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June 3

Strength of a linear relationship

If you have some data with (x,y) coordinates and you plot it on a scatter plot, is the strength of the linear relationship the correlation between x and y, or is it the value of the slope of the best fitted linear regression line? Widener (talk) 07:16, 3 June 2011 (UTC)[reply]

Correlation and dependence says "The Pearson correlation coefficient indicates the strength of a linear relationship between two variables". This makes sense, because the correlation coefficient is unaffected by rescaling of the variables, whereas the slope of the regression line does depend on the scale of the variables, so it is somewhat arbitrary. Another way to think of this is to see that the slope of the regression line actually has units (which depend on the units in which you have measured the dependent and independent variables), but a useful mesaure of relationship strength should be a dimensionless quantity. Gandalf61 (talk) 09:25, 3 June 2011 (UTC)[reply]

I seem to recall that if x is regressed on y, then y regressed on x, the product of the two slopes is the correlation coefficient. So in this sense, the slope(s) of the best fitted regression line(s) do give the strength of the linear relationship - but it's really the correlation coefficient which does so.→86.132.165.117 (talk) 21:24, 3 June 2011 (UTC)[reply]

That is complete nonsense. The product of their slopes is not generally between −1 and 1, and is not dimensionless. The correlation is a dimensionless number between −1 and 1. Michael Hardy (talk) 15:49, 4 June 2011 (UTC)[reply]

I think the product of the slopes of the two regression lines must be the square of the correlation coefficient. Gandalf61 (talk) 08:23, 4 June 2011 (UTC)[reply]

Gandalf, that is also nonsense. The square of the correlation is obviously dimensionless; the product of the slopes is not. Michael Hardy (talk) 15:50, 4 June 2011 (UTC)[reply]

Agreed, my memory was correct but incomplete - for negative correlation, both slopes and the cc would be negative, but the product would be positive. It can be seen from this that it's only if all the points lie on a straight line that the regression lines coincide, and so have slopes reciprocal to each other.→86.132.165.117 (talk) 15:44, 4 June 2011 (UTC)[reply]

The slope says nothing about the strength of the relationship. You can have the same slope when the correlation is 0.01 or when it's 0.99. Michael Hardy (talk) 15:47, 4 June 2011 (UTC)[reply]

Agreed. But the slopes of the two regression lines have recriprocal dimensions - one is dim(y)/dim(x), the other is dim(x)/dim(y) - so their product is dimensionless. Slope of one regression line is:

\beta _{1}={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}}

slope of the other regression line is:

\beta _{2}={\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}

and their product is:

\beta _{1}\beta _{2}={\frac {\left(\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})\right)^{2}}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}\sum _{i=1}^{n}(y_{i}-{\bar {y}})^{2}}}=r^{2}

Gandalf61 (talk) 18:06, 4 June 2011 (UTC)[reply]

If I didn't know something, I'd be more courteous (and wary) than to describe someone else's statement as complete nonsense, in bold for emphasis - these things sometimes come back to bite you. My statement was only slight nonsense, as I'd forgotten the squaring, but Gandalf's were spot on.→86.132.165.117 (talk) 20:25, 4 June 2011 (UTC)[reply]

Don't hold back anon! Spit out the truth: Michael Hardy is a cunt. --72.179.51.84 (talk) 21:25, 4 June 2011 (UTC)[reply]

The second statement, by Gandalf, was right: the product of the slopes is dimensionless and is the square of the correlation. Haste makes waste: I saw the first attempt and was hasty thereafter. Michael Hardy (talk) 22:00, 4 June 2011 (UTC)[reply]

Look up Beta_(finance) and Pearson_product-moment_correlation_coefficient. The regression coefficient beta is the covariance between the two variables divided by the variance of the independent variable. The linear correlation determined by Pearson's r is the covariance divided by the product of the standard deviations. — Preceding unsigned comment added by 99.100.92.26 (talk) 14:06, 5 June 2011 (UTC)[reply]

That old earth curvature chestnut again, but with a difference

Turn the page if you've already been bored to death with this. But if you haven't, you may be able to help us non mathematicians (who came up with a variety of answers you wouldn't believe). Two tall people are standing at sea level, facing each other across the water. Their eyes are exactly six feet above sea level. How far apart are they when they can just see each other's eyes, but not the bridge of the noses? OK, it's a fabulously clear day, flat calm sea, and they have fabulously keen eyesight. We're interested in the distance, not the possibility/impossibility of being able to view each other's eyes. Moriori (talk) 21:41, 3 June 2011 (UTC)[reply]

The distance between their eyes is twice the distance to the horizon. PrimeHunter (talk) 22:50, 3 June 2011 (UTC)[reply]

The same approach answers the question of whether the highest summits in Scotland (Ben Nevis), England (Scafell Pike) and Wales (Snowdon) are visible from each other, assuming perfect visibility and the absence of anything in between - only the last two are.→86.132.165.117 (talk) 16:12, 4 June 2011 (UTC)[reply]