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Hi Randy:

I am working on an audit engagement and encountered the following scenario.

1 of my sample of 25 in a population of 250 has a deviation from the expected behavior.My sample was selected with 0 deviations expected. Our firm guidance says the sample should be expanded by 15 to 40 to evaluate the control as effective. In other words, 1 error out of 40 appear to be acceptable. It also suggests 2 errors out of 60 samples is acceptable to conclude control is effective. I am trying to understand the underlying concept and rationale. The confidence interval is 90 percent. I would appreciate if you could explain the inputs/rationale that is driving these numbers. Thanks a lot

Hi Mark, Here's what I think is going on.

Recommending a sample size of 40 from an original size of 25 in order to achieve a confidence interval of 90 percent involves a so-called z-score

z90 = ∆p/{p(1-p)/n}^1/2 = 1.645

where

1.645 = value of z-score (= number of standard deviations) that gives a one-sided probability of 95% of values less than z

- this z-score gives a 2-sided (central) probability of 90%, as desired

- this assumes the population distribution of defects is close to normal, which is not a bad assumption in this case

- this statistic is used all the time

p = probability of a sample being defective

- your question states that errors (defects?) of 1-out-of-40, or 1/40 = 0.025, is OK, so I'll take p = 0.025

∆p = deviation of measured probability from expected probability

- here, I assume the measured probability is the one from your first sampling, p = 1/25 = 0.04.

- I'm not sure this is strictly kosher since the the expected number of deviations is so small (i.e., 1), but whatever

- your question says that the expected number of deviations = 0, therefore ∆p = 0.04

- I'm not sure an expected value of precisely 0 is legal but I suppose assuming it is << 0.04 works

{p(1-p)/n}^1/2 = standard deviation of estimate of p

- from assuming a binomial distribution with mean = p

n = number of elements in the sample (sample size)

The goal here is to use the z90 eq to specify a sample size to achieve the 90% level. So, rearranging

n = (z90/∆p)^2･p(1-p)

= (1.645/0.04)^2･(0.025)･(0.975) = 41.2 ~ 40.

So this is probably where the sample size of 40 comes from, although using p = 1/40 and then solving for n = 40 is a little disconcerting.

Your question also states that 2-out-of-60 errors are also acceptable. If we try to calculate p by assuming a binomial distribution and setting probability of 2-out-0f-60, P(2|60), equal to the probability of 1-out-of-40, P(1|40) ,then we get

P(1|40) = {40!/(39!1!)}･p(1-p)^39 = probability of getting 1 error in 40 samples for given p

P(2|60) = {60!/(58!2!)･p^2･(1-p)^58 = probability of getting 2 errors out of 60 samples.

Setting these equal to each other gives

(40)･p(1-p)^39 = (1770)･p^2･(1-p)^58

40/1770 = p･(1-p)^19

ln(40/1770) - lnp = 19･ln(1-p)

This is an implicit equation for p which has to be "solved" graphically by looking at the intersection of the left and right hand sides. This, unfortunately, does not give a clear result: the 2 curves approach each other near 0.05 but do not cross at a well defined point => inconclusive. So I'm not sure where this information fits in. If you can get some information from your "firm guidance" regarding this, please pass it on.

Hope this helps.

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