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प्रश्न
Obtain trend values for data in Problem 19 using 3-yearly moving averages.
उत्तर
Construct the following table for finding 3-yearly moving averages:
| Year t |
Yield (in '000 tonnes) yt |
3–yearly moving total | 3–yearly moving averages trend value |
| 1959 | 0 | – | – |
| 1960 | 1 | 3 | 1 |
| 1961 | 2 | 6 | 2 |
| 1962 | 3 | 6 | 2 |
| 1963 | 1 | 4 | 1.3333 |
| 1964 | 0 | 5 | 1.6667 |
| 1965 | 4 | 5 | 1.6667 |
| 1966 | 1 | 7 | 2.3333 |
| 1967 | 2 | 13 | 4.3333 |
| 1968 | 10 | – | – |
Notes
Answers given in the textbook for trend values are 1.4, 1.4, 2, 1.8, 1.6, 3.4. However, as per our calculation they are 1, 2, 2, 1.3333, 1.6667, 1.6667, 2.3333, 4.3333.
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0 | 0 | 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 8 | 9 | 10 |
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| 1980 | 10 | – 3 | 9 | – 30 |
| 1985 | 7 | – 2 | 4 | – 14 |
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The normal equations are
Σy = na + bΣx
As, Σx = 0, a = `square`
Also, Σxy = aΣx + bΣx2
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∴ The equation of trend line is y = `square`
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| Year | IMR | 3 yearly moving total |
3-yearly moving average (trend value) |
| 1980 | 10 | – | – |
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| 1990 | 5 | 16 | `square` |
| 1995 | 4 | 12 | 4 |
| 2000 | 3 | 8 | `square` |
| 2005 | 1 | `square` | 1.33 |
| 2010 | 0 | – | – |
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| Year | Production | 4 yearly moving total |
4 yearly centered total |
4 yearly centered moving average (trend values) |
| 2006 | 19 | – | – | |
| `square` | ||||
| 2007 | 20 | – | `square` | |
| 72 | ||||
| 2008 | 17 | 142 | 17.75 | |
| 70 | ||||
| 2009 | 16 | `square` | 17 | |
| `square` | ||||
| 2010 | 17 | 133 | `square` | |
| 67 | ||||
| 2011 | 16 | `square` | `square` | |
| `square` | ||||
| 2012 | 18 | 140 | 17.5 | |
| 72 | ||||
| 2013 | 17 | 147 | 18.375 | |
| 75 | ||||
| 2014 | 21 | – | – | |
| – | ||||
| 2015 | 19 | – | – |
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| 1931 | 1 | 1937 | 8 |
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| 1934 | 2 | 1940 | 1 |
| 1935 | 3 | 1941 | 4 |
| 1936 | 2 |
Complete the following activity to fit a trend line by method of least squares:
