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प्रश्न
The equations of two regression lines are 10x − 4y = 80 and 10y − 9x = − 40 Find:
- `bar x and bar y`
- bYX and bXY
- If var (Y) = 36, obtain var (X)
- r
उत्तर
(i) Given equations of regression are
10x − 4y = 80
i.e., 5x − 2y = 40 .....(i)
and 10y − 9x = −40
i.e., − 9x + 10y = −40 .....(ii)
By 5 × (i) + (ii), we get
25x − 10y = 200
− 9x + 10y = − 40
16x = 160
∴ x = 10
Substituting x = 10 in (i), we get
5(10) − 2y = 40
∴ 50 − 2y = 40
∴ −2y = 40 − 50
∴ −2y = − 10
∴ y = 5
Since the point of intersection of two regression lines is `(bar x, bar y)`, `bar x = 10 and bar y = 5`
(ii) Let 10y − 9x = −40 be the regression equation of Y on X.
∴ The equation becomes 10Y = 9X − 40
i.e., Y = `9/10X − 40/10`
Comparing it with Y = bYX X + a, we get
`b_(YX) = 9/10 = 0.9`
Now, the other equation 10x − 4y = 80 be the regression equation of X on Y.
∴ The equation becomes 10X = 4Y + 80
i.e., X = `4/10 Y + 80/10`
i.e., X = `2/5 Y + 8`
Comparing it with X = bXY Y + a', we get
`b_(XY) = 2/5 = 0.4`
(iii) Given, Var (Y) = 36, i.e., `sigma_Y^2` = 36
∴ σY = 6
Since `b_(XY) = r xx sigma_X/sigma_Y`
`2/5 = 0.6 xx sigma_X/6`
∴ `2/5 = 0.1 xx sigma_X`
∴ `2/(5 xx 0.1) = sigma_X`
∴ `sigma_X` = 4
∴ `sigma_X^2 = 16` i.e., Var(X) = 16
(iv) r = `+-sqrt(b_(XY) *b_(YX)`
`= +-sqrt(2/5 xx 9/10) +- sqrt(9/25)`
`= +- 3/5`
`= +- 0.6`
Since bYX and bXY are positive,
r is also positive.
∴ r = 0.6
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| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
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| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
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