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Validate 99.5% reliability 90% confidence  Sample size calculation
If I state a component is 99.5% reliable at 90% confidence, I can calculate the sample size needed to validate the R&C target is 460 samples. I would expect to see .5% cumulative failures in the field for the life of the product. Sample size = ln(1C)/ln(R)
Someone has stated if the test is performed using the 95th percentile usage profile (more harsh usage profile than the median user), then I can validate the same reliability using a smaller sample size. For example, I can use a sample size of 22, to meet 99.5% reliability with 90% confidence if I use a 95th percentile usage profile. I expect .5% cumulative failures in the field for the life of the product?
Are the previous statements correct? Can I reduce the sample size from 460 to 22, and validate 99.5% reliability of the population by adjusting the usage profile?
Calculation:
I test to R90/C90, I need 22 samples.
Population Reliability = 1(195th percentile user)(1Reliability)
=1(1.95)(1.9)
=R99.5%

8th December 2008, 06:14 AM


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Re: Validate 99.5% reliability 90% confidence  Sample size calculation
Can someone help here?
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12th December 2008, 05:03 PM

Willy Grunfeld
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Re: Validate 99.5% reliability 90% confidence  Sample size calculation
Someone has stated if the test is performed using the 95th percentile usage profile (more harsh usage profile than the median user), then I can validate the same reliability using a smaller sample size.
I never heard of 95th percentile usage profile and don't know what is meant by that. It doesn't sound right to me. The word harsher sugests you are referring to some accelaration factor. Other than temperature which can be used as an acceleration factor and for which the relationships are known I don't know of any acceleration factor that can be used readily.
Assuming that you are referring to accelerated testing, the test time can be reduced or conversely the sample size if the test time remains unchanged. I wouldn't take that formula as correct unless you have the full reference how is it derived. ooks suspicious to me
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12th December 2008, 05:18 PM

Willy Grunfeld
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Re: Validate 99.5% reliability 90% confidence  Sample size calculation
"If I state a component is 99.5% reliable at 90% confidence, I can calculate the sample size needed to validate the R&C target is 460 samples. I would expect to see .5% cumulative failures in the field for the life of the product. Sample size = ln(1C)/ln(R)"
99.5% reliable means there is a 0.995 probability it will fail at time t. You can't "expect" =calculate the number of failures without knowing the failure rate and the time.
90% confidence is just what it sounds that there is a 90 %probability that the actual reliability will indeed be 99.5% and a 10% chance that it won't.
I don't know where did you take that formula to calculate the sample size....you can't talk about sample size without taking time into consideration. The assumption (for constant failure rate) is that 100 parts operating for 1,000 hours is the same as 1000 parts operating for 100 hours.
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23rd December 2008, 10:53 AM

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Re: Validate 99.5% reliability 90% confidence  Sample size calculation
w_grunfeld, Thanks for your reply!
The sample size equation comes from the number of samples for the lower binomial confidence bound, for a zero failure test. The R&C are life targets, and the zero failure test is for 1 life. I should have clarified.
Thanks again for your response!

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