Question



2. For the data plotted below, draw (visually) a best-fit line. Then write down an equation for the best-fit line you have drawn. Finally, extrapolate (i.e. pred value when the independent variable is at 3.5. It is not necessary to calculate the least-squares best-fit line! lict) the dependent variable 3.5 2.5 1.5 0.5 0 0.5 1 1.522.5 3 3.5 3.5 T 2.5 1.5 0.5 1.5 2.5
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Answer #1

I have taken printout of the image and have drawn a best fit line.

0.5 0.5 12.5 25 2.5 1.5 05 1,5 2.5 3.5

The equation of the first line which passes at (2.5, 3) and (2, 2.5)

y - 3 = (3 - 2.5) / (2.5 - 2) (x - 2.5)

=> y - 3 = (x - 2.5)

=> y = x + 0.5

The equation of the second line which passes at (0.5, 3.5) and (3, 0.75)

y - 3.5 = (3.5 - 0.75) / (0.5 - 3) (x - 0.5)

=> y - 3.5 = -1.1 (x - 0.5)

=> y - 3.5 = -1.1x + 0.55

=> y = -1.1x + 4.05

For x = 3.5

Estimate of dependent variable in the first plot is,

y = 3.5 + 0.5 = 4

Estimate of dependent variable in the second plot is,

y = -1.1 * 3.5 + 4.05 = 0.2

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