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प्रश्न
In a partially destroyed laboratory record of an analysis of regression data, the following data are legible:
Variance of X = 9
Regression equations:
8x − 10y + 66 = 0
and 40x − 18y = 214.
Find on the basis of above information
- The mean values of X and Y.
- Correlation coefficient between X and Y.
- Standard deviation of Y.
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उत्तर
Given, `sigma_"X"^2 = 9`
∴ σX = 3
(i) The two regression equations are
8x - 10y + 66 = 0
i.e., 8x - 10y = - 66 ...(i)
and 40x - 18y = 214 ....(ii)
By 5 × (i) - (ii), we get
40x - 50y = - 330
40x - 18y = 214
(-) (+) (-)
- 32y = - 544
∴ y = `544/32 = 17`
Substituting y = 17 in (i), we get
8x - 10 × 17 = - 66
∴ 8x - 170 = - 66
∴ 8x = - 66 + 170
∴ 8x = 104
∴ x = `104/8 = 13`
Since the point of intersection of two regression lines is `(bar x, bar y)`,
`bar x` = mean value of X = 13, and
`bar y` = mean value of X = 17.
(ii) Let 8x - 10y + 66 = 0 be the regression equation of Y on X.
∴ The equation becomes 10Y = 8X + 66
i.e., Y = `8/10 "X" + 66/10`
i.e., Y = `4/5 "X" + 33/5`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 4/5`
Now, the other equation, i.e., 40x - 18y = 214 is the regression equation of X on Y.
∴ The equation becomes X = `18/40 "Y" + 214/40`
i.e., X = `9/20 "Y" + 107/20`
Comparing it with X = bXY Y + a', we get
`"b"_"XY" = 9/20`
r = `+- sqrt("b"_"XY" * "b"_"YX")`
∴ r = `+- sqrt(9/20 xx 4/5) = +- sqrt(9/25) = +- 3/5 = +- 0.6`
Since bYX and bXY both are positive,
r is also positive.
∴ r = 0.6
(iii) `"b"_"YX" = "r" sigma_"Y"/sigma_"X"`
∴ `4/5 = 0.6 xx sigma_"Y"/3`
∴ `4/5 = sigma_"Y"/5`
∴ `sigma_"Y" = 4`
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| x | y | `x - barx` | `y - bary` | `(x - barx)(y - bary)` | `(x - barx)^2` | `(y - bary)^2` |
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| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
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Mean of x = `barx = square`
Mean of y = `bary = square`
bxy = `square/square`
byx = `square/square`
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∴ Regression equation of y on x is `square`
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∴ `bary = square`
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∴ byx = `square/square`
∴ byx = `square/square`
∴ r = `square`
