ADS-79D-HDBK Downloaded from https://www.everyspec.com If a null hypothesis H 0 is formed by assuming R(x) , the probability of not encountering a fault up to a CI value of x , is less than the required reliability of 90%, then N , the minimum number of fault-free items with a CI value = x needed to reject H 0 at the 1-C significance level, is given by: (Equation 10) Where: C = the required one-sided lower confidence bound to reliability x = the CI value to which we test If we substitute Equation 6 , the Weibull reliability equation, into Equation 10 , then the minimum sample size when faulted CI values follow a Weibull distribution is given by:
(Equation 11) Finally, recall Equation 9 , where η is expressed in terms of the threshold CI value and its associated reliability. Substituting Equation 9 into Equation 11 and simplifying gives the minimum number of fault-free items with a CI value = x needed to demonstrate the required reliability R(X 0 ) at X 0 with the required confidence bound C : (Equation 12) There are several noteworthy things about Equation 12 : a. As mentioned earlier, for CBM applications, β should always be greater than one ( β > 1). If β is found to be less than or equal to one, then we can conclude that the associated CI is a poor indicator of component health. b.
The CI value we are testing each item to, x , must always be greater than or equal to our threshold X 0 . c. When x = X 0 , Equation 12 reduces to Equation 10 , which provides a sample size based on the binomial distribution (with N successes out of N trials) that is independent of the CI distribution. d. Most importantly, this method relies on N false positives to demonstrate the required reliability at X 0 . And when x , the CI value we test to, is allowed to increase above our threshold X 0 , the required sample size N decreases. 168