Tolerance Interval for Standard Uncertainty

#1
Hello all,

Here's my question. I'm reading an old work document written back in the 80s. The purpose is to calculate a 95/95 one-sided tolerance interval for a parameter of interest. To do this, they use the following equation:

x + 1.645*S

where

S = sqrt( (dF/dx * zx * sx)^2 + (dF/dy * zy * sy)^2 + ... )

It is assumed that each parameter, i.e., x, y, ..., follows a normal distribution.

Note that each term has a corresponding z applied to it for a 95% confidence interval. The entire thing is then multiplied again outside the square root by 1.645. Does this give a 95/95 one-sided tolerance interval, or is this closer to a confidence interval?

I've seen a variety of explanations online that explain calculating the tolerance interval as x+KS where S is only based on one parameter, and hence one sample size (n). The k-factor is then looked up based on n. However, in the above scenario, each component of S has a different n. I can't find anything that addresses this scenario.

I'm an engineer that hasn't had to use statistics in almost 20 years, please be gentle with the response. And please let me know if this question is better suited for a different forum.
 

fed2

Active Member
#2
formula for S has delta method written all over it. guessing that n is 'degree of freedom' sort of in this case. charts given with 'n' are most likely designed for usual standard deviation.

use t-test.
 
#3
Hi fed2, thanks for the response.

So, if I understand your suggestion, you're saying to use the student-t values to look up zx, zy, etc, right? I believe this is what they are already doing to come up with the zx, zy, ... values that are applied inside the square root. But I'm wondering if this approach, along with the 1.645 outside the square root, truly gives a tolerance interval? Or is it some sort of confidence interval instead?
 
#5
So I'm not 100% clear on that yet, but I think it's the mean of F determined when calculating all of the standard deviations (sx, sy, etc.).
 

fed2

Active Member
#6
is there any chance x is an estimate of a population percentile? there is a close relationship between confidence intervals on percentiles and tolerance intervals. i have found a remarkable proof of this that will not fit in this post.
 
#7
Hi,

So I've been reading the document to try and get a better grasp for answering your question. Turns out X is 1. Let me explain. The purpose of the calculation is to put an upper limit (via a 95/95 tolerance interval) on the relative change of a parameter. So, if the value "1.645*S" (from my original post) comes to a value of 0.04, their conclusion is that applying a 4% multiplier to my measured value will bound the effects of all of the various uncertainties with a one-sided 95/95 tolerance interval.

All of their "dF/dx * z * x" components are relative values so that the final "1.645*S" is also relative (it makes my head spin reading it). So, if "1.645*S" is 0.04, then the final multiplier is "1+0.04=1.04". Hence why X is a value of 1.

So, in their calculation, dF is a relative change, dx is relative, and s is relative. Then, when you combine all of these relative components together and add them to 1, they give a maximum possible relative change.
 
#9
Any way you could retype the link to the "Tolerance_Limits_..." document? The link seems to be broken when I click on it.

Also, maybe I've stumbled across an answer? As you know, my original concern was that when reading about tolerance intervals, they use a special "k-factor" instead of the more common z factor that I've historically seen. For example, here is an old document that determined k-factors for use in tolerance intervals. However, the document I'm reading doesn't use k factors. First it applies a 95 confidence interval to each of the standard deviations inside the square root (as explained here), and then applies a 1.645 on the outside of the square root. I wanted to try the k-factor approach, but the k-factor that you select is dependent on the number of samples n. But, in this case, each term under the square root has a different n.

Well today I stumbled across this which provides a method to calculate an approximate degree of freedom for a linear combination of variances (which is what I'm doing under the square root, right?). So now I'm thinking I can use this approach to come up with a single value for n, and then use that n to look up my 95/95 k-factor. Thoughts?