Basically 12 people out of 31 said they felt their care was a 6 or better and I’m curious as to how the probability that someone would say that is calculated
In: 0
If you want to get just a little bit more sophisticated, you could provide an interval in which you are fairly confident that the probability lies.
Very rough estimate: assume the distribution for your belief of the probability is Normal centered at 0.39. Then assume that the width of this Normal distribution is sigma ~ 1/sqrt(N) ~ 1/sqrt(31) ~ 0.18. So, a rough guess would say the probability is 11% < p < 57%.
For this problem, the precise posterior distribution for the probability is a [beta](https://en.wikipedia.org/wiki/Beta_distribution) with alpha = n_successes = 12 and beta = n_fails = 19. Using the formula for the standard deviation of a beta rv, I find sigma = 9%. Thus, I would say that a good estimate for the interval containing the true probability is 30% < p < 48%. (This is called a credible interval btw.)
I included the rough estimate first to show that the width of the interval will get smaller as the number of samples gets bigger via sigma ~ 1/sqrt(N). The more precise estimate using the formula for a beta is what you have given the number of samples you have currently collected.
Latest Answers