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Question
In the regression equation of Y on X, byx represents slope of the line.
Options
True
False
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Solution
This statement is True.
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RELATED QUESTIONS
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For the following bivariate data obtain the equations of two regression lines:
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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
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 regression equation of x on y is 40x – 18y = 214 ......(i)
The regression equation of y on x is 8x – 10y + 66 = 0 ......(ii)
Solving equations (i) and (ii),
`barx = square`
`bary = square`
∴ byx = `square/square`
∴ bxy = `square/square`
∴ r = `square`
Given variance of x = 9
∴ byx = `square/square`
∴ `sigma_y = square`
The management of a large furniture store would like to determine sales (in thousands of ₹) (X) on a given day on the basis of number of people (Y) that visited the store on that day. The necessary records were kept, and a random sample of ten days was selected for the study. The summary results were as follows:
`sumx_i = 370 , sumy_i = 580, sumx_i^2 = 17200 , sumy_i^2 = 41640, sumx_iy_i = 11500, n = 10`
Complete the following activity to find, the equation of line of regression of Y on X and X on Y for the following data:
Given:`n=8,sum(x_i-barx)^2=36,sum(y_i-bary)^2=40,sum(x_i-barx)(y_i-bary)=24`
Solution:
Given:`n=8,sum(x_i-barx)=36,sum(y_i-bary)^2=40,sum(x_i-barx)(y_i-bary)=24`
∴ `b_(yx)=(sum(x_i-barx)(y_i-bary))/(sum(x_i-barx)^2)=square`
∴ `b_(xy)=(sum(x_i-barx)(y_i-bary))/(sum(y_i-bary)^2)=square`
∴ regression equation of Y on :
`y-bary=b_(yx)(x-barx)` `y-bary=square(x-barx)`
`x-barx=b_(xy)(y-bary)` `x-barx=square(y-bary)`
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`
