http://en.wikipedia.org/wi

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Posted on 2006-05-01

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.

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.

3 Comments

There are some more formula here:

http://en.wikipedia.org/wiki/Standard_deviation

http://en.wikipedia.org/wi

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

This is from the formula on the page http://en.wikipedia.org/wi

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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