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I have a randomly selected group of 25 children from an unknown population size, and I am trying to determine if the student population as a whole are capable of long division. Each student is represented by an ID number S1 through to S25 and they are given a long division problem to solve and the result is stored as a 1 for success and 0 for failure (example and link to full sized example table below).

Student ID   Pass/Fail
S1           1
S2           1
S3           1
S4           1
S5           1
S6           0
S7           0

Google Drive Link to Example Data

The data looks like most students are capable and long division will not have to be re-taught but how can I statistically test that the skew towards pass (1) is greater than if pass and fail results were generated randomly?

I know it appears simple but I am new to statistics.

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    $\begingroup$ In my opinion, the problem with this is you need to (quantitatively speaking) define what "capable" means (usually in a quantitative sense). Maybe I'm wrong, though, and someone else will know something that I don't. $\endgroup$
    – Clarinetist
    Commented Mar 7, 2016 at 16:22
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    $\begingroup$ In order to have a testable hypothesis, you need to provide a quantitative criterion for "as a whole" or "most." Does that mean more than half? If not, exactly what does it mean? $\endgroup$
    – whuber ♦
    Commented Mar 7, 2016 at 16:31
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    $\begingroup$ If your hypothesis is that all students know long division, then a single zero in your dataset disproves it. It's difficult to justify the meaningfulness of a model in which students would be correct with 50% probability: getting a long division answer right isn't a guessing game. Arguably, correct answers are unlikely to be accidentally correct. $\endgroup$
    – whuber ♦
    Commented Mar 7, 2016 at 17:00
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    $\begingroup$ Let $p$ = the true proportion of students who can do long division. You want to test the hypothesis $p = .50$ versus the alternative that $p > .50$. However, it seems that if for example $.55$ proportion of students are capable of long division, this seems like a rather small number when the goal is to teach this concept. So you might want to figure out a higher number that is satisfying to you, and then test that hypothesis. $\endgroup$
    – Greenparker
    Commented Mar 7, 2016 at 17:01
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    $\begingroup$ Although what @Greenparker says is valid, be very careful not to misinterpret it! If you develop your hypothesis by examining these data, then you cannot turn around and test that hypothesis with the same data. Normally, your hypothesis is related to a meaningful, quantitative question. You should be able to construct one before you ever see the data. $\endgroup$
    – whuber ♦
    Commented Mar 7, 2016 at 20:34

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How can I statistically test that the skew towards pass (1) is greater than if pass and fail results were generated randomly?

Testing this hypothesis would lend itself to a binominal test comparing the number of students passing, out of total students tested, with the default proportion 0.5.

But is this really what you want to know? It is probably not a relevant question whether the proportion of students getting the correct answer is different than 50%. And as @whuber pointed out, there is no reason to suspect that for this kind of task that incapable students would get the correct answer 50% of the time. (This would be the case if it were a multiple choice question with two options.)

My suggestion would be to calculate the proportion of students passing, and calculate a confidence interval about this proportion. This result would probably be meaningful for what you are trying to determine.

When calculating the confidence interval, be sure to use a method appropriate for proportions.

The following is the R code to find the confidence interval for the Example Data in the question. There are other methods for confidence intervals for proportions.

Pass.Fail = c(1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1) 
Pass = sum(Pass.Fail)
Total = length(Pass.Fail)
binom.test(Pass, Total, 0.5)

This gives the results: 19 pass out of 25, for a proportion of 0.76, and a confidence interval of about 0.55 to 0.91.

This still allows you to compare to proportions of interest, based on whether the confidence interval overlaps a given proportion. So in this case you can say statistically that the proportion of students capable of getting the correct answer is likely greater than 0.50, but you cannot say it is likely greater than 0.75. Likewise, the proportion of students is likely less than 1.00.

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