Linear Regression on a String
Octave, 29 26 24 20 bytes
@(s)s/[!!s;1:nnz(s)]

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We have the model
y= intercept *x^0 + slope * x
 = intercept * 1  + slope * x

Here y is the  ASCII value of string s
To find parameters intercept and slope we can form the following equation:
s = [intercept slope] * [1 X]

so
[intercept slope] = s/[1 x]

!!s converts a string to a vector of ones with the same length as the string.
The vector of ones is used for estimation of the intercept.
1:nnz(s) is range of values from 1 to number of elements of the string used as x.
Previous answer
@(s)ols(s'+0,[!!s;1:nnz(s)]')

For test paste the following code into Octave Online
(@(s)ols(s'+0,[!!s;1:nnz(s)]'))('meta.codegolf.stackexchange.com')

A function that accepts a string as input and applies ordinary least squares estimation of model y = x*b + e
The first argument of ols is y that for it we transpose the string s and add with number 0 to get its ASCII code.