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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.
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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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| x | y | `x - barx` | `y - bary` | `(x - barx)(y - bary)` | `(x - barx)^2` | `(y - bary)^2` |
| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
| 2 | 7 | – 1 | – 2 | `square` | 1 | 4 |
| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
| Total = 15 | Total = 45 | Total = 0 | Total = 0 | Total = `square` | Total = 10 | Total = 40 |
Mean of x = `barx = square`
Mean of y = `bary = square`
bxy = `square/square`
byx = `square/square`
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∴ Regression equation x on y is `square`
Regression equation of y on x is `(y - bary) = "b"_(yx) (x - barx)`
∴ Regression equation of y on x is `square`
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Mean of y = 28
Regression coefficient of y on x = – 1.2
Regression coefficient of x on y = – 0.3
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| x | y | |
| Mean | 53 | 142 |
| Variance | 130 | 165 |
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`sum(x - overlinex)^2` = 1200, `sum(y - overliney)^2` = 300, `sum(x - overlinex)(y - overliney)` = – 250
Find:
- byx
- bxy
- Correlation coefficient between x and y.
