Failure data too small

[Wardoyo]

Currently we are always find unreasonalble result when calculate small data using weibull method. Some of our case like follows:
We have three failure 6595, 7191, 13013, and we have 21 suspension data 18,193, 10,159, 11,101, 13,230, 18,521, 18,397, 14,328, 16,310, 11,539, 17,156, 15,911, 14,799, 11,101, 6,160, 16,481, 11,611, 16,304, 16,310, 18,507, 233, 15,911.

We periodically review our component life to identify the appropriate component life for our budget, we choose B50 as our reasonable budget.
If we find the data like this the result always too high. It's critical we can't use unreasonable result, it made our budget unreasonable also. Other wise if we neglect the suspension data the result will too small and the budget will unreasonable also

Is there any suggestion method to treat the data like this?
Thanks In advance happy new year for all

[Tarik El-Azzouzi]

Can you tell me what analysis method are you using (MLE, regression..), what distribution are you assuming, the results you are obtaining and whether you are using conficence bounds.

you also check these references for analysing censored data http://www.weibull.com/LifeDataWeb/right_censored_suspended_data.htm

[Wardoyo]

We use Weibull distribution, the method we use is both RRX and MLE, and we still can't accept the result from these two method. So far the result can exceed 3-4 times our expected value so we don't have any idea t use confidence bound...

[Tarik El-Azzouzi]

I entered your data into Weibull++
Using Rank Regression on X, I obtain the following:
Beta = 2.336225046
Eta = 26280.60151
B50 = 2.2465E4

Using MLE, I obtain the following:
Beta = 1.746394194
Eta = 45733.69094
B50 = 3.7076E+4

Let me know if my results don’t match yours.

It could be that under the current status your product doesn't meet goal. If you are very surprised with your results, can you describe how you obtained this data, how the test was performed and how the data was collected.

[Wardoyo]

Tarik,
For your info, from these data we get B50 36,184. The value mean hours, so far we confidence the component life is about 20,000 hours. Some other case (other component) may result too high than this case. Sometime we get about 100,000 hours other, ot above. To reduce the value we try to use Weibull 1 parameter, but it must be use Beta 12 or 15 so the result more reasonable. Analyse this way may confusing when we have to assume Beta, How much the Beta must be placed.

[Wardoyo]

In the machine we have hour meter to measure the component life in hours. Our mechanic always take the component hours when they remove the component from the machine. And the data always keep in the our system. For component run on data we have the tools to measure the current machine hours so the current component hours can be measure by compare the current machine hours and machine hours when the component installed.
Your result using RRX is more reasonable, does it mean if the data too small using RRX method is recommended?

[Tarik El-Azzouzi]

1- I can’t give you suggestions about what Beta value to use if you assume a 1-parameter Weibill. You can use historical data/engineering judgment. Since you have enough data in this data set, why do you want to you use a 1-parameter Weibull? If you want to involve prior knowledge, I suggest you use a Bayesian approach (supported in Weibull++) which would give you a way to use prior knowledge and what the new data is telling you.

2-Regression is normally preferred over MLE in the case of small data sets. MLE is usually the preference in cases of heavily censored data. However, these are not fixed rules that always apply, you need to assess each method depending on the type of data you have.

3-I have a question. When you say that you measure the hours on the unit when you check it…do you actually know when the unit exactly failed? Or do you only know that it failed some time after the last inspection and the current inspection?
In the first case you can use exact data, in the second, you should use interval data.