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Question
For the following data, find the regression line of Y on X
| X | 1 | 2 | 3 |
| Y | 2 | 1 | 6 |
Hence find the most likely value of y when x = 4.
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Solution
| X = xi | Y = yi | `bb("x"_"i"^2)` | xi yi |
| 1 | 2 | 1 | 2 |
| 2 | 1 | 4 | 2 |
| 3 | 6 | 9 | 18 |
| 6 | 9 | 14 | 22 |
From the table, we have
n = 3, ∑ xi = 6, ∑ yi = 9, `sum "x"_"i"^2 = 14`, ∑ xi yi = 22
`bar x = (sum x_i)/"n" = 6/3 = 2`
`bar y = (sum y_i)/"n" = 9/3 = 3`
Now, `"b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar"x"^2)`
`= (22 - 3xx2xx3)/(14 - 3(2)^2)`
`= (22 - 18)/(14 - 12)`
`= 4/2`
= 2
The regression equation of Y on X is,
`(y - bary) = "b"_("YX") (x - barx)`
y − 3 = 2(x − 2)
y − 3 = 2x − 4
y = 2x − 4 + 3
y = 2x − 1 is the regression equation of Y on X. ...(1)
When x = 4, y = ?
Substituting x = 4 in equation (1)
y = 2(4) − 1
= 8 − 1
y = 7
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