Advertisements
Advertisements
प्रश्न
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.
उत्तर
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
APPEARS IN
संबंधित प्रश्न
The following data give the height in inches (X) and the weight in lb. (Y) of a random sample of 10 students from a large group of students of age 17 years:
X | 61 | 68 | 68 | 64 | 65 | 70 | 63 | 62 | 64 | 67 |
Y | 112 | 123 | 130 | 115 | 110 | 125 | 100 | 113 | 116 | 125 |
Estimate weight of the student of a height 69 inches.
Obtain the two regression lines from the following data N = 20, ∑X = 80, ∑Y = 40, ∑X2 = 1680, ∑Y2 = 320 and ∑XY = 480.
The following data relate to advertisement expenditure (in lakh of rupees) and their corresponding sales (in crores of rupees)
Advertisement expenditure | 40 | 50 | 38 | 60 | 65 | 50 | 35 |
Sales | 38 | 60 | 55 | 70 | 60 | 48 | 30 |
Estimate the sales corresponding to advertising expenditure of ₹ 30 lakh.
You are given the following data:
Details | X | Y |
Arithmetic Mean | 36 | 85 |
Standard Deviation | 11 | 8 |
If the Correlation coefficient between X and Y is 0.66, then find
- the two regression coefficients,
- the most likely value of Y when X = 10.
The regression coefficient of X on Y
The regression coefficient of Y on X
When one regression coefficient is negative, the other would be
If the regression coefficient of Y on X is 2, then the regression coefficient of X on Y is
The lines of regression intersect at the point
The term regression was introduced by