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Question
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
Solution
In the given problem, n = 12 (even), two middle t − values are 1976 and 1977, h = 1
u = `("t" - "mean of two middle values")/("h"/2)`
= `("t" - 1976.5)/(1/2)`
= 2(t − 1976.5)
We obtain the following table:
Year t |
Poduction yt |
u = 2(t − 1976.5 | u2 | uyt | Trend Value |
1971 | 1 | – 11 | 121 | – 11 | 0.1535 |
1972 | 0 | – 9 | 81 | – 0 | 0.7165 |
1973 | 1 | – 7 | 49 | – 7 | 1.2795 |
1974 | 2 | – 5 | 25 | – 10 | 1.8425 |
1975 | 3 | – 3 | 9 | – 9 | 2.4055 |
1976 | 2 | – 1 | 1 | – 2 | 2.9685 |
1977 | 4 | 1 | 1 | 4 | 3.5315 |
1978 | 6 | 3 | 9 | 18 | 4.0945 |
1979 | 5 | 5 | 25 | 25 | 4.6575 |
1980 | 1 | 7 | 49 | 7 | 5.2205 |
1981 | 4 | 9 | 81 | 36 | 5.7835 |
1982 | 10 | 11 | 121 | 110 | 6.3465 |
Total | 39 | 0 | 572 | 161 |
From the table, n = 12, ∑yt = 39, ∑u = 0, ∑u2 = 572, ∑uyt = 161
The two normal equations are:
∑yt = na' + b'∑u and ∑uyt = a'∑u + b'∑u2
∴ 39 = 12a' + b'(0) ......(i)
and 161 = a'(0) + b'(572) ......(ii)
From (i), a′ = `39/12` = 3.25
From (ii), b′ = `161/572` = 0.2815
∴ The equation of the trend line is yt = a′ + b′u
i.e., yt = 3.25+ 0.2815 u,
where u = 2(t − 1976.5)
RELATED QUESTIONS
Obtain the trend line for the above data using 5 yearly moving averages.
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.
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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?
Fill in the blank :
The complicated but efficient method of measuring trend of time series is _______.
State whether the following is True or False :
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State whether the following is True or False :
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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 :
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.
Solve the following problem :
Fit a trend line to the data in Problem 7 by the method of least squares.
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.
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 :
Following table shows the number of traffic fatalities (in a state) resulting from drunken driving for years 1975 to 1983.
Year | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |
No. of deaths | 0 | 6 | 3 | 8 | 2 | 9 | 4 | 5 | 10 |
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Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |
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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.
Choose the correct alternative:
Moving averages are useful in identifying ______.
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 |
Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010
Year | 1980 | 1985 | 1990 | 1995 |
IMR | 10 | 7 | 5 | 4 |
Year | 2000 | 2005 | 2010 | |
IMR | 3 | 1 | 0 |
Fit a trend line by the method of least squares
Solution: Let us fit equation of trend line for above data.
Let the equation of trend line be y = a + bx .....(i)
Here n = 7(odd), middle year is `square` and h = 5
Year | IMR (y) | x | x2 | x.y |
1980 | 10 | – 3 | 9 | – 30 |
1985 | 7 | – 2 | 4 | – 14 |
1990 | 5 | – 1 | 1 | – 5 |
1995 | 4 | 0 | 0 | 0 |
2000 | 3 | 1 | 1 | 3 |
2005 | 1 | 2 | 4 | 2 |
2010 | 0 | 3 | 9 | 0 |
Total | 30 | 0 | 28 | – 44 |
The normal equations are
Σy = na + bΣx
As, Σx = 0, a = `square`
Also, Σxy = aΣx + bΣx2
As, Σx = 0, b =`square`
∴ The equation of trend line is 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 | – | – |
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.
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`