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प्रश्न
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.
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 graph.
APPEARS IN
संबंधित प्रश्न
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 |
Obtain the trend values for the above data using 3-yearly moving averages.
Choose the correct alternative :
We can use regression line for past data to forecast future data. We then use the line which_______.
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.
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 :
Obtain trend values for data in Problem 13 using 4-yearly moving averages.
Solve the following problem :
Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010.
Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |
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.
State whether the following statement is True or False:
The secular trend component of time series represents irregular variations
State whether the following statement is True or False:
Least squares method of finding trend is very simple and does not involve any 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 |
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
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.
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 |
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`