Tamil Nadu Board of Secondary EducationHSC Commerce Class 11th

# The heights (in cm.) of a group of fathers and sons are given below: Heights of fathers: 158 166 163 165 167 170 167 172 177 181 Heights of Sons: 163 158 167 170 160 180 170 175 172 175 - Business Mathematics and Statistics

Sum

The heights (in cm.) of a group of fathers and sons are given below:

 Heights of fathers: 158 166 163 165 167 170 167 172 177 181 Heights of Sons: 163 158 167 170 160 180 170 175 172 175

Find the lines of regression and estimate the height of the son when the height of the father is 164 cm.

#### Solution

 Heights of fathers(X) Heights of Sons(Y) dx = X − 168 dy = Y − 169 dx2 dy2 dxdy 158 163 − 10 − 6 100 36 − 60 166 158 − 2 − 11 4 121 22 163 167 − 5 − 2 25 4 10 165 170 − 3 1 9 1 − 3 167 160 − 1 − 9 1 81 9 170 180 2 11 4 121 22 167 170 − 1 1 1 1 −1 172 175 4 6 16 36 24 177 172 9 3 81 9 27 181 175 13 6 169 36 78 1686 1690 6 0 410 446 248

N = 10, ∑X = 1686, ∑Y = 1690, ∑dx2 = 410, ∑y2 = 446, ∑dxdy = 248, bar"X" = 1686/10 = 168.6, bar"Y" = 1690/10 = 169

bxy = ("N"sum"dxdy" - (sum"dx")(sum"dy"))/("N"sum"dy"^2 - (sum"dy")^2)

= (10(248) - 6(0))/((10)(446) - 0^2)

= 2480/4460

= 0.556

Regression equation of X on Y

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

X – 168.6 = 0.556 (Y – 169)

X – 168.6 = 0.556Y – 93.964

X = 0.556Y – 93.964 + 168.6

X = 0.556Y + 76.636

X = 0.556Y + 74.64

byx = ("N"sum"dxdy" - (sum"dx")(sum"dy"))/("N"sum"dx"^2 - (sum"dx")^2)

= (10(248) - 0)/(10(410) - 6^2)

= 2480/(4100 - 36)

= 2480/4064

= 0.610

Regression equation of Y on X

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

Y − 169 = 0.610 (X − 168.6)

Y – 169 = 0.610X – 102.846

Y = 0.610X – 102.846 + 169

Y = 0.610X + 66.154 ………(1)

To get son’s height (Y) when the father height is X = 164 cm.

Put X = 164 cm in equation (1) we get

Son’s height = 0.610 × 164 + 66.154

= 100.04 + 66.154

= 166.19 cm

Concept: Regression Analysis
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