yihfenchen
September 1st, 2000, 03:34 PM
[Originally Posted: 3/24/00-- Transferred by ReliaSoft Moderator]
In your FAQ section, you discuss how the required number of units are calculated. Here you define P to be the required confidence level in %. In equation format, P is equal to Prob(X>=k), where X is a binomial random variable and k is the number of desired successes. I think P should be 1-Prob(X>=k). For example, using your notation assume k=N. Then P = Z^N. Solving for N you get N =k) and solving for N you get N>= ceiling(ln(1-P)/ln(Z)). Now, let P= 95% and Z= 96%. Using the first equation, N = 74. The second value mathes the value given by DRT. So, P should be defined to be the probability of observing at most k-1 successes or the probability of observing at least N-k+1 failures. Is this correct or am I interpreting the documentaion in the FAQ incorrectly.
In your FAQ section, you discuss how the required number of units are calculated. Here you define P to be the required confidence level in %. In equation format, P is equal to Prob(X>=k), where X is a binomial random variable and k is the number of desired successes. I think P should be 1-Prob(X>=k). For example, using your notation assume k=N. Then P = Z^N. Solving for N you get N =k) and solving for N you get N>= ceiling(ln(1-P)/ln(Z)). Now, let P= 95% and Z= 96%. Using the first equation, N = 74. The second value mathes the value given by DRT. So, P should be defined to be the probability of observing at most k-1 successes or the probability of observing at least N-k+1 failures. Is this correct or am I interpreting the documentaion in the FAQ incorrectly.