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For the following bivariate data obtain the equations of two regression lines: - Mathematics and Statistics

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प्रश्न

For the following bivariate data obtain the equations of two regression lines:

X 1 2 3 4 5
Y 5 7 9 11 13

For the following bivariate data obtain the equation of regression line of Y on X.

X 1 2 3 4 5
Y 5 7 9 11 13
योग
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उत्तर

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

∴ X = 0.5 Y - 1.5

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अध्याय 3: Linear Regression - Exercise 3.1 [पृष्ठ ४२]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 3 Linear Regression
Exercise 3.1 | Q 8 | पृष्ठ ४२

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From the data of 20 pairs of observations on X and Y, following results are obtained.

`barx` = 199, `bary` = 94,

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∴ byx = `square/square`

∴ bxy = `square/square`

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∴ byx = `square/square`

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∴ `b_(yx)=(sum(x_i-barx)(y_i-bary))/(sum(x_i-barx)^2)=square`

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∴ When x = 5

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∴ Y − 12 = −4

∴ Y = `square`


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