether an event ("1") happend by chance $p$ or not would be an binomial test. In R this can be done via the binom.test function. 
So if 30 participants voted "1" and 35 voted "0" the test of votings happening bei 50:50 chance ist
> binom.test(c(30, 35), p=.05)

     Exact binomial test

data:  c(30, 35)
number of successes = 30, number of trials = 65, p-value < 2.2e-16
alternative hypothesis: true probability of success is not equal to 0.05
95 percent confidence interval:
 0.3370209 0.5896749
sample estimates:
probability of success 
         0.4615385 

This is an exact test that does not rely on approximations as opposed to $\chi^2$ tests, which seem still more reasonable then a t-test.
