>-----Original Message-----
>From: John Pearson [mailto:xero@CMC.NET]
>Sent: 31 March 2004 06:56
>To: PICLIST@MITVMA.MIT.EDU
>Subject: [OT:] linear regression analysis
>
>
>I am not very good at math. Please assume I am younger than 9
>or older than 40.
>
>I want my mpu to be able to create values for a linear
>regression equation from 4 values that are entered, example:
>
>When A = 6, B = 2.1
>When A = 30, B = 0
>
>How do I come up with the equation to corrolate B from A
>(besides using analysis software). Hopefully I can program my
>mpu to determine the equation needed as I change the values.
>
>Thank you
>John

A straight line has the form y = mx + c where m is the slope and c is the
offset.

m = dy/dx (i.e. change in y relative to the change in x)

m =  (0-2.1)/(30-6) = -2.1/24 = -0.0875 (this is value of the slope)

c = the value of y when x = 0.  After finding the slope you simply need to
plug one of the original sets of values back into the equation:

y = mx + c (m is known at this stage)

y = (-0.0875x) + c

Rearange to get c:

c = y - (-0.0875x)

Put one of the original sets of values in

c = 2.1 - (-0.0875 * 6) = 2.625

The equation of your straight line is therefore

y = -0.0875x + 2.625


FWIW this is not difficult to do on a PIC, even without using floating point
math.

Regards

Mike




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