Berton,
   
  Thanks for your inupt. The 'nist' link you mentioned was one of the reasons for my confusion and how it is implemented in R. As for now I am assuming predict function with 'prediction' option will provide me tolerance/prediction interval. Is this a proper assumption?
   
  TIA for your help.
  Sachin
  

Berton Gunter <gunter.berton@gene.com> wrote:
  Peter et. al.:
> 
> With those definitions (which are hardly universal), tolerance
> intervals are the same as prediction intervals with k == m == 1, which
> is what R provides.
> 
> 

I don't believe this is the case. See also:

http://www.itl.nist.gov/div898/handbook/prc/section2/prc263.htm

This **is** fairly standard, I believe. For example, see the venerable
classic text (INTRO TO MATH STAT) by Hogg and Craig.

To be clear, since I may also be misinterpreting, what I understand/mean is:

Peter's definition of a "tolerance/prediction interval" is a random interval
that with a prespecified confidence contain a future predicted value;

The definition I understand to be a random interval that with a prespecified
confidence will contain a prespecfied proportion of the distribution of
future values. ..e.g. a "95%/90%" tolerance interval will with 95%
confidence contain 90% of future values (and one may well ask, "which
90%"?).

Whether this is a useful idea is another issue: the parametric version is
extremely sensitive (as one might imagine) to the assumption of exact
normality; the nonparametric version relies on order statistics and is more
robust. I believe it is nontrivial and perhaps ambiguous to extend the
concept from the usual fixed distribution to the linear regression case. I
seem to recall some papers on this, perhaps in JASA, in the past few years.

As always, I welcome correction of any errors or misunderstandings herein.

Cheers to all,

Bert Gunter



 		
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