Re: quality control
- From: Richard Ulrich <Rich.Ulrich@xxxxxxxxxxx>
- Date: Tue, 14 Nov 2006 01:28:41 -0500
On Mon, 13 Nov 2006 10:04:03 -0600, David Winsemius
<doe_snot@xxxxxxxxxxx> wrote:
Richard Ulrich <Rich.Ulrich@xxxxxxxxxxx> wrote in news:vnsfl2d685snrulj63m9iel784mhgnjop8@xxxxxxx:
On 12 Nov 2006 17:00:04 -0800, "Frank" <deps_bear@xxxxxxxxx> wrote:
If I know a product fails .01% of the time and I have 1500 items I'm
running through a process. How many items do I need to check with,
say, 99% confidence that all the items are built correctly.
How many failures do you expect? Almost always, zero.
This is dealing with exact probabilities. For a higher failure
rate, you might want to look at the p of success, and raise
to a power, e.g., (.9999)^n . For the tiny p of 0.01%,
the figuring can be pretty much additive
You want to have only so many items *unchecked* that there
will be, on the average, only 1 bad item in 100 samplings --
so that 99 times out 100, there will be none.
You expect 1 failure in 10,000. One hundred samplings
that each fail to test 100 items will meet that condition.
So you need to check 1400 of each 1500.
How did you go from 100 samplings of size 100 to the number 1400?
What I got was 100 samplings of 100, with a long-term average
of 1 defect per 10,000. The 1500 and subsequently 1400 are
largely irrelevant to the problem, as I construed it. I agree that
it is not necessarily a good way to devise a real-life test of
defects. As you point out, it is even more difficult to assure
that there are *no* defects, than what you get with this strong
assumption. Basically, you might as well go with 100% testing.
If the long term rate is 1 per 10,000, then any sample of 100
has a (very close to) 99% chance of being clean. That's
easy arithmetic, following stern logic.
However you did it, you are then claiming that after examining 1400
items that you are 99% confident that there are no defective items in
the remaining 100 items, when the past experience indicated that the
failure rate was 0.0001 (so it was quite unlikely that you would have
observed _any_ even if all 1500 were examined? Why not stop at 1300?
Seems to me that you might have incorrectly inverted the problem to one
of reducing the size of the population at risk.
The best answer to a binomial acceptance sampling problem in a finite
population that I have seen recently was given by Ted Harding in
Medstats:
<http://groups.google.com/group/MedStats/msg/65e62e78dfe9b3f0?dmode=source&hl=en>
When I apply his method I get a confidence level of .9333 for an
expected rate of 1/1500 and an observed of zero after 1400 were sampled
and that is a much higher predicted rate (at least on a ratio scale)
than the OP specified. (code in the R system.)
I took the expected rate of 1/10,000 as guaranteed --
I don't see how you fit that into a 2x2 table, which is
what the hypergeometric describes, e.g., Fisher's Exact Test.
*Do* the entries of phyper ( ) represent a 2x2 table in
some fashion? For problems of finite samples, with larger
numbers of errors, the counts can be construed as
numbers (okay-seen, defects-seen; yet-to-see-okay,
defects-yet-to-see) -- for a table with cells (A,B; C,D).
That is a problem with the sample providing the rates.
N<-1500; M<-1; x<-0;[1] 0.9333333
k<-1400; 1-phyper(x,M,N-M,k)
So even with a higher prior than specified, you cannot get close to 99%.
If you assume a prior of 2/1500, you can get to 0.9956 after sampling
1400.
Huh? A higher prior *what*? I don't get this. A higher defect
rate should not have a higher confidence of more zeros
(Even after 1350, you can get to 0.99006 with the rate of
0.000667.) I see no way that with an expected as low as 0.0001 and
sample size of 1500 that you can anywhere near the confidence level
specified.
--
Rich Ulrich, wpilib@xxxxxxxx
http://www.pitt.edu/~wpilib/index.html
.
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