Confidence level and reliability level relationship

[thomastong]

Hi,

Could anyone help to tell me where I can see the introduction for the relationship between confidence level and reliability level?

Someone told me that 50% confidence level means nothing. Is that right? But I always see some tests with that confidence level.

What confidence level and reliability level used in prediction for a product are appropriate? Is there a standard instruction to follow? If have, please do me a favor to tell me.

Thanks!

[tarik]

You can start by reading the following reference: http://www.weibull.com/LifeDataWeb/confidence_bounds.htm.

[adam]

Thomas,

It is a correct statement. 50% confidence doesn't mean much. It means that the true reliability could be higher or lower than the estimated value with a 50% probability (same as flipping a coin).

Regarding what reliability and confidence to use, there is no standard value. It depends on the product, and the risks that you (or your company) are willing to take. The only general rule is that for safety items a high reliability high confidence value is used.

[Rui Assis]

Although there is no standard, a connection between reliability and confidence level is well established since long.

The theorem of survival according to Bayes linking reliability and confidence level is given by: R = (1 – C)^[1/(n+1)], where: R – Reliability; C – Confidence level; n – Sample size. This expression can be rewritten as follows: C = 1 – (R)^(n+1). From the first expression we conclude that for a given C, R increases with n. From the second expression we conclude that for a given R, C increases with n.

Example: The failure behaviour of an item follows a Weibull with eta = 25.000 hours and beta = 2.

a) What are the reliability and the confidence level if one unit tested for 12.000 hours without failing?

R = EXP[-(12.000/25.000)^2] = 0,794
C = 1 – 0,794^(1+1) = 0,369

We can conclude that reliability is 79,4% reliable at a confidence level of 36,9%.

b) What are the new values of the reliability and the confidence level if the test is extended up to 20.000 hours?

R = EXP[-(20.000 / 25.000)^2] = 0,527
C = 1 – 0,527^(1+1) = 0,722

As expected, the reliability decreases and the confidence level of the result increases. One can, of course, reach higher levels of confidence faster by increasing the sample size n.

I hope this helps.

[tarik]

I haven’t seen that relationship before. Can you give me a reference for the relationship you presented? It reminds me of the Non-parameteric test design relationship, please read the following material (http://www.weibull.com/LifeDataWeb/test_design.htm under Nonparametric Test Design), this is a non-parametric method in which you don’t have to assume a distribution, and given three of the inputs of CL, Rtest, n (number of units) and nf (number of failures) you can solve for the forth.

[Rui Assis]

The analysis is based on the Weibull distribution and the success-run theorem. The success-run theorem (Bayes´ formula) is a nonparametric equation given by: R = (1 – C)^[1/(n+1)], where: R – Reliability at a confidence level C; C – Confidence level; n – Sample size. In this case, the characteristic life is not needed.

Please refer to Charles Lipson and Narendra J. Sheth, “Statistical Design and Analysis of Engineering Experiments”, McGraw Hill Kogakusha, Tokyo, 1973, pages 178-181.

[tarik]

I went over the example you provided. I have difficulty understanding why would a demonstration test be needed at all if the parameters of the product are already known ‘for sure’. The parameters automatically give you the reliability. In reliability testing, if you use the Parametric Binomial approach, an assumed distribution is required. If you assume a Weibull for example, a beta, not both parameters, value is required. If both parameters were required, then there is no point in testing when you already have a model for the product.
If you are not sure about the parameters, the uncertainty in the parameters can only be obtained from looking at the actual test data from which the parameters were obtained.

Another comment about that example. I tried modeling the data in the example. The example says that there is 1 suspension at 12,000h. If you enter this data into Weibull++, you will be required to assume a beta value and a confidence level that you want your results to be calculated at, in other words a 1-parameter Weibull model is used. This is the common approach that people use to model data that is only made of suspensions (or assume an exponential distribution). So if I enter the beta=2 as the example states and a confidence level of 36.8%. The estimated eta is in fact 1.7715E4 and R(12000)=63.2% unlike what is in the example.

[Rui Assis]

First of all, let me tell you that I am not familiar with product testing. What I know comes mostly from what I read in the Lipson and Sheth book and from attending the Reliasoft forum and just for the sake of curiosity as this is not my professional field. I work as a teacher on industrial engineering at the university – where I teach simulation techniques, economy engineering, operations management and maintenance management. I work also as a consultant engineer in the field of maintenance of process industries (power plants mainly) and develop software for decision making processes.

Now with regard to your first point, I think you are right when you say it is no use testing when you know both parameters beforehand. My opinion is that the text is misleading and shouldn’t say “…, with eta = 25,000…” leaving the impression that eta is known “for sure”. I think it should say “…with an estimated (predicted) eta of 25,000…” instead. Later this value should be revaluated I think. Let me tell you a bit of my experience: Perhaps this is a similar procedure than the one I follow whenever I am faced with the result of an inspection. As matter of fact, in power plants the estimation of remaining life of critical components play a very important role. In these cases one needs to inspect critical parts on a timely basis. When a part is known to follow a Weibull failure distribution, you start by determining a schedule for inspections based on the knowledge of the parameters (beta for sure as it is solely dependent on the mechanism of failure, and a presumed eta), considering an acceptable value of conditional reliability between any inspection moment and the next. Later on, after each inspection takes place resulting negative, you have to adjust the schedule for the inspections to come based on the calculated prior conditional probability of failure, on the quality (accuracy) of the test and on the level of trust/confidence the inspector himself has on the result. Then you apply the Baye´s theorem to account for this and determine a new (adjusted) value of eta which will be used to reschedule future inspections. You can read more in CRTD – Vol. 20-3 (p. 70-77) from ASME. Because this is a common practise in what inspections are concerned, I think (suspect) that the same rationale can be extended to product testing. If you have just one or two prototypes of a very expensive item, you can easily estimate the parameter beta, as it depends on the physics of failure alone, but the same doesn’t apply to eta, as it depends on the environment and operating conditions. Am I right? So, you have to look for it.

Regarding your last point, I am afraid that, because of the same reasons I explained above and despite I have Weibull++ Reliasoft software with me, I don´t know why both answers don´t coincide. I just can say that I feel uncomfortable about that and I am going to give it some more thought. Please let me know if you come to any conclusion.

Lipson & Sheth say, at the end of the example, in page 181, that “...the characteristic life is not really needed…”. I can’t see exactly how but, at least, it conforms approximately to what I said about the inspections.

Thank you for your interest and for the time spent reading these lines.

[tarik]

Thanks for your reply. As for your question regarding estimating eta, usually people are more comfortable estimating a value for beta, based on failure mechanisms, then to estimate eta, there are other ways to estimate the two parameters (for ex, based on two failure probably, at different times, estimates or a combination of failure probably and a parameter estimate). I suggest you also consider the BlockSim software and the new RCM++ software, they can help a great deal in your maintenance and inspection planning and cost analysis.