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

Posted on 2006-05-01
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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.
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Question by:sjaguar13

LVL 18

Accepted Solution

There are some more formula here:
http://en.wikipedia.org/wiki/Standard_deviation
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LVL 18

Expert Comment

The answer can be got from :

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

This is from the formula on the page http://en.wikipedia.org/wiki/Standard_deviation for obtaining sample standard deviation.

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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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