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Question
From the two regression equations, find r, `bar x and bar y`. 4y = 9x + 15 and 25x = 4y + 17
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Solution
Given regression equations are
4y = 9x + 15
i.e., - 9x + 4y = 15 ....(i)
and 25x = 4y + 17
i.e., 25x - 4y = 17 ...(ii)
Adding equations (i) and (ii), we get
- 9x + 4y = 15
25x - 4y = 17
16x = 32
∴ x = 2
Substituting x = 2 in (i), we get
- 9(2) + 4y = 15
∴ - 18 + 4y = 15
∴ 4y = 33
∴ y = 8.25
Since the point of intersection of two regression lines is `(bar x, bar y), bar x = 2 and bar y = 8.25`
Let 4y = 9x + 15 be the regression equation of Y on X.
∴ The equation becomes Y = `9/2 "X" + 15/4`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 9/4 = 2.25`
Now, the other equation, i.e., 25x = 4y + 17 is the regression equation of X on Y.
∴ The equation becomes X = `4/25 "Y" + 17/25`
Comparing it with X = bXY Y + a', we get
`"b"_"XY" = 4/25 = 0.16`
r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt (0.16 xx 2.25)`
`= +- sqrt0.36 = +- 0.6`
Since bYX and bXY are positive,
r is also positive.
∴ r = 0.6
∴ `bar x = 2 and bar y = 8.25` and r = 0.6
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Mean of x = 53
Mean of y = 28
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Regression coefficient of x on y = – 0.3
a. r = `square`
b. When x = 50,
`y - square = square (50 - square)`
∴ y = `square`
c. When y = 25,
`x - square = square (25 - square)`
∴ x = `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
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`2 square - 5 bary + 60` = 0
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`sigma_x : sigma_y` = 3 : 2
∴ byx = `square/square`
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| 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`
