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Question
The equations of two regression lines are x − 4y = 5 and 16y − x = 64. Find means of X and Y. Also, find correlation coefficient between X and Y.
Solution
Given equations of regression lines are
x - 4y = 5 …(i)
16y - x = 64
i.e., - x + 16y = 64 …(ii)
Adding (i) and (ii), we get
x - 4y = 5
- x + 16y = 64
12y = 69
∴ y = `69/12 = 5.75`
Substituting y = 5.75 in (i), we get
x - 4(5.75) = 5
∴ x - 23 = 5
∴ x = 5 + 23 = 28
Since the point of intersection of two regression lines is `(bar x, bar y)`,
∴ `bar x = 28 and bar y = 5.75`
Let, x - 4y = 5 be the regression equation of X on Y
∴ The equation becomes X = 4Y + 5
Comparing it with X = bXY Y + a', we get
bXY = 4
Now, the other equation i.e. 16y - x = 64 is regression equation of Y on X
∴ The equation becomes 16Y = X + 64
i.e., Y = `1/16 "X" + 64/16`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 1/16`
r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt(4 xx 1/16) = +- sqrt(1/4) = +- 1/2 = +- 0.5`
Since bXY and bYX both are positive,
r is also positive.
∴ r = 0.5
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byx = `square/square`
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Regression equation of y on x is `(y - bary) = "b"_(yx) (x - barx)`
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b. When x = 50,
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