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प्रश्न
Solve the following problem :
Fit a trend line to data in Problem 16 by the method of least squares.
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उत्तर
In the given problem, n = 7 (odd), middle t – value is 1995, h = 5
u = `"t - middle value"/"h" = ("t" - 1995)/(1)`
We obtain the following table.
| Year t |
Infant mortality rate yt |
u = `("t" - 1995)/(5)` | u2 | uyt | Trend Value |
| 1980 | 10 | –3 | 9 | –30 | 8.9999 |
| 1985 | 7 | –2 | 4 | –14 | 7.4285 |
| 1990 | 5 | –1 | 1 | –5 | 5.8571 |
| 1995 | 4 | 0 | 0 | 0 | 4.2857 |
| 2000 | 3 | 1 | 1 | 3 | 2.7143 |
| 2005 | 1 | 2 | 4 | 2 | 1.1429 |
| 2010 | 0 | 3 | 9 | 0 | –0.4285 |
| Total | 30 | 0 | 28 | –44 |
From the table, n = 7, `sumy_"t" = 10, sumu = 0, sumu^2 = 28,sumuy_"t" = – 44`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 30 = 7a' + b'(0) ...(i) and
– 44 = a'(0) + b'(28) ...(ii)
From (i), a' = `(30)/(7)` = 4.2857
From (ii), b' = `(-44)/(28)` = 1.5714
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.2857 – 1.5714 u, where u = `("t" - 1995)/(5)`.
APPEARS IN
संबंधित प्रश्न
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 :
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 complicated but efficient method of measuring trend of time series is _______.
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 :
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 |
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 :
The percentage of girls’ enrollment in total enrollment for years 1960-2005 is shown in the following table.
| Year | 1960 | 1965 | 1970 | 1975 | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 |
| Percentage | 0 | 3 | 3 | 4 | 4 | 5 | 6 | 8 | 8 | 10 |
Fit a trend line to the above data by graphical method.
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.
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 16 using 3-yearly moving averages.
Obtain trend values for data in Problem 19 using 3-yearly moving averages.
The simplest 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
State whether the following statement is True or False:
Least squares method of finding trend is very simple and does not involve any calculations
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
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 |
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 | – | – |
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`
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:
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`
