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प्रश्न
The regression equation of x on y is 40x – 18y = 214 ......(i)
The regression equation of y on x is 8x – 10y + 66 = 0 ......(ii)
Solving equations (i) and (ii),
`barx = square`
`bary = square`
∴ byx = `square/square`
∴ bxy = `square/square`
∴ r = `square`
Given variance of x = 9
∴ byx = `square/square`
∴ `sigma_y = square`
उत्तर
The regression equation of x on y is 40x – 18y = 214 ......(i)
The regression equation of y on x is 8x – 10y + 66 = 0 ......(ii)
Solving equations (i) and (ii),
By 5 × (ii) – (i), we get
40x – 50y = – 330
40x – 18y = 214
– + –
– 32y = – 544
∴ y = `544/32` = 17
Substituting y = 17 in (i), we get
8x – 10 × 17 = – 66
∴ 8x – 170 = – 66
∴ 8x = – 66 + 170
∴ 8x = 104
∴ x = `104/8` = 13
`barx = 13
`bary` = 17
8x – 10y + 66 = 0 is regression equation of y on x.
⇒ 10y = 8x + 66
⇒ y = `4/5 x + 6.6
⇒ byx = `4/5`
40x – 18y = 214 is regression equation of x on y.
⇒ 40x = 18y + 214
⇒ x = `9/20 y + 5.35`
∴ bxy = `9/20`
r = `+- sqrt("b"_(xy) "b"_(yx))`
= `+ sqrt((9/20 xx 4/5))`
= `+ sqrt(9/25)`
= `+ 3/5`
= 0.6
Given variance of x = 9 ⇒ σx2 = 9 ⇒ σx = 3
∴ byx = `("r"sigma_y)/sigma_x`
∴ `4/5 = (0.6 xx sigma_y)/3`
∴ σy = `(4 xx 3)/(5 xx 0.6)`
∴ σy = 4
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