This test provides for the demonstration of maintainability when the requirement is stated in terms of both a required critical percentile value ( T_{1} ) and a design goal value ( T_{0} ) [or when the requirement is stated in terms of a required percentile value ( T_{1}) and a design goal value ( T_{0}) 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 lognormal distribution. The variance, ^{2}, of the logarithms of the maintenance times is known from prior information or reasonably precise estimates can be obtained.
HYPOTHESES
H_{0} : (1p)th percentile, X_{P} = T_{0} (Equation B10)
or P[X > T_{0} ] = p
H_{1} : (1p)th percentile, X_{P} = T_{1 } (Equation B11)
or P[X > T_{0}] = p, ( T_{1} > T_{0} )
Illustration:

H_{0} : 95th percentile = X_{P} = X_{.05} = T_{0} =
1.5 hours
ln T_{0} = 0.4055
H_{1} : 95th percentile = X_{P} = X_{.05} =
T_{1} = 2 hours
ln T_{1} =
0.6932 
SAMPLE SIZE  To meet specified a and b risks, the sample size to be used is given by the formula
where:
^{2} is a prior estimate of ^{2}, the true variance of the logarithms of the
maintenance times.
Z_{P} is the standardized normal deviate corresponding to the (1p)th percentile.
DECISION PROCEDURE  Compute:
Reject otherwise.
DISCUSSION  This test is based upon the fact that under the lognormal assumption, the (1p)th percentile value is given by X_{P} = e^{( + ZP )}. Taking logarithms gives lnX_{P} = + Z_{P}, and using maximum likelihood estimates for the normal parameters and , the (1p)th percentile maximum likelihood estimate is ln_{P} = + Z_{P} . ln X_{P} is approximately normal. To meet the producer's risk requirements, a critical value X* is chosen for the sample estimate of the (1p)th percentile X_{P}. Note = is an estimate for .
Example  The following hypotheses are to be tested at = _{} =
.10.
H_{0} : 95th percentile = X_{.05} = 1.5 hours = T_{0}; ln T_{0} =
.4055
H_{1} : 95th percentile = X_{.05} = 2.0 hours = T_{0}; ln T_{0} =
.6932