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प्रश्न
If the two regression lines for a bivariate data are 2x = y + 15 (x on y) and 4y = 3x + 25 (y on x), find
- `bar x`,
- `bar y`,
- bYX
- bXY
- r [Given `sqrt0.375` = 0.61]
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उत्तर
Given,
regression equation of X on Y is 2x = y + 15,
i.e., 2x - y = 15 …(i)
regression equation of Y on X is 4y = 3x + 25,
i.e., - 3x + 4y = 25 …(ii)
By (i) × 4 + (ii) we get
8x - 4y = 60
+ - 3x + 4y = 25
5x = 85
∴ x = 17
Substituting the value of x = 17 in (i), we get
2(17) – y = 15
∴ 34 y = 15
∴ y = 34 – 15 = 19
Since the point of intersection of two regression lines is `(bar x, bar y)`,
(i) `bar x` = 17
(ii) `bar y` = 19
(iii) Regression equation of Y on X is 4y = 3x + 25
i.e., 4Y = 3X + 25
i.e., Y = `3/4 "X" + 25/4`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 3/4`
(iv) Regression equation of X on Y is 2x = y + 15
i.e., 2X = Y + 15
i.e., X = `"Y"/2 + 15/2`
Comparing it with X = bXY Y + a' we get,
`"b"_"XY" = 1/2`
(v) r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt((1/2) * (3/4)) = +- sqrt0.375 = +- 0.61`
Since bYX and bXY both are positive,
r is also positive.
∴ r = 0.61
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| x | y | `x - barx` | `y - bary` | `(x - barx)(y - bary)` | `(x - barx)^2` | `(y - bary)^2` |
| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
| 2 | 7 | – 1 | – 2 | `square` | 1 | 4 |
| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
| Total = 15 | Total = 45 | Total = 0 | Total = 0 | Total = `square` | Total = 10 | Total = 40 |
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`
