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प्रश्न
The equations of the two lines of regression are 3x + 2y − 26 = 0 and 6x + y − 31 = 0 Find
- Means of X and Y
- Correlation coefficient between X and Y
- Estimate of Y for X = 2
- var (X) if var (Y) = 36
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उत्तर
(i) Given equations of regression are
3x + 2y - 26 = 0
i.e., 3x + 2y = 26 ….(i)
and 6x + y - 31 = 0
i.e., 6x + y = 31 ….(ii)
By (i) - 2 × (ii), we get
3x + 2y = 26
12x + 2y = 62
- - -
- 9x = - 36
∴ x = `(-36)/-9 = 4`
Substituting x = 4 in (ii), we get
6 × 4 + y = 31
∴ 24 + y = 31
∴ y = 31 - 24
∴ y = 7
Since the point of intersection of two regression lines is `(bar x, bar y)`,
`bar x` = mean of X = 4, and
`bar y` = mean of Y = 7
(ii) Let 3x + 2y 26 = 0 be the regression equation of Y on X.
∴ The equation becomes 2Y = 3X + 26
i.e., Y = `(-3)/2"X" + 26/2`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = (- 3)/2`
Now, the other equation 6x + y - 31 = 0 is the regression equation of X on Y.
∴ The equation becomes 6X = - Y + 31
i.e., X = `(-1)/6 "Y" + 31/6`
Comparing it with X = bXY Y+ a', we get
`"b"_"XY" = (-1)/6`
∴ r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt((- 1)/6 xx (- 3)/2) = +- sqrt(1/4) = +- 1/2 = +- 0.5`
Since the values of bXY and bYX are negative,
r is also negative.
∴ r = - 0.5
(iii) The regression equation of Y on X is
Y = `(- 3)/2 "X" + 26/2`
For X = 2, we get
Y = `(- 3)/2 xx 2 + 26/2 = - 3 + 13 = 10`
(iv) Given, Var (Y) = 36, i.e., `sigma_"Y"^2` = 36
∴ `sigma_"Y" = 6`
Since `"b"_"XY" = "r" xx sigma_"X"/sigma_"Y"`
`(- 1)/6 = - 0.5 xx sigma_"X"/6`
∴ `sigma_"X" = (-6)/(- 6 xx 0.5) = 2`
∴ `sigma_"X"^2` = Var(X) = 4
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