Need statistics help - calc. margin of error

[Barry1492]

I'll try to simplify the terms.

I have a total population of 167 widget machines that run 24/7. Each keep a running error log. I've sampled 25 widget machines for 14 days worth or information and these are the results:

3 machines produced errors during the 14 day period.
There were a total of 8 errors among the 25 machines.

ie 12% with an error, .3 errors per machine


Can I calculate a margin of error given this information? If so, how?

 

[tomwalz]

Sure.

Basic math.

Won't do you any good. You can't fix an "error". You can fix an out of spec part if you know how it is our of spec. But "error" is a meaningless term.

About half the problems on here are the result on unclear specifications.

Tell the prof that you would never manufacture to specifications that use the term "error" without defining it so you figured this was a trick question.

Might work.

Thomas J. Walz
Carbide Processors, Inc.

http://www.carbideprocessors.com

Good engineering starts with a Grainger Catalog.

 

[Barry1492]

Hmmm....I seem to be having a hard time explaining myself (ran into this problem on another forum).

I'm an engineer working in diagnostics. We have flags that come up on our machines as "errors." I checked the data base that maintains the logs of all 169 machines in Europe for the last month. I got the information that I needed, but the database has been known to be incomplete.

So I logged directly onto the machine to check their error logs to see if the numbers are similar. However, I don;t have the time to log onto 169 individual machines, so I am sampling and trying to extrapolate the data across the entire population. The reason being, if the error occurs often enough, then R&D will put in a fix. If not, we'll just let the user work around it while we work on bigger fish to fry.

So, given a population of 169 and a sample of 25, I'm looking for the margin of error on the percentage of machines recording errors, and/or the average number of errors per machine.