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प्रश्न
Solve the following problem :
The following table shows the production of pig-iron and ferro- alloys (‘000 metric tonnes)
| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
Fit a trend line to the above data by graphical method.
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उत्तर
Taking year on X-axis and production on Y-axis, we plot the points for production corresponding to years. Joining these points, we get the graph of time series, we fit a trend line as shown in the figure.
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संबंधित प्रश्न
Fit a trend line to the data in Problem 4 above by the method of least squares. Also, obtain the trend value for the index of industrial production for the year 1987.
Obtain the trend values for the data in using 4-yearly centered moving averages.
| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 |
| Index | 0 | 2 | 3 | 3 | 2 | 4 | 5 | 6 | 7 | 10 |
Choose the correct alternative :
Which of the following is a major problem for forecasting, especially when using the method of least squares?
State whether the following is True or False :
Moving average method of finding trend is very complicated and involves several calculations.
State whether the following is True or False :
Least squares method of finding trend is very simple and does not involve any calculations.
Fit a trend line to the following data by the method of least squares.
| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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.
Solve the following problem :
Obtain trend values for data in Problem 10 using 3-yearly moving averages.
Solve the following problem :
Following table shows the number of traffic fatalities (in a state) resulting from drunken driving for years 1975 to 1983.
| Year | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |
| No. of deaths | 0 | 6 | 3 | 8 | 2 | 9 | 4 | 5 | 10 |
Fit a trend line to the above data by graphical method.
Solve the following problem :
Fit a trend line to data in Problem 16 by the method of least squares.
Obtain trend values for data in Problem 19 using 3-yearly moving averages.
The simplest method of measuring trend of time series is ______
State whether the following statement is True or False:
Least squares method of finding trend is very simple and does not involve any calculations
The following table gives the production of steel (in millions of tons) for years 1976 to 1986.
| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 |
| Production | 0 | 4 | 4 | 2 | 6 | 8 | 5 | 9 | 4 | 10 | 10 |
Obtain the trend value for the year 1990
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.
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 | – | – |
Following table shows the amount of sugar production (in lakh tonnes) for the years 1931 to 1941:
| 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 |
Complete the following activity to fit a trend line by method of least squares:
Fit a trend line to the following data by the method of least square :
| Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
| IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |
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`
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`
