```Date: Mar 17, 2013 4:44 PM
Author: RGVickson@shaw.ca
Subject: Re: Estimate failure rate: Variable degree of freedom in chi-square

On Saturday, March 16, 2013 10:27:21 PM UTC-7, Paul wrote:> I've found conflicting information about the degrees of freedom to use> > in the chi-square distribution when estimating failure rate from the> > number of failures seen over a specified period of time.  To be sure,> > the lower MTBF (upper failure rate) always uses 2n+2, where n is the> > number of failures.  However, the upper MTBF (lower failure rate) is> > shown as using both 2n and 2n+2, depending on the source.  I haven't> > found an online explanation of exactly how the chi-square distribution> > enters into the calculation (other than http://www.weibull.com/hotwire/issue116/relbasics116.htm,> > which I'm still chewing on).  So I haven't been able to determine> > whether 2n or 2n+2 is correct from first principles at this point.> > Based on the reasoning in the above weibull.com page, however, I am> > inclined to believe that the degrees of freedom should be 2n because> > we're talking about the two tails of the *same* distribution for upper> > and lower limits.  But this leaves the mystery of why 2n+2 shows up> > frequently.  Is the reason for this straightforward enough to explain> > via this newsgroup?I wanted to reply to David Jones's response below, but my browser is forcing me to reply instead to you.With regard to David's point (1): _given_ n observed outcomes in (0,T) the outcome times are UNIFORMLY distributed in (0,T); that is, the individual arrival times are the n order statistics of the distribution U(0,T). Therefore, the observed inter-arrival times are NOT exponential and are NOT independent. (This is a fundamental and well-known property of Poisson processes.)
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