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प्रश्न
Solve the following problem :
Following data shows the number of boxes of cereal sold in years 1977 to 1984.
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
| No. of boxes in ten thousand | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
Fit a trend line to the above data by graphical method.
उत्तर
Taking year on X-axis and number of boxes on Y-axis, we plot the points for number of boxes corresponding to years. Joining these points, we get the graph of the time series. We fit the trend line as shown in the following graph.
APPEARS IN
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.
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.
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 | 1970 | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production (Million Barrels) |
0 | 0 | 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 8 | 9 | 10 |
i. Obtain trend values for the above data using 5-yearly moving averages.
ii. Plot the original time series and trend values obtained above on the same graph.
Choose the correct alternative :
We can use regression line for past data to forecast future data. We then use the line which_______.
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?
State whether the following is True or False :
Moving average method of finding trend is very complicated and involves several calculations.
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 :
Obtain trend values for data in Problem 16 using 3-yearly moving averages.
Obtain trend values for data in Problem 19 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.
State whether the following statement is True or False:
Moving average method of finding trend is very complicated and involves several calculations
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 |
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 complicated but efficient method of measuring trend of time series is ______.
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`
