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1P-Weibull with Zero Failure Data
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This example validates the calculations for a 1-parameter Weibull with zero failure data in Weibull++ standard folios.
Reference Case
The data set from Table 8.2 on page 196 of the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998 is used.
Data
| Number in State
|
State F or S
|
Time to Failure
|
| 10 |
S |
500
|
| 12 |
S |
1000
|
| 8 |
S |
1500
|
| 9 |
S |
2000
|
| 7 |
S |
2500
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| 9 |
S |
3000
|
| 6 |
S |
3500
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| 3 |
S |
4000
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Result
The formulas for calculating the [math]\displaystyle{ \eta \,\! }[/math] at a given confidence level of [math]\displaystyle{ 1 - \alpha\,\! }[/math] is on page 195.
- [math]\displaystyle{ \hat{\eta}_{L} = \left (\frac{2\sum_{i=1}^{n} t^{\beta}_{i}}{X^{2}_{(1-\alpha ;2)}}\right ) ^{\beta} }[/math]
The 95% lower confidence bound on [math]\displaystyle{ \eta \,\! }[/math] when [math]\displaystyle{ \beta = 2\,\! }[/math] is:
- [math]\displaystyle{ \hat{\eta}_{L} = \left (\frac{2\sum_{i=1}^{n} t^{\beta}_{i}}{X^{2}_{(1-\alpha ;2)}} \right )^{\beta} = 10250\,\! }[/math]
Results in Weibull++
The following picture shows the result in Weibull++:
