Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 4 Time Series Ex 4.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1

Question 1.

The following data gives the production of bleaching powder (in ‘000 tonnes) for the years 1962 to 1972.

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

Solution:

Question 2.

Use the method of least squares to fit a trend line to the data in problem 1 above. Also, obtain the trend value for the year 1975.

Solution:

n = 11, let the trend line the

y = a + bu ……..(I)

Σy = na + bΣu ……..(i)

Σuy = aΣu + bΣu^{2} ………(ii)

Substituting the values of Σy, Σu, Σuy, & Σu^{2}, we get

46 = 11a + 0

∴ a = 4.18 And

114 = 0 + b(110)

∴ b = 1.04

By (I) the equation of the trends line is

y = 4.18 + 1.04u

Where u = t – 1967 ……..(iii)

For the year 1975 we have u = 8

Substituting in (iii) we get

Y= 4.18 + 1.04(8) = 12.5

Trend value for the year 1975 is 12.5 (in ‘000 tonnes).

Question 3.

Obtain the trend line for the above data using 5 yearly moving averages.

Solution:

Question 4.

The following table shows the index of industrial production for the period from 1976 to 1985, using the year 1976 as the base year.

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

Solution:

Question 5.

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.

Solution:

u = \(\frac{t-1980.5}{\frac{1}{2}}\), n = 10, Σu = 0, Σy = 42, Σu^{2} = 330, Σuy = 148

Let the trend line be y = a + bu ……(i)

where u = \(\frac{t-1980.5}{\frac{1}{2}}\)

i.e. u = 2t – 3961

Σy = na + bΣu ……(ii)

Σuy = aΣu + bΣu^{2} ……….(iii)

Substituting the values of Σy, n, Σu, Σuy & Σu^{2} We get

42 = 10a + 0

∴ a = 4.2 and

148 = 0 + 5.330

∴ b = 0.4485

∴ by (i) the equation of the trends line is

Y = 4.2 + 0.4485u ………(iv)

where u = 2t – 3961

For the year 1987,

u = 13 by (iv) we have

Y = 4.2 + 0.4485(13) = 10.0305

∴ The trend value for the year 1987 is 10.0305

Question 6.

Obtain the trend values for the data in problem 4 using 4-yearly centered moving averages.

Solution:

Question 7.

The following table gives the production of steel (in millions of tonnes) for the years 1976 to 1986.

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

Solution:

Question 8.

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.

Solution:

u = \(\frac{t-1981}{1}\), n = 10, Σu = 0, ΣY = 62, Σu^{2} = 110, Σuy = 87

Let the equation of the trend line be

Y = a + bu

where u = t – 1981 ……(i)

ΣY = na + bΣu ………(ii)

Σuy = aΣu + bΣu^{2} ………(iii)

Substituting the values of Σy, n, Σu, Σuy, Σu^{2} in (ii) & (iii)

62 = 11a + 0

∴ a = 5.6364 And

87 = 0 + 5(110)

∴ b = 0.7909

∴ by (i) equation of the trend line is y = 5.6364 + 0.7909u

Where u = t – 1981

For the year 1990,

u = 9

∴ y = 5.6364 + 0.7909(9)

∴ y = 12.7545 (in million tonnes)

Question 9.

Obtain the trend values for the above data using 3-yearly moving averages.

Solution:

Question 10.

The following table shows the production of gasoline in the U.S.A. for the years 1962 to 1976.

(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.

Solution:

(i)

(ii)