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Question
Solve the following problem :
The percentage of girls’ enrollment in total enrollment for years 1960-2005 is shown in the following table.
Year | 1960 | 1965 | 1970 | 1975 | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 |
Percentage | 0 | 3 | 3 | 4 | 4 | 5 | 6 | 8 | 8 | 10 |
Fit a trend line to the above data by graphical method.
Solution
Taking year on X-axis and percentage of enrolment on Y-axis, we plot the points for enrolment corresponding to years. Joining these points, we get the graph of time series. We fit the trend line as shown in the following graph.
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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`