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| ===Biasing and Unbiasing of Beta===
| | #REDIRECT [[Crow-AMSAA - NHPP]] |
| Eqn. (6) returns the biased estimate of <math>\beta </math> . The unbiased estimate of <math>\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):
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| ::<math>\bar{\beta }=\frac{N-1}{N}\hat{\beta }</math>
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| <br>
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| For failure terminated data (meaning that the test ends after a specified test time):
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| <br>
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| ::<math>\bar{\beta }=\frac{N-2}{N}\hat{\beta }</math>
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| <br>
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| '''Example 1'''
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| <br>
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| 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.
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| <br>
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| <br>
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| <center>Table 5.1 - Developmental test data for two identical systems </center>
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| {|style= align="center" border="1"
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| !Failure Number
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| !Failed Unit
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| !Test Time Unit 1(hr)
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| !Test Time Unit 2(hr)
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| !Total Test Time(hr)
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| !<math>ln{(T)}</math>
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| |-
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| |1|| 1|| 1.0|| 1.7|| 2.7|| 0.99325
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| |-
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| |2|| 1|| 7.3|| 3.0|| 10.3|| 2.33214
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| |-
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| |3|| 2|| 8.7|| 3.8|| 12.5|| 2.52573
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| |-
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| |4|| 2|| 23.3|| 7.3|| 30.6|| 3.42100
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| |-
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| |5|| 2|| 46.4|| 10.6|| 57.0|| 4.04305
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| |-
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| |6|| 1|| 50.1|| 11.2|| 61.3|| 4.11578
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| |-
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| |7|| 1|| 57.8|| 22.2|| 80.0|| 4.38203
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| |-
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| |8|| 2|| 82.1|| 27.4|| 109.5|| 4.69592
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| |-
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| |9|| 2|| 86.6|| 38.4|| 125.0||4.82831
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| |-
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| |10|| 1|| 87.0|| 41.6|| 128.6|| 4.85671
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| |-
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| |11|| 2|| 98.7|| 45.1|| 143.8|| 4.96842
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| |-
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| |12|| 1|| 102.2|| 65.7|| 167.9|| 5.12337
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| |-
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| |13|| 1|| 139.2 ||90.0||229.2|| 5.43459
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| |-
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| |14|| 1|| 166.6|| 130.1|| 296.7|| 5.69272
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| |-
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| |15|| 2|| 180.8|| 139.8 ||320.6||5.77019
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| |-
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| |16|| 1|| 181.3|| 146.9|| 328.2|| 5.79362
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| |-
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| |17|| 2|| 207.9|| 158.3 ||366.2||5.90318
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| |-
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| |18|| 2|| 209.8|| 186.9|| 396.7|| 5.98318
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| |-
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| |19|| 2|| 226.9|| 194.2|| 421.1|| 6.04287
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| |-
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| |20|| 1|| 232.2|| 206.0|| 438.2|| 6.08268
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| |-
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| |21|| 2|| 267.5|| 233.7|| 501.2|| 6.21701
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| |-
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| |22|| 2|| 330.1|| 289.9|| 620.0|| 6.42972
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| |}
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|
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| '''Solution'''
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| <br>
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| For the failure terminated test, using Eqn. (amsaa6):
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| ::<math>\widehat{\beta }=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}}</math>
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| :where:
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| <br>
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| ::<math>\underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355</math>
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| <br>
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| :Then:
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| <br>
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| ::<math>\widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142</math>
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| <br>
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| From Eqn. (amsaa5):
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| <br>
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| ::<math>\widehat{\lambda }=\frac{22}{{{620}^{0.6142}}}=0.4239</math>
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| <br>
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| Therefore, <math>{{\lambda }_{i}}(T)</math> becomes:
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| <br>
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| ::<math>\begin{align}
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| & {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\
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| & = & 0.0217906\frac{\text{failures}}{\text{hr}}
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| \end{align}</math>
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| <br>
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| Figure 4fig81 shows the plot of the failure rate. If no further changes are made, the estimated MTBF is <math>\tfrac{1}{0.0217906}</math> or 46 hr.
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| <br>
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| <br>
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| [[Image:rga5.1.png|thumb|center|400px|Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.]]
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| <br>
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