## Sample size for cycle testing based on reliability/confidence

## Sample size for cycle testing based on reliability/confidence

(OP)

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.

## RE: Sample size for cycle testing based on reliability/confidence

Cheers

Greg Locock

## RE: Sample size for cycle testing based on reliability/confidence

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

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## RE: Sample size for cycle testing based on reliability/confidence

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

## RE: Sample size for cycle testing based on reliability/confidence

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

## RE: Sample size for cycle testing based on reliability/confidence

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.

## RE: Sample size for cycle testing based on reliability/confidence

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

http

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.

## RE: Sample size for cycle testing based on reliability/confidence

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

## RE: Sample size for cycle testing based on reliability/confidence

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

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

## RE: Sample size for cycle testing based on reliability/confidence

## RE: Sample size for cycle testing based on reliability/confidence

Cheers

Greg Locock

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

## RE: Sample size for cycle testing based on reliability/confidence

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%