J
James Oldham
Question regarding the following formula for fraction defective of a population (from Juran Quality Handbook):
n=p(1-p)*(Z/E)^2
If you choose E=.05, does this imply the population fraction defective of the sample will be within .05% or 5% of the actual fraction non-conforming?
n=p(1-p)*(Z/E)^2
If you choose E=.05, does this imply the population fraction defective of the sample will be within .05% or 5% of the actual fraction non-conforming?