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प्रश्न
Two lines of regression are 10x + 3y − 62 = 0 and 6x + 5y − 50 = 0. Identify the regression of x on y. Hence find `bar x, bar y` and r.
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उत्तर
Given, two lines of regression are
10x + 3y – 62 = 0
i.e., 10x + 3y = 62 …(i)
and 6x + 5y – 50 = 0
i.e., 6x + 5y = 50 …(ii)
By (i) × 5 - (ii) × 3, we get
50x + 15y = 310
18x + 15y = 150
- - -
32x = 160
∴ x = 5
Substituting x = 5 in (i) we get,
10(5) + 3y = 62
∴ 50 + 3y = 62
∴ 3y = 62 - 50 = 12
∴ y = 4
Since the point of intersection of two regression lines is `(bar x, bar y)`,
`bar x = 5 and bar y = 4`
Now,
Let 10x + 3y - 62 = 0 be the regression equation of X on Y.
∴ The equation becomes 10x = –3y + 62
i.e., 10X = –3Y + 62
i.e., X = `- 3/10 "Y" + 62/10`
Comparing it with X = bXY Y + a, we get
∴ `"b"_"XY" = - 3/10`
Now, other equation 6x + 5y – 50 = 0 be the regression equation of Y on X.
∴ The equation becomes 5y = – 6x + 50
i.e., 5Y = – 6X + 50
i.e., Y = `- 6/5 "x" + 50/5`
Comparing it with Y = bYX X + a', we get
`"b"_"YX" = - 6/5`
Now, `"b"_"YX" * "b"_"XY" = (- 3/10)(- 6/5) = 9/25`
i.e., bXY . bYX < 1
∴ Assumption of regression equations is true.
∴ r = `+-sqrt("b"_"XY" * "b"_"YX") = +-sqrt(9/25) = +- 3/5`
Since bYX and bXY both are negative,
r is negative.
∴ r = `- 3/5 = - 0.6`
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