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Find the equation $y=\beta_{0}+\beta_{1} x$ of the least-squares line that best fits the given data points.$(1,0),(2,1),(4,2),(5,3)$

$y=-0.6+0.7 x$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 6

Applications to Linear Models

Vectors

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Harvey Mudd College

University of Nottingham

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we know that the problem of finding the regression line is essentially a problem off the square solution. So we're given the data, which are the four points. So we formed the Medics Axe, which has four rows as many as the data we have. And on the first column, we have all ones because these corresponds to the constant term better. Zero. And the second column we put the ax is off the points they were given. So 124 and five and then for the constant term we have the wise of the points. So 0123 Let's call these metrics. Why? So since we need to solve the least square solution, we know that we need to form the normal equation for it. So let's start computing X transpose x a simple computation. It's a two by two matrix with entries 4 12 12 46 And then we also need to compute X transpose. Why? Which is a two by two matrix? Well, to buy ones are column vector with entries six and 25. Now that we have these data, we can just compute the least court solution because you know that are solution Beata It's going to be even by x transpose x inverse that multiplies X transpose y and we have completed these medicines in the previous page. So x transpose x inverse is one over. It's the terminal which turned out to be 40 then 46 4 months storm on a cell phone on the second and I are gonna and then the other term in green is just X transpose y, which was 6 25 Now it is just initial matrix multiplication. So one over 40 and the resolve to year is minus 24 28. After the vote, anybody 40 we finally got at zero point was minus 0.60 point seven and this is the least squares beata And so these are the conditions of the regression line. So we know the constant term bit zero. He's minus 0.6 and the slope beta one is 0.7

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