Re: August is tied for 5TH warmest in 125 years

From: charliew2 (charliew2_at_ev1.net)
Date: 10/04/04


Date: Mon, 4 Oct 2004 12:50:29 -0500


Roger Coppock <rcoppock@adnc.com> wrote in message
news:25516292.0409301759.33b879f9@posting.google.com...
> These globally averaged temperature data come from NASA:
> http://www.giss.nasa.gov/data/update/gistemp/GLB.Ts+dSST.txt
> They represent the results of tens of millions of readings
> taken at thousands of land stations and ships around the globe
> over the last 125 years. Yes, the land data are corrected for
> the urban heat island effect. The sea data do not need to be.
> There are few urban centers in the sea.
>
> The Mean August temperature over the last 125 years is 13.966 C.
> The Variance is 0.04142.
> The Standard Deviation is 0.2035.
>
> Rxy 0.795544 Rxy^2 0.632891
> TEMP = 13.682823 + (0.004487 * (YEAR-1879))
> Degrees of Freedom = 123 F = 212.05004
> Confidence of nonzero correlation = approximately
> 0.999999999999999999999999999 (27 nines), which is darn close to 100%!
>
> The month of August in the year 2004,
> is linearly projected to be 14.244,
> yet it was 14.38. <- Above linear projection.
> The sum of the residuals is 12.337094
>
> Exponential least squares fit:
> TEMP = 13.685377 * e^(.0003200 * (YEAR-1879))
> The sum of the residuals is 12.288896
>

(cut)

Once again, Roger, you need to extrapolate your linear and exponential
curves on the same graph. If you do, you'll convince yourself that you
can't distinguish between the two extrapolations for at least 2000 years,
which means that it is WAY too early to assume an exponential fit. The
slightly lower sum of residuals for the exponential fit does not justify its
use.

However, let me rephrase my statement. Do some statistics on the sum of
residuals, and let me know if the difference between 12.337094 and 12.288896
is statistically significant in any way. My guess is that they aren't or
you wouldn't be using so many significant digits (i.e., both numbers round
to 12.3).



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