×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Terminated testing

## Terminated testing

(OP)
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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Terminated testing

(OP)
Amazing what a walk round the carpark will do. My brains must be in my feet. Forget it.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Terminated testing

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

### RE: Terminated testing

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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

### RE: Terminated testing

(OP)
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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Terminated testing

Thanks!

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

### RE: Terminated testing

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

### RE: Terminated testing

(OP)
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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Terminated testing

(OP)
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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!