Standard Deviation using the Sumation of X and Summation of X-Squared?

Posted on 2006-05-01
Last Modified: 2012-06-21
If you had a set of X values and a set of Y values, how do you get the standard deviation of the difference?

Give the n=16, the summation of the differences = 108, and the summation of the differences squared is 1746, the mean of the differences is 6.75. How would you get the standard deviation (it should come out to 8.234)?

 I tried 1746-6.75^2/16 and that didn't come out quite right. Then I started playing around by squaring and dividing stuff. I know I somehow need to get to about 68, so that when I do the square root, I get back 8.234. I would need 1020 if I divide by n-1 (15) or 1088 if I divide by 16. Can't seem to get it, though, and everything I looked up assumes you have the list of numbers and can do the whole (summation of (difference - average difference)^2)/N.
Question by:sjaguar13
    LVL 18

    Accepted Solution

    There are some more formula here:
    LVL 18

    Expert Comment

    The answer can be got from :

    Sqrt((1746 -(108^2/16))/15)

    This is from the formula on the page for obtaining sample standard deviation.

    LVL 22

    Assisted Solution

    Forget the fact that this data results from the difference between two random variables, that's just a red herring.  You have a random variable Z, you have 16 values for it, their sum is 108 and the sum of their squares is 1746.

    n = 16
    mean of Z is <Z>, which we know is 108/16
    sum of squares is 1746, so <Z^2> = 1746/16

    standard deviation of Z = sqrt( <(Z - <Z>)^2> * n/(n-1) )
    (Z - <Z>)^2 = Z^2 - 2*Z*<Z> + <Z>^2
    <(Z - <Z>)^2> = <Z^2> - <2*Z*<Z>> + <<Z>^2>
    2*<Z> can be factored out of the middle term, since it is a constant
    The last term is constant, so it is its own mean.
                          = <Z^2> - 2*<Z>*<Z> + <Z>^2
                          = <Z^2> - <Z>^2

    standard deviation of Z = sqrt((<Z^2> - <Z>^2) * n/(n-1))
        = sqrt((1746/16 - (108/16)^2) * 16/15)
        = sqrt((1746 - (108^2)/16) / 15), same as JR2003

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