daveman
Diamond Member
Why don't you see if you can find some data to refute this, cause I'm getting tired of posting it:
An unsourced imaged, with no indication where it came from?
Nothing to refute. It's not proof of anything.
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Why don't you see if you can find some data to refute this, cause I'm getting tired of posting it:
It was 19 degrees above freezing in the Mid Atlantic on July 5th. Global cooling?
Probably something to do with our superior and paranoia-free grasp of reality.
[MENTION=30473]flacaltenn[/MENTION]
I know. A simple task to demomstrate your grasp of science. This is easy. You can use Excel for it.
Get the data on global mean temps that Watt used in his graph thqt the OP is talking about. Do a linear linear regression, make a graph like the one the OP claims as proof, then tell us what the confidence interval for the slope is. That done, explain why the linear regression and graph are meaningless.
Prove you aren't a moron without an original thought
You won't. You can't.
I can't.. I won't.. Because it's not neccessary to abuse mathematics like you do..
NOW -- you've gone from BABBLING to ABUSE of mathematics. Why the GRAPH is meaningless? It's not. It WILL ALWAYS be a plot of US (not Global -- dummy) temperatures from 2005 to 2014.. Not matter how dirty you talk to it --- YOU can't change the meaning of the graph or call it meaningless. Secondly, there is no REQUIREMENT to test a TREND LINE. It will always be the BEST linear approximation to the chart. Lowest error bar fit to the equation F(X) = aX + b. Plenty of data points to make that determination. Knock yourself and weep and wring your hands about the variance of US Temperatures over time.. It won't change a damned thing.. It IS -- what it is... It's not a linear function that's for sure. So it won't fit.
Now why it's plotted to BEGIN in 2005 is mighty suspicious.. I suppose if you can "cherry-pick" a cooling trend line by doing so --- SOMEONE might think that's meaningful, because it allows them to state that the linear trend in temps since 2005 indicate a slight cooling for the United States. Guess they didn't want to start that graph EARLIER for a reason..
Enough honesty for ya Clyde?? Enough original thought? You just go on faking your math chops and pretending that the normal variance of temperature data sets has some significance in it that allows you to DISMISS a simple graph and trend line calc..
[MENTION=30473]flacaltenn[/MENTION]
I know. A simple task to demomstrate your grasp of science. This is easy. You can use Excel for it.
Get the data on global mean temps that Watt used in his graph thqt the OP is talking about. Do a linear linear regression, make a graph like the one the OP claims as proof, then tell us what the confidence interval for the slope is. That done, explain why the linear regression and graph are meaningless.
Prove you aren't a moron without an original thought
You won't. You can't.
Why don't you see if you can find some data to refute this, cause I'm getting tired of posting it:
oh ................ oh ............. oh ...... !!!!!!!!!
Well.....this sure sucks for the AGW alarmists!!!
Government data shows conclusively........... Government Data Show U.S. in Decade-Long Cooling - Forbes
USA in decade long cOoLInG
How stoopid do you look if you are an alarmist k00k in 2014? For 15 years, people have been saying this is a fucking hoax yet the obsessed nutters clung to their fairy tale!!!
ghey
Why don't you see if you can find some data to refute this, cause I'm getting tired of posting it:
Probably something to do with our superior and paranoia-free grasp of reality.
I'm wondering why you deniers bother. You can keep loudly proclaiming the glorious victory of the denier cult here, but the world just goes on laughing its ass off at you all.
Given that, how long do you plan to keep up the crazy? Oh wait, I know the answer. You'll be crazy for as long as your political cult orders you to be crazy. It's not like any of you can think independently.
A safe assumption is that the newer dataset is more accurate. That would be the paranoia-free position.
The difference between the two, as explained by the people who did the adjusting, is that such adjustments correct for errors and differences such as time-of day, location changes, analog to digital and general magic ;-),
A safe assumption is that the newer dataset is more accurate. That would be the paranoia-free position.
The difference between the two, as explained by the people who did the adjusting, is that such adjustments correct for errors and differences such as time-of day, location changes, analog to digital and general magic ;-),
Why go back and alter data from over 50 years ago? What is the reason for altering it?
[MENTION=30473]flacaltenn[/MENTION]
I know. A simple task to demomstrate your grasp of science. This is easy. You can use Excel for it.
Get the data on global mean temps that Watt used in his graph thqt the OP is talking about. Do a linear linear regression, make a graph like the one the OP claims as proof, then tell us what the confidence interval for the slope is. That done, explain why the linear regression and graph are meaningless.
Prove you aren't a moron without an original thought
You won't. You can't.
I can't.. I won't.. Because it's not neccessary to abuse mathematics like you do..
NOW -- you've gone from BABBLING to ABUSE of mathematics. Why the GRAPH is meaningless? It's not. It WILL ALWAYS be a plot of US (not Global -- dummy) temperatures from 2005 to 2014.. Not matter how dirty you talk to it --- YOU can't change the meaning of the graph or call it meaningless. Secondly, there is no REQUIREMENT to test a TREND LINE. It will always be the BEST linear approximation to the chart. Lowest error bar fit to the equation F(X) = aX + b. Plenty of data points to make that determination. Knock yourself and weep and wring your hands about the variance of US Temperatures over time.. It won't change a damned thing.. It IS -- what it is... It's not a linear function that's for sure. So it won't fit.
Now why it's plotted to BEGIN in 2005 is mighty suspicious.. I suppose if you can "cherry-pick" a cooling trend line by doing so --- SOMEONE might think that's meaningful, because it allows them to state that the linear trend in temps since 2005 indicate a slight cooling for the United States. Guess they didn't want to start that graph EARLIER for a reason..
Enough honesty for ya Clyde?? Enough original thought? You just go on faking your math chops and pretending that the normal variance of temperature data sets has some significance in it that allows you to DISMISS a simple graph and trend line calc..
It's real simple. Every regression requires the test for statistical significance for it to have any meaning. When the regression is run, then p-value is also provided. It's just real basic stuff. Not doing it, not presenting with the graph is because of a) a lack of understanding of the fundamentals or b) it doesn't support the the bs.
Linear regression ALWAYS REQUIRES that the coefficients be tested significance. The reason is obvious. The complete equation that represents the data is F(X) = aX + b + e. e is the error, the diffence between the model and the data. The error term has a variance. Can you explain what the mean of the error is?
So, seeing as you are lacking the basic knowlede, I will explain it for you. There are two coeficients that linear regression calculates. These are the slope of the line and the constant. Whe these are calculated, the program also calculates the p-value. The p-value is used to determine if the coefficients are statistically significant. These are comared to the alpha which sets the level of statistical significance that is desired for the test to be interpreted as acceptable. The alpha valuesbate the cuttoff points for acceptance of the coefficient. Typical alphas that are used set the cuttoff value at 90%, 95%, and 99%.
Without th p-value, the linear trend line has no meaning.
The probem with regession slopes near zero is that the p-value is nearly always to large and the coeffcient meaningless.
I can guarantee from inspection that the p-value, for the graph in question, is to large for the regression line to have any meaning. There are not enough data points and the variance is to great.
But you don't have to take my word for it. You can do it yourself. But you won't, and Watt didn't because the result will invalidate what you want to believe. This is what you always do.
Here is a good page that explains the confidence interval for the slope
Regression Slope: Confidence Interval