Chapter 4 Time Series Ex 4.1 Solutions

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).

u = , n = 10, Σu = 0, Σy = 42, Σu2 = 330, Σuy = 148 Let the trend line be y = a + bu ……(i) where u = 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

u = , 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)

(i) (ii)