Advertisements
Advertisements
Question
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
APPEARS IN
RELATED QUESTIONS
Obtain the two regression lines from the following data N = 20, ∑X = 80, ∑Y = 40, ∑X2 = 1680, ∑Y2 = 320 and ∑XY = 480.
Find the equation of the regression line of Y on X, if the observations (Xi, Yi) are the following (1, 4) (2, 8) (3, 2) (4, 12) (5, 10) (6, 14) (7, 16) (8, 6) (9, 18).
A survey was conducted to study the relationship between expenditure on accommodation (X) and expenditure on Food and Entertainment (Y) and the following results were obtained:
Details | Mean | SD |
Expenditure on Accommodation (₹) | 178 | 63.15 |
Expenditure on Food and Entertainment (₹) | 47.8 | 22.98 |
Coefficient of Correlation | 0.43 |
Write down the regression equation and estimate the expenditure on Food and Entertainment, if the expenditure on accommodation is ₹ 200.
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
The following data pertains to the marks in subjects A and B in a certain examination. Mean marks in A = 39.5, Mean marks in B = 47.5 standard deviation of marks in A = 10.8 and Standard deviation of marks in B = 16.8. coefficient of correlation between marks in A and marks in B is 0.42. Give the estimate of marks in B for the candidate who secured 52 marks in A.
Find the line regression of Y on X
X | 1 | 2 | 3 | 4 | 5 | 8 | 10 |
Y | 9 | 8 | 10 | 12 | 14 | 16 | 15 |
The following information is given.
Details | X (in ₹) | Y (in ₹) |
Arithmetic Mean | 6 | 8 |
Standard Deviation | 5 | `40/3` |
Coefficient of correlation between X and Y is `8/15`. Find
- The regression Coefficient of Y on X
- The most likely value of Y when X = ₹ 100.