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Question
Solve the following problem :
The following table shows the production of pig-iron and ferro- alloys (‘000 metric tonnes)
Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
Fit a trend line to the above data by graphical method.
Solution
Taking year on X-axis and production on Y-axis, we plot the points for production corresponding to years. Joining these points, we get the graph of time series, we fit a trend line as shown in the figure.
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Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
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Year | 1980 | 1985 | 1990 | 1995 |
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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`
Obtain trend values for data, using 3-yearly moving averages
Solution:
Year | IMR | 3 yearly moving total |
3-yearly moving average (trend value) |
1980 | 10 | – | – |
1985 | 7 | `square` | 7.33 |
1990 | 5 | 16 | `square` |
1995 | 4 | 12 | 4 |
2000 | 3 | 8 | `square` |
2005 | 1 | `square` | 1.33 |
2010 | 0 | – | – |