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

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.
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q1
Fit a trend line by graphical method to the above data.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q1.1

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:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q2.1
n = 11, let the trend line the
y = a + bu ……..(I)
Σy = na + bΣu ……..(i)
Σuy = aΣu + bΣu2 ………(ii)
Substituting the values of Σy, Σu, Σuy, & Σu2, 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).

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

Question 3.
Obtain the trend line for the above data using 5 yearly moving averages.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q3.1

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.
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q4
Fit a trend line to the above data by graphical method.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q4.1

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:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q5
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q5.1
u = \(\frac{t-1980.5}{\frac{1}{2}}\), n = 10, Σu = 0, Σy = 42, Σu2 = 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 + bΣu2 ……….(iii)
Substituting the values of Σy, n, Σu, Σuy & Σu2 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

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

Question 6.
Obtain the trend values for the data in problem 4 using 4-yearly centered moving averages.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q6

Question 7.
The following table gives the production of steel (in millions of tonnes) for the years 1976 to 1986.
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q7
Fit a trend line to the above data by the graphical method.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q7.1

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:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q8
u = \(\frac{t-1981}{1}\), n = 10, Σu = 0, ΣY = 62, Σu2 = 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 + bΣu2 ………(iii)
Substituting the values of Σy, n, Σu, Σuy, Σu2 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)

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

Question 9.
Obtain the trend values for the above data using 3-yearly moving averages.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q9

Question 10.
The following table shows the production of gasoline in the U.S.A. for the years 1962 to 1976.
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q10
(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)
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q10.1
(ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1 Q10.2

Leave a Comment