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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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| 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`
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∴ y = `square`
| x | y | xy | x2 | y2 |
| 6 | 9 | 54 | 36 | 81 |
| 2 | 11 | 22 | 4 | 121 |
| 10 | 5 | 50 | 100 | 25 |
| 4 | 8 | 32 | 16 | 64 |
| 8 | 7 | `square` | 64 | 49 |
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