From the data of 7 pairs of observations on X and Y, following results are obtained. ∑(xi - 70) = - 35, ∑(yi - 60) = - 7, ∑(xi - 70)2 = 2989, ∑(yi - 60)2 = 476, ∑(xi - 70)(yi - 60) = 1064 [Given: `sqrt0.7884` = 0.8879] Obtain The line of regression of Y on X. The line regression of X on Y. The correlation coefficient between X and Y.

From the data of 7 pairs of observations on X and Y, following results are obtained.

Question

From the data of 7 pairs of observations on X and Y, following results are obtained.

∑(x_i - 70) = - 35, ∑(y_i - 60) = - 7,

∑(x_i - 70)² = 2989, ∑(y_i - 60)² = 476,

∑(x_i - 70)(y_i - 60) = 1064

[Given: `sqrt0.7884` = 0.8879]

Obtain

The line of regression of Y on X.
The line regression of X on Y.
The correlation coefficient between X and Y.

Sum

Solution

Given: n =7, ∑(x_i - 70) = - 35, ∑(y_i - 60) = - 7,

∑(x_i - 70)² = 2989, ∑(y_i - 60)² = 476,

∑(x_i - 70)(y_i - 60) = 1064

Let u_i = x_i - 70 and v_i = y_i - 60

∴ ∑ u_i = - 35, ∑ v_i = - 7

`sum "u"_"i"^2 = 2989, sum "v"_"i"^2 = 479`

∑ u_i v_i = 1064

∴ `bar "u" = (sum "u"_"i")/"n" = (-35)/7 = - 5`

∴ `bar "v" = (sum "v"_"i")/"n" = (-7)/7 = - 1`

Now, `sigma_"u"^2 = (sum "u"_"i"^2)/"n" - (bar"u")^2`

`= 2989/7 - (- 5)^2` = 427 - 25 = 402

and `sigma_"v"^2 = (sum "v"_"i"^2)/"n" - (bar"v")^2`

`= 476/7 - (- 1)^2 = 68 - 1 = 67`

cov(u, v) = `(sum "u"_"i" "v"_"i")/"n" - bar"uv"`

`= 1064/7 - (- 5)(- 1)` = 152 - 5 = 147

Since the regression coefficients are independent of change of origin,

b_YX = b_VU and b_XY = b_UV

∴ b_YX = b_VU = `("cov" ("u", "v"))/sigma_"U"^2 = 147/402 = 0.36`

and b_XY = b_UV= `("cov" ("u", "v"))/sigma_"V"^2 = 147/67 = 2.19`

Also, `bar x = bar u` + 70 = - 5 + 70 = 65

and `bar y = bar v` + 60 = - 1 + 60 = 59

(i) The line of regression of Y on X is

`("Y" - bar y) = "b"_"YX" ("X" - bar x)`

∴ (Y - 59) = (0.36)(X - 65)

∴ Y - 59 = 0.36X - 23.4

∴ Y = 0.36X + 59 - 23.4

∴ Y = 0.36X + 35.6

(ii) The line of regression of X on Y is

`("X" - bar x) = "b"_"XY" ("Y" - bar y)`

∴ (X - 65) = (2.19)(Y - 59)

∴ X - 65 = 2.19Y - 129.21

∴ X = 2.19Y + 65 - 129.21

∴ X = 2.19Y - 64.21

(iii) r = `+-sqrt("b"_"YX" * "b"_"XY")`

`= +- sqrt((0.36)(2.19))`

`= +- sqrt0.7884 = +- 0.8879`

Since b_YX and b_XY both are positive,

r is also positive.

∴ r = 0.8879

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Properties of Regression Coefficients

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Q 3 Q 2 Q 4

Chapter 3: Linear Regression - Exercise 3.2 [Page 47]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board

Chapter 3 Linear Regression
Exercise 3.2 | Q 3 | Page 47

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