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1P-Weibull MLE Solution for Multiple Right Censored Data
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This example compares the calculation for a 1-parameter Weibull MLE solution with right censored data.
Reference Case
The data set in Table C.5 on page 633 in 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
|
288 |
S |
50
|
148 |
S |
150
|
1 |
F |
230
|
124 |
S |
250
|
1 |
F |
334
|
111 |
S |
350
|
1 |
F |
423
|
106 |
S |
450
|
99 |
S |
550
|
110 |
S |
650
|
114 |
S |
750
|
119 |
S |
850
|
127 |
S |
950
|
1 |
F |
990
|
1 |
F |
1009
|
123 |
S |
1050
|
93 |
S |
1150
|
47 |
S |
1250
|
41 |
S |
1350
|
27 |
S |
1450
|
1 |
F |
1510
|
11 |
S |
1550
|
6 |
S |
1650
|
1 |
S |
1850
|
2 |
S |
2050
|
Result
The formulas for calculating the ML [math]\displaystyle{ \eta\,\! }[/math] and the standard error of [math]\displaystyle{ \eta\,\! }[/math] are given on page 193.
- [math]\displaystyle{ \hat{\eta}=\left (\frac{\sum^{n}_{i=1}t^{\beta}_{i}}{r} \right)^{\frac{1}{\beta}}\,\! }[/math] and [math]\displaystyle{ se_{\hat{\eta}}=\frac{\hat{\eta}}{\beta}\sqrt{\frac{1}{r}}\,\! }[/math]
where [math]\displaystyle{ \beta\,\! }[/math] is given, [math]\displaystyle{ t_{i}\,\! }[/math] is the time for the ith observation, r is the number of failures. Appling this equation, we get the following results:
- [math]\displaystyle{ \hat{\eta}=\left (\frac{\sum^{n}_{i=1}t^{\beta}_{i}}{r} \right)^{\frac{1}{\beta}} = 12320.33\,\! }[/math] and [math]\displaystyle{ se_{\hat{\eta}}=\frac{\hat{\eta}}{\beta}\sqrt{\frac{1}{r}} = 2514.88\,\! }[/math]
Results in Weibull++
The standard deviation of eta is 2.529845E+07^0.5 = 5029.756 (SYN-I-2958).