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Question
Obtain trend values for data in Problem 19 using 3-yearly moving averages.
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Solution
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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| 1975 | 0 | - 4 | 16 | 0 |
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| `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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| Number of accidents | 39 | 18 | 21 | 28 | 27 | 27 | 23 | 25 | 22 |
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We take origin to 18, we get, the number of accidents as follows:
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| 2008 | 21 | 1 | -4 | 16 | -84 |
| 2009 | 0 | 2 | -3 | 9 | 0 |
| 2010 | 3 | 3 | -2 | 4 | -6 |
| 2011 | 10 | 4 | -1 | 1 | -10 |
| 2012 | 9 | 5 | 0 | 0 | 0 |
| 2013 | 9 | 6 | 1 | 1 | 9 |
| 2014 | 5 | 7 | 2 | 4 | 10 |
| 2015 | 7 | 8 | 3 | 9 | 21 |
| 2016 | 4 | 9 | 4 | 16 | 16 |
| `sumx_t=68` | - | `sumu=0` | `sumu^2=60` | `square` |
The equation of trend 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=68,sumu=0,sumu^2=60,sumux_t=-44`
Putting these values in normal equations, we get
68 = 9a' + b'(0) ...(3)
∴ a' = `square`
-44 = a'(0) + b'(60) ...(4)
∴ b' = `square`
The equation of trend line is given by
xt = `square`
