Statistical analysis of binary outcome I am trying to figure out the best statistical test for a data set I've collected. I presented a group of adults (n=9) with a 21 different auditory cues. For each cue I asked them to choose the corresponding word from list of four words, one which was "correct" and three "incorrect." I then had them take the test again (I would have run more trials, but I didn't have enough time).  
Mainly I want to test if their answers show that they recognized the cue and assigned meaning to it, or if they chose it by chance. But I'm getting caught up in the details and it has been a while since I took a stats course. 
I want to compare the number of correct choices to the theoretical amount of correct choices (1/4). Should I use a binomial test or Chi-squared? Or do you recommend something totally different? Also should I look at each patient's individual data or the aggregate? 
I also wanted to analyze the amount "recognized" for each patient -  the amount of times the patient chose the same word for both the test and retest (despite if it was right or wrong). This is where I am really lost - could I simply compare the total "recognized" to the expected 1/4? (I got 1/4 by looking at the probability of selecting the correct twice [1/4*1/4=1/16] and selecting the same wrong twice [3/4*1/4= 3/16])? 
I'm sure there is a simple solution, but I may be overthinking this. Any help is much appreciated! 
 A: Suppose you want to take both the test and re-test into account.
On the first test a subject who is just guessing performs $r = 21$ trials with
success probability $p = 1/4$.
Again on the re-test the subject does the same.
So the subject's total score is on the two tests is
a binomial random variable $X$ with $t = 2n = 42$ trials
and success probability $p = 1/4$ on each. Thus
the average (or expected) value is $\mu = E(X) = tp = 42/4 = 10.5,$
variance $V(X) = tp(1-p) = 126/16 = 7.875$, and $\sigma = SD(X) = \sqrt{7.875} = 2.8062.$ Of course, we expect that
subjects who are not just guessing will tend to get higher
scores.
For $n = 9$ subjects you would find the average score $\bar X.$ Then
$$Var(\bar X) = \sigma^2/n = 7.875/9 =  0.875$$ and $SD(\bar X) = 0.9354.$
Your null hypothesis is $H_0: \mu = 10.5$ and your alternative
hypothesis is that $H_a: \mu > 10.5.$ 
The average score $\bar X$ is close enough to normal that you can
reject the null hypothesis at the 5% level if 
$Z = (\bar X - 10.5)/0.9354 \ge 1.645$ or $\bar X \ge 12.$ 
Notes: (a) This is assuming that the 42 trials for each subject are
independent, this includes the assumption that the first test
and re-test are independent. (b) We also assume that the nine subjects are chosen at random
for the population of interest. 
(c) Because the standard deviation of $\bar X$ under the null
hypothesis is know, this is a Z-test. (d) Similar tests could
be done for individual subject (use his/her $X$ instead of $\bar X$
with the appropriate $\sigma$). Also, you could do a similar
procedure for the first test only (use $n = 21$ instead of $t = 42$).
(e) Just $X$ for a given subject is nearly normal, so there is
no problem assuming that $\bar X$ is normal.
