Showing 90% Confidence in 99.99% Correct "Threshold Number" | Statistics Help

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we need to show with 90% confidence that the probability of the "treshhold number" being correct is 99.99%. "So we want P=0.9999 with C=0.90". we can consider 5 trails without put { 4, 7, 12, 9, 11}..we can say the treshhold # is the "2nd trail = 7"... i tried doing it using t alpha but i failed to get it... i haven't taken statistcs in years..thanks for help folks
 
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is it that hard??
 
sstudent said:
is it that hard??

Yes. I don't understand what you are talking about. Can you give a reference for the background to this problem?
 


Ok, now can you explain what your problem is? You have five trials? You want to say the probability of something is 0.9999? The probability of what exactly?
 


the probabilty of the treshhod # being correct, which is the 2nd trail..thanks
 


sstudent said:
the probabilty of the treshhod # being correct, which is the 2nd trail..thanks

This isn't helping. You are just repeating what I don't understand. What do the trials represent? Are they samples from some distribution? What does it mean for one of them to be 'correct'. What is correct? What's a 'threshold'? Can you define all of these things?
 


the trails are just sample from some distributaion, for the treshhold # u can say its just a numer, so we are trying to prove the the output of the samples larger than 7 "treshhold" is 99.99% with 90%confience..i hope that makes it more clear
 


I believe you are talking about the tolerance interval. In this problem, your confidence level is 95% and the threshold interval for capturing at least 99.99% i.e. % of these samples falling in the tolerance interval is:
\bar{x} \pm (tolerance)s , where tolerance is your tolerance critical value.

Your statistics book should have a tolerance critical value table.
 
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