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Question
The complicated but efficient method of measuring trend of time series is ______.
Options
graphical method
method of moving average
method of least squares
method of addition
Solution
The complicated but efficient method of measuring trend of time series is method of least squares.
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RELATED QUESTIONS
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 |
Fit a trend line to the data in Problem 7 by the method of least squares. Also, obtain the trend value for the year 1990.
Choose the correct alternative :
We can use regression line for past data to forecast future data. We then use the line which_______.
Fill in the blank :
The method of measuring trend of time series using only averages is _______
State whether the following is True or False :
Graphical 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.
State whether the following is True or False :
All the three methods of measuring trend will always give the same results.
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.
Solve the following problem :
Following table shows the amount of sugar production (in lac tonnes) for the years 1971 to 1982.
| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 | 3 | 6 | 5 | 1 | 4 | 10 |
Fit a trend line to the above data by graphical method.
Solve the following problem :
Fit a trend line to data in Problem 4 by the method of least squares.
Obtain trend values for the following data using 4-yearly centered moving averages.
| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 3 | 6 | 5 | 1 | 4 | 10 |
Solve the following problem :
Obtain trend values for the data in Problem 7 using 4-yearly moving averages.
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 |
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.
Solve the following problem :
Obtain trend values for data in Problem 16 using 3-yearly moving averages.
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
Use the method of least squares to fit a trend line to the data given below. Also, obtain the trend value for the year 1975.
| 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 | 8 | 9 | 9 | 8 | 7 | 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.
Obtain trend values for data, using 3-yearly moving averages
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 | – | – |
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`
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`
