Simple and Multiple Linear Regression Analysis Sample Assignment
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Questions: Give σx = 3 and Regression equation 8X – 10Y + 66 = 0; 40X – 18Y = 214.
Find Out (i) The mean value of X and Y, (ii) Coefficient of correlation between X and Y and (iii) Standard deviation of Y.
Sol: (i) We know, the two regression lines intersect each other at the mean values of X and Y. Therefore, these values can be obtained by solving these two regression equations 8X – 10Y + 66 = 0; 40 X – 18Y = 214
or 8X – 10Y = – 6
40 X – 18Y = 214
Multiplying equations (i) by 5, we have
40X – 50Y = –330
40 X – 18Y = 214
_-_ + ___ -
-32 Y = -544
or Y = +544 = 17
+32
Y = 17
By substituting the value of Y in equation (i),
8X – 10(17) = – 66
or 8X = – 66 + 170
or 8X = 104
X = 13 or
X = 13
(ii) In order to determine the correlation coefficient, we will have to find out regression coefficients. Here, we do not know which of the two regression equation (i) as regression equation X on Y.
Given equation is
8X – 10Y + 66 = 0
Or 8X – 10Y = – 66
or X = 10Y/8 – 66/8
or X = 10/8 (Y – 6.6)
.: bxy = 10/8 = +1.25.
From equation (ii)
40X – 18Y = 214
or 18Y = 40X – 214
or Y = 40X/18 – 214/18
.: byx = 40/18 + 2.22.
Since the product of these two regression coefficients (i.e., bxy x byx = 1.25 X 2.2) is exceeding one, our assumption is wrong. Therefore, first equation is Y on X and the second X on Y.
.: 8X – 10Y + 66 = 0
or 10Y = 8X +66 = 0
or Y = 8X + 66
10
or Y = 8/10 (X + 52.8)
.: Y = 8/10 + .8
From equation (ii)
40X – 18Y = 214
40X = 214 + 18Y
X = 18Y + 214
40 40
or X = 18/40Y + 5.35
.: bxy = 18/40 = +0.45
.: r = √(bxy x byx)
.: σy = 4
Simple and Multiple Linear Regression Analysis Sample Assignment
Questions: Give σx = 3 and Regression equation 8X – 10Y + 66 = 0; 40X – 18Y = 214.
Find Out (i) The mean value of X and Y, (ii) Coefficient of correlation between X and Y and (iii) Standard deviation of Y.
Sol: (i) We know, the two regression lines intersect each other at the mean values of X and Y. Therefore, these values can be obtained by solving these two regression equations 8X – 10Y + 66 = 0; 40 X – 18Y = 214
or 8X – 10Y = – 6
40 X – 18Y = 214
Multiplying equations (i) by 5, we have
40X – 50Y = –330
40 X – 18Y = 214
_-_ + ___ -
-32 Y = -544
or Y = +544 = 17
+32
Y = 17
By substituting the value of Y in equation (i),
8X – 10(17) = – 66
or 8X = – 66 + 170
or 8X = 104
X = 13 or
X = 13
(ii) In order to determine the correlation coefficient, we will have to find out regression coefficients. Here, we do not know which of the two regression equation (i) as regression equation X on Y.
Given equation is
8X – 10Y + 66 = 0
Or 8X – 10Y = – 66
or X = 10Y/8 – 66/8
or X = 10/8 (Y – 6.6)
.: bxy = 10/8 = +1.25.
From equation (ii)
40X – 18Y = 214
or 18Y = 40X – 214
or Y = 40X/18 – 214/18
.: byx = 40/18 + 2.22.
Since the product of these two regression coefficients (i.e., bxy x byx = 1.25 X 2.2) is exceeding one, our assumption is wrong. Therefore, first equation is Y on X and the second X on Y.
.: 8X – 10Y + 66 = 0
or 10Y = 8X +66 = 0
or Y = 8X + 66
10
or Y = 8/10 (X + 52.8)
.: Y = 8/10 + .8
From equation (ii)
40X – 18Y = 214
40X = 214 + 18Y
X = 18Y + 214
40 40
or X = 18/40Y + 5.35
.: bxy = 18/40 = +0.45
.: r = √(bxy x byx)
.: σy = 4
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