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
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39% is a reasonable estimate absent other information. There are more sophisticated statistical approaches you could use here if you knew something about the people rating you or what you “expect” the rating to be, but those tools require quite a bit more information and can be tricky to use.
39% is the best guess without any more info but it’s impossible to know the true probability from just a sample. it could be much less or much more than 39%, and the 31 people sampled just had an unusual amount of people rating 6+.
39% is a reasonable estimate absent other information. There are more sophisticated statistical approaches you could use here if you knew something about the people rating you or what you “expect” the rating to be, but those tools require quite a bit more information and can be tricky to use.
39% is the best guess without any more info but it’s impossible to know the true probability from just a sample. it could be much less or much more than 39%, and the 31 people sampled just had an unusual amount of people rating 6+.
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.
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.
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