This test provides for the demonstration of maintainability when the requirement is stated in terms of both a required mean value (µ_{1}) and a design goal value (µ_{0}) (or when the requirement is stated in terms of a required mean value (µ_{1}) and a design goal value (µ_{0}) is chosen by the contractor). The test plan is subdivided into two basic procedures, identified herein as Test Plan A and Test Plan B. Test A makes use of the lognormal assumption for determining the sample size, whereas Test B does not. Both tests are fixed sample tests, (minimum sample size of 30), which employ the Central Limit Theorem and the asymptotic normality of the sample mean for their development.
ASSUMPTIONS
Test A 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.
Test B No specific assumption concerning the distribution of maintenance times are necessary. The variance d^{2} of the maintenance times is known from prior information or reasonably precise estimates can be obtained.
Hypotheses
H_{0} : Mean = µ_{0} 

(Equation B2) 
H_{1} : Mean = µ_{1}, ( µ_{1} > µ_{0 }) 

(Equation B3) 
Illustration: 
H_{0} : µ_{0} =
30 minutes 

H_{1} : µ_{1} =
45 minutes 
Note that µ_{0} is normally the specified maintainability index value, and that µ_{1} is typically the maximum acceptable value of the specified index.
SAMPLE SIZE  For a test with producer's risk and consumer's risk _{}, the sample size for Test A is given by:
where ^{2} is a prior estimate of the variance of the maintenance times and Z_{} a and Z_{ }are standardized normal deviates. The sample size for Test B is given by:
where ^{2} is a prior estimate of the variance of the maintenance times. Z_{} and Z_{} are standardized normal deviates.
Decision Procedure  Obtain a random sample of n maintenance times, X_{1}, X_{2}, . . . , X_{n}, and compute the sample mean,
and the sample variance
Test A: Accept if 

Test B: Accept if 

Reject otherwise
Discussion  By the central limit theorem, the sample mean is approximately normal for large n with mean E(X) and variance Var (). In Test A, under the lognormal assumption Var = d^{2} where d^{2} = e^{( 2 + )}( e^{2} 1 ) = µ^{2 }(
e^{2} 1 ) Thus the sample size, N, can be computed using a prior estimate of ^{2}. In Test B, a a prior estimate of d^{2} is assumed to be available to calculate the sample size. A Critical value C is chosen such that µ_{0} + Z_{} = C = µ_{1}  Z_{} . If µ = µ_{0} , then P( > C) = and if µ = µ1, then P( C) = .
Example  It is desired to test the hypothesis that the mean corrective maintenance time is equal to30 minutes against the alternate hypothesis that the mean is 45 minutes with = = 0.05.
Then 
H_{0} : µ_{0} = 30 minutes. 

H_{1} : µ_{1} =
45 minutes. 
Test A: Under the lognormal assumption with prior estimate of ^{2} = 0.6, the sample size using equation B4 is:
n_{c} = 

Test B: Under the distributionfree case with a prior estimate of ^{2} = 900, (or = 30), the sample size using equation B5 is:
Operating Characteristic (OC) Curve The OC curve for Test B for this example is given in Figure B4. It gives the probability of acceptance for values of the mean maintenance time from 20 to 60 minutes. The OC curve for Test A for this example is given in Figure B3. It gives the probability of acceptance for various values of the mean maintenance time. Thus, if the true value of m is 40 minutes, then the probability that a demonstration will end in acceptance is 0.21 as seen from Figure B3.
Figure B3  OC Curve for Test A
Figure B4  OC Curve for Test B