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Time-to-Failure Data

Fielded Systems Data

Goodness-of-Fit Tests for the Crow-AMSAA Model

Goodness-of-Fit Tests for Repairable Systems

Appendix B: Common Beta Hypothesis Tests

The Common Beta Hypothesis (CBH) Test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet. As shown by Crow [17], suppose that Κ number of systems are under test. Each system has an intensity function given by:

MATH (4)

where q = 1, ... ,Κ . You can compare the intensity functions of each of the systems by comparing the βq 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 β1 = β2 =...= βΚ. Let $\tilde{\beta}_{q}$ denote the conditional maximum likelihood estimate of βq, which is given by:

MATH (5)

where:

Then for each system, assume that:

MATH

are conditionally distributed as independent Chi-Squared random variables with 2Mq degrees of freedom. When Κ = 2, you can test the null hypothesis, Ho, using the following statistic:

MATH (6)

If Ho is true, then F equals MATH and conditionally has an F-distribution with (2M1,2M2) degrees of freedom. The critical value, F, can then be determined by referring to the Chi-Squared tables. Now, if Κ ≥ 2, then the likelihood ratio procedure [17] can be used to test the hypothesis β1 = β2 =...= βΚ. Consider the following statistic:

MATH

where:

Also, let:

MATH

Calculate the statistic D, such that:

MATH (7)

The statistic D is approximately distributed as a Chi-Squared random variable with (Κ - 1) degrees of freedom. Then after calculating D, refer to the Chi-Squared tables with (Κ - 1) degrees of freedom to determine the critical points. Ho is true if the statistic D falls between the critical points.

Common Beta Hypothesis Example

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 MATH, test the hypothesis that β1 = β2 while assuming a significance level equal to 0.05. Calculate $\tilde{\beta}_{1}$ and $\tilde{\beta}_{2}$ using Eqn. (5). Therefore:

MATH

Then MATH. Using Eqn. (6) calculate the statistic F with a significance level of 0.05.

MATH

Since 1.2408 < 2.0980 we fail to reject the null hypothesis that β1 = β2 at the 5% significance level.

Now suppose instead it is desired to test the hypothesis that β1 = β2 = β3. Calculate the statistic D using Eqn. (7).

MATH

Using the Chi-Square tables with Κ - 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, we fail to reject the null hypothesis that β1 = β2 = β3 at the 5% significance level.