I notice my organization does not follow to NIST standards and I seek for help from the community:

All the while, my company has been using the following; and I am often confused on the validity:

The binomial equation used to predict the probability of a specific number of r defects or failures in n samples is:

P(r ≤ k) = k pr (1 – p)(n – r)

∑ n!

r!(n – r)!

r=0

Where: p = lot tolerance percent defective.

n = sample size.

r = number defective.

P(r ≤ k) = probability of k or fewer defective or failed units in a sample size of n units.

Confidence Level = CL = 1 – P(r ≤ k)

To obtain a 95% Confidence Level with 90% reliability, for 0 failure allowed, the required sample size is:

0.95 = 1 – P(r ≤ k)

P(r ≤ k) = 0.05

0.05 = k (0.1)0 (1 – 0.1)(n – 0)

∑ n!

0!(n – 0)!

r=0

= 0.9n

n = 29 (Round up)

However; from what I learn from MINITAB, the sample size calculation should use margin of error.

I googled abit and saw this: http://www.mddionline.com/article/sa...error-approach

This based on NIST.

n = (1.96 sq) x (pq) / (d sq)

where n=sample size, p=percent conforming, q=1-p, d=margin of error

If I plug in my company requirement (95% CI and 90% reliability) into the NIST eqn:

n=29, p=10%=0.1, q= 1-0.1 =0.9

29 = 1.962 x 0.1 x 0.9 / (E2)

E ≈ 0.109

E ≈ 10%

Is this correct?

Q1) Is my company protocol using binomial equation valid?

Q2) How to relate my company protocol to the NIST equation?

Appreciate the help from the community, many thanks in advance!!!

Regards

GP ]]>