The following code calculates the regression line slope.

One thought was to have the last point be the YIntercept, but

not sure how -- or -- if the formula for the slope needs to be modified to have the previous point be the YIntercept?

Given the formula Y = mx + b one would think that

setting b = YIntercept = last data point, this would give the correct result.

Any other solutions appreciated.

Code:

`Private Function fLFit()`

'IN PROCESS -- 20180319

'Linear Regression

Dim i As Integer

Dim n As Integer '# Data Pts.

Dim SX As Double 'Sum X

Dim SY As Double 'Sum Y

Dim SX2 As Double 'Sum X Squared

Dim SXY As Double 'Sum X * Y

Dim SY As Double 'Sum Y

Dim Num As Double 'Numerator of Fraction

Dim Dem As Double 'Denominator of Fraction

n = CInt(msngInput1)

SX = 0

SY = 0

For i = 1 To n

SX = SX + X(i)

SX2 = SX2 + X(i) ^ 2

SY = SY + Y(i)

SY2 = SY2 + Y(i) ^ 2

SXY = SXY + X(i) * Y(i)

Next i

'Calculate Slope

Num = (n * SXY) - (SX * SY)

Dem = (n * SX2) - (SX * SX)

Slope = Num / Dem

End Function