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The following results were obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men. X Y Mean 50 140 Variance 150 165 and ∑(xi-x¯)(yi-y¯)=1120. Find the prediction of

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Question

The following results were obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.

  X Y
Mean 50 140
Variance 150 165

and `sum (x_i - bar x)(y_i - bar y) = 1120`. Find the prediction of blood pressure of a man of age 40 years.

Sum
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Solution

Given, X = Age, Y = Systolic blood pressure,

n = 10, `bar x = 50","  bar y = 140`,

`sigma_X^2 = 150,  sigma_Y^2 = 165` and

`sum (x_i - bar x)(y_i - bar y) = 1120`

Since Var(X) = `(sum (x_i - bar x)^2)/n`,

`sigma_x^2 = (sum (x_i - bar x)^2)/n`

∴ `150 = (sum (x_i - bar x)^2)/10`

∴ `sum (x_i - bar x)^2 = 1500`

Now, `b_(YX) = (sum (x_i - bar x)(y_i - bar y))/(sum (x_i - bar x)^2) = 1120/1500 = 0.7`

∴ The regression equation of systolic blood pressure of the men (Y) on their age (X) is 

`(Y - bar y) = b_(YX) (X - bar x)`

∴ (Y − 140) = 0.7(X − 50)

∴ Y − 140 = 0.7X − 35

∴ Y = 0.7X − 35 + 140

∴ Y = 0.7X + 105

For X = 40,

Y = 0.7(40) + 105

Y = 28 + 105

Y = 133

∴ The man of age 40 years has a systolic blood pressure of 133.

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Properties of Regression Coefficients
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Chapter 3: Linear Regression - Miscellaneous Exercise 3 [Page 54]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 3 Linear Regression
Miscellaneous Exercise 3 | Q 4.1 | Page 54

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