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I'm getting confused by this and was wondering if someone can enlighten me:

I have a random sample consisting of 50 percentages. Each percentage can take on any value between 0% and 100% inclusive and is suppose to represent the quality of a file produced by a specific team. The desire is to provide inferential statistics - most notably basic hypothesis testing and confidence intervals regarding the quality of files produced by this team.

I am fairly confident the binomial distribution is not suitable as I have a vector of proportions, rather than a vector of passes/fails. The following thread Distribution for percentage data seems to confirm this. However, the solution gung suggested relates to continuous proportions. In my scenario, each percentage is formed by taking the proportion of correct answers from 15 questions. So I don't have continuous proportions but rather discrete proportions. In addition, it is possible to obtain a percentage of 100% or 0% (because all 15 questions were correct or were incorrect).

This makes me wary about using a beta distribution.

I should add that each of the 15 questions is binary. So effectively each percentage is the mean of a binomial variable that takes on the value 1 (correct) or 0 (correct) across 15 trials. This is similar (I think) to the following post Statistical tests when each variable in a sample is a percentage though I admit I struggled to grasp the underlying message from the post.

Should I still work with the beta distribution?

Does each percentage being the mean of a binomial variable allow me to consider the sampling distribution of the percentages as binomial?

Is it easier/better to simply work with the count of correct questions instead? So I would have a sample of 50 integers that can take values between 0-15.

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  • $\begingroup$ Do I understand you well that you have 50 percentages, and that each of these 50 percentages is the fraction of 'positive' answers on 15 questions ? Furthermore, can you assume the the chance of answering yes on a question is the same for each question ? $\endgroup$ – user83346 Aug 18 '15 at 15:31
  • $\begingroup$ Hi @fcoppens - that understanding is correct. The chance of getting a correct answer on each question (across all teams) is not the same: some are answered correct most of the time, others less so. This is because each question captures a different element of what is expected in a high quality file. $\endgroup$ – Nick2015 Aug 18 '15 at 15:39
  • $\begingroup$ We have to be precise here, can you give an answer of two different questions ? Because you talk about a correct answer, so that means that you can, for each of these 15 questions say what the correct answer is ? However, the probability of getting a correct answer on question one is different from the probability of getting a correct answer to questions 2 ? $\endgroup$ – user83346 Aug 18 '15 at 15:57
  • $\begingroup$ Sorry for the confusion - each question has a binary response (either 'yes' or 'no'). For each question, I know which of 'yes' and 'no' is correct and therefore which is incorrect. But, as you say, the probability of getting a correct answer on question 1 is different to question 2,3,4,...15. $\endgroup$ – Nick2015 Aug 18 '15 at 16:04
  • $\begingroup$ Sorry to ask so many questions, but I have two additional ones: (1) do you know the probabilities of correctness for each question or do you have an idea about their order of magnitude (is 50pct 1pct 0.1pct ...) and (2) what do you want to make inferences about , the probabibility of a correct answer for each question or the average correctness of each of the teams ? $\endgroup$ – user83346 Aug 18 '15 at 17:25
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All your proportions $X_i$ is the proportion of "yes" (say) from 15 questions. Assuming each question is independent from the others, we have that $$ 15\cdot X_i \sim \text{bin}(15,p) $$ From that, for example, you might calculate confidence intervals for $p$, say that results $(p_l, p_u)$. Those confidence intervals then can be interpreted as confidence intervals for he expected value of your proportions. So, yes, effectively you are working with the counts, not the proportions.

EDIT

I see now from the comments above that we cannot assume equal $p$'s among the 15 questions, so a binomial distribution is not appropriate. As a first approximation, then, you could try a beta-binomial distribution. You could search this site for beta-binomial to get some hints: https://stats.stackexchange.com/search?q=betabinomial

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