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Forum: Reliability & Maintainability Questions and Answers

Topic: Reliability & Maintainability Questions and Answers

Topic Posted by: Reliability & Maintainability Forum (src_forum@alionscience.com )
Organization: System Reliability Center
Date Posted: Mon Aug 31 12:47:36 US/Eastern 1998

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Original Message:

Posted by: Prasad (hariprasad@telco.co.in )
Date posted: Wed Jun 18 5:04:47 US/Eastern 2003
Subject: Sample size
Message:
Hi, I am working in an automobile company. Can you please help me out, what is the minimum sample size required for Reliability testing? Since there is time and cost factors are involved, How many samples should i take for testing? Is there any standard for deciding sample size?


Reply:

Subject: Sample Size
Reply Posted by: J.L.Romeu (jromeu@alionscience.com )
Organization: RAC
Date Posted: Thu Jun 19 16:04:29 US/Eastern 2003
Message:
It is known that, if the time to failure (t) is Exponential (with mean = ) and the test is failure terminated, then (2T / ) is distributed as a Chi Square with 2n degrees of freedom (2n) But if the test is time truncated (i.e. stopped arbitrarily at any time) then: (2T / ) is distributed as a Chi Square with 2n+2 degrees of freedom (2n+2) and this is the relationship which fundaments the sample size requirement. Kececioglu (1) recommends that the device's MTTF goal (denoted mG) be taken to be equal to mL1, the lower, one-sided confidence limit on the mean. In this case, for pre specified values of the risk , of sample size n and of total test time T, we have: mL1 = mG = 2T / n that is, the mean will be, with confidence (1-), larger than mL1 when the test time is T, the accepted risk is  and the sample size is n. With some algebra, and for the case where no failures are observed during testing (i.e. DF =2*0+2 = 2) we get: T = mG * ( 1/2 * 2 ) = mG * ( 1/2 * 4.605) = 2.3025 mG (for =0.1) T = mG * ( 1/2 * 2 ) = mG * ( 1/2 * 5.991) = 2.9955 mG (for =0.05) We can reverse this statement and obtain an estimate of the required sample size N, for the same constraints above expressed and prescribed test time td: T = 2.9955 mG = N* td; (for =0.05) Where the prescribed test time td, based on delivery schedule, is multiplied by the number of samples N to obtain the total test time T. With some additional algebra: N = 2.3025* mG / td ; for Confidence Limit of 90% N = 2.9955* mG / td ; for Confidence Limit of 95% This formula will solve the case where no failures are expected and you test for a pre specified time td. In general, the situation may be more complex than that, but this formula will provide a first approximation. Let's see an example: Assume prescribed test time td, is 100 hours and that the target MTTF, mG is 10000 hours. For a risk =0.05 (Confidence Limit of 95%) we obtain a sample size N: N = 2.9955* mG / td = 2.9955 * 10000 / 100 = 299.6  300 You will need 300 samples on test, for these prescribed 100 hours. Bibliography: 1. Keccecioglu, D. Reliability and Life Testing Handbook (page 200, Volume 1). Prentice Hall. 1993


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