The following results are obtained for a bivariate data.particularsxyN

1.	The following results are obtained for a bivariate data.particularsxyNo of observation8Mean100100The sum of square of deviation taken from mean130145The sum of product of deviation taken from mean115Obtain the regression line of Y on X.
Answer» Here, n = 8; x̄ = 100; ȳ = 100; Σ(x – x̄)² = 130; Σ(y – ȳ)² = 145 and Σ(x – x̄) (y – ȳ) =115 are given. \(Now, b = \frac{Σ(x−\bar x)(y−\bar y)}{Σ(x−\bar x)^2}\) \(∴ b = \frac{115}{130}\) = 0.88 a = ȳ – bx̄ ∴ a = 100 – 0.88 (100) = 100 – 88 = 12 Regression line of Y on X is ŷ = a + bx Putting a = 12 and b = 0.88 in ŷ = a + bx, the regression line of Y on X obtained is ŷ = 12 + 0.88x.

The following results are obtained for a bivariate data.particularsxyNo of observation8Mean100100The sum of square of deviation taken from mean130145The sum of product of deviation taken from mean115Obtain the regression line of Y on X.

Answer»

Here, n = 8;

x̄ = 100; ȳ = 100;

Σ(x – x̄)² = 130;

Σ(y – ȳ)² = 145 and

Σ(x – x̄) (y – ȳ) =115 are given.

\(Now, b = \frac{Σ(x−\bar x)(y−\bar y)}{Σ(x−\bar x)^2}\)

\(∴ b = \frac{115}{130}\)

= 0.88

a = ȳ – bx̄
∴ a = 100 – 0.88 (100)
= 100 – 88
= 12

Regression line of Y on X is ŷ = a + bx

Putting a = 12 and b = 0.88 in ŷ = a + bx,

the regression line of Y on X obtained is ŷ = 12 + 0.88x.