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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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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
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 = 5, ∑ 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" = (sumx_i y_i − n bar x bar y)/(sum x_i^2 − n barx^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 the regression equation of X on Y,

`"b"_"XY" = (sumx_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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संबंधित प्रश्न

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Find the equation of the regression line of y on x, if the observations (x, y) are as follows : 
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Hence find co-ordinates of corner points of the common region.


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

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

`sum(x_i - barx)^2` = 1200, `sum(y_i - bary)^2` = 300,

`sum(x_i - bar x)(y_i - bar y)` = –250

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Solving equations (i) and (ii),

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∴ r = `square`

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

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Following is the data that the Manager refers to:

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