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Question
From the data given below:
| Marks in Economics: | 25 | 28 | 35 | 32 | 31 | 36 | 29 | 38 | 34 | 32 |
| Marks in Statistics: | 43 | 46 | 49 | 41 | 36 | 32 | 31 | 30 | 33 | 39 |
Find
- The two regression equations,
- The coefficient of correlation between marks in Economics and Statistics,
- The mostly likely marks in Statistics when the marks in Economics is 30.
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Solution
| Marks in Economics (X) | Marks in Statistics (Y) | x = `"X" - bar"X"` | y = `"Y" - bar"Y"` | x2 | y2 | xy |
| 25 | 43 | − 7 | 5 | 49 | 25 | − 35 |
| 28 | 46 | − 4 | 8 | 16 | 64 | − 32 |
| 35 | 49 | 3 | 11 | 9 | 121 | 33 |
| 32 | 41 | 0 | 3 | 0 | 9 | 0 |
| 31 | 36 | − 1 | − 2 | 1 | 4 | 2 |
| 36 | 32 | 4 | − 6 | 16 | 36 | − 24 |
| 29 | 31 | − 3 | − 7 | 9 | 49 | 21 |
| 38 | 30 | 6 | − 8 | 36 | 64 | − 48 |
| 34 | 33 | 2 | − 5 | 4 | 25 | − 10 |
| 32 | 39 | 0 | 1 | 0 | 1 | 0 |
| 320 | 380 | 0 | 0 | 140 | 398 | − 93 |
N = 10, ∑X = 320, ∑Y = 280, ∑x2 = 140, ∑y2 = 398, ∑xy = − 93, `bar"X" = 320/100` = 32, `bar"Y" = 380/100` = 38
(a) Regression equation of X on Y.
bxy = `"r"(sigma_"x")/(sigma_"y") = (sum"xy")/(sum"y"^2) = (-93)/398` = − 0.234
`"X" - bar"X" = "b"_"xy"("Y" - bar"Y")`
X − 32 = − 0.234(Y − 38)
X = − 0.234Y + 8.892 + 32
X = − 0.234Y + 40.892
Regression equation of Y on X.
`"Y" - bar"Y" = "b"_"xy"("X" - bar"X")`
byx = `"r"(sigma_"x")/(sigma_"y") = (sum"xy")/(sum"y"^2) = (-93)/140` = − 0.664
Y − 38 = − 0.664(X − 32)
Y = − 0.664X + 21.248 + 38
Y = − 0.664X + 59.248
(b) Coefficient of correlation (r) = `±sqrt("b"_"xy" xx "b"_"yx")`
= `sqrt((-0.234)(-0.664))`
= − 0.394
(c) When X = 30, Y = ?
Y = − 0.664(30) + 59.248
= − 19.92 + 59.248
= 39.328
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