2
$\begingroup$

This is new statistics space for me, so please excuse my ignorance.

I have some data (N=180) that isn't normally distributed (verified in Minitab, P<0.005) and I want to use the data to get a +/- 4 sigma specification limit. Basically, I want to say:

"We can expect 99.993666% of parts to have a measured dimension between DATA_LSL and DATA_USL."

None of the distributions in Minitab fit with good confidence, and so I found that sometimes you can do bootstrapping, which is something like re-sampling your data to get an idea of the uncertainty in some statistic based on 1000+ random re-samples. I think I got that right, but I may be mistaken.

Does anyone know how to bootstrap the standard deviation in Minitab? I found a macro that does it for the mean, and I can put in whatever confidence interval that I want. I think that only tells me about the mean, though, and not the standard deviation.

There are some papers that talk about bootstrapping for LSL/USL, but nothing that I can find that's an example/macro in Minitab.

Any help or advice would be greatly appreciated :)

Thanks, Adam Hartman Mechanical Engineer - Flextronics

stat_stackexchange.15.arrakis@spamgourmet.com

$\endgroup$
0
1
$\begingroup$

Edit: The standard deviation won't give you the limits you seek. In particular 4 sd's will not give you a 99.993666% interval.

You don't even have Chebyshev (which only would give you an 93.75% interval) ... because it needs the population standard deviation

Accurate/reproducible bootstrapping of tail quantiles without strong (effectively parametric) assumptions requires a really large original sample... Straight out maxima and minima don't perform well under the standard sort of nonparametric bootstrap (see the reference), let alone trying to bootstrap way past the sample extrema (however that works!).

I would suggest not trying it with a sample size many times greater than $1/p$, where $p$ is the proportion in the tail - at least ten times, perhaps more, might be feasible. For your data, that suggests you need an original sample size of over 300,000, not 180. A more sophisticated approach - maybe an $m$-out-of-$n$ bootstrap, say - might let you do something where $n$ is around $1/p$ ... which is 31576, if I did it right.

If you want to get away with $n=180$, you you're pretty much going to need some strong parametric assumptions. Even then, the assumptions will be critical - even a small change in assumptions will produce very large changes in the estimates.

--

Bickel, P.J. and Freedman, D. (1981). Some asymptotic theory for the bootstrap.
Ann. Stat. 9, 1196- 1217.

See also this answer:

It will not work for estimating extremes regardless of the population distribution.

$\endgroup$
3
  • $\begingroup$ I understand about half of what you said :) To estimate the standard deviation of my non-normal data, I will need either 30k or 300k measured data points? Can I estimate anything with 180 data points? I'm worried that I've explained the problem wrong. $\endgroup$
    – AHartman
    Mar 5 '13 at 11:53
  • $\begingroup$ "To estimate the standard deviation of my non-normal data, I will need either 30k or 300k measured data points" -- Your question was about constructing a "99.993666%" interval. If a standard deviation gave you the limits, it would be different and you could get away with less. It doesn't. Even the Chebyshev theorem doesn't save you. Neither sample standard deviation nor population standard deviation will get you that interval you asked for. $\endgroup$
    – Glen_b
    Mar 5 '13 at 13:41
  • $\begingroup$ @AHartman I edited my answer to clarify stuff I should have included before, but it may be that some clarification in the question would help. Specifically, what is the problem you're really trying to solve? You can get an estimate of of the s.d., and you should be able to bootstrap that, but I had understood that you were ultimately trying to make a particular probability statement... and the estimate of sigma won't really do that for you as it stands. $\endgroup$
    – Glen_b
    Mar 5 '13 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.