This test provides for the demonstration of maintainability when the requirement is stated in terms of both a required critical percentile value ( T1 ) and a design goal value ( T0 ) [or when the requirement is stated in terms of a required percentile value ( T1) and a design goal value ( T0) is chosen by the system developer]. If the critical percentile is set at 50 percent, then this test method is a test of the median. The test is a fixed sample size test. The decision criterion is based upon the asymptotic normality of the maximum likelihood estimate of the percentile value.
ASSUMPTIONS - Maintenance times can be adequately described by a log-normal distribution. The variance, 2, of the logarithms of the maintenance times is known from prior information or reasonably precise estimates can be obtained.
H0 : (1-p)th percentile, XP = T0 (Equation B-10)
or P[X > T0 ] = p
H1 : (1-p)th percentile, XP = T1 (Equation B-11)
or P[X > T0] = p, ( T1 > T0 )
||H0 : 95th percentile = XP = X.05 = T0 =
ln T0 = 0.4055
H1 : 95th percentile = XP = X.05 =
T1 = 2 hours
ln T1 =
SAMPLE SIZE - To meet specified a and b risks, the sample size to be used is given by the formula
2 is a prior estimate of 2, the true variance of the logarithms of the
ZP is the standardized normal deviate corresponding to the (1-p)th percentile.
DECISION PROCEDURE - Compute:
DISCUSSION - This test is based upon the fact that under the log-normal assumption, the (1-p)th percentile value is given by XP = e( + ZP ). Taking logarithms gives lnXP = + ZP, and using maximum likelihood estimates for the normal parameters and , the (1-p)th percentile maximum likelihood estimate is lnP = + ZP . ln XP is approximately normal. To meet the producer's risk requirements, a critical value X* is chosen for the sample estimate of the (1-p)th percentile XP. Note = is an estimate for .
Example - The following hypotheses are to be tested at = =
H0 : 95th percentile = X.05 = 1.5 hours = T0; ln T0 =
H1 : 95th percentile = X.05 = 2.0 hours = T0; ln T0 =