Template:Biasing and unbiasing of beta camsaa: Difference between revisions

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[[Image:rga5.1.png|thumb|center|400px|Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.]]
[[Image:rga5.1.png|center|500px|Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.]]
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Revision as of 17:09, 19 April 2012

Biasing and Unbiasing of Beta

Eqn. (6) returns the biased estimate of [math]\displaystyle{ \beta }[/math] . The unbiased estimate of [math]\displaystyle{ \beta }[/math] can be calculated by using the following relationships. For time terminated data (meaning that the test ends after a specified number of failures):

[math]\displaystyle{ \bar{\beta }=\frac{N-1}{N}\hat{\beta } }[/math]


For failure terminated data (meaning that the test ends after a specified test time):


[math]\displaystyle{ \bar{\beta }=\frac{N-2}{N}\hat{\beta } }[/math]


Example 1
Two prototypes of a system were tested simultaneously with design changes incorporated during the test. Table 5.1 presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.

Table 5.1 - Developmental test data for two identical systems
Failure Number Failed Unit Test Time Unit 1(hr) Test Time Unit 2(hr) Total Test Time(hr) [math]\displaystyle{ ln{(T)} }[/math]
1 1 1.0 1.7 2.7 0.99325
2 1 7.3 3.0 10.3 2.33214
3 2 8.7 3.8 12.5 2.52573
4 2 23.3 7.3 30.6 3.42100
5 2 46.4 10.6 57.0 4.04305
6 1 50.1 11.2 61.3 4.11578
7 1 57.8 22.2 80.0 4.38203
8 2 82.1 27.4 109.5 4.69592
9 2 86.6 38.4 125.0 4.82831
10 1 87.0 41.6 128.6 4.85671
11 2 98.7 45.1 143.8 4.96842
12 1 102.2 65.7 167.9 5.12337
13 1 139.2 90.0 229.2 5.43459
14 1 166.6 130.1 296.7 5.69272
15 2 180.8 139.8 320.6 5.77019
16 1 181.3 146.9 328.2 5.79362
17 2 207.9 158.3 366.2 5.90318
18 2 209.8 186.9 396.7 5.98318
19 2 226.9 194.2 421.1 6.04287
20 1 232.2 206.0 438.2 6.08268
21 2 267.5 233.7 501.2 6.21701
22 2 330.1 289.9 620.0 6.42972

Solution
For the failure terminated test, using Eqn. (amsaa6):

[math]\displaystyle{ \widehat{\beta }=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}} }[/math]
where:


[math]\displaystyle{ \underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355 }[/math]


Then:


[math]\displaystyle{ \widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142 }[/math]


From Eqn. (amsaa5):


[math]\displaystyle{ \widehat{\lambda }=\frac{22}{{{620}^{0.6142}}}=0.4239 }[/math]


Therefore, [math]\displaystyle{ {{\lambda }_{i}}(T) }[/math] becomes:


[math]\displaystyle{ \begin{align} & {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\ & = & 0.0217906\frac{\text{failures}}{\text{hr}} \end{align} }[/math]


Figure 4fig81 shows the plot of the failure rate. If no further changes are made, the estimated MTBF is [math]\displaystyle{ \tfrac{1}{0.0217906} }[/math] or 46 hr.