Advertisements
Advertisements
प्रश्न
Solve the following problem :
Obtain trend values for data in Problem 13 using 4-yearly moving averages.
उत्तर
Construct the following table for finding 4-yearly centred moving averages.
Year t |
No. of deaths yt |
4–yearly moving average | 4–yearly moving Averages | 2 unit moving total | 4–yearly centred moving averages trend value |
1975 | 0 | ||||
1976 | 6 | ||||
17 | 4.25 | ||||
1977 | 3 | 9 | 4.5 | ||
19 | 4.75 | ||||
1978 | 8 | 10.25 | 5.125 | ||
22 | 5.5 | ||||
1979 | 2 | 11.25 | 5.625 | ||
23 | 5.75 | ||||
19880 | 9 | 1075 | 5.375 | ||
20 | 5 | ||||
1981 | 4 | 12 | 6 | ||
28 | 7 | ||||
1982 | 5 | ||||
1983 | 10 |
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.
Choose the correct alternative :
Which of the following is a major problem for forecasting, especially when using the method of least squares?
The simplest method of measuring trend of time series is ______.
Solve the following problem :
Obtain trend values for the following data using 5-yearly moving averages.
Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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.
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 :
Fit a trend line to data in Problem 13 by the method of least squares.
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.
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.
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 |
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
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 |
Fit equation of trend line for the data given below.
Year | Production (y) | x | x2 | xy |
2006 | 19 | – 9 | 81 | – 171 |
2007 | 20 | – 7 | 49 | – 140 |
2008 | 14 | – 5 | 25 | – 70 |
2009 | 16 | – 3 | 9 | – 48 |
2010 | 17 | – 1 | 1 | – 17 |
2011 | 16 | 1 | 1 | 16 |
2012 | 18 | 3 | 9 | 54 |
2013 | 17 | 5 | 25 | 85 |
2014 | 21 | 7 | 49 | 147 |
2015 | 19 | 9 | 81 | 171 |
Total | 177 | 0 | 330 | 27 |
Let the equation of trend line be y = a + bx .....(i)
Here n = `square` (even), two middle years are `square` and 2011, and h = `square`
The normal equations are Σy = na + bΣx
As Σx = 0, a = `square`
Also, Σxy = aΣx + bΣx2
As Σx = 0, b = `square`
Substitute values of a and b in equation (i) the equation of trend line is `square`
To find trend value for the year 2016, put x = `square` in the above equation.
y = `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`