huangyhg

terminated testing
if you test 22 items from a very large batch to a given number of cycles, with no failures, then you have demonstrated with 90% confidence that each item in your batch has at least 90% probability of reaching that number of cycles.
can someone with greater statistical knowledge than i, give me a hint how to calculate this? i can do it by a monte carlo method, i guess, but what's the proper way?
cheers
greg locock
amazing what a walk round the carpark will do. my brains must be in my feet. forget it.
cheers
greg locock
hello greglock, i have many times a dialog like this:
hi,tom, could you tell me what's the familly name of mary?
which mary, there are five of them?
mary jones of course!
aaah, that one...let me think for a while....eeeeh
regards
m777182
greg,
could you please share the answer you "discovered"? i know the three numbers you mentioned, and i know a little about gaussian distributions, student-t test, and other statistics tidbits, but a concise answer from a respected author would be appreciated. thanks.
regards,
cory
i'll dig out the exact formula tomorrow.
the right way of thinking is that we need no failures.
roughly speaking
p(0 failures)=1-sigma [n=1-22] (p(n failures))
where
p(1 failure) =22*.1*.9^21
p(2 failures)= 2c22 *.01*.9^20
p(3 failures)= 3c22 *.001*.9^19
etc
well as you can see it becomes a permutations and combinations problem - that was the clue. it's called a negative binomial, from memory. i'll enter the correct formula tomorrow - sorry, i don't know the derivation.
cheers
greg locock
thanks!
regards,
cory
are you trying to use your 22 tested samples, all of which passed the test, to deduce the likely proportion of "failures" in your population? if so, then presumably your problem stems from the fact that your sampling encountered zero failures.
there are some texts that present tables of confidence bounds for this (almost degenerate) case of sampling a population whose distribution is taken to be binomial. try:
"principles and procedures of statistics", by r.g.d.steel and j.h.torrie;
or
"statistical tables for use with binomial samples - contingency tests, confidence limits, and sample size estimates", by maitland, herrera and sutcliffe, published by the department of medical statistics, new york university college of medicine.
(this comes from some work notes i made nine years ago. i do not have the actual references, merely some photocopied pages. hence the lack of proper bibliographic data.)
hth
here's the answer i came up with (i notice it doesn't include confidence limits)
n=number of trials until you have r successes
p=probability of success in one trial
c=(n-1)!/((r-1)!*(n-r)!)
q=1-p
p=c*p^r*q^(n-r)
probability of r successes from n trials =1-p
in the case above n=r=22, p=0.9,
with the result 1-p=0.9
cheers
greg locock
ooops
sorry, i display (yet again) my slightly dodgy understanding of statistical confidence.
1-p is the probability that the observation i have made is consistent with n r and p, that is, it is my confidence. he said, confidently.
cheers
greg locock

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借用达朗贝尔的名言：前进吧，你会得到信心!
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