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प्रश्न
Mean of x = 53
Mean of y = 28
Regression coefficient of y on x = – 1.2
Regression coefficient of x on y = – 0.3
a. r = `square`
b. When x = 50,
`y - square = square (50 - square)`
∴ y = `square`
c. When y = 25,
`x - square = square (25 - square)`
∴ x = `square`
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उत्तर
Mean of x = 53
Mean of y = 28
Regression coefficient of y on x = – 1.2
Regression coefficient of x on y = – 0.3
a. r = `+- sqrt("b"_(xy)*"b"_(yx))`
= `+- sqrt((-0.3)(-1.2))`
= `+- 0.6`
Since bYX and bXY both are – negative,
r is also negative.
∴ r = – 0.6
b. When x = 50,
`(y - bary) = "b"_(yx) (x- barx)`
∴ `(y - 28) = - 1.2 (50 - 53)`
∴ y = 28 – 60 + 63.6
∴ y = 31.6
c. When y = 25,
`(x - 53) = - 0.3 (25 - 28)`
∴ X = 53 – 7.5 + 8.4
∴ X = 53.9
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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`
Mean of x = 25
Mean of y = 20
`sigma_x` = 4
`sigma_y` = 3
r = 0.5
byx = `square`
bxy = `square`
when x = 10,
`y - square = square (10 - square)`
∴ y = `square`
The regression equation of y on x is 2x – 5y + 60 = 0
Mean of x = 18
`2 square - 5 bary + 60` = 0
∴ `bary = square`
`sigma_x : sigma_y` = 3 : 2
∴ byx = `square/square`
∴ byx = `square/square`
∴ r = `square`
|bxy + byz| ≥ ______.
