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Just as the normal distribution has the 68–95–99.7 rule with 68% of the data within +- 1 standard deviation and so on, does the binomial distribution too has something like that. Or does its being a discrete distribution, make it difficult to have a similar rule.

Actually, I wanted to calculate the number of trials required (binomial distribution case) for a given maximum error (+-0.1) and confidence level (95%) for different probabilities from 0:0.05:1 using the formula: max. error = standard error * no. of standard deviations for 95% confidence (as shown in http://en.wikipedia.org/wiki/Checking_whether_a_coin_is_fair: but instead of a normal approximation, i want to use the binomial distribution).

I would like to use the Clopper-Pearson (CP) interval, but the CP interval gives me the probability range, can I somehow use it to find the number of trials required when I fix the interval to +-0.1.

Any help is highly appreciated. Thanks.

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I guess I have found the number of samples required for 95% confidence interval with a given precision which incorporates the Clopper-Pearson (CP) aka exact method. Also I don't think the binomial distribution has any rule close to the 68-95-99.7 rule of the normal distribution (didn't find anything closer, that's why :-)

The CP interval is asymmetric about the estimated proportion unlike the Wald interval (normal approximation) or the Wilson-score interval, except at mid-p values of around 0.5. Hence, if the CP interval is 0.2 it cannot be used as estimated proportion +-0.1 as I was looking for in the question I posted.

However, the following journal paper:

Krishnamoorthy, K. and Peng, J. 2007. Some properties of the exact and score methods for binomial proportion and sample size calculation. Communications in Statistics—Simulation and Computation 36: 1171–1186.

does calculate the sample size for constructing 95% confidence interval with given precision of 0.01 to 0.2 for both the score and exact method (Table 2 in the mentioned paper) which comes to something like this: n for 0.1 precision with p=0.15:0.05:0.5 is 56,68,79,88,94,99,102,103 trials for CP interval; and which in fact is very close to trials calculated using normal approximation. Here 0.1 precision means upperbound - lowerbound = 2*0.1 and not 0.1 alone.

I hope this answers the above question and wish that someone finds the answer helpful too :-)

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