Hi M_1:
I have been thinking about your question some more, and I came up with an interpretation that feels right to me. It it based on the following
"Were this procedure to be repeated on multiple samples, the calculated confidence interval (which would differ for each sample) would encompass the true population parameter 90% of the time"
from
With a Gaussian distribution, which is symmetrical about the mean, it is reasonable to find a +/- x, where x is a multiple of the standard deviation, such that the probability that a repeat of the experiment will result in a new calculated mean m that will be inside the range m
0 +/- x (where m
0 is the old mean) a specified percentage of the time. That is, for example, if x = 2 x standard deviation:
(1) Probability of m ∈ {m0 - x, m0 + x} > 95%.
So, we want to make a similar statement for the experiment that resulted in 99 h and 1 t. However, the result for this experiment is at the tail of the binomial distribution, which is far from Gaussian, and even from being symmetrical. So instead of (1) we want something like
(2) Probability of m ∈ {Min, 100} > 95%.
To find the value of Min, calculate the binomial probability distribution distribution for the range of integers 100, 99, 98, . . ., Min so that
(3) P100 + P99 + ... + PMin ≥ 95%.
To calculate these values:
(4) P100 = 0.99100
(5) Pk = Pk+1 (1/0.99) (k+1) / (100-k)
My guess is that Min will be about 94.
One more observation. This calculation is not exactly what the quoted interpretation above says. However, I think it will be a good approximation of what it says.
Regards,
Buzz