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Question
Solve the following problem :
Fit a trend line to data by the method of least squares.
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
| Number of boxes (in ten thousands) | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
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Solution
In the given problem, n = 8 (even), two middle t – values are 1980 and 1981, h – 1
u = `"t - mean of two middle values"/("h"/2) = ("t" - 1980.5)/(1/2)` = 2(t – 1980.5)
We obtain the following table.
| Year t |
No. of boxes (in ten thousands) yt |
u = 2(t – 1980.5) | u2 | uyt | Trend Value |
| 1977 | 1 | –7 | 49 | –7 | 1.5836 |
| 1978 | 0 | –5 | 25 | 0 | 2.5240 |
| 1979 | 3 | –3 | 9 | –9 | 3.4644 |
| 1980 | 8 | –1 | 1 | –8 | 4.4048 |
| 1981 | 10 | 1 | 1 | 10 | 5.3452 |
| 1982 | 4 | 3 | 9 | 12 | 6.2856 |
| 1983 | 5 | 5 | 25 | 25 | 7.2260 |
| 1984 | 8 | 7 | 49 | 56 | 8.1664 |
| Total | 39 | 0 | 168 | 79 |
From the table, n = 8, `sumy_"t" = 39, sumu = 0, sumu^2 = 168,sumuy_"t" = 79`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 39 = 8a' + b'(0) ...(i) and
79 = a'(0) + b'(168) ...(ii)
From (i), a' = `(39)/(8)` = 4.875
From (ii), b' = `(79)/(168)` = 0.4702
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.875 + 0.4702 u, where u = 2(t – 1980.5).
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RELATED QUESTIONS
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| 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 |
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| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
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| Production | 4 | 6 | 5 | 1 | 4 | 10 |
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| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
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| Production | 4 | 6 | 5 | 1 | 4 | 10 |
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 |
- Obtain trend values for the above data using 5-yearly moving averages.
- Plot the original time series and trend values obtained above on the same graph.
Fit equation of trend line for the data given below.
| Year | Production (y) | x | x2 | xy |
| 2006 | 19 | – 9 | 81 | – 171 |
| 2007 | 20 | – 7 | 49 | – 140 |
| 2008 | 14 | – 5 | 25 | – 70 |
| 2009 | 16 | – 3 | 9 | – 48 |
| 2010 | 17 | – 1 | 1 | – 17 |
| 2011 | 16 | 1 | 1 | 16 |
| 2012 | 18 | 3 | 9 | 54 |
| 2013 | 17 | 5 | 25 | 85 |
| 2014 | 21 | 7 | 49 | 147 |
| 2015 | 19 | 9 | 81 | 171 |
| Total | 177 | 0 | 330 | 27 |
Let the equation of trend line be y = a + bx .....(i)
Here n = `square` (even), two middle years are `square` and 2011, and h = `square`
The normal equations are Σy = na + bΣx
As Σx = 0, a = `square`
Also, Σxy = aΣx + bΣx2
As Σx = 0, b = `square`
Substitute values of a and b in equation (i) the equation of trend line is `square`
To find trend value for the year 2016, put x = `square` in the above equation.
y = `square`
Complete the table using 4 yearly moving average method.
| 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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The following table shows gross capital information (in Crore ₹) for years 1966 to 1975:
| Years | 1966 | 1967 | 1968 | 1969 | 1970 |
| Gross Capital information | 20 | 25 | 25 | 30 | 35 |
| Years | 1971 | 1972 | 1973 | 1974 | 1975 |
| Gross Capital information | 30 | 45 | 40 | 55 | 65 |
Obtain trend values using 5-yearly moving values.
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 |
| 1978 | 8 | - 1 | 1 | - 8 |
| 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`
