Just to clear the water, there are two formulas for Pearson's r:
Deviation formula:r = \(\frac{S_{xy}}{\sqrt{S_{xx}\times S_{yy}}}\) (covariance divided by sqrt of var X and var Y)
Z score formula:r = \(\frac{1}{n-1}\Sigma \left[Z_xZ_y\right]=\frac{1}{n-1}\Sigma \left[\left(\frac{x-\overline{x}}{SD_x}\right)\left(\frac{y-\overline{y}}{SD_y}\right)\right]\) (n=observations; SD_x is standard deviation or x; SD_y is standard deviation of y)
When we square r \(\left(r^2\right)\), it represents the amount of covariation compared to the amount of total variation. For example, if r=0.8, squaring it gets you 0.64. Then X explains 64% of the variability in Y, and vice versa.
The standard statistic is based on the z-scores, so the statistic will not change if you change the units of x and y.
To find the equation's slope for y = a + bx (where slope is b and a is y-intercept), you use the formula:
b = \(r\left(\frac{SD_y}{SD_x}\right)=\frac{S_{xy}}{SS_x}\)
a = (mean of y's) - b * (mean of x's)
An unstandardized statistic, which is what you're looking for will change if you change the units of X and Y.
What bio_man did was wrong. He found the equation for the standardized regression line using the formulas:
\(Z_Y'=\beta \left(Z_X\right)\)
remember \(Z_X\) = \(\Sigma \left(\frac{x-\overline{x}}{SD_x}\right)\)
where \(\beta =r\)
And no y-intercept for standardized equations
Hope you didn't submit it
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