Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Summation Simplification Help?

Hello -

I am studying statistics and am trying to understand how the below formula:

Sxy = Summation (XY) - (Summation (X))(Summation (Y))/n

Sxx = Summation (X^2) - (Summation X)^2/n

translates to the following:

Sxy/Sxx = [Summation (X - Xbar)(Y - Ybar)]/[Summation (X - Xbar)^2]

Please help....

Also, is there a website that clearly explains the summation rules?

2 Answers

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  • 1 decade ago
    Favorite Answer

    Summations are "symbol" ( +, -, /, and * are also "symbols") which are use to indicate long additions ( eg. 30 or more numbers). You need to treat them like they are: one number which you may know or not (in which case is a constant (eg. "a"). In the formula simplify as follow:

    Sum (XY) - (Sum (X)*Sum (Y)) /n

    ---------------------------------------------

    Sum (x^2)*Sum(x)^2 / n

    n is eliminated....

    xbar = mean (eg. x/n)

    (Sum (x)*Sum (y)) /n ) = Sum (xbar*ybar)

    Sum (x-xbar)*(y-ybar)

    ------------------------------

    Sum (x - xbar)^2

    good luck!

  • 1 decade ago

    Xbar is the average of all the values of x.

    If there are n values of x, then Xbar = (1/n) Σx ... ( 1 )

    => Σ(X - Xbar)^2

    = Σ[X^2 - 2X*Xbar + Xbar^2]

    = ΣX^2 - 2Xbar*ΣX + [ΣXbar]^2

    [ 2Xbar being constant can be taken out of summation.]

    = ΣX^2 - (2/n)[ΣX]^2 + (1/n)[ΣX]^2 [putting the value of Xbar from (1)]

    = ΣX^2 - (1/n)[ΣX]^2 = Sxx

    Thus, Sxx = Σ(X - Xbar)^2 ... ( 2 )

    Now, Σ(X - Xbar)(Y - Ybar)

    = Σ(XY - Xbar*Y - Ybar*X + Xbar*Ybar)

    = Σ(XY) - Xbar*ΣY - Ybar*ΣX + Xbar*Ybar*Σ1

    = Σ(XY) - (1/n)ΣX*ΣY - (1/n)ΣY*ΣX + (1/n)ΣX*(1/n)ΣY*(n)

    (because Σ1 = n)

    = Σ(XY) - (1/n)ΣXY - (1/n)ΣYX + (1/n)ΣXY

    = Σ(XY) - (1/n)ΣXY = Sxy

    Thus, Sxy = Σ(X - Xbar)(Y - Ybar) ... ( 3 )

    From eqns. ( 2 ) and ( 3 ),

    Sxy/Sxx = [Summation (X - Xbar)(Y - Ybar)]/[Summation (X - Xbar)^2]

    Hope, I have been able to explain all steps clearly.

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