RELIABILITY TRENDS TEST
Name:
Type:
Purpose:
Compute tests to determine whether or not the repair times of
a system show significant trend. The trend may be either an
improvement or a degradation.
Specifically, it computes the following three tests
 The reverse arrangement test.
 The military handbook test.
 The Laplace test.
Description:
Determining if repair times exhibit a trend is necessary for
choosing an appropriate reliability model. These formal tests
for trend in repair times are a compliment to the Duane plot
for detecting trend.
Each of these tests will be described separately.
 The Reverse Arrangement Test
Given r repairs, T_{1}, T_{2}, ....,
T_{r}, the interarrival times
I_{2}=R_{2}T_{1},
I_{3}=T_{3}T_{2}, ....,
I_{r}=T_{r}T_{r1} and the
censoring time T_{end} > T_{r}, we
calculate how many instances we have of a later
interarrival time being strictly greater
than an earlier interarrival time. These are called
reversals. Too many reversals indicates a significant
improving trend and too few reversals indicates a
significant degradation trend. More formally,
 Count a reversal every time I_{j} <
I_{k} for some j and k with j < k.
 Compute the total number of reversals, R.
 For r repair times, the maximum possible number of
reversals is r(r1)/2.
 If there are no trends, the expected number of
reversals is r(r1)/4.
 For r > 12, the following approximation can be
used to determine if the number of reversals is
statistically significant.
The test statistic is compared to a standard
normal distribution. That is, if z > 1.96, we
have statistical significance at the 95% level.
For r <= 12, tabled values are used.
The advantage of this test is that it is simple and it
makes no assumptions about a model for the possible trend.
 The Military Handbook Test
Given r repairs, T_{1}, T_{2}, ....,
T_{r} and the censoring time
T_{end} > T_{r}, we calculate the test
statistic
This test statistic follows a chisquare distribution with
2*r degrees of freedom.
This test is recommended for the case when the choice is
between no trend and a nonhomogeneous Poisson process
(NHPP) power law (Duane) model.
 The Laplace Test
Given r repairs, T_{1}, T_{2}, ....,
T_{r} and the censoring time
T_{end} > T_{r}, we calculate the test
statistic
This test statistic follows a standard normal distribution.
This test is recommended for the case when the choice is
between no trend and a nonhomogeneous Poisson process
(NHPP) exponential law model.
If you have internet access, you can enter the following
command to see a discussion of the power law model
WEB HANDBOOK NHPP POWER LAW
Syntax:
RELIABILITY TREND TEST <y>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Examples:
RELIABILITY TREND TEST Y
RELIABILITY TREND TEST Y SUBSET TAG > 0
Note:
These commands require a censoring time. Enter the following
command before the RELIABILITY TREND TEST command to
specify the censoring time:
Default:
Synonyms:
Related Commands:
DUANE PLOT

= Generate a Duane plot.

POWER LAW RANDOM NUMBERS

= Generate random number from a NHPP power law model.

INTERARRIVAL TIME

= Compute interarrival times of a variable.

Applications:
Implementation Date:
Program:
REVERSE ARRANGEMENTS TEST:
NUMBER OF FAILURE TIMES = 37
OBSERVED NUMBER OF REVERSALS = 215
MAXIMUM POSSIBLE NUMBER OF REVERSALS = 666
EXPECTED NUMBER OF REVERSALS = 333
VALUE OF TEST STATISTIC = 21.40790
90% CRITICAL VALUE FOR NO IMPROVEMENT = 1.282
95% CRITICAL VALUE FOR NO IMPROVEMEMT = 1.645
99% CRITICAL VALUE FOR NO IMPROVEMENT = 2.33
MILITARY HANDBOOK TEST:
NUMBER OF FAILURE TIMES = 37
CHISQUARE TEST STATISTIC VALUE = 35.72037
CHISQUARE TEST STATISTIC CDF VALUE = 0.4925163E04
IMPROVEMENT TEST
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
NO TREND (0.000,0.900) ACCEPT
NO TREND (0.000,0.950) ACCEPT
NO TREND (0.000,0.990) ACCEPT
DEGRADATION TEST
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
NO TREND (0.100,1) REJECT
NO TREND (0.050,1) REJECT
NO TREND (0.010,1) REJECT
LAPLACE TEST:
NUMBER OF FAILURE TIMES = 37
NORMAL TEST STATISTIC VALUE = 3.416969
NORMAL TEST STATISTIC CDF VALUE = 0.9996833
IMPROVEMENT TEST
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
NO TREND (0.000,0.900) REJECT
NO TREND (0.000,0.950) REJECT
NO TREND (0.000,0.990) REJECT
DEGRADATION TEST
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
NO TREND (0.100,1) ACCEPT
NO TREND (0.050,1) ACCEPT
NO TREND (0.010,1) ACCEPT
