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Re: [Phys-l] calibration

On Aug 6, 2007, at 12:36 PM, Jeffrey Schnick wrote:

Check out the statistics applet at:
<http://espse.ed.psu.edu/edpsych/faculty/rhale/Statistics/statlets/ free/calib.htm>
Using it (and treating the resistivity as the dependent variable in the calibration plot), using the data you provided, I get, for r=62 nano-ohm meters, with the confidence level for the prediction interval set to 68.3%:
p = (33.1 +/- 7.4) %
________________________________

From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of Ludwik Kowalski
Sent: Sun 8/5/2007 5:54 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] calibration

Aug 5, 2007, at 2:23 PM, Folkerts, Timothy J wrote:
Finding the prediction interval for a linear regression is a little
more complicated than just finding the St Dev of the various points.
It may be convenient to define the uncertainty of p as the standard
deviation of such fluctuations, but it is not the standard practice.
You should perhaps check a stats book for more info on the topic.

A quick run of the data through Minitab suggests that the +/-1 stdev
prediction interval (68.3% confidence level) for r = 0.62 is p =
30.25 +/- 6.5. This is half again as large as the interval you list.
The size of the prediction interval also changes across the range of
the graph. . . .

Jeff,
Thanks for the shared reference:

http://www.chem.utoronto.ca/coursenotes/analsci/StatsTutorial/ ErrRegr.html

I like it, except for one detail. The predicting variable y (what is measured is fluorescence) was plotted vertically while the inferred variable (concentration) is plotted horizontally. Their formula to calculate standard deviation of the inferred variable contains the slope of the line, b.

My predicting (independent) variable, r, is plotted horizontally while the inferred (dependent) variable is plotted vertically. That seems to be more logical. But that is not the main point. I think that what they call slope b (in the formula for the standard deviation of the predicted concentration) should be replaced by 1/b', where b' is the slope of my line. I will try to do this later and see if our results agree. What was the numerical value of b that you used in their formula? _______________________________________________________
Ludwik Kowalski, a retired physicist
5 Horizon Road, apt. 2702, Fort Lee, NJ, 07024, USA
Also an amateur journalist at http://csam.montclair.edu/~kowalski/cf/