Question #349509

In a statistical study relating to the prices (in T) of two shares, X and Y, the following two regression lines were found as 8X - 10Y + 70 = 0 and 20X - 9Y - 65 = 0. The standard deviation of X = 3, then find (i) the values of X and Y, (ii) r(X, Y), and (iii) standard deviation of Y.


1
Expert's answer
2022-06-14T12:07:48-0400

(i) The given equation of the lines of regression are


8X10Y+70=08X - 10Y + 70 = 020X9Y65=020X - 9Y - 65 = 0


40X50Y+350=040X - 50Y + 350 = 040X18Y130=040X - 18Y - 130 = 0

32Y+480=0-32Y+480=0

Y=15Y=15

8X10(15)+70=08X - 10(15) + 70 = 0

X=10X=10

X=10,Y=15X=10, Y=15


(ii) Let the equation 8X10Y+70=08X - 10Y + 70 = 0 be the regression equation of YY on X.X. Then


Y=0.8X+7Y=0.8X+7

Comparing it with Y=bYXX+a,Y=b_{YX}X+a, we get bYX=0.8.b_{YX}=0.8.

Let the equation 20X9Y65=020X - 9Y - 65 = 0 be the regression equation of XX on Y.Y. Then


X=0.45X+3.25X=0.45X+3.25

Comparing it with X=bXYY+a,X=b_{XY}Y+a', we get bXY=0.45.b_{XY}=0.45.


r=±bXYbYXr=\pm\sqrt{b_{XY}\cdot b_{YX}}

Since bXY,bYXb_{XY},b_{YX} both are positive, then rr is also positive:


r=0.80.45=0.6r=\sqrt{0.8\cdot 0.45}=0.6

(iii)


bYX=rσYσXb_{YX}=r\dfrac{\sigma_Y}{\sigma_X}

Then


σY=bYXσXr\sigma_Y=b_{YX}\dfrac{\sigma_X}{r}

σY=0.8(30.6)=4\sigma_Y=0.8(\dfrac{3}{0.6})=4


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