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From the following data obtain the equation of two regression lines:

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Question

From the following data obtain the equation of two regression lines:

X 6 2 10 4 8
Y 9 11 5 8 7
Sum
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Solution

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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Types of Linear Regression
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Chapter 3: Linear Regression - Exercise 3.1 [Page 42]

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