# Confidence Interval for Attendance

I have a survey where one of the questions is "How likely are you to attend this virtual event?" Responses range from 1 to 5 (least likely to most likely). I would like to know what the 90% confidence interval is for the number of people who responded to the survey that will attend.

My instinct is to use something along the lines of a "binomial proportion confidence interval" but it essentially sounds like I have 5 different coins so I'm a little stuck on combining the 5 resulting intervals. I am fine assuming independence when it comes to any individual attending or not attending.

I'm also trying to encode this in a spreadsheet so although I could come to an answer via running some monte carlo experiments, I'd rather not have to go that route.

• You probably need no reminding that what people say they do may have little bearing on what they actually do. Perhaps a more serious problem is that "least" and "most" likely, and therefore everything in between, are not quantitative. This makes it impossible to translate your survey results into any kind of estimate of attendance. It's hopeless to look for a confidence interval for a number you can't even estimate.
– whuber
Aug 20, 2020 at 21:15

Besides the assumption of independence that you already mentioned, you would probably also need to make the assumption that the behavior of different people is identical ( the other i in iid) conditioned on their "attendance".
In this case, one way to model the survey responses for both of the groups separately is with a multinomial model, the natural extension of the binomial model that you mentioned. (There might be other approaches that make use of the ordinal nature of the 5 boxes, see e.g. ordinal regression). In the multinomial model the probability mass function for the numbers, $$x$$ ($$x \in \mathbb{N}^5 \land \sum_i^5 x_i = n$$), of $$n$$ people checking the corresponding of the 5 boxes is given by $$p(x | \theta) = \frac{\Gamma(\sum_i x_i + 1)}{\prod_i \Gamma(x_i+1)} \prod_{i=1}^5 \theta_i^{x_i}$$ where $$\theta\in (0,1)^5 \land \sum_{i=1}^5\theta_i = 1$$, i.e. $$\theta_i$$ is the probability that a person from this group ticks box $$i$$. As @whuber pointed out in a comment, if you do not know this probability and don't have data (e.g. form previous years) that link the survey responses to actual attendance there is no point in calculating anything and you start guessing right away.
However if you do know these probabilities for both attenders (A) and non-attenders (Q), since you don't know how many attenders you have nor who the attenders are, you would still need to model them at the same time: $$p(x|\theta^A, n_A, \theta^Q, n_Q) = \sum_{\{(x^A, x^Q)| x^A+x^Q=x\land \sum_i^5x^A_i=n^A\land \sum_i^5x^Q_i=n^Q\}} p(x^A|\theta^A)p(x^Q|\theta^Q)$$