# Linear Algebra - Vector Spaces Quiz 6 (Least Squares)

In the least squares problem we find that

$A^T Ax = A^T b$

and

$c_1 = \frac{\sum y_i}{n} , c_2 = \frac{\sum x_i y_i }{\sum x_i^2}$

where $y = c_1 + c_2 x$

I'm assuming this linear equation represents some kind of fitted line for the data. However, a previous step in the quiz involved setting all the $x_i$ coordinates to $x_i - \bar{x}$ to simplify finding $c_1$ and $c_2$.

So I went ahead and tried this method on a simple set I made up of 20 pieces of data of coordinates $(x,y)$:

$\{ (1,4), (3,5) , (4,3), (4,9), (6,2), (6,8), (8,6), (10,9), (11,7), (12, 5), (13,11), (14,7), (15,9), (16,12), (17,8), (18,10), (18,13), (19,14), (20,13), (20,14)\}$

A little spread out but I expect basically a positive linear relationship with y as a function of x.

Now I find with my calculator stat functions that $\bar{x} = 11.75$, so I set all $x_i$ to $x_i - 11.75$:

$X = \{ -10.75, -8.75, -7.75, -7.75, -5.75, -3.75, -1.75, -0.75, 0.25, 1.25, 2.25, 3.25, 4.25, 5.25, 6.25, 6.25, 7.25, 8.25, 8.25 \}$

My calculations from here:

$c_1 = \frac{\sum y_i}{n} = \frac{169}{20} = 8.45$ $c_2 = \frac{\sum x_i y_i}{\sum x_i^2} = \frac{340.25}{725.75} \approx 0.46883$

I wind up with a line equation that looks like

$y = 0.46883x + 8.45$

If I try and plot this, the line sits way outside any of my data. Did I miss some crucial step? Do the two constants change in subtracting the mean of X to begin with? Or am I misunderstanding the meaning of $c_1$ and $c_2$? As far as I can tell the quiz doesn't seem to suggest one way or another. Note by Jeff Folster
11 months, 3 weeks ago

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