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प्रश्न
From the following data obtain the equation of two regression lines:
X | 6 | 2 | 10 | 4 | 8 |
Y | 9 | 11 | 5 | 8 | 7 |
उत्तर
X = xi | Y = yi | `"x"_"i"^2` | `"y"_"i"^2` | xi yi |
6 | 9 | 36 | 81 | 54 |
2 | 11 | 4 | 121 | 22 |
10 | 5 | 100 | 25 | 50 |
4 | 8 | 16 | 64 | 32 |
8 | 7 | 64 | 49 | 56 |
30 | 40 | 220 | 340 | 214 |
From the table, we have
n = 5, ∑ xi = 30, ∑ yi = 40, `sum "x"_"i"^2 = 220`, `sum "y"_"i"^2 = 340,` ∑ xi yi = 214
`bar x = (sum x_i)/"n" = 30/5 = 6`
`bar y = (sum y_i)/"n" = 40/5 = 8`
Now, for regression equation of Y on X,
`"b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar"x"^2)`
`= (214 - 5xx 6 xx 8)/(220 - 5(6)^2) = (214 - 240)/(220 - 180) = (-26)/40` = - 0.65
Also, `"a" = bar y - "b"_"YX" bar x`
= 8 - (- 0.65)(6) = 8 + 3.9 = 11.9
The regression equation of Y on X is
Y = a + bYX X
∴ Y = 11.9 - 0.65X
Now, for regression equation of X on Y,
`"b"_"XY" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "y"_"i"^2 - "n" bar"y"^2)`
`= (214 - 5 xx 6 xx 8)/(340 - 5(8)^2) = (214 - 240)/(340 - 320) = (-26)/20` = - 1.3
Also, `"a"' = bar x - "b"_"XY" bar y`
= 6 - (- 1.3)8 = 6 + 10.4 = 16.4
The regression equation of Y on X is
X = a' + bXY Y
∴ X = 16.4 - 1.3Y
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