Expert: Josh - 4/3/2006

Question
Can you folks help me with a little bit of math.  This has to do
with a specification of printing papers  call "Opacity".

I have had 4 different samples of paper tested and obtained their
"opacity" %.

Sheet#1 - 91.79
Sheet#2 - 91.36
Sheet#3 - 89.78
Sheet#4 - 89.17

These are based on a scale of 1-100.  with 1% being transparent
(no opacity)  and 100  being 100% opaque.
Paper for printing generally fall into a scale of 80-98%.

Looking at the above test results, one could say they are all
pretty close - ( sheet 4 is 2.85% less opaque than sheet #1)
however, that is looking at them in a scale of 1-100. MY
PROBLEM IS HOW TO RELATE THESE FIGURES TO EACH OTHER ON
A SCALE OF 80-98.

If I were to select Sheet #2 as my benchmark  (average opacity
91.36)
Knowing that the typical range of opacity for printing papers is
on a scale of 80-98  (not a scale of 0-100)
Please help me with:

Sheet #1 (average 91.79)  is  WHAT % more opaque than the
benchmark #2 ?
Sheet #3 (average 89.78)  is  WHAT % less opaque then the
benchmark #2 ?
Sheet #4 (average 89.17)  is  WHAT % less opaque then the
benchmark #2 ?

AND  the sheet I will be using this year has an opacity of 92 -
again on a scale of 80-98...
Sheet #4 (average 89.17)  is WHAT % less opaque then the new
sheet (92)?

THANK-YOU FOR YOUR HELP.


Answer
Hi Dave,

Consider the interval [80,98] (i.e., from 80 to 98). The nominal range is R=18. Under this new scale, a 1% change change is represented by a difference of 0.18.

If you choose #2 as the bench mark and use x(2)=91.36 to denote the opacity of number 2, then

#1 is 100*[x(1)-x(2)]/R = 100*(91.79-91.36)/18 % more opaque than #2, relative to the effective range of opacity expected from printing papers.

Likewise, for #3, it is 100*|x(3)-x(2)|/R = |100*(89.78-91.36)/18| % less opaque than #2. Note: I've used |.| to denote the absolute value, changing a negative quantity to a positive quantity.

Let the opacity for the brand you'll be using be x(5)=92.
Again, #4 is less opaque than #5 by
100*|x(5)-x(4)|/R %, where x(5)=92, x(4)=89.17 and R=18.

When presenting these figures, it's important that you point out the effective range is limited to the interval [80,98] based on a priori expectation. Paragraph one serves as your point of reference.

Cheers.