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Correlation and gradient

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    I have have a set of data pairs (x, y);
    (1, a)
    (2, b)
    (3, c)
    (4, d)
    (5, e)
    (6, f)
    (7, g)

    The least squares regression line for the this set is y=3x-12

    Determine the new gradient of this line if the original set of scores has been transformed to;

    (6, a+3)
    (12, b+3)
    (18, c+3)
    (24, d+3)
    (30, e+3)
    (36, f+3)
    (42, g+3)

    i.e. the x scores have been multiplied by 6, and the +3 has been added to the y scores.
    Now from my statistical tables book; I have the formula;
    [tex]m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }}[/tex]

    how can I find the new gradient?

    The answer says;
    [tex]\begin{array}{l}
    m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }} = \frac{{ \times 6}}{{ \times 36}} = \times \frac{1}{6} \\
    Hence\;gradient\;is\;now\;3 \times \frac{1}{6} = \frac{1}{2} \\
    \end{array}[/tex]

    I don't really understand how this process works and don't want to assume anything that is wrong

    thanks in advance
    Steven
     
  2. jcsd
  3. Nov 4, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm no expert on statistics but I know that the "gradient" of a line is just its slope. I would have done this, ignoring all the statistical stuff, by arguing that y is now yold+ 3 and x is now 6xold so that yold= y- 3 and xold= x/6. Since you are told that yold= 3xold- 12, you now have y-3= 3(x/6)- 12 or simply y= x/2- 15. The slope (gradient) of that line is 1/2.
     
  4. Nov 4, 2007 #3
    Cheers for that HallsofIvy!

    seems like the obvious thing to do looking back lol

    thanks!
     
  5. Mar 13, 2009 #4
    hi steven10137,
    can you please tell me the name of the book from where you read this formula
    of gradient and correlation.
     
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