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A product being designed will have a failure-free warranty requirement of 1 year and the Engineers are planning a series of tests to estimate our warranty exposure.

Based on the binomial confidence interval analysis procedure described in section 7.2.4.1 of the NIST/SEMATECH e-Handbook of Statistical Methods I have computed the following in advance of the test. I used the exact intervals for small numbers of failures.

  • If out of a random sample of 20 units, 4 units fail the 1 year warranty simulation test, I believe we can say that we are 90% confident that the population defect rate will lie between 7.1% and 40.1%
  • If out of a random sample of 20 units, 1 unit fails the 1 year warranty simulation test, I believe we can say that we are 90% confident that the population defect rate will lie between 0.26% and 21.6%
  • If out of a random sample of 20 units, 0 units fail the 1 year warranty simulation test, I believe we can say that we are 90% confident that the population defect rate will be 13.9% or less.
  • If out of a random sample of 1 unit, zero units fail the 1 year warranty simulation test, I believe we can say that we are 90% confident that the population defect rate will be 95% or less.

Now assuming no failure occurs at the 1 year point in the test(s), continuing the test(s) beyond the 1 year warranty period with a some corresponding benefit in confidence would be desirable as the test units are costly to build. Can the above analysis procedure be extended to answer the following questions when the simulation test time exceeds the actual warranty period?

  1. At the 90% confidence level, what are the upper and lower limits of the population defect rate if a 1.5 year warranty simulation test is run on 1 unit and that unit does not fail?

  2. At the 90% confidence level, what are the limits of the population defect rate if 2 units are warranty simulated. One unit fails at 1.5 years and the other fails at 4 years?

  3. Can Monte Carlo be used here to help answer/validate these questions?

Many thanks for any guidance.

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  • $\begingroup$ I changed the title to make the question clearer/more attractive in the main list, please check if I did not inadvertently changed the meaning of your question. $\endgroup$ – Gala Jul 24 '13 at 15:21
  • $\begingroup$ Gael, thanks for making the edits, I think they add clarity as this question is at the intersection of applied statistics, reliability engineering and decision making. Also thanks for adding the hyper-link to the e-handbook . . . I'm a newbie at this site. $\endgroup$ – Steve Jul 24 '13 at 15:29
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With respect to the underlying statistical technique, whether the duration is shorter or longer than the warranty seems immaterial. You are just making inference on a proportion and you can indeed use the same technique for another point in time (your question 1).

That said, the time points need to be specified in advance. You should not collect data continuously and then choose the point at which you compute a confidence interval based on these data. For example, you should not:

  • Simulate two years of product life, notice that two units fail after 20 months and compute a confidence interval at 19 months purporting to show that reliability is high at this point in time.
  • Compute many tests/confidence intervals (say one every month or more) and just pick one that suits you (say the last one with an upper bound under a particular threshold or something like that).

In both cases, you are increasing the error level by (implicitly or explicitly) running many tests and the confidence intervals will be misleading.

A better way to approach this if you want something else than a confidence interval for a pre-specified time point (and an answer to your question 2) is to model the time-to-failure instead of the proportion of defective/failure rate. Survival analysis techniques are also relevant.

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  • $\begingroup$ thank you for the reply. I’m assuming that you’ve confirmed my 4 sample calculations or at least 1 of them allowing us to be on the “same computational page”. I’m encouraged when you say that I can use the same technique for another point in time but I don’t see quite how to do that in regards to the two questions I posted. The technique is able to handle varying “points in time” but I believe only if those points in time are equivalent. My questions involve nonequivalent time points. [I've reached the character limit, more to come] $\endgroup$ – Steve Jul 24 '13 at 17:51
  • $\begingroup$ The technique as you know takes 3 inputs and produces 2 outputs: Input 1: Confidence Level (only interested in 0.90 for all analyses) Input 2: The Sample Size Input 3: Number of defectives or test failures (this has to assume a test duration that is consistent with the interpretation of the outputs) Output 1: The 90% Lower Confidence Limit associated with the time duration embodied in Input #3 Output 2: The 90% Upper Confidence Limit associated with the time duration embodied in Input #3 [More to come] $\endgroup$ – Steve Jul 24 '13 at 17:52
  • $\begingroup$ I have trust in the internal consistency of my 4 sample calculations above because for each calculation the time durations assumed in Input #3 (number of failures) is equivalent to that required by the interpretation of the two outputs, namely 1 year; that is a simulated 1 year test duration and a 1 year warranty requirement. However, the questions I have involve the case where the two time durations are not equal as in my Question 1.[More to come] $\endgroup$ – Steve Jul 24 '13 at 17:54
  • $\begingroup$ Here we have the warranty requirement (associated with the confidence limit outputs) as remaining at 1 year but the time duration applied during the test (and associated with the inputs) is 1.5 years. How can the technique deal with this time inconsistency? Don’t we have apples and oranges? If I’m not seeing something here can you run the numbers and show your results (inputs and outputs) for my Questions 2 & 3 ? That way I should be able to reproduce them. Many thanks. $\endgroup$ – Steve Jul 24 '13 at 17:55
  • $\begingroup$ Typo in my last posting, sorry. Not "Questions 2 & 3" but Questions 1 & 2 . . . . $\endgroup$ – Steve Jul 24 '13 at 18:50

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