1. ## No Standard Deviation

Hello statistical geniuses,

I have 450 items and I want to test a sample and be sure with 95% confidence that if all of my tested items are functioning properly, that 95% of my total population will be functioning properly. Is this possible? Thanks a lot for the help.

2. ## Re: No Standard Deviation

Why do you want to use a sample when you have the population? Normally you sample because you can't obtain the whole population.

I don't understand what you mean by "functioning properly" or what this has to do with standard deviation.

3. ## Re: No Standard Deviation

Originally Posted by noetsi
Why do you want to use a sample when you have the population? Normally you sample because you can't obtain the whole population.
Just because you can make a list of the items in the population doesn't mean you have the measurements on those observations.

I don't understand what you mean by "functioning properly" or what this has to do with standard deviation.
My guess is that "functioning properly" is just what it sounds like - does the item do what it is supposed to do. It's a yes/no response. My guess is that the OP is somewhat familiar with sample size calculations for continuous data but doesn't have experience dealing with binary data and wants input on how to do a sample size calculation for this type of scenario.

4. ## Re: No Standard Deviation

Hi, sorry for the confusion. I have the entire population, but given the amount of time and work involved, I don't want to have to test all 450 items. I was just curious if there was a way to test a sample, say 50, and if all 50 work properly then I can be X% confident that all 450 will work properly. Thanks very much for your help, maybe it's not possible.

5. ## Re: No Standard Deviation

"functioning properly" is a yes or no response, sorry if my original post was a bit cryptic.

6. ## Re: No Standard Deviation

hi,
this would be something called a success run test.

The probability of an item picked from your population functioning properly will be R = (1-CL)**(1/N) where CL is your confidence interval and N is the number of tested units, provided all N functioned properly.

Regards
rogojel

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