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#1
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Excel: Can I force a linear trendline through the origin?
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#2
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Yes, open at the Option tab when you make (or format) a trendline
There is a text box to set intercept to any value (including 0) best wishes -- Bernard V Liengme www.stfx.ca/people/bliengme remove caps from email "Bill" wrote in message ... |
#3
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I have done that, but it is not working. Perhaps it is because I am plotting
a log-log graph and therefore there is no actual zero value for the y-axis. But I need the line to pass through the origin, and because the data covers several decades, it needs to be plotted as log-log. Any ideas? Bill "Bernard Liengme" wrote: Yes, open at the Option tab when you make (or format) a trendline There is a text box to set intercept to any value (including 0) best wishes -- Bernard V Liengme www.stfx.ca/people/bliengme remove caps from email "Bill" wrote in message ... |
#4
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Try this it may work
1) Display the equation of the line. 2) use the equation to calculate the value y by taking x as 0.0000001 or any value 3) add these values (x,y) to your series as new points you can then hide the marker if you want. Hope this helps "Bill" wrote: I have done that, but it is not working. Perhaps it is because I am plotting a log-log graph and therefore there is no actual zero value for the y-axis. But I need the line to pass through the origin, and because the data covers several decades, it needs to be plotted as log-log. Any ideas? Bill "Bernard Liengme" wrote: Yes, open at the Option tab when you make (or format) a trendline There is a text box to set intercept to any value (including 0) best wishes -- Bernard V Liengme www.stfx.ca/people/bliengme remove caps from email "Bill" wrote in message ... |
#5
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Hi,
If the plot 'looks' linear in the log-log plot, then the x,y-data are NOT linear (and y =a* x^m). calculate the logarithms of the x and y values in new columns and make a plot of log y vs log x, and get a trendline (which you can force to pass through the origin; i.e., you are forcing "a" to be equal to 1). However, note that the slope of this line is indeed the exponent "m" in the equation, y = x^m. Formatting the axes of a graph to logarithmic scales only changes the visual appearance of the graph, but does not actually transform the x,y-data to their logarithm values. If you still want to stick to the y vs x plot with log-log scales(and not the log y vs log x plot), you can still get a trendline that 'looks' linear by selecing "Power" and not "linear" for the trendline type. The trendline equation will show up as y = a*x^m; but you can't force "a" to become equal to 1. The latter can be done using Excel's 'Solver' utility. Regards, B. R. Ramachandran "Bill" wrote: I have done that, but it is not working. Perhaps it is because I am plotting a log-log graph and therefore there is no actual zero value for the y-axis. But I need the line to pass through the origin, and because the data covers several decades, it needs to be plotted as log-log. Any ideas? Bill "Bernard Liengme" wrote: Yes, open at the Option tab when you make (or format) a trendline There is a text box to set intercept to any value (including 0) best wishes -- Bernard V Liengme www.stfx.ca/people/bliengme remove caps from email "Bill" wrote in message ... |
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