Re: Comparisons using RMS Error and Correlation
- From: "Reef Fish" <large_nassua_grouper@xxxxxxxxx>
- Date: 28 Nov 2006 23:14:37 -0800
Nicholas Kinar wrote:
Reef Fish Bob,
First and foremost, this is a really nice solution, and I thank you for
pointing me in the right direction.
Thanks you, Nicholas. This is probably the first time I've erased the
solution I had already typed, because I realized it was such a good
exercise to understand those relations in a regression! And there are
so many variations of the same theme you can ask and provide answers
that are not obvious even to the expert.
If I were still teaching, that theme would have provided many of my
exam question (Open Book) which require comprehension but no
memorazation of formulas will help unless the students can THINK. :-)
I am very glad that you worked it out and appreciated the solution.
I am very glad that I have been given
the opportunity to swim with your school of fish.
Thank you again. It is my pleasure. I assume you recognize the
paraphrase of this expression that originated from dog lovers,
"The more I know people, the more I like fish". :-)
-- Reef Fish Bob.
Nicholas Kinar wrote:
I have two datasets that have been collected in the context of snow
science.
Both datasets contain esimates of Snow Water Equilivant (SWE), which is a
measure of the amount of water that remains when snow on the ground
surface
in a particular area is melted down, thereby giving a measurement of the
equilivant depth of water (i.e. 20 mm). Dataset A consists of
observations
made with Instrument #1, whereas Dataset B consists of observations made
with Instrument #2. To compare the datasets, I plot dataset A on the
x-axis
and dataset B on the y-axis, and then calculate the correlation
coefficient
to determine how well the relationship between the datasets can be
described
by a linear model. I do this for two separate research sites, Site #1
and
Site #2. Thus, I have one graph of the correlations for Site #1, and
another graph of the correlations for Site #2. Suppose that the
correlation
coefficient is r^2 = 0.86 for Site #1, and the correlation coefficient is
r^2= 0.79 for Site #2. However, when I calculate the Root Mean Squared
Error for both sites, I find that RMS = 12.65 mm for Site #1, and RMS =
6.23
mm for Site #2. Why is the RMS greater for Site #1 (which has a higher
correlation), whereas the RMS is lower for Site #2 (which has a lower
correlation)?
This is an easy one since the difference in RMS (assume you mean
the square root of the MSE) depends on the sample size for each
regression/correlation and the variance of Y for each Site.
Regardless the values of the r^2, their corresponding RMS can
be larger or smaller depending on the variances of the Ys in the
two Sites.
If we assume n1 = n2, and put in your values for r^2 and RMS,
you'll find the var(Y) for Site 1 is 6.1844 var(Y) for Site 2.
I erased my derivation because it's such a nice little problem for
simple regression that I'll leave it as an exercise, giving only the
solution above.
Very nice.
BTW, given your values for r, the difference in the RMS depends
ONLY on the variances of Y in the two sites, and nothing else
(assuming equal sample sizes in each Site).
-- Reef Fish Bob.
Thank you ever so much for your time and assistance.
Nicholas
Reef Fish Bob, thank you for taking the time to answer my question. This is
greatly appreciated!
Nicholas
.
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