Weibull parameter estimate for inequality data

Hi all,

Is it possible to estimate weibull fit parameters using inequality data.

Say my data is:

1,  7,  9,  4,  8,  3,  >9,  11,  14,   >10.

If I remove the ineqality data say, A = [1 7 9 4 8 3 11 14];

and estimate the parameter using

wblfit(A);

I get some values but that's incorrect.

Is there anyway I can include inequality data into my code. Any help is appreciated.

[dpb]

See the documentation for wblfit -- it includes the optional censoring argument variable to indicate which values are censored observations.

[Deepti Ranjan Majhi]

Thank you @dpb for your reply. Yes I tried with censoring argument but the I am still not getting the correct answer.

Is there anything else we can try.

[dpb]

I dunno,

>> wblfit([1,7,9,4,8,3,9,11,14,10],[zeros(1,6) 1 0 0 1])
ans =
   8.516791692357502   2.061950820041830
>> figure, plot([0:60],wblpdf([0:60],w(1),w(2))) 
>>

surely looks pretty reasonable to me given the data..

What do you think the answer should be and how do you determine this is wrong?

[Deepti Ranjan Majhi]

Hi @dpb, thank you for your promt reply.

My actually data:

124, >124, >124, 404, 404, 314, 342, >342, 635

and the corresponding answer is 458.6 and 2.265.

If I try with wblfit:

>> wblfit([124 124 124 404 404 314 342 342 635],[0 1 1 0 0 0 0 1 0])
ans =
  354.0105    2.0894

There is a large difference in both the results.

Is there is any other way to handle this problem.

[dpb]

"and the corresponding answer is 458.6 and 2.265."

And how do you arrive at that?

It's been alongtime™ since I read Lawless on MLE estimation for the censored case, but it's the reference in the MATLAB doc and I'd have pretty high confidence The Mathworks will have implemented the algorithm correctly.

I've got the reference on the shelf in the basement that I can go dig out, but where do you get the other values you claim are correct for comparison?

Hmmm...I see that wblfit returns same estimates regardless of whether use the censoring input or not -- I was thinking there was some adjustment made although it's indeterminate just what those values would be so the estimation problem is theoretically intractable.

If just ignore the truncated observations (which tend towards the lower end), then

 wblfit([124 404 404 314 342 635])
ans =
  416.7561    2.6509
>> 

which is closer to your claimed values.

But, still, how do you come by the above values as being "right" from the given data set?

[dpb]

As for the Q? of "what else can we do?", it would be to write your own MLE function and estimation code -- the Q? is how to derive the proper MLE -- the classical definitions of Type I and Type II censoring are that for Type I censoring one observes N units for a fixed time, T and counts the number of failures, R; the remaining (N-R) units all survived at least T.

Type II censoring is one in which one runs the N units for an indeterminate time until a predetermined number of failures occur; then the test time is T(R).

It's not clear precisely how your test data were arrived at; doesn't appear to precisely fit either of those two models so would need more details to work out what it (the MLE estimator) would end up being.

For some rudiments, see

https://www.itl.nist.gov/div898/handbook/apr/section1/apr131.htm

https://www.itl.nist.gov/div898/handbook/apr/section4/apr412.htm

from the NIST handbook; those are just a couple of the more pertinent sections from the overall reliability chapters.