Question #279920

Section: 1 of 1 Question: 3 of 7 Marks forthls Question: 10 a



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An analysis of the data of the partially destroyed dairy are



variance x=8 and the least square regression lines for the



random variables are 3x 4y = 20, 12x — 8y = 62. Find



the following basis of above infonnation:



(i) Mean value of x and y.



(ii) Correlation co-efficient between x and y.



(iii) SD of x and y.



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Expert's answer
2021-12-15T17:31:27-0500

i)i)

Since two regression lines always intersect at a point (xˉ,yˉ)(\bar{x},\bar{y}), representing mean values of the values x and y as shown below:

3x+4y=203x+4y=20

12x8y=6212x-8y=62

Multiplying the first equation by 4 and subtracting from the second,

24y=18    yˉ=0.7524y=18\implies\bar{y}=0.75

From the first equation

xˉ=(204yˉ)3=173=5.67\bar{x}={(20-4\bar{y})\over3}={17\over3}=5.67

Therefore, the mean value of x and y are 5.67 and 0.75 respectively.


ii)ii)

To find the given regression equations in such a way that the coefficient of dependent variable is less than one at least in one equation. 

So, 3x+4y=20    4y=203x    y=534x3x+4y=20\implies4y=20-3x\implies y=5-{3\over4}x

That is, myx=34=34|m_{yx}|=|-{3\over4}|={3\over4}

and

12x8y=62    12x=62+8y    x=6212+812y12x-8y=62\implies12x=62+8y\implies x={62\over12}+{8\over12}y

Thus, mxy=812m_{xy}={8\over 12}

Hence coefficient of correlation rr between xx and yy is given by: 

r=mxymyx=34812=12=0.7071(4dp)r=\sqrt{m_{xy}*m_{yx}}=\sqrt{{3\over4}*{8\over12}}=\sqrt{1\over2}=0.7071(4dp)  


iii)iii)

 Since the variance of x is 8, therefore, its standard deviation is,

σx=8=2.8284(4dp)\sigma_x=\sqrt{8}=2.8284(4dp)

To determine the standard deviation of y, consider the formula

σy=myxσxr=(0.752.82840.7071)=3\sigma_y={m_{yx}\sigma_x\over r}=({0.75*2.8284\over0.7071})=3

Therefore, the standard deviation for x and y are 2.8284 and 3 respectively.


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