For the following bivariate data obtain the equations of two regression lines:
| X | 1 | 2 | 3 | 4 | 5 |
| Y | 5 | 7 | 9 | 11 | 13 |
Solution
| X = xi | Y = yi | `"x"_"i"^2` | `"y"_"i"^2` | xi yi |
| 1 | 5 | 1 | 25 | 5 |
| 2 | 7 | 4 | 49 | 14 |
| 3 | 9 | 9 | 81 | 27 |
| 4 | 11 | 16 | 121 | 44 |
| 5 | 13 | 25 | 169 | 65 |
| 15 | 45 | 55 | 445 | 155 |
From the table, we have
n = 6, ∑ xi = 15, ∑ yi = 45, `sum "x"_"i"^2 = 55`, `sum "y"_"i"^2 = 445`, ∑ xi yi = 155
`bar x = (sum x_i)/"n" = 15/5 = 3`
`bar y = (sum y_i)/"n" = 45/5 = 9`
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)`
`= (155 - 5 xx 3 xx 9)/(55 - 5(3)^2) = (155 - 135)/(55 - 45) = 20/10 = 2`
Also, `"a" = bar y - "b"_"XY" bar x` = 9 - 2(3) = 9 - 6 = 3
The regression analysis of Y on X is
Y = a + bYX X
∴ Y = 3 + 2X
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)`
`= (155 - 5xx3xx9)/(445 - 5(9)^2) = (155 - 135)/(445 - 405) = 20/40 = 0.5`
Also, `"a"' = bar x - "b"_"XY" bar y`
= 3 - (0.5)(9) = 3 - 4.5 = - 1.5
The regression equation of X on Y is
X = a' + bXY Y
∴ X = - 1.5 + 0.5Y
