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in Statistics by (57.2k points)

Find a parabolic trend of the form Y = ax + bx + cx2 for the following time series.

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Let x and y be the year and students strength.

Here X̄ = middle most year = 2012

From the normal equation: na + b Σx + c Σx2 = Ey

5a + b.0 + c.10 = 53; 5a + 10c = 53 ………. (1)

From: a Σx + b Σx2 + c Σx3 – Σxy; a.0 + b.10 + c.0 = 0

10 b = 0 

∴ b = 0.

And from: a Σx2 + b Σx3 + c Σx4 = Σx2y; a. 10 + b.0 + c.34 = 120

i. e., 10a + 34c = 120 …….(2)

From (1) and (2)

∴ c = 1. put c = 1 in (1) we get 5a +10 (1) = 53, 5a = 53 – 10

a = \(\frac{43}{5}\) = 8.6.

The fitted parabolic trend equation is: ŷ = a + bx + cx2

ŷ = 8.6 + 0x + 1x2 ; ŷ = 8.6 + x2.

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