# Demonstration Reliability Test Formula with allowance for 'x' Failures

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#### Disenchanted9000

Hi all,

I am in search of a formula that will provide the number of test units required to demonstrate a reliability figure with allowance for 'x' failures.

I know of this formula which allows for zero failures:

N=(ln(1-confidence)/ln(R)), where N is the number of units (round up this answer), R is the reliability from 0-1. Confidence level is also 0-1. Straight forward.

Then, I found this formula which is confusing and also solves for confidence.

______________f_____n!
(1-confidence)= Σ ----------- x (1-R)^i x R^(n-i)
_____________i=0_ i! x (n-i)!

Yikes! Brain hurting....

Need a similar formula to solve for number of units.

Thanks. Robert

#### Miner

##### Forum Moderator
Look starting at page 167 (p. 175 of the pdf file) of Sample Size Calculations Practical Methods for Engineers and Scientists.

The text shows a worked out problem then follows up by demonstrating how to obtain the same solution in Minitab.

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#### Disenchanted9000

Thanks v9991. I think the link to "src.alionscience.com/toolbox/oneshotcalc" is what I was looking for. Will try in Excel.

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#### Disenchanted9000

Thanks Miner. I have a read. Lots of examples to work through.