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Question
Solve the following problem :
Fit a trend line to the data in Problem 7 by the method of least squares.
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Solution
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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| `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`
