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The line of best fit. Need help?

  1. Feb 22, 2008 #1
    Well, i am taking a first course in elementary statistics, so we do not actually prove anything at all. So i was wonderign how does one determine the line of best fit?
    y=mx+b, where m is the slope of the line of best fit,
    I know that the slope is equal to

    m=SS(xy)/SS(x), where SS(x) is the sum of square of x, while SS(xy) the sum of the squares of x,y. also

    b=[SUM(y)-m*SUM(x)]/n but i have no idea how one would come up with these expressions.
    Can somebody show a proof for this, or just point me to the right direction?
  2. jcsd
  3. Feb 22, 2008 #2


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    The proof reqires a knowledge of elementary calculus. Basic idea is to assume a straight line fit with m and b unknown. Set up the expression for the sum of the squares of the distances of the points from the line. Find m and b which minimize the expression.
  4. Feb 27, 2008 #3


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    Let μ and β be the least squares estimators of m and b in y = mx + b.

    The estimated equation is then y[t] = β + μ x[t] + u[t] where u is the residual error and t indexes "data row" (e.g., observation).

    u[t] = y[t] - (β + μ x[t])

    u[t]^2 = (y[t] - (β + μ x[t]))^2

    Sum over t:
    Σt u[t]^2 = Σt (y[t] - (β + μ x[t]))^2

    Now minimize with respect to μ and β by differentiating Σt u[t]^2 with respect to μ and β separately, setting each derivative to zero, then solving for μ and β that satisfy these two equations simultaneously:

    ∂Σt u[t]^2/∂μ = ∂Σt u[t]^2/∂β = 0.
    Last edited: Feb 27, 2008
  5. Feb 27, 2008 #4
    Thankyou for your replies! I totally forgot to let you know that i had managed to get to the result i wanted, my approach was almost identical with what u did, with the exception of notation, but i got to the result!

    thnx both of you!
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