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For 5 observations of pairs of (X, Y) of variables X and Y the following results are obtained. ∑X = 15, ∑Y = 25, ∑X2 = 55, ∑Y2 = 135, ∑XY = 83. - Business Mathematics and Statistics

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प्रश्न

For 5 observations of pairs of (X, Y) of variables X and Y the following results are obtained. ∑X = 15, ∑Y = 25, ∑X2 = 55, ∑Y2 = 135, ∑XY = 83. Find the equation of the lines of regression and estimate the values of X and Y if Y = 8; X = 12.

बेरीज

उत्तर

N = 5, ΣX = 15, ΣY = 25, ΣX2 = 55, ΣY2 = 135, ΣXY = 83, `bar"X" = 15/5` = 3, `bar"Y" = 25/5` = 5.

byx = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"(sum"X"^2) - (sum"X")^2)`

= `(5(83) - (15)(25))/(5(55) - (15)^2)`

= `(415 - 375)/(275 - 225)`

= `40/50`

= 0.8

Regression line of Y on X:

`"Y" - bar"Y" = "b"_"yx"("X" - bar"X")`

Y – 5 = 0.8(X – 3)

Y = 0.8X – 2.4 + 5

Y = 0.8X + 2.6

When X = 12, Y = 0.8X + 2.6

Y = (0.8)12 + 2.6

= 9.6 + 2.6

= 12.2

bxy = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"(sum"Y"^2) - (sum"Y")^2)`

= `(5(83) - (15)(25))/(5(135) - (25)^2)`

= `(415 - 375)/(675 - 625)`

= `40/50`

= 0.8

Regression line of X on Y:

`"X" - bar"X" = "b"_"xy"("Y" - bar"Y")`

X – 3 = 0.8(Y – 5)

X = 0.8Y – 4 + 3

X = 0.8Y – 1

When Y = 8, X = 0.8Y – 1

X = (0.8)8 – 1

= 6.4 – 1

= 5.4

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Regression Analysis
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पाठ 9: Correlation and Regression Analysis - Exercise 9.2 [पृष्ठ २२७]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
पाठ 9 Correlation and Regression Analysis
Exercise 9.2 | Q 10 | पृष्ठ २२७

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