For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.

प्रश्न

बेरीज

उत्तर

Let 10y = 3x + 170 be the regression equation of Y on X.

∴ The equation becomes 10y = 3x + 170

i.e., Y = `3/10 "X" + 170/10`

Comparing it with Y = b_YX X + a, we get

`"b"_"YX" = 3/10`

Now other equation 5x + 70 = 6y be the regression equation of X on Y.

∴ The equation becomes 5x = 6y – 70

i.e., X = `6/5 "Y" - 70/5`

Comparing it with X = b_XY Y + a', we get

`"b"_"XY" = 6/5`

∴ r = `+-sqrt("b"_"XY" * "b"_"YX") = +-sqrt(6/5 xx 3/10) = +- sqrt(9/25) +- 3/5`

Since b_YX and b_XY both are positive,

r is positive.

∴ r = `3/5` = 0.6

Now, two correlated lines of regression are

10y = 3x + 170

i.e., - 3x + 10y = 170 …(i)

and 5x + 70 = 6y

i.e., 5x - 6y = –70 …(ii)

By (i) × 5 + (ii) × 3, we get

- 15x + 50y = 850
+ 15x - 18y = - 210
32y = 640

∴ y = 20

Substituting y = 20 in equation (i), we get

- 3x +10(20) = 170

∴ - 3x + 200 = 170

∴ 3x = 200 - 170

∴ x = 10

Since the point of intersection of two regression lines is `(bar x, bar y)`,

`bar x` = mean value of X = 10, and

`bar y` = mean value of Y = 20.

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Properties of Regression Coefficients

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Q 12 Q 11 Q 13

पाठ 3: Linear Regression - Exercise 3.3 [पृष्ठ ५०]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board

पाठ 3 Linear Regression
Exercise 3.3 | Q 12 | पृष्ठ ५०

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