1P-Weibull MLE Solution for Multiple Right Censored Data

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1P-Weibull MLE Solution for Multiple Right Censored Data

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 6.324612E+06.

1PMLE multiple right censored.png