By Steve Goddard
We are all familiar with the GISS graph below, showing how the world has warmed since 1880.
http://data.giss.nasa.gov/gistemp/graphs/Fig.A2.lrg.gif
The GISS map below shows the geographic details of how they believe the planet has warmed. It uses 1200 km smoothing, a technique which allows them to generate data where they have none – based on the idea that temperatures don’t vary much over 1200 km. It seems “reasonable enough” to use the Monaco weather forecast to make picnic plans in Birmingham, England. Similarly we could assume that the weather and climate in Portland, Oregon can be inferred from that of Death Valley.
The map below uses 250 km smoothing, which allows us to see a little better where they actually have trend data from 1880-2009.
I took the two maps above, projected them on to a sphere representing the earth, and made them blink back and forth between 250 km and 1200 km smoothing. The Arctic is particularly impressive. GISS has determined that the Arctic is warming rapidly across vast distances where they have no 250 km data (pink.)
A way to prove there’s no data in the region for yourself is by using the GISTEMP Map locator at http://data.giss.nasa.gov/gistemp/station_data/
If we choose 90N 0E (North Pole) as the center point for finding nearby stations:
We find that the closest station from the North Pole is Alert, NWT, 834 km (518 miles) away. That’s about the distance from Montreal to Washington DC. Is the temperature data in Montreal valid for applying to Washington DC.?
Even worse, there’s no data in GISTEMP for Alert NWT since 1991. Funny though, you can get current data right now, today, from Weather Underground, right here. WUWT?
Here’s the METAR report for Alert, NWT from today
METAR CYLT 261900Z 31007KT 10SM OVC020 01/M00 A2967 RMK ST8 LAST OBS/NEXT 270600 UTC SLP051
The next closest GISTEMP station is Nord, ADS at 935 km (580 miles) away.
Most Arctic stations used in GISTEMP are 1000 km (621 miles) or more away from the North Pole. That is about the distance from Chicago to Atlanta. Again would you use climate records from Atlanta to gauge what is happening in Chicago?
Note the area between Svalbard and the North Pole in the globe below. There is no data in the 250 km 1880-2009 trend map indicating that region has warmed significantly, yet GISS 1200 km 1880-2009 has it warming 2-4° C. Same story for northern Greenland, the Beaufort Sea, etc. There’s a lot of holes in the polar data that has been interpolated.
The GISS Arctic (non) data has been widely misinterpreted. Below is a good example:
Monitoring Greenland’s melting
The ten warmest years since 1880 have all taken place within the 12-year period of 1997–2008, according to the NASA Goddard Institute for Space Studies (GISS) surface temperature analysis. The Arctic has been subject to exceptionally warm conditions and is showing an extraordinary response to increasing temperatures. The changes in polar ice have the potential to profoundly affect Earth’s climate; in 2007, sea-ice extent reached a historical minimum, as a consequence of warm and clear sky conditions.
If we look at the only two long-term stations which GISS does have in Greenland, it becomes clear that there has been nothing extraordinary or record breaking about the last 12 years (other than one probably errant data point.) The 1930s were warmer in Greenland.
Similarly, GISS has essentially no 250 km 1880-2009 data in the interior of Africa, yet has managed to generate a detailed profile across the entire continent for that same time period. In the process of doing this, they “disappeared” a cold spot in what is now Zimbabwe.
Same story for Asia.
Same story for South America. Note how they moved a cold area from Argentina to Bolivia, and created an imaginary hot spot in Brazil.
Pay no attention to that man behind the curtain.
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Buffoon replied, “Lack of data means the trend isn’t valid. Are you saying there isn’t a lack of data? I don’t get what you’re arguing.”
While there is insufficient data to create trends from the dataset with 250km radius smoothing, there is apparently enough data with to do it from the data with 1200km radius smoothing.
As I explained to Steven upthread:
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To create the trend maps, GISS uses cells where at least 66% of the data exists. Refer to their map making webpage:
http://data.giss.nasa.gov/gistemp/maps/
The note toward the bottom of the page reads, “’Trends’ are not reported unless >66% of the needed records are available.”
Since the maps with 250km radius smoothing have much less data from which to create trends (than the maps using 1200km radius smoothing), the trends will be different.
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Buffoon, presenting examples to me why this is right or wrong doesn’t serve any purpose (unless you’re trying to educate me), because it is GISS who established the 1200km radius smoothing and GISS who set the threshold on data availability for trends for their data.
I’ve noted to you the problems I find with the 1200km radius smoothing in an earlier comment.
What many took as an argument between Steven and me was simply a lack of communication–I was simply trying to explain to Steven how and why, regardless of whether or not it was correct. In other words, GISS set the “rules” for how this was done and I was simply trying to explain those to Steven.
Regards
Frank K. says:
July 27, 2010 at 5:36 pm
I have studied this paper in the past, particularly Figure 3, and the ONLY region where the correlation is any good in 44.4 N and above. Everywhere else, especially the Southern Hemisphere, it’s marginal or complete crud! But hey, it’s climate science – anything goes…
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Thank you for checking out the paper. I am not surprised at your finding.
70% of this planet is water and you only have to look at how the ENSO patterns change to see how utterly bogus the concept of a 1200KM smoothing is . This is especially when coupled with a claimed 0.1C accuracy as implied by all the media releases.
RW says:
July 29, 2010 at 6:17 am
“It seems to me that you have very little knowledge of statistics. You found a web page that made a number of mistaken claims about the Pearson coefficient, and you seem to think that permits you to make statements like this. The web page you found was wrong; those four statements about the Pearson coefficient are wrong…..
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You are working with the text-only light edition of “H.Lohninger: Teach/Me Data Analysis, Springer-Verlag, Berlin-New York-Tokyo, 1999. ISBN 3-540-14743-8
The correlation coefficient r (also called Pearson’s product moment correlation after Karl Pearson) is calculated by ….
Assumptions:
* linear relationship between x and y
* continuous random variables
* both variables must be normally distributed
* x and y must be independent of each other
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This is from a statistical program manual and illustrates the fast and loose use of Pearson correlation that I dislike so much because it is done without any real knowledge of the person typing on the keys:
Chapter Seven: Correlation and Regression
The most widely used bivariate test is the Pearson correlation. It is intended to be used when both variables are measured at either the interval or ratio level, and each variable is normally distributed. However, sometimes we do violate these assumptions. If you do a histogram of both EDUC, chapter 4, and PRESTG80, you will notice that neither is actually normally distributed. Furthermore, if you noted that PRESTG80 is really an ordinal measure, not an interval one, you would be correct. Nevertheless, most analysts would use the Pearson correlation because the variables are close to being normally distributed, the ordinal variable has many ranks, and because the Pearson correlation is the one they are used to. SPSS includes another correlation test, Spearman’s rho, that is designed to analyze variables that are not normally distributed, or are ranked, as is PRESTG80
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::This article is about the correlation coefficient between two random variables.
If we have a series of n measurements of X and Y written as xi and yi where i = 1, 2, …, n, then the Pearson product-moment correlation coefficient can be used to estimate the correlation of X and Y . The Pearson coefficient is
[formula]
also known as the “sample correlation coefficient”. It is especially important if X and Y are both normally distributed. The Pearson correlation coefficient is then the best estimate of the correlation of X and Y .
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Chapter 12 Correlation:
Pearson correlation coefficient page 12-2………………… 2007 A. Karpinski
1. Pearson product moment correlation coefficient…
3. Correlational assumptions
• The assumptions of this of the test for a correlation are that both X and Y are
normally distributed (Actually X and Y must jointly follow a bivariate
normal distribution).
o No other assumptions are required, but remember:
• The correlation coefficient is very sensitive to outliers
• The correlation coefficient only detects linear relationships….
o The covariance only describes linear relationships between X and Y.
o There may be a non-linear relationship between X and Y, but rXY will only
capture linear relationships. rXY will not be useful in measuring non-
linear relationships between X and Y.
o The correlation coefficient is quite sensitive to outliers
• Anytime you report a correlation, you should examine the scatterplot
between those two variables.
⇒ To check for outliers
⇒ To make sure that the relationship between the variables is a
linear relationship
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Pearson product-moment correlation coefficient
In statistics, the Pearson product-moment correlation coefficient (sometimes referred to as the MCV or PMCC) (r) is a common measure of the correlation between two variables X and Y. When measured in a population the Pearson Product Moment correlation is designated by the Greek letter rho (ρ). When computed in a sample, it is designated by the letter “r”….
The statistic is defined as the sum of the products of the standard scores of the two measures divided by the degrees of freedom: [Formula]
Note that this formula assumes the Z scores are calculated using sample standard deviations which are calculated using n − 1 in the denominator. When using population standard deviations, divide by n instead.
The result obtained is equivalent to dividing the covariance between the two variables by the product of their standard deviations…..
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http://en.wikipedia.org/wiki/Bimodal_distribution
Bimodal distributions are a commonly-used example of how summary statistics such as the mean, median, and standard deviation can be deceptive when used on an arbitrary distribution. For example, in the distribution in Figure 1, the mean and median would be about zero, even though zero is not a typical value. The standard deviation is also larger than deviation of each normal distribution.
Elementary Concepts in Statistics
Are All Test Statistics Normally Distributed?
Not all, but most of them are either based on the normal distribution directly or on distributions that are related to and can be derived from normal, such as t, F, or Chi-square. Typically, these tests require that the variables analyzed are themselves normally distributed in the population, that is, they meet the so-called “normality assumption.” Many observed variables actually are normally distributed, which is another reason why the normal distribution represents a “general feature” of empirical reality. The problem may occur when we try to use a normal distribution-based test to analyze data from variables that are themselves not normally distributed…..
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(This is why my Stat. teacher said to PLOT THE DATA!)
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RW says:
July 29, 2010 at 6:17 am
This is not correct at all. The distribution of the data is to a large extent irrelevant.
It’s poor form to throw around phrases like “People with very little knowledge of statistics…”, even if you yourself understand statistics. It’s much worse when you’re making basic errors.
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I think your statements are a very good illustration of why I made the statement about a little knowledge in statistics being dangerous. And that is without thirty years of watching snakeoil salesmen/consultants selling high priced stat programs to novices with promises of magically improved production performance.
Oh and you never even caught the trap I set to find out if anyone here had any real knowledge of Statistics….
GISS 250km data is no longer available (the page opens, but no graphical representations and all download links give an error). If I were i a US taxpayer I’d ask them where the data went myself.
Gail Combs – the internet’s a big place. No matter how wrong the idea, you’ll easily be able to find 20 people who will support it. Your statements about the Pearson coefficient are not outrageously wrong, but they are wrong nonetheless. The maths behind the Pearson correlation coefficient does not assume or require that the data be normally distributed, and the coefficient gives meaningful results for non-normal data. I gave you an example that showed this quite clearly.
The wider point is that the correlation between temperature anomalies at widely spaced locations doesn’t go away if you use a different correlation statistic. It is a physical fact.