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Question
For the following bivariate data obtain the equations of two regression lines:
| X | 1 | 2 | 3 | 4 | 5 |
| Y | 5 | 7 | 9 | 11 | 13 |
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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 = 5, ∑ 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" = (sumx_i y_i − n bar x bar y)/(sum x_i^2 − n barx^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 the regression equation of X on Y,
`"b"_"XY" = (sumx_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
∴ X = 0.5 Y − 1.5
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