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Question
Two samples from bivariate populations have 15 observations each. The sample mean of X and Y are 25 and 18 respectively. The corresponding sum of squares of deviations from means are 136 and 148. The sum of product of deviations from respective means is 122. Obtain the equation of line of regression of X on Y.
Solution
Given n = 15, `barx = 25, bary = 18`
`Σ ( x - bar x )^2 = 136. Σ ( y - bary )^2 = 148`
`Σ ( x - bar x )^2( y - bary )^2 = 122`
we have `b_xy = [ Σ( x - barx )( y - bary )]/[ Σ ( y - bary)^2]`
= `122/148` = 0.8243
Regression line of x on y is
`x - barx = b_(xy) ( y - bary )`
⇒ x - 25 = 0.8243 ( y - 18 )
⇒ x = 0.8243 ( y - 18 ) + 25
⇒ x = 0.8243y - 14.8374 + 25
∴ x = 0.8243y + 10.1626
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