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प्रश्न
State whether the following statement is True or False:
y = 5 + 2.8x and x = 3 + 0.5y be the regression lines of y on x and x on y respectively, then byx = – 0.5
पर्याय
True
False
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उत्तर
False
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संबंधित प्रश्न
The equations given of the two regression lines are 2x + 3y - 6 = 0 and 5x + 7y - 12 = 0.
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Find the feasible solution for the following system of linear inequations:
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If Σx1 = 56 Σy1 = 56, Σ`x_1^2` = 478,
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From the two regression equations y = 4x – 5 and 3x = 2y + 5, find `bar x and bar y`.
The equations of the two lines of regression are 3x + 2y − 26 = 0 and 6x + y − 31 = 0 Find
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- var (X) if var (Y) = 36
Find the equation of the line of regression of Y on X for the following data:
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In the regression equation of Y on X, byx represents slope of the line.
Choose the correct alternative:
The slope of the line of regression of y on x is called the ______
Choose the correct alternative:
u = `(x - 20)/5` and v = `(y - 30)/4`, then bxy =
Choose the correct alternative:
y = 5 – 2.8x and x = 3 – 0.5 y be the regression lines, then the value of byx is
State whether the following statement is True or False:
If equation of regression lines are 3x + 2y – 26 = 0 and 6x + y – 31= 0, then mean of X is 7
State whether the following statement is True or False:
bxy is the slope of regression line of y on x
If the regression equations are 8x – 10y + 66 = 0 and 40x – 18y = 214, the mean value of y is ______
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The equations of the two lines of regression are 6x + y − 31 = 0 and 3x + 2y – 26 = 0. Identify the regression lines
If n = 5, Σx = Σy = 20, Σx2 = Σy2 = 90, Σxy = 76 Find the regression equation of x on y
If `bar"X"` = 40, `bar"Y"` = 6, σx = 10, σy = 1.5 and r = 0.9 for the two sets of data X and Y, then the regression line of X on Y will be:
For certain bivariate data on 5 pairs of observations given:
∑x = 20, ∑y = 20, ∑x2 = 90, ∑y2 = 90, ∑xy = 76 then bxy = ______.
For a bivariate data `barx = 10`, `bary = 12`, V(X) = 9, σy = 4 and r = 0.6
Estimate y when x = 5
Solution: Line of regression of Y on X is
`"Y" - bary = square ("X" - barx)`
∴ Y − 12 = `r.(σ_y)/(σ_x)("X" - 10)`
∴ Y − 12 = `0.6 xx 4/square ("X" - 10)`
∴ When x = 5
Y − 12 = `square(5 - 10)`
∴ Y − 12 = −4
∴ Y = `square`
