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प्रश्न
The two regression equations are 5x − 6y + 90 = 0 and 15x − 8y − 130 = 0. Find `bar x, bar y`, r.
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उत्तर
Given, the two regression equations are
5x − 6y + 90 = 0
i.e., 5x − 6y = −90 ...(i)
and 15x − 8y − 130 = 0
i.e., 15x − 8y = 130 ...(ii)
By (i) × 3 – (ii), we get
15x − 18y = −270
15x − 8y = 130
− + −
− 10y = −400
∴ y = 40
Substituting y = 40 in (i), we get
5x − 6(40) = −90
∴ 5x − 240 = −90
∴ 5x = −90 + 240
∴ 5x = 150
∴ x = 30
Since the point of intersection of two regression lines is `(bar x, bar y)`.
∴ `bar x` = 30 and `bar y` = 40
Now, let 5x – 6y + 90 = 0 be the regression equation of Y on X.
∴ The equation becomes 6Y = 5X + 90
i.e., Y = `5/6 X + 90/6`
Comparing it with Y = bYX X + a, we get
∴ `b_(YX) = 5/6`
Now, other equation 15x – 8y – 130 = 0 be the regression equation of X on Y.
∴ The equation becomes 15X = 8Y + 130
i.e., X = `8/15 Y + 130/15`
Comparing it with X = bXY Y + a', we get
∴ `b_(XY) = 8/15`
∴ r = `+-sqrt(b_(XY) * b_(YX))`
= `+- sqrt(8/15 * 5/6)`
= `+- sqrt(4/9)`
= `+- 2/3`
Since bYX and bXY both are positive, r is positive.
∴ r = `2/3`
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| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
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| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
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Mean of x = `barx = square`
Mean of y = `bary = square`
bxy = `square/square`
byx = `square/square`
Regression equation of x on y is `(x - barx) = "b"_(xy) (y - bary)`
∴ Regression equation x on y is `square`
Regression equation of y on x is `(y - bary) = "b"_(yx) (x - barx)`
∴ Regression equation of y on x is `square`
