For example, say we have three coins. Due to several tests, we have 95% confidence intervals for X (the heads percentage) of each coin. Coin A is 0-5%, coin B is 1-9%, and coin C is 40-70%.
[For the curious, these CI's were created using inverse binomial CDF, based a small sample of ~30 flips of the coin. Not surprisingly, we have no clue about the composition of the coin-bag these coins were drawn from.]
Can we say anything about the number of heads we would expect to see if we flipped all three? Or if we chose one coin randomly and flipped it? Based on the computed CI's alone? What if we can also use the info that led us to those CI's?
Edit: let me tell you much more about my problem.
We have 50 coins, all of which we suspect are very low probability heads (well under 5%). We do 30 trials, where each trial consists of flipping all the coins and recording all the results. The value we seek is a CI on the average number of heads in a trial.
One approach would be to record the 30 "numbers of heads", and do a standard normal approximation. The problem is that many of these 30 numbers are 0 and 1,
so while we get the correct mean, I fear the resulting CI is way tighter than it has any right to be.
What I would like to do is calculate a CI for each coin, using it's 30 binary outcomes. I have calculated that for 0 heads out of 30, this leads to a 95% CI of 0-11.5% (with a mean of 0, of course). I did this by substituting values for P in "cumulative distribution function(1) of binomial distribution n=30 p=P" in Wolfram Alpha until the result was around 0.025. For 1 heads out of 30, I get a CI of 0-17.2%
Once I have done this, can I add up the CI's in any way?
Best Answer
There are multiple methods to calculate the binomial confidence intervals, and in case of your small (~30) sample size this may mean notable differences.
Can we say anything about the number of heads we would expect to see if we flipped all three? Assuming that the normal approximation was used and was appropriate to get the three confidence intervals we can go on with the normal approximation: expected number of heads $E=\frac{0+5}{2} + \frac{1+9}{2} + \frac{40+70}{2}$, $CI = E ± ½ \sqrt{(5-0)^2 + (9-1)^2 + (70-40)^2}$, see sum of normally distributed random variables
Or if we chose one coin randomly and flipped it? If randomly meant choosing one of these 3 coins with equal probabilities than we expect ⅓ times the expected value and a CI of $\frac{1}{\sqrt{3}}$ times the width of the result of the previous question.
What if we can also use the info that led us to those CI's? I that info could mean all or some of your assumptions before setting up the experiment, the experimental conditions and the observations. In that case you could formulate the full conditional probability, tell a bit more about the coins, tackle the small sample size better, particularly giving more accurate CIs for the previous questions.