# Use the method of least squares to fit a trend line to the data in Problem 6 below. Also, obtain the trend value for the year 1975 - Mathematics and Statistics

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Use the method of least squares to fit a trend line to the data given below. Also, obtain the trend value for the year 1975.

 Year 1962 1963 1964 1965 1966 1967 1968 1969 Production (million barrels) 0 0 1 1 2 3 4 5 Year 1970 1971 1972 1973 1974 1975 1976 Production (million barrels) 6 8 9 9 8 7 10

#### Solution

In the given problem, n = 15 (odd), middle t – values is 1969, h = 1

u = ("t" - "middle value")/"h"

= ("t" - 1969)/1

= t – 1969

We obtain the following table:

 Year t Productionyt u = t − 1969 u2 uyt Trend Value 1962 0 − 49 0 − 0.6 1963 0 − 6 36 0 0.2 1964 1 − 5 25 − 5 1 1965 1 − 4 16 − 4 1.8 1966 2 − 3 9 − 6 2.6 1967 3 − 2 4 − 6 3.4 1968 4 − 1 1 − 4 4.2 1969 5 0 0 0 5 1970 6 1 1 6 5.8 1971 8 2 4 16 6.6 1972 9 3 9 27 7.4 1973 9 4 16 36 8. 1974 8 5 25 40 9 1975 9 6 36 54 9.8 1976 10 7 49 70 10.6 Total 75 0 280 224

From the table, n = 15, ∑yt = 75, ∑u = 0, ∑u2 = 280, ∑uyt = 224

The two normal equations are:

∑yt = na' + b'∑u and ∑uyt = a' ∑u + b'∑u2

∴ 75 = 15a' + b'(0)   ......(i)

and

224 = a′(0) + b′(280)  .....(ii)

From (i), a′ = 75/15 = 5

From (ii), b′= 224/280 = 0.8

∴ The equation of the trend line is yt = a′ + b′u

i.e., yt = 5 + 0.8 u, where u = t – 1969

Now, for t = 1975, u = 1975 – 1969 = 6

∴  yt = 5 + 0.8 × 6 = 9.8

Concept: Measurement of Secular Trend
Is there an error in this question or solution?
Chapter 2.4: Time Series - Q.4
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