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Correlation formula

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janne

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Jan 6, 2002, 10:17:21 PM1/6/02
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I have a correlation formula I don't get to work. And we must use this
formula on the test. Let me give you an example: Let's say X and Y
are:
x y
1 68
2 91
3 102
3 107
4 105
4 114
5 115
6 127
_ ___
28 829
_ _
X is =3.5 and Y is =103.625

Now to my problem. Look at the formula in this URL:
http://www.jannesgallery.com/corr.html.
How do I do the first (X-X(with a line above))? I have tried to take
_
X-X
1-3.5=2.5
2-3.5=-1.5
3-3.5=-0.5
3-3.5=-0.5
4-3.5=0.5
4-3.5=0.5
5-3.5=1.5
6-3.5=2.5
________
0

As you see the answer is zero. What do I do wrong? and the same with
Y-Y(with a line above). It turns out to be zero. Please help me to tell
how I should do.

Janne

Patrick FitzGerald

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Jan 6, 2002, 10:51:48 PM1/6/02
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On Mon, 07 Jan 2002 03:17:21 GMT, janne <jan.h....@telia.com>
wrote:


>As you see the answer is zero. What do I do wrong? and the same with
>Y-Y(with a line above). It turns out to be zero. Please help me to tell
>how I should do.
>
>

Study your formula more closely


What you have to is to take each x value subtract the x mean and
multiply it by its corresponding y value with the y mean subtracted
and add all those rsults to gether

THere are better formulas for calculating r, go to your library and
consult a statistic text


Patrick


Patrick

JJ Diamond

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Jan 25, 2002, 2:20:34 PM1/25/02
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this comment is probably too late to be of use, BUT remember that
correlation is about covariability. our interest is not just how X
varies about its mean, or Y about its mean, but how the PAIRS of X's
and Y's vary together. since the mean is the balancing point for a
distribution, the sum of deviations of each X from the mean will
always equal zero. so patrick is absolutely correct in asking you to
look at the PRODUCT of the deviations for each observation. this is
the essence of covariation - how X and Y vary together. lastly,
realize that you are calculating a measure of linear relationship. a
correlation of the type you are calculating can be quite low if there
is a relationship, but it is not linear.


patr...@netaccess.co.nz (Patrick FitzGerald) wrote in message news:<3c391a8e...@news.netaccess.co.nz>...

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