For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y − 5x + 180 = 0. - Mathematics and Statistics

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For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y − 5x + 180 = 0.  The variance of marks in statistics is `(9/16)^"th"` of the variance of marks in accountancy. Find the correlation coefficient between marks in two subjects.

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Solution

Given, n = 50, X = marks in Statistics,

Y = marks in Accountancy,

Regression equation of X on Y is

3y – 5x + 180 = 0,

`bar y = 44,  sigma_X^2 = 9/16 sigma_Y^2`

Now, 3y – 5x + 180 = 0 is the regression equation of X on Y.

∴ The equation becomes 5X = 3Y + 180

i.e., X = `3/5` Y + `180/5`

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

`b_(XY) = 3/5, a = 180/5` = 36

a = `barx - b_(XY)  bary`

∴ 36 = `bar x - 3/5 xx 44`

∴ 36 = `bar x` – 26.4

∴ `bar x` = 36 + 26.4 = 62.4

Also, `sigma_X^2 = 9/16 sigma_Y^2`

∴ `sigma_X^2/sigma_Y^2 = 9/16`

∴ `sigma_X/sigma_Y = 3/4`

`b_(XY) = r xx sigma_X/sigma_Y`

∴ `3/5 = r xx 3/4`

∴ `3/5 xx 4/3` = r

∴ r = `4/5` = 0.8

∴ Mean marks in statistics `(barx)` are 62.4 and correlation coefficient (r) between marks in the two subjects is 0.8.

Concept: Properties of Regression Coefficients
  Is there an error in this question or solution?
Chapter 3: Linear Regression - Exercise 3.3 [Page 50]

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