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Question
For a certain bivariate data of a group of 10 students, the following information gives the internal marks obtained in English (X) and Hindi (Y):
| X | Y | |
| Mean | 13 | 17 |
| Standard Deviation | 3 | 2 |
If r = 0.6, Estimate x when y = 16 and y when x = 10
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Solution
Given, `barx` = 13, `bary` = 17, `sigma_x` = 3, `sigma_y` = 2, r = 0.6
byx = `"r" sigma_y/sigma_x = 0.6 xx 2/3` = 0.4
bxy = `"r" sigma_x/sigma_y = 0.6 xx 3/2` = 0.9
The regression equation of X on Y is given by `("X" - barx) = "b"_(xy) ("Y" - bary)`
(X – 13) = 0.9(Y – 17)
X – 13 = 0.9Y – 15.3
X = 0.9Y – 15.3 + 13
X = – 2.3 + 0.9Y ......(i)
For Y = 16, from equation (i) we get
X = – 2.3 + (0.9)(16)
= – 2.3 + 14.4
= 12.1
The regression equation of Y on X is given by `("Y" - bary) = "b"_(yx) ("X" - barx)`
(Y – 17) = 0.4(X – 13)
Y – 17 = 0.4X – 5.2
Y = 0.4X – 5.2 + 17
Y = 11.8 + 0.4X .....(ii)
For X = 10, from equation (ii) we get
Y = 11.8 + 0.4(10)
= 11.8 + 4
= 15.8
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|
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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`
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∴ 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`
| 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 |
| Total = 30 | Total = 40 | Total = `square` | Total = 220 | Total = `square` |
bxy = `square/square`
byx = `square/square`
∴ Regression equation of x on y is `square`
∴ Regression equation of y on x is `square`
If byx > 1 then bxy is _______.
