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प्रश्न
Choose the correct alternative :
What is a disadvantage of the graphical method of determining a trend line?
पर्याय
Provides quick approximations
Is subject to human error
Provides accurate forecasts
Is too difficult to calculate
उत्तर
Is subject to human error.
APPEARS IN
संबंधित प्रश्न
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.
Fill in the blank :
The complicated but efficient method of measuring trend of time series is _______.
State whether the following is True or False :
All the three methods of measuring trend will always give the same results.
Fit a trend line to the following data by the method of least squares.
Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 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 10 using 3-yearly moving averages.
Solve the following problem :
Fit a trend line to data in Problem 13 by the method of least squares.
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 :
Fit a trend line to data in Problem 16 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:
Moving average method of finding trend is very complicated and involves several calculations
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`
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 |
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`