Question 1 (5 marks)

a) –

b)  Additive: = −

Multiplicative:

(0.5 mark / 0 mark)

(0.5 mark / 0 mark)

(4 marks)

(1 mark)

Question 2 (10 marks)

a)  Each  of  values  gives  estimate  of  average  monthly  temperature  for  different  months (January – December)

(1 mark)

b)  Coefficient of T in Model 2 is not significant (p-value > 0.05). Model 1 has smaller AIC and SIC.

Thus, Model 1 should be used.

(2 marks)

c) 0 : 1  = 2  = ⋯ = 11  = 12

1 : 0 is not true (1 / 1)

(37. 1230−15.4378)/11

=                                                       = 21.0702      (1 / 1)

Reject the null hypothesis.          (0.5 / 0.5)

There is seasonal variation.        (1 / 1)

(4 marks)

d)  MAPE = 2.1616. The average absolute errors is 2.1616% of the observed values.      (1)

ACF1 = 0.4512. Correlation between two successive errors is 0.4512, considerably is not strong.                                                                                                                             (1)

Theil’s U = 1.6157. Since the value is greater than 1, naïve method produces better      forecasts                                                                                                                        (1)

(3 marks)

Question 3 (7.5 marks)

a)  Both ADF tests have p-value of the t-stat smaller than 0.05. ADF test results suggest that the level of the series is stationary.      (0.75)

Both KPSS tests have LM-stat smaller than critical value at 5% significance value. KPSS

test results suggest that the level of the series is stationary.       (0.75)

Overall decision; differencing is not needed.      (0.5)

(2 marks)

b)  Since the level is stationary, identify possible models by inspecting correlogram of level .

Possible models:

1) ARMA(0,6)(2,0)12

ACF: Spikes at lags 1 to 6, cut off to zero.

Exponential decay at lags 12, 24, 36.

PACF: Assuming true PAC exponential decay at lags 1, 2,3,… .

Spikes at lags 12, 24, and cut off to zero at lags 36, 48,…

(2.5 marks)

2) ARMA(6,0)(2,0)12

ACF: Assuming true AC exponential decay at lags 1,2,3,… .

Exponential decay at lags 12, 24, 36.

PACF: Spikes at lags 1 to 6, cut off to zero.

Spikes at lags 12, 24, and cut off to zero at lags 36, 48,…

(2.5  marks)

# Other appropriate models are also acceptable.

c)  Since the level is stationary and ARMA models can be applied on levels, decision on inclusion of a constant term can be made based on the time plot of the series.

(0.5 mark)

Question 4 (7.5 marks)

a)  Both ADF tests have p-value of the t-stat greater than 0.05. ADF test results suggest that the level of the series is non-stationary.      (0.75)

Both KPSS tests have LM-stat smaller than critical value at 5% significance value. KPSS

test results suggest that the level of the series is stationary.       (0.75)

Overall decision; differencing is needed. Since the series has seasonal variation, seasonal

differencing is preferable.      (0.5)

(2 marks)

b)  Since the level is considered as non-stationary, identify possible models by inspecting correlogram of the seasonally differenced series.

Possible models:

1) ARMA(0,5)(0,1)12

ACF: Spikes at lags 1 to 5, cut off to zero.

Spike at lags 12, and cut off to zero at lags 24, 36,… .

PACF: PAC exponential decay at lags 1, 2, 3,… .

Assuming the true PAC exponential decay at lags 12, 24, 36

(2.5 marks)

2) ARMA(2,0)(3,0)12

ACF: Exponential decay at lags 1, 2, 3,… .

Assuming true AC exponential decay at lags 12, 24, 36 …

PACF: Spikes at lags 1 and 2, cut off to zero.

Spikes at lags 12 and 36 and cut off to zero at lags 48, 60 …

(2.5 marks)

# Other appropriate models are also acceptable.

c)  Since the level is non-stationary and ARMA models cannot be applied on level, decision on inclusion of a constant term cannot be made based on the time plot of the series.

(0.5 mark)

Question 5 (10 marks)

a)  All inverted AR roots are inside unit circle, the estimated process is stationary   (1)

c)  (1 − 1 − 2 2 )(1 − 12 ) = (1 + Θ1 12) (  )1

= 1 −1 + 2 −2 + −12 − 1 −13 − 2 −14 + + Θ 1 −12 (  )2

d)  January 2016:

193  = 25.4787      (1.5)

February 2016:

194  = 25.6107      (1.5)

January 2017:

25.4787 ± 1.96(√0.0663) = (24.9740, 25.9834)      (  )1

25.6107 ± 1.96((√0.0754) = (25.0725, 26.1489)        (1)