1P-Weibull MLE Solution for Multiple Right Censored Data

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1P-Weibull MLE Solution for Multiple Right Censored Data

This example validates the calculations for a 1-parameter Weibull MLE solution with right censored data in Weibull++ standard folios.


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 variance of eta is 6.324612E+06. The standard deviation is 2514.88.

1PMLE multiple right censored.png