Estimation of parameters and interpretation of the results

[DILIPAN. A]

when analysing the data shown below, with a qualifying degree of fit using RRX, I find that two parameter weibull model results in Rho of 0.9704, while three parameter fit gives 0.9982 which don't seem to be significantly different, But the B10 are 458.4961 and 37.7946 resp. which are sig. different. Further the failure rate trend is decreasing for Weibull 2 and increasing for webull three. How does one interpret and select a model in such cases. Request y'r help.

NS------LI----STATE----SET (NS- No. in state, LI
_________________________ last inspected, SET-
27------0-------F-------1 state end time)
2628----0-------S-------1
2-------1-------F-------2
1002----1-------S-------2
3-------2-------F-------3
1554----2-------S-------3
3-------3-------F-------4
768-----3-------S-------4
3-------4-------F-------5
1110----4-------S-------5
5-------5-------F-------6
1102----5-------S-------6
11------6-------F-------7
1813----6-------S-------7
8-------7-------F-------8
1488----7-------S-------8
7-------8-------F-------9
1493----8-------S-------9
5-------9-------F-------10
1725----9-------S-------10
9-------10------F-------11
1693----10------S-------11
3-------11------F-------12
1578----11------S-------12
8-------12------F-------13
2104----12------S-------13
2-------13------F-------14
1380----13------S-------14
0-------14------F-------15
1083----14------S-------15
0-------15------F-------16
1078----15------S-------16
0-------16------F-------17
340-----16------S-------17
0-------17------F-------18
80------17------S-------18

1. I tried using MLE method (as per options provided in software), however there is no big difference in our results. Pl clarify.

2. I tried grouping data into two sets and did the analysis separately (took the last 3 months failure data separately and did analysis, there is no correlation in predictions - we took Weeks in service rather than MIS for better analysis.
Given below is the WIS data :
NS LI STATE SET (NS- No. in state,LI-
_________________________ last inspected, SET
10------0-------F-------1 -State End Time)
596-----0-------S-------1
8-------1-------F-------2
594-----1-------S-------2
3-------2-------F-------3
594-----2-------S-------3
2-------3-------F-------4
595-----3-------S-------4
0-------4-------F-------5
251-----4-------S-------5
1-------5-------F-------6
248-----5-------S-------6
0-------6-------F-------7
251-----6-------S-------7
1-------7-------F-------8
251-----7-------S-------8
0-------8-------F-------9
388-----8-------S-------9
0-------9-------F-------10
389-----9-------S-------10
0-------10------F-------11
389-----10------S-------11
0-------11------F-------12
389-----11------S-------12

I appreciate your earliest advise/help in this,

[Dr. Dave]

I asked the poster to e-mail me the data set and then to call me. The modeling choice here involves a decision as to whether or not a failure free period makes sense, and what failure rate behavior is expected a priori. This data set appears to be warranty data, and we will also check that it is correctly tabulated.

If we get a good resolution, we'll post the results of our discussion.

Best wishes to all,

Dr. Dave Olwell

[DILIPAN. A]

Dr.Dave Olwell, The data we posted is warranty data, and we don't think this is a case of failure free period or failures before time zero. I will e-mail, u the data in excel format for your review. Kindly advise on the best fit and how one needs to proceed furthur. Thanks.

[Dr. Dave]

I looked at the data using Weibull++6.0. The best fit appears to be a 3P Weibull, using RRX with the RS ranking method.

A mixed Weibull also appears to fit well, but there are real issues with extrapolating using that model.

The characteristic of interest seems to be B10 life. The 3P model gives a 90% LCB of B10 as 35.1188 using likelihood ratio bounds, and a very small chance (0.03%) that failure time <0. The MW model gives a B10 time of 48.5, but due to numerical issues with the all-interval data, will not return an LCB.

Without knowing more about the physics of failure or the production issues, I'd go with the more conservative answer using the 3P Weibull.

Best wishes,

Dr. Dave

PS Paid commercial announcement: we do this type of analysis with lots more customer dialog at the consulting arm of ReliaSoft, ReliaSoft Professional Services.

[Herjit Gill]

I am using Maximum Likelihood to analyze my data. I have got only a few failures and the failures and suspensions are not spread uniformly. Am I using the correct parameter estimation. Is there also anyway I can use the correlation coefficent when using the MLE?
Any help would be greatly appreciated.

[RS Support]

MLE is better when random suspensions are present in my opinion. As for the correlation coefficient, under MLE analysis it is not really the best measure (but you could use it – and Weibull 6 provides this value even when using MLE). If what you are trying to do is ascertain the adequacy of fit among different distributions for the same data set, then use the value of the likelihood function value provided when using MLE.

[S Goodman]

I have just started looking at failure data using the Weibul distributions. I have been reading some reference materiel which states a shape parameter above 6 or if the plot is not linear but curves then the alpha, beta parameters are not valid. I am looking for more information on valid data sets for Weibull distributions - any recommendations?

[Pantelis]

First, these are really two questions. Furthermore saying that the parameters are not valid is incorrect – the parameters are always valid (if computed correctly) -- the question is “Is the model appropriate for the data set? -- or is a 2-parameter Weibull the right model for my failure data?”

Now with respect to the first question, “If the shape parameter is > 6 is a 2-parameter Weibull an inappropriate model?” This statement is Incorrect – you cannot judge model appropriateness based on the value of some arbitrary cutoff point (i.e. 6) for the model parameters.

Now with respect to the second question. If the data does not follow a linear trend then yes, this is an indication that the current model may be inappropriate and that other life-models should be utilized. It should be noted that a 3-parameter Weibull can be used if the data follows a smooth curve. See http://www.weibull.com/LifeDataWeb/three_parameter_weibull_regression.htm.

Hope this helps.