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प्रश्न
Solve the following problem :
Fit a trend line to the data in Problem 7 by the method of least squares.
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उत्तर
In the given problem, n = 10 even), middle t – value is 1980 and 1985, h = 5
u = `"t - mean of two middle values"/("h"/2) = ("t" - 1985.5)/(5/2) = (2("t" – 1985.5))/(5)`
We obtain the following table.
| Year t |
Percentage of enrolment yt |
u = `(2("t" – 1982.5))/(5)` | u2 | uyt | Trend Value |
| 1960 | 0 | –9 | 81 | 0 | 0.8187 |
| 1965 | 3 | –7 | 49 | –21 | 1.7701 |
| 1970 | 3 | –5 | 25 | –15 | 2.7215 |
| 1975 | 4 | –3 | 9 | –12 | 3.6729 |
| 1980 | 4 | –1 | 1 | –4 | 4.6243 |
| 1985 | 5 | 1 | 1 | 5 | 5.5757 |
| 1990 | 6 | 3 | 9 | 18 | 6.5271 |
| 1995 | 8 | 5 | 25 | 40 | 7.4785 |
| 2000 | 8 | 7 | 49 | 56 | 8.4299 |
| 2005 | 10 | 9 | 81 | 90 | 9.3813 |
| Total | 51 | 0 | 330 | 157 |
From the table, n = 10, `sumy_"t" = 51, sumu = 0, sumu^2 = 330,sumuy_"t" = 157`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 51 = 10a' + b'(0) ...(i) and
157 = a'(0) + b'(330) ...(ii)
From (i), a' = `(51)/(10)` = 5.1
From (ii), b' = `(157)/(330)` = 0.4757
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 5.1 + 0.4757 u, where u = `(2("t" – 1985.5))/(5)`.
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| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
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| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
| No. of boxes in ten thousand | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
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| Year | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |
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| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 |
| Production | 0 | 4 | 4 | 2 | 6 | 8 | 5 | 9 | 4 | 10 | 10 |
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| 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 |
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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 | – | – |
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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| Year | Production | Year | Production |
| 1931 | 1 | 1937 | 8 |
| 1932 | 0 | 1938 | 6 |
| 1933 | 1 | 1939 | 5 |
| 1934 | 2 | 1940 | 1 |
| 1935 | 3 | 1941 | 4 |
| 1936 | 2 |
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| 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.
The publisher of a magazine wants to determine the rate of increase in the number of subscribers. The following table shows the subscription information for eight consecutive years:
| Years | 1976 | 1977 | 1978 | 1979 |
| No. of subscribers (in millions) |
12 | 11 | 19 | 17 |
| Years | 1980 | 1981 | 1982 | 1983 |
| No. of subscribers (in millions) |
19 | 18 | 20 | 23 |
Fit a trend line by graphical method.
Following table gives the number of road accidents (in thousands) due to overspeeding in Maharashtra for 9 years. Complete the following activity to find the trend by the method of least squares.
| Year | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 |
| Number of accidents | 39 | 18 | 21 | 28 | 27 | 27 | 23 | 25 | 22 |
Solution:
We take origin to 18, we get, the number of accidents as follows:
| Year | Number of accidents xt | t | u = t - 5 | u2 | u.xt |
| 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`
