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Question
State whether the following statement is True or False:
Moving average method of finding trend is very complicated and involves several calculations
Options
True
False
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Solution
False
RELATED QUESTIONS
Choose the correct alternative :
Which of the following is a major problem for forecasting, especially when using the method of least squares?
Choose the correct alternative :
What is a disadvantage of the graphical method of determining a trend line?
The simplest method of measuring trend of time series is ______.
Fill in the blank :
The method of measuring trend of time series using only averages is _______
Solve the following problem :
Fit a trend line to data in Problem 4 by the method of least squares.
Solve the following problem :
Fit a trend line to the data in Problem 7 by the method of least squares.
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 :
Obtain trend values for data in Problem 16 using 3-yearly moving averages.
Solve the following problem :
Following tables shows the wheat yield (‘000 tonnes) in India for years 1959 to 1968.
| Year | 1959 | 1960 | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 |
| Yield | 0 | 1 | 2 | 3 | 1 | 0 | 4 | 1 | 2 | 10 |
Fit a trend line to the above data by the method of least squares.
The complicated but efficient method of measuring trend of time series is ______
State whether the following statement is True or False:
The secular trend component of time series represents irregular variations
Following table shows the amount of sugar production (in lac tons) for the years 1971 to 1982
| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 4 | 6 | 5 | 1 | 4 | 10 |
Fit a trend line by the method of least squares
Obtain trend values for data, using 4-yearly centred moving averages
| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 4 | 6 | 5 | 1 | 4 | 10 |
Obtain the trend values for the data, using 3-yearly moving averages
| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 |
| Production | 0 | 4 | 4 | 2 | 6 | 8 |
| Year | 1982 | 1983 | 1984 | 1985 | 1986 | |
| Production | 5 | 9 | 4 | 10 | 10 |
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 | – | – |
Obtain the trend values for the following data using 5 yearly moving averages:
| Year | 2000 | 2001 | 2002 | 2003 | 2004 |
| Production xi |
10 | 15 | 20 | 25 | 30 |
| Year | 2005 | 2006 | 2007 | 2008 | 2009 |
| Production xi |
35 | 40 | 45 | 50 | 55 |
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:
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.
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.
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`
