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Nonlinear Regression Using Excel's Data Analysis Tool | Rate this: (2.5/5 from 2 votes) |
I’m trying to understand how to make use of the equation suggested by excel when fitting a trendline through the following set of data
and selecting the polynomial curve as the best fit. The regression returns the following coefficients: Y=-0.0022x^2 +8.8713x-8919.6 So to evaluate Y when x=2010 I would calculate: Y=-0.0022*(2010)^2 +8.8713*2010-8919.6 Which results in Y being equal to 23.493 which is non sensible and doesn’t match the Y value the graph would imply for the year 2010 (something around 27%). Can somebody please tell me what I’m doing wrong? | ||||||||||||||||
Posted by Catrina on |
Replies - Displaying 1 to 3 of 3 | Order Replies By: Most Recent | Chronological | Highest Rated |
Rate this: (3/5 from 2 votes) the software take x as 1,2,3,4,5,6 not as you expecet as 2004,2005......2009 then if you aplay x=7 (which should be 2010) the result will be close to what you expect | |
All the best - m.saad | |
Posted by msaad on |
Rate this: (3/5 from 1 vote) | |
Posted by Catrina on |
Rate this: (4/5 from 2 votes) Regards,Rachel Gomez | |
rachel | |
Posted by RachelGomez161999 on |
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