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The following table shows the production of gasoline in U.S.A. for the years 1962 to 1976.i. Obtain trend values for the above data using 5-yearly moving averages. ii. Plot the original time series - Mathematics and Statistics

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Question

The following table shows the production of gasoline in U.S.A. for the years 1962 to 1976.

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

i. Obtain trend values for the above data using 5-yearly moving averages.
ii. Plot the original time series and trend values obtained above on the same graph.

Sum
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Solution

i.

Year
t
Production (millions of barrels) yt 5–yearly
moving total
5–yearly moving averages trend value
1962 0
1963 0
1964 1 4 0.8
1965 1 7 1.4
1966 2 11 2.2
1967 3 15 3
1968 4 20 4
1969 5 25 5
1970 6 30 6
1971 7 35 7
1972 8 38 7.6
1973 9 41 8.2
1974 8 44 8.8
1975 9
1976 10

ii.
Taking year on X-axis and production trend on Y-axis, we plot the points for production corresponding to years to get the graph of time series and plot the points for trend values corresponding to years to get the graph of trend as shown in the adjoining figure. Production: 5 yearly moving Average: ---------------

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Measurement of Secular Trend
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Chapter 4: Time Series - Exercise 4.1 [Page 67]

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The following table shows the production of gasoline in U.S.A. for the years 1962 to 1976.

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 7 8 9 8 9 10  
  1. Obtain trend values for the above data using 5-yearly moving averages.
  2. Plot the original time series and trend values obtained above on the same graph.

Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010

Year 1980 1985 1990 1995
IMR 10 7 5 4
Year 2000 2005 2010  
IMR 3 1 0  

Fit a trend line by the method of least squares

Solution: Let us fit equation of trend line for above data.

Let the equation of trend line be y = a + bx   .....(i)

Here n = 7(odd), middle year is `square` and h = 5

Year IMR (y) x x2 x.y
1980 10 – 3 9 – 30
1985 7 – 2 4 – 14
1990 5 – 1 1 – 5
1995 4 0 0 0
2000 3 1 1 3
2005 1 2 4 2
2010 0 3 9 0
Total 30 0 28 – 44

The normal equations are

Σy = na + bΣx

As, Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx2

As, Σx = 0, b =`square`

∴ The equation of trend line is y = `square`


Complete the following activity to fit a trend line to the following data by the method of least squares.

Year 1975 1976 1977 1978 1979 1980 1981 1982 1983
Number of deaths 0 6 3 8 2 9 4 5 10

Solution:

Here n = 9. We transform year t to u by taking u = t - 1979. We construct the following table for calculation :

Year t Number of deaths xt u = t - 1979 u2 uxt
1975 0 - 4 16 0
1976 6 - 3 9 - 18
1977 3 - 2 4 - 6
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1979 2 0 0 0
1980 9 1 1 9
1981 4 2 4 8
1982 5 3 9 15
1983 10 4 16 40
  `sumx_t` =47 `sumu`=0 `sumu^2=60` `square`

The equation of trend line is xt= a' + b'u.

The normal equations are,

`sumx_t = na^' + b^' sumu`              ...(1)

`sumux_t = a^'sumu + b^'sumu^2`      ...(2)

Here, n = 9, `sumx_t = 47, sumu= 0, sumu^2 = 60`

By putting these values in normal equations, we get

47 = 9a' + b' (0)       ...(3)

40 = a'(0) + b'(60)      ...(4)

From equation (3), we get a' = `square`

From equation (4), we get b' = `square`

∴ the equation of trend line is xt = `square`


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