Working out the equation for coordinates on a graph

[Saints-94]

I have a series of data points for X and Y points on a graph. The data is quite random and I am trying to work out a trend line so I can then form an equation for the line. How would I go about working out the equation for the data below.
(0, 580)
(6.7, 495)
(13.4, 445)
(18.7, 365)
(22.8, 350)
(27, 340)

 

[DrClaude]

Unless there is an underlying model that would suggest otherwise, I would consider that to be a straight line and do a linear least-square fit.

 

[Saints-94]

I'm not sure how I can apply that to the data that I have. Where would I start with working out an equation?

 

[DrClaude]

You need to first calculate the averages, $\bar{x}$ and $\bar{y}$. Then use eqs. (16) and (20) in the link I gave you to get $\mathrm{ss}_{xx}$ and $\mathrm{ss}_{xy}$. You then get the slope from eq. (27) and the intercept from eq. (28).

 

[Saints-94]

I was expecting to get a trend line that looked like an exponential curve. Is it possible to work out an equation that would give me an exponential curve?

 

[DrClaude]

I was expecting to get a trend line that looked like an exponential curve. Is it possible to work out an equation that would give me an exponential curve?

The problem with the non-linear case is that you can't get a direct answer. It becomes a multidimensional minimization problem.

The best is to use existing software to do this. Most graphing programs can do this, as well as other software such as Matlab.

 

[Saints-94]

Ok, thanks. I have the Matlab software, but am unsure how to programme my data.

 

[DrClaude]

https://www.mathworks.com/help/curvefit/linear-and-nonlinear-regression.html