Maharashtra State BoardHSC Commerce 12th Board Exam
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For the following bivariate data obtain the equations of two regression lines: - Mathematics and Statistics

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Sum

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 = 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

Concept: Types of Linear Regression
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