"Looking at the data in Exh A, I have calculated an R^2 value for the pre-inflection of 0.712. However, calculations of the R^2 value for the post-inflection points runs from a miserable 0.416 down to a disastrous value of 0.18."
A person who contended this would be a complete idiot, and I would never listen to another thing they said. Th…
"Looking at the data in Exh A, I have calculated an R^2 value for the pre-inflection of 0.712. However, calculations of the R^2 value for the post-inflection points runs from a miserable 0.416 down to a disastrous value of 0.18."
A person who contended this would be a complete idiot, and I would never listen to another thing they said. These are 'Variations Against Trend (Baseline)' charts - they have an inherent 'bias to R^2 fit'. Any lower R^2 or departure from fit indicates a SIGNAL, not the lack of one.
I think you are missing this, and the four points I made.
You just used the word Trend in your quotes. The trend line represents the linear interpolation of the data points. This is true no matter what signal you are trying to convey. Point 1 is an excellent example of where a regression is called for. Point 2 is what you are trying to convey, or signal. No problem.
In closing, if you leave the trend line out(post-inflection), you would have no concern about the R^2 value, and the non-statistics among us can make and see their own best fit. Your 'aha!' moment. If/when you are going to put a linear regression trend line(post-inflection) of the data points to prove your signal direction and magnitude, leaving out an R^2 value - even if it is worse than pre-inflection is a mistake. As noted, we expect the regression to get worse. But no matter the delta R^2, from pre and post-inflection, either do or don't for both.
Trend and 'linear fit' are not congruent, despite your insistence that they are. There exists such a thing called Non-Linear optimization (my Major is in this). It uses the Golden Section method to conduct regression/trend analysis.
As you can see below, gravity (no inflection, single function, and indeed a TREND) produces a trend in the motion of a planet, but if you place an R^2 on it the fit is going to be horrible. Does that mean that gravity does not exist!!!!!!!!!!?????
This is what you are saying - that gravity does not exist, because it doesn't fit an R^2 well.
"Looking at the data in Exh A, I have calculated an R^2 value for the pre-inflection of 0.712. However, calculations of the R^2 value for the post-inflection points runs from a miserable 0.416 down to a disastrous value of 0.18."
A person who contended this would be a complete idiot, and I would never listen to another thing they said. These are 'Variations Against Trend (Baseline)' charts - they have an inherent 'bias to R^2 fit'. Any lower R^2 or departure from fit indicates a SIGNAL, not the lack of one.
I think you are missing this, and the four points I made.
You just used the word Trend in your quotes. The trend line represents the linear interpolation of the data points. This is true no matter what signal you are trying to convey. Point 1 is an excellent example of where a regression is called for. Point 2 is what you are trying to convey, or signal. No problem.
In closing, if you leave the trend line out(post-inflection), you would have no concern about the R^2 value, and the non-statistics among us can make and see their own best fit. Your 'aha!' moment. If/when you are going to put a linear regression trend line(post-inflection) of the data points to prove your signal direction and magnitude, leaving out an R^2 value - even if it is worse than pre-inflection is a mistake. As noted, we expect the regression to get worse. But no matter the delta R^2, from pre and post-inflection, either do or don't for both.
Haveaniceday; doc :-)
Trend and 'linear fit' are not congruent, despite your insistence that they are. There exists such a thing called Non-Linear optimization (my Major is in this). It uses the Golden Section method to conduct regression/trend analysis.
As you can see below, gravity (no inflection, single function, and indeed a TREND) produces a trend in the motion of a planet, but if you place an R^2 on it the fit is going to be horrible. Does that mean that gravity does not exist!!!!!!!!!!?????
This is what you are saying - that gravity does not exist, because it doesn't fit an R^2 well.
https://theethicalskeptic.com/wp-content/uploads/2022/08/Gravity-does-not-exist.png