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प्रश्न
Find the line regression of Y on X
X | 1 | 2 | 3 | 4 | 5 | 8 | 10 |
Y | 9 | 8 | 10 | 12 | 14 | 16 | 15 |
उत्तर
X | Y | dx = X − 5 | dy = Y − 12 | dx2 | dy2 | dxdy |
1 | 9 | − 4 | − 3 | 16 | 9 | 12 |
2 | 8 | − 3 | − 4 | 9 | 16 | 12 |
3 | 10 | − 2 | − 2 | 4 | 4 | 4 |
4 | 12 | − 1 | 0 | 1 | 0 | 0 |
5 | 14 | 0 | 2 | 0 | 4 | 0 |
8 | 16 | 3 | 4 | 9 | 16 | 12 |
10 | 15 | 5 | 3 | 25 | 9 | 15 |
33 | 84 | − 2 | 0 | 64 | 58 | 55 |
`bar"X" = (sum"X")/"n" = 33/7` = 4.71
`bar"Y" - (sum"Y")/"n" = 84/7` = 12
Regression coefficient
byx = `("N"sum"dxdy" - (sum"dx")(sum"dy"))/("N"sum"dx"^2 - (sum"dx")^2)`
= `(7(55) - (-2)(0))/(7(64) - (-2)^2)`
= `385/444`
= 0.867
∴ Regression line of Y on X is
`"Y" - bar"Y" = "b"_"yx"("X" - bar"X")`
Y − 12 = 0.867 (X − 4.71)
Y − 12 = 0.867X − 4.084
Y = 0.867X + 7.916
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