How to compute Cronbach's alpha with only one measurement per subject? Consider the following scenario. I ran an experiment with 200 trials, each with a different stimulus. Each subject did exactly 200 trials. The subject responds with a single number anywhere between 5 and 50. The correct answer also ranges between 5 - 50. For each subject that did the experiment, I compute a single value, $V$, for that person. This value $V$ uses computations that use the expected answer and the observed answer for each trial. That is all the experiment does. It allows me to find $V$ for a subject.
I was asked to find Cronbach's alpha for this particular experiment. How do I do it? I see the formula on Wikipedia, however the denominator is $K-1$ and in my case, $K=1$, so how would I find Cronbach's Alpha? Should I find a single Cronbach's alpha value for the experiment as a whole, or is each subject getting an alpha value? I couldn't find many good online resources to learn about Cronbach's alpha value, so if anyone has any good suggestions, I would love to see links to places I can learn this stuff. Thanks!
I will be using R, and I have a CSV file where column 1 is the subject, column 2 is the expected answer, and column 3 is the subjects answer. There are 200 rows.
 A: The formula for Cronbach's alpha is:
$$
\alpha =\frac{K}{K-1} ( 1-\frac{\sum_{i=1}^{K}\sigma^2_{Y_i}}{\sigma^2_X})
$$
Here, K is the number of different items you administered to each subject. Sometimes items are different questionnaire items designed to measure the same underlying construct. In your case, it sounds like each item is a separate run of the experiment. 
In order to calculate Cronbach's alpha, you need to put your data in "wide format" (as Michelle mentioned). This means that each of the 200 measurements needs to have its own column/variable. So your columns would be SubjectID, Answer1, Answer2, ... , Answer 200.
I'm not sure where your expected answer column comes in. Cronbach's alpha is used to test the consistency of answers to each other, not to some true value, because the true value being measured is latent (i.e., unknown). I suppose you could calculate the correlation between the mean of the answers and the expected answer to see how well people did. Or calculate the times they answered exactly the way you expected and call that their score. 
As has been suggested in the comments, it doesn't seem to make sense to calculate alpha on the raw scores. If they are only meaningful in relation to the expected answer (i.e., as indicative of how well a subject does on this particular test), you need to use adjusted scores that actually quantify how well a subject did. If an answer of 50 with an expected answer of 25 doesn't mean the same thing as an answer of 50 with an expected answer of 45, then calculating alpha on the raw scores is meaningless. 
To calculate Cronbach's alpha using R, read the CSV file into a dataframe, reformat into wide format, then run cronbach.alpha on only the answer columns (assuming your columns for subject and values are called SubjectID and Score):
x.long <- read.csv(file="myfile.csv")
library(reshape2)
x.wide <- dcast(x.long, SubjectID ~ Score)
library(ltm)
cronbach.alpha(x.wide[,-1]) # remove SubjectID in first column

A: Update with proviso: it is impossible to answer the question without knowing the detail of what occurred in the experiment, as the risk of providing incorrect advice is very high. Depending on what actually happened in the experiment, it is possible that both of my suggestions below are incorrect. This update is from my perspective that "software is stupid" - so long as the input is numbers (in most cases) you'll get an answer on any analysis you do, the problem being that the analysis may not have been appropriate, and therefore the analysis "answer" is wrong. This assumes that the desired ultimate outcome is to provide an answer that is appropriate to the method and data, and therefore may recommend a solution that is different to the one asked in the original question.
Information on Cronbach's alpha is here and any standard statistics package will calculate this for you. You will need to use the individual trial data to do the calculation so K=200. Assuming you're going to be using R or similar, you'll need the data in wide format. If you believe that you may have training or acclimatisation effects in your data, you may wish to consider a split half reliability alternative, for which you will still need all the individual data points. The point of doing this is to tell you whether V is an appropriate summary statistic, given your data.
Is there a reason why you think that something like a mean and standard deviation won't be informative? Internal reliability is normally performed when you don't administer an identical stimulus repeatedly to a person (e.g. when they answer a questionnaire, because a questionnaire doesn't consist of the same question repeated a number of times). If you're presenting them with the same stimulus repeatedly, I'm not sure why you would wish to use a reliability measure. I'm now not convinced that a mean and SD is the right approach either, however depending on what the experiment actually did with the subjects an analysis based on the difference (maybe relative rather than absolute given the response range) between the expected answer and the subject's answer could be a way forward.
