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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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 leas - Mathematics and Statistics

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प्रश्न

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`

तक्ता

उत्तर

Let the equation of trend line be y = a + bx   .....(i)

Here n = 7(odd), middle year is 1995 and h = 5

x = `("t" - "middle year")/"h"`

= `("t" - 1995)/5`

We obtain the following table:

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

From the table, n = 7, Σyt = 30, Σx = 0, Σx2 = 28, Σxyt = – 44

The normal equations are

Σy = na + bΣx

∴ 30 = 7a + bΣx

As, Σx = 0, a = `30/7` = 4.2857

Also, Σxy = aΣx + bΣx

∴ – 44 = aΣx + b′(28)

As, Σx = 0, b =`(-44)/28` =  – 1.5714

∴ The equation of trend line is y = a + bx

∴ The equation of trend line is y = 4.2857 – 1.5714x

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Measurement of Secular Trend
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2.4: Time Series - Q.5

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

Fill in the blank :

The method of measuring trend of time series using only averages is _______


Fill in the blank :

The complicated but efficient 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 :

Fit a trend line to the data in Problem 7 by the method of least squares.


Solve the following problem :

Obtain trend values for the data in Problem 7 using 4-yearly moving averages.


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 :

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

Fit a trend line to the above data by graphical method.


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.


Choose the correct alternative:

Moving averages are useful in identifying ______.


State whether the following statement is True or False:

The secular trend component of time series represents irregular variations


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


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

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`


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