Appendix B: Hypothesis TestsAppendix B: Hypothesis TestsThis chapter includes the following sections:
The Common Beta Hypothesis (CBH) test is applicable to the following data types: Multiple Systems (Unknown Equivalent Times), Repairable and Fleet. As shown by Crow [17], suppose that K number of systems are under test. Each system has an intensity function given by Eqn. 1.
(1)
where q
= 1, ... , K. You can
compare the intensity functions of each of the systems by comparing the
of each system.When conducting an analysis
of data consisting of multiple systems, you expect that each of the systems
performed in a similar manner. In particular, you would expect the interarrival
rate of the failures across the systems to be fairly consistent. Therefore,
the CBH test tests the hypothesis, Ho,
such that 
. Let 
denote the conditional maximum likelihood estimate of 
is given by:
(2)
where:
K = 1
Mq
= Nq if data
on the qth system is time
terminated or Mq = (Nq
- 1) if data on the qth system is failure
terminated (Nq is the number of failures
on the qth system)
Xiq is the ith time-to-failure on the qth system
Then for each system, assume that:
are conditionally distributed as independent Chi-Square random variables with 2Mq degrees of freedom. When K = 2, you can test the null hypothesis, Ho, using the following statistic.
(3)
If Ho is true, then F
equals 
and has conditionally an F-distribution
with (2M1, 2M2)
degrees of freedom. The critical value, F, can then
be determined by referring to the Chi-Square tables. Now, if 
, then the likelihood ratio procedure [17] can be used to test the hypothesis 
. Consider the following statistic:
where:
Also, let:
Calculate the statistic D, such that:
(4)
The statistic D is approximately distributed as a Chi-Square random variable with (K - 1) degrees of freedom. Then after calculating D, refer to the Chi-Square tables with (K - 1) degrees of freedom to determine the critical points. Ho is true if the statistic D falls between the critical points.
Consider the data in Table B.1.
Table B.1 - Repairable system data
|
|
System 1 |
System 2 |
System 3 |
|
Start |
0 |
0 |
0 |
|
End |
2000 |
2000 |
2000 |
|
Failures |
1.2 |
1.4 |
0.3 |
|
|
55.6 |
35 |
32.6 |
|
|
72.7 |
46.8 |
33.4 |
|
|
111.9 |
65.9 |
241.7 |
|
|
121.9 |
181.1 |
396.2 |
|
|
303.6 |
712.6 |
444.4 |
|
|
326.9 |
1005.7 |
480.8 |
|
|
1568.4 |
1029.9 |
588.9 |
|
|
1913.5 |
1675.7 |
1043.9 |
|
|
|
1787.5 |
1136.1 |
|
|
|
1867 |
1288.1 |
|
|
|
|
1408.1 |
|
|
|
|
1439.4 |
|
|
|
|
1604.8 |
Given that the intensity function for the qth
system is 
, test the hypothesis that 
while assuming a significance level equal
to 0.05. Calculate 
and 
using Eqn. 2. Therefore:
Then 
. Using Eqn. 3 calculate the statistic
F
with a significance level of 0.05.
Since 1.2408 < 2.0980 the hypothesis is accepted that 
at the 5% significance level.
Now suppose instead it is desired to test the hypothesis that 
. Calculate the statistic D using Eqn.
4.
Using the Chi-Square tables with K -
1 = 2 degrees of freedom, the critical values at the 2.5
and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since 0.1026 <
D < 5.9915,
the hypothesis is accepted that 
at the
5% significance level.
The Laplace Trend test tests the hypothesis that a trend does not exist within the data. The Laplace Trend test is applicable to the following data types: Multiple Systems (Unknown Equivalent Times), Repairable and Fleet. The Laplace Trend test can determine whether the system is deteriorating, improving, or if there is no trend at all. Calculate the test statistic, U, using the following equation.
(5)
where:
T = total operating time (termination time)
Xi = age of the system
at the ith
successive failure
N = total number of failures
The test statistic U
is approximately a standard normal random variable. The critical value
is read from the Standard Normal tables with a given significance level,
.
Consider once again the data in Table B.1. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic U for System 1 using Eqn. 5.
From the Standard Normal tables with a significance level of 0.10, the
critical value is equal to 1.645. If -1.645 <
U
< 1.645 then accept the hypothesis of no trend.
However, since U
< 1.645 then an improving trend exists within
System 1. If U
> 1.645
then a deteriorating trend would exist.
Table B.2 displays the critical values for the Cramér von Mises goodness-of-fit
test given the sample size, M, and the
significance level, 
.
Table B.2 - Critical values for Cramér von Mises test
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