Sample size for cycle testing based on reliability/confidence 3

[mjduffy]

I have been asked by my client to test a rear trunk assembly to a specific number of cycles, to a specified reliability and confidence. How do I determine how many samples to be tested? These assemblies are not tested to failure and there is no failure data available. I'm reviewing my stats text book but nothing's jumping out at me. If anyone can help, note that I'm knowledgable in engineering math in general, but a statistics ignoramous.

 

[GregLocock]

I hate suspended tests (ie tests that are terminated before failure). Without an estimate of the standard deviation (or whatever) I don't see how you can do it. Ask their Supplier Technical Assistance guy to help you, evil grin.




Cheers

Greg Locock

 

[patprimmer]

I studied stats many years ago, but I still remember to support Greg's statement. You need to test to failure, then do standard deviations then you can test for degrees of confidence based on sample size and variation.

As I remember, if you test say 10 specimens at a time, then test 10 times to give a total of 100 specimens, you get a higher degree of confidence than one test of 100 specimens

Regards
pat pprimmer@acay.com.au
eng-tips, by professional engineers for professional engineers
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[GregLocock]

I've just thought of an example.

Suppose you had a million components, and you tested them for 10 cycles each, and they all passed.

What have you learned about their ability to reach 20000 cycles?

Absolutely nothing.


Cheers

Greg Locock

 

[Worried]

I too am in the situation you describe.There is a website called "alionscience.com" which has a calculator for "one-shot devices" that calculates any one of sample size, reliability, confidence and number of failures. (Tell it any three, and it will calculate the fourth.) There is also a good discussion paper on the site about sampling plans, which applies to this situation. I also have an nth generation photocopy of "Larson's Binomial Nomograph" which does the same thing, and agrees with the calculated results. My problem is that I do not know anything about the derivation of this graph, so when I tell someone that 95% reliability @ 50% confidence requires 14 samples with no failures, they don't believe me, and I can't do the proof.
Given that your post is 6 months old as I write this, I'm hoping that you are now an expert and can help me!

Thanks
Dave Parent

 

[NickE]

Just to add IIRC the standard deviation and number of smples is governed by an inverse root law. IE:

Sigma (stddev) is proportional to one over the root of the number of samples.

Since at some point the root of n increases much slower than n, above some number n(o) there is no way to reduce variation by testing more.

 

[GregLocock]

Worried, is your problem that you have no tests to failure? Or that you don't know how to prove that a given sample size is too small to be helpful?

The first I cannot help with. The second is in any stats book beyond highschool, I think.

http://mathworld.wolfram.com/BinomialDistribution.html

for a rather mathematical approach

http://en.wikipedia.org/wiki/Binomial_distribution

for a readable approach



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Worried]

Greg;

I apologise for being unclear in my original post. My problem is that when asked to specify how many items should be tested, to generate 95% Reliabiltiy @ 50% Confidence, (as an example,) I can quote a number from various calculators or graphs, but cannot justify the number with the appropriate reasoning and/or math. Naturally, designers and managers want to prove out a new design with the minimum number of expensive prototypes. My response to the above example would "14, if all the individuals pass the test." This usually causes people to say "That's too many. I'll test two."
All I want to do is educate myself as to how the above-mentioned graphs are created. Your suggested sources are very helpful in that regard.

Thanks
Dave Parent

 

[GregLocock]

We call that test to bogey, and you are right, it takes an awful lot of samples to generate a reasonable confidence (I don't think 50% confidence is going to satisfy anyone!). Far better to test one part to failure or some ridicuous multiple of the bogey and then consider your options.

In your example you would have 50% confidence in 95% reliability at 14 passes out of 14 tests because

(hopefully) 0.95^14 = 0.5 (actually 0.4876)

That is, in order to get 14 passes the first one has to pass, which has a 95% probability, and the second one has to pass. also 95%, so the probability of them both passing is 0.95^2

etc etc.

Much more difficult to work out for other confidences.





Cheers

Greg Locock

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[buckeye1160]

Contact MGA testing lab. They are an industry leader in automotive testing.

 

[GregLocock]

Reinvented statistics have they?

Cheers

Greg Locock

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[ivymike]

If I understand the problem correctly, then what you have is a sample N of parts which have not failed in X cycles, and you want to be able to say that you have Y confidence that Z (95) percent of the parts produced will make it past X cycles.

In other words, you want to check whether the cumulative probability of failure from 0 to X cycles is less than (1-Z) 5%, and you need to know how many not-failed samples you need at X hours to be able to say with Y confidence that the above is the case.

Without breaking out a stats book, it seems to me that Greg's approach is the right one. You want to have less than (1-Y) chance that your lack of failures happened randomly, which means that you need to test enough parts that (Z)^N<(1-Y).

If you tested 59 parts to X hours and saw no failures, then you could say that there was only 4.85% chance that this would happen randomly if the worst-allowable failure rate were assumed to be the actual failure rate. 5% failure rate gives 95% passing and 95%^59 =
4.85%.

With 59 passes, you could also say that you're 99.8% sure that the failure rate at X hours is less than 10%. 1-((90%)^59) = 99.8%