II. Probability and Statistics for Reliability
B. Statistical inference
2. Statistical interval estimates (Evaluate)
Compute confidence intervals, tolerance intervals, etc., and draw conclusions from the results.
The common metric, MTBF, has confidence intervals.
Additional References
Confidence Intervals for MTBF (article)
Perils of MTBF (NoMTBF.com article)
Quick Quiz
1-21. A randomly failing component is tested to X failures and a total of N observations are made. How many degrees of freedom are used in calculating confidence limits on the mean?
(A) 2X + 2
(B) 2X
(C) N
(D) N − 1
(B) 2X
There are a few clues in the question that help narrow down the problem to the correct formula to examine. First, “randomly” implies the exponential distribution. The testing terminated when it reached “X failures” suggesting a failure terminated test design. It calls× for the calculation of the “mean” “confidence interval”.
MTBF is the mean of the exponential distribution and the lower one-sided confidence interval is
$$ \frac{2T}{\chi _{\left( \alpha ,\text{ }2r \right)}^{2}}\le \theta $$
where 2r (or in terms of this problem 2X) is the degrees of freedom.
I cannot get the same answer in the Type 1 censoring example.
For lower bound:
T = 4 units x 250 hr = 1000 hrs
α = .10
r= 4
Is this correct?
I think you are right, the answer noted in the lecture is 128.97 and I’m now finding 125.17… T = 4 x 250, r = 4 alpha=0.1 making chi-squared value found in table of 15.987…. thus 2x 1000 / 15.978 = 125.17
Thanks for the catch, not sure what I did wrong the first time.
Cheers,
Fred
Good! We agree, except my chi-square .1, 10 reads 15.987. I think you flipped the digits. Still close enough by any measure.
Yep, moving too fast…. and fixed now.
Cheers,
Fred