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1P-Weibull with Zero Failure Data
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This example compares the calculation for a 1-parameter Weibull with zero failure data.
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
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State F or S
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Time to Failure
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10 |
S |
500
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12 |
S |
1000
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8 |
S |
1500
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9 |
S |
2000
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7 |
S |
2500
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9 |
S |
3000
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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 results in Weibull++: