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A Look at Some Sample Solutions From Our Writers
Are we still having doubts? Well, then, here’s a simple linear regression solution from a higher/applied mathematics assignment crafted by our mathematics coursework experts.
Problem →
Consider the following observation pairs (xi, yi)
(-1,6), (0,3), (1,2), (2,-1)
- Find the approximate regression line.
y’ = ß’0 + ß’1 xi
based on the observed data.
- For each xi, compute the fitted value of yi using
yi’ = ß’0 + ß’1 xi
- Find out the residuals
ei = yi- y’i
- Find R-squared (the coefficient of determination)
Solution
- We have
x’ = (-1+0+1+12)/4= 0.5
y’= (6+3+2-1)/4= 2.5
sxx= (-1-0.5)2 + (0-0.5)2 + (1-0.5)2 + (2-0.5)2 = 5 → this is the sample corrected sum of squares
sxy= (-1 - 0.5) (6 - 2.5) + (0 - 0.5)(3 - 2.5) + (1- 0.5)(2 - 2.5) + (2 - 0.5)(-1 - 0.25) = -11 →this is the sum of the products of differences between x values and differences between y values
So, we have
ß’1 = sxy / sxx = -11/5= -2.2
ß’0 = 2.5 – (-2.2)(0.5)= 3.6
- The fitted values are given by
yi’ = 3.6 -2.2 xi
So, we have
y’1 = 5.8, y’2= 3.6, y’3= 1.4, y’4= -0.8
- The residuals obtained are as follows →
e1 = y1 - y1’ = 6 - 5.8 = 0.2
e2 = y2 - y2’ = 3 - 3.6 = -0.6
e3 = y3 - y3’ =2 - 1.4 = 0.6
e4 = y4 - y4’ = -1 - (-0.8) = -0.2
- Finally, we find R-squared or the coefficient of determination as follows →
syy = (6 - 2.5)2 + (3 - 2.5)2 + (2 - 2.5)2 +(-1 - 2.5)2 = 25
R2 = (-11)2/ (5*25) ~~ 0.968
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