Day 6: Correlation and Regression Lines #1
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sarcalumenin 9 years ago Can someone explain why: r^2 = bxy × byx?
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neild799 9 years ago Usually, we define
b_xy = r * sy/sx
where sy is the standard deviation of the y values and sx is the standard deviation of the x values.
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Conversely, we could say that:
b_yx = r * sx/sy
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Rewrite both:
r = b_xy * sx/sy
r = b_yx * sy/sx
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Then:
r^2 = r*r = (b_xy * sx/sy)*(b_yx * sy/sx) = b_xy*b_yx
as desired.
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Sepestious 9 years ago Im confused....isnt bxy -3/4 and byx -4/3? So if r is given by the root of the product of those two, shouldnt it be equal to 1?
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AbhishekVermaIIT 9 years ago Actually byx is also -3/4 (not -4/3). Also, 'r' should be negative since both bxy and byx are negative !
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Sepestious 9 years ago Thank you! I figured it out
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tagore15 9 years ago What am I missing here?
cor(x,y)^2=-4/3*-3/4=1 cor(x,y)=-1
but it is not accepted this answer. Is it something related to output format?
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rashmisethia89 9 years ago Even i am confused. Aren't you calculating the slope of the two lines? According the problem for the line of regression of y on x, slope should be -3/4 and for line of regression of x on y, the slope should be -3/4 also. Am I wrong?
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AbhishekVermaIIT 9 years ago I guess you are missing the fact that sign of
r
would be same as that ofbxy
andbyx
. If you are submitting "k" as answer, try submitting "-k" !
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AbhishekVermaIIT 9 years ago The slope (for both the lines) is -3/4. Correct that !
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