ARIMA-MODEL BY"IBM spss statistics 24"

  1. The first data set contains observations adapted from a series provided by a large U.S. corporation. There are 90 weekly observations showing the percentage of the time that parts for an industrial product are available when needed. The data can be found on Blackboard with “parts” as the pre-fix. Use the observation 1 to 85 for model building and assessing model goodness of fit for estimation purposes. Use the observation 86 to 90 for assessing model goodness of fit for forecasting purposes.

  1. Fit a White Noise model to the original data and based on the results of Residuals ACF and Residuals PACF, select an initial model. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied? In efforts to approach White Noise series, improve your initial model by fitting another model based on the initial model results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied?


ARIMA-MODEL BY"IBM spss statistics 24" 1

ARIMA-MODEL BY"IBM spss statistics 24" 2

The sample ACF and PACF functions of original data seem to suggest that the data follow ARMA processes. Based on the principle of parsimony, we will start fitting an AR(1) process to the data. The summaries of the fit and diagnostics are listed below:


Model Description

Model Type

Model ID

Percentage

Model_1

ARIMA(1,0,0)





The Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Percentage-Model_1

0

.184

2.050

2.025

1.657

27.308

17

.054

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Percentage-Model_1

Percentage

No Transformation

Constant

81.982

.387

211.660

.000

AR

Lag 1

.431

.100

4.302

.000



Forecast

Model

86

87

88

89

90

Percentage-Model_1

Forecast

83.33

82.56

82.23

82.09

82.03

UCL

87.40

87.00

86.73

86.60

86.54

LCL

79.25

78.12

77.73

77.58

77.51


ARIMA-MODEL BY"IBM spss statistics 24" 3

Testing for significance of models parameters:

1/ State the null and alternate hypotheses.

H0: Ø1 = 0

H1: Ø1 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value = 0 (there is a small chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that the first parameter of AR(1) is contributing significantly into explaining the variation of the response variable (the percentage of the time that parts for an industrial product are available when needed)

Checking the stationarity conditions:

│<1


│0.431│<1 so the model is stationary.


Checking invertibility conditions:

AR processes are always invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 4

The normality assumption seems to be satisfied. However, there are two observation at the bottom and the top of the graph above may decrease the P.V of the goodness of fit test.



Constant variance of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 5



The constant variance of the residual terms seems to be satisfied since there is no megaphone shape is observed in the graph.










Independence of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 6

The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model is close to White-Noise Series but it has not reached it yet; which suggest fitting more complex models.









The sample ACF and PACF functions of residuals of AR(1) model seem to suggest that the data follow ARMA processes. we should try fitting an MA(1) process to the data. However, I chose to fit AR(2) process first then fit MA(1) and ARMA(1,1) later in the analysis. The summaries of the fit and diagnostics are listed below:


Model Description

Model Type

Model ID

Percentage

Model_1

ARIMA(2,0,0)



Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Percentage-Model_1

0

.253

1.973

1.910

1.561

17.523

16

.353

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Percentage-Model_1

Percentage

No Transformation

Constant

82.042

.528

155.528

.000

AR

Lag 1

.305

.105

2.897

.005

Lag 2

.299

.108

2.776

.007



Forecast

Model

86

87

88

89

90

Percentage-Model_1

Forecast

84.07

83.57

83.12

82.83

82.60

UCL

87.99

87.67

87.49

87.28

87.11

LCL

80.15

79.48

78.74

78.38

78.10


ARIMA-MODEL BY"IBM spss statistics 24" 7


Testing for significance of models parameters:

1/ State the null and alternate hypotheses.

H0: Øi = 0

H1: Øi ≠ 0 , i = 1,2

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø1) = 0.005 (there is a small chance that the null hypothesis is true)

P-value (Ø2) = 0.007 (there is a small chance that the null hypothesis is true)


3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that the first and second parameters of AR(2) are contributing significantly into explaining the variation of the response variable (the percentage of the time that parts for an industrial product are available when needed)

Checking the stationarity conditions:

│<1

< 1

< 1


│0.299│<1

< 1

< 1

So the model is stationary.


Checking invertibility conditions:

AR processes are always invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 8


The normality assumption seems to be satisfied.


















Constant variance of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 9


The constant variance of the residual term seems to be satisfied since there is no megaphone shape is observed in the graph.











Independence of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 10

The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model has reached White-Noise Series

  1. Select an alternative model (one that is different from your initial model) fit a White Noise model to the original data and based on the results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied? In efforts to approach Whit Noise series, improve your alternative model by fitting another model based on the alternative model results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied?

ARIMA-MODEL BY"IBM spss statistics 24" 11

The sample ACF and PACF functions of original data seem to suggest that the data follow ARMA processes. Based on the principle of parsimony, we will start fitting an MA(1) process to the data. The summaries of the fit and diagnostics are listed below:




Model Description

Model Type

Model ID

Percentage

Model_1

ARIMA(0,0,1)



Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Percentage-Model_1

0

.122

2.126

2.118

1.734

49.784

17

.000

0




ARIMA Model Parameters

Estimate

SE

t

Sig.

Percentage-Model_1

Percentage

No Transformation

Constant

81.969

.300

272.836

.000

MA

Lag 1

-.307

.105

-2.923

.004



Forecast

Model

86

87

88

89

90

Percentage-Model_1

Forecast

82.64

81.97

81.97

81.97

81.97

UCL

86.87

86.39

86.39

86.39

86.39

LCL

78.41

77.55

77.55

77.55

77.55


ARIMA-MODEL BY"IBM spss statistics 24" 12


Testing for significance of model parameters:

1/ State the null and alternate hypotheses.

H0: θ1 = 0

H1: θ1 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value = 0.004 (there is a small chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that the first parameter of MA(1) is contributing significantly into explaining the variation of the response variable (the percentage of the time that parts for an industrial product are available when needed)

Checking the stationarity conditions:

MA processes are always stationary.


Checking invertibility conditions:

│<1


│-.307│<1 so the model is invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 13


The normality assumption seems to be satisfied. However, there are two observation at the bottom and the top of the graph above may decrease the P.V of the goodness of fit test.



Constant variance of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 14


The constant variance of the residual term seems to be satisfied since there is no megaphone shape is observed in the graph.


Independence of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 15


The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model is close to White-Noise Series but it has not reached it yet; which suggest fitting more complex models.









The sample ACF and PACF functions of residuals of MA(1) model seem to suggest that the data follow ARMA processes. we should try fitting an ARMA(1,1) process to the data. The summaries of the fit and diagnostics are listed below:



Model Description

Model Type

Model ID

Percentage

Model_1

ARIMA(1,0,1)



Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Percentage-Model_1

0

.300

1.910

1.869

1.527

13.924

16

.604

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Percentage-Model_1

Percentage

No Transformation

Constant

82.154

.830

98.928

.000

AR

Lag 1

.920

.070

13.188

.000

MA

Lag 1

.654

.130

5.015

.000



Forecast

Model

86

87

88

89

90

Percentage-Model_1

Forecast

83.94

83.79

83.66

83.54

83.43

UCL

87.72

87.71

87.69

87.65

87.62

LCL

80.15

79.88

79.64

79.43

79.24


ARIMA-MODEL BY"IBM spss statistics 24" 16


Testing for significance of model parameters:

1/ State the null and alternate hypotheses.

H0: Ø1 = 0

H1: Ø1 ≠ 0

H0: θ1 = 0

H1: θ1 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø1) = 0 (there is a small chance that the null hypothesis is true)

P-value (θ1) = 0 (there is a small chance that the null hypothesis is true)


3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that both parameters of ARMA(1,1) are contributing significantly into explaining the variation of the response variable (the percentage of the time that parts for an industrial product are available when needed)

Checking the stationarity conditions:


│<1


│0.920│<1 so the model is stationary.


Checking invertibility conditions:


│<1


│0.654│<1 so the model is invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 17


The normality assumption seems to be moderately violated.


















Constant variance of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 18


The constant variance of the residual term seems to be satisfied since there is no megaphone shape is observed in the graph.











Independence of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 19

The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model has reached White-Noise Series

c. Based on the results obtained from parts a and b, fill in the tables below to evaluate the potential models. Which model do you prefer for estimation purposes? Which model do you prefer for forecasting purposes?

Checking Model Assumptions (Residuals)

Model/ Measure

ARIMA (1,0,0)

AEIMA (2,0,0)

ARIMA (0,0,1)

ARIMA (1,0,1)

Normality

Satisfied

Satisfied

Satisfied

Moderately violated

Constant Variance

Satisfied

Satisfied

Satisfied

Satisfied

Independence

Moderately violated

Moderately violated

Moderately violated

Moderately violated

White noise

violated

Satisfied

violated

Satisfied

Model Stationarity

Satisfied

Satisfied

Satisfied

Satisfied

Model Invertibility

Satisfied

Satisfied

Satisfied

Satisfied

The highlighted model does not suggest further refinement, while the other models suggest some refinements.

Goodness of fit for (Estimation)

Model/ Measure

ARIMA (1,0,0)

AEIMA (2,0,0)

ARIMA (0,0,1)

ARIMA (1,0,1)

RMSE

2.050

1.973

2.126

1.910

MAE

1.657

1.561

1.734

1.527

MAPE

2.025

1.910

2.118

1.869

Number of Parameters (Complexity)

Number of Significant Parameters

The highlighted model is the best model for estimation purposes.

Goodness of fit for (Forecasting)

Model/ Measure

ARIMA (1,0,0)

AEIMA (2,0,0)

ARIMA (0,0,1)

ARIMA (1,0,1)

RMSE

3.706869299

3.297838686

3.908943591

3.09327658

MAE

3.148

2.618

3.444

2.504

MAPE

0.036641804

0.030578392

0.040101984

0.029427493

The highlighted model is the best model for forecasting purposes.

  1. Use the entire data set and based on your best model for forecasting, forecast the observations: 91, 92, 93, 94, and 95. Provide both a point estimate and a confidence interval for each forecasted observation.

Forecast

Model

91

92

93

94

95

Percentage-Model_1

Forecast

84.44

84.29

84.16

84.03

83.92

UCL

88.45

88.43

88.39

88.36

88.31

LCL

80.43

80.16

79.92

79.71

79.52



1. The data set contains the monthly total (in thousands) of persons unemployed in Canada from January 1956 to December 1975. There are a total of 240 observations. The data can be found on Blackboard with “Unemployment-Canada” as the title. Use the observation 1 to 230 for model building and assessing model goodness of fit for estimation purposes. Use the observation 231 to 240 for assessing model goodness of fit for forecasting purposes.

  1. a. Fit a White Noise model to the original data and based on the results of Residuals ACF and Residuals PACF, select an initial model for both seasonal components and non-seasonal components. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied? In efforts to approach White Noise series, improve your initial model by fitting another model both seasonal components and non-seasonal components based on the initial model results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied?

  2. Instead of doing part (a), you could run a several models aiming to reach White Noise series. Examine each model Residuals ACF and Residuals PACF.


ARIMA-MODEL BY"IBM spss statistics 24" 20



Model Description

Model Type

Model ID

Unemp

Model_1

ARIMA(0,0,0)(0,0,0)




Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Unemp-Model_1

0

-2.168E-19

144.154

37.342

119.948

920.567

18

.000

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Unemp-Model_1

Unemp

No Transformation

Constant

401.561

9.505

42.246

.000



ARIMA-MODEL BY"IBM spss statistics 24" 21




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(0,0,0)(1,0,0)


ARIMA-MODEL BY"IBM spss statistics 24" 22




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(1,0,0)(1,0,0)


ARIMA-MODEL BY"IBM spss statistics 24" 23




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(1,0,1)(1,0,0)


ARIMA-MODEL BY"IBM spss statistics 24" 24




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(1,0,1)(1,0,1)


ARIMA-MODEL BY"IBM spss statistics 24" 25




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,1)(1,0,1)


ARIMA-MODEL BY"IBM spss statistics 24" 26




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,2)(1,0,1)


ARIMA-MODEL BY"IBM spss statistics 24" 27




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,2)(2,0,1)


ARIMA-MODEL BY"IBM spss statistics 24" 28




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,2)(2,0,2)


ARIMA-MODEL BY"IBM spss statistics 24" 29




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,1,2)(2,0,2)


ARIMA-MODEL BY"IBM spss statistics 24" 30




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,2)(2,1,2)


ARIMA-MODEL BY"IBM spss statistics 24" 31




Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,1,2)(2,1,2)


ARIMA-MODEL BY"IBM spss statistics 24" 32




  1. Do this part if you chose to do part (a) only. Select an alternative model (one that is different from your initial model) fit a White Noise model to the original data and based on the results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied? In efforts to approach Whit Noise series, improve your alternative model by fitting another model based on the alternative model results of Residuals ACF and Residuals PACF. Estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied?

  1. Do this part if you chose to do part (b) only. Select one of the best models you have from part (b) as initial model. For further diagnosis, estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied? Then select another good model you have from part (b) as alternative model. For further diagnosis, estimate the model parameters. Check the stationarity and invertibility conditions. Are the parameter estimates significant? Are the model assumptions satisfied?



Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,1)(1,0,1)



Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Unemployment-Model_1

0

.965

27.290

5.950

20.340

29.929

13

.005

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Unemployment-Model_1

Unemployment

No Transformation

Constant

437.249

305.087

1.433

.153

AR

Lag 1

1.609

.225

7.140

.000

Lag 2

-.635

.217

-2.925

.004

MA

Lag 1

.461

.259

1.778

.077

AR, Seasonal

Lag 1

.977

.013

77.173

.000

MA, Seasonal

Lag 1

.516

.077

6.689

.000



Forecast

Model

Mar 1919

Apr 1919

May 1919

Jun 1919

Jul 1919

Aug 1919

Sep 1919

Oct 1919

Nov 1919

Dec 1919

Unemployment-Model_1

Forecast

820.26

790.10

730.27

701.01

671.28

631.90

599.36

593.84

632.02

691.27

UCL

867.92

862.66

823.04

810.57

794.84

767.11

744.26

746.79

791.67

856.50

LCL

772.61

717.53

637.49

591.44

547.72

496.69

454.46

440.88

472.36

526.04

For each model, forecasts start after the last non-missing in the range of the requested estimation period, and end at the last period for which non-missing values of all the predictors are available or at the end date of the requested forecast period, whichever is earlier.


ARIMA-MODEL BY"IBM spss statistics 24" 33


Testing for significance of model parameters (Seasonal Components) ARIMA (1,0,1)12:

1/ State the null and alternate hypotheses.

H0: Ø12 = 0

H1: Ø12 ≠ 0

H0: θ12 = 0

H1: θ12 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø12) = 0 (there is a small chance that the null hypothesis is true)

P-value (θ12) = 0 (there is a small chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that both parameters of ARIMA(1,0,1)12 are contributing significantly into explaining the seasonal variation of the response variable (the monthly total (in thousands) of persons unemployed in Canada from January 1956 to December 1975)

Checking the stationarity conditions:


│<1


│0.977│<1 so the seasonal model is stationary.


Checking invertibility conditions:


│<1


│0.516│<1 so the seasonal model is invertible.


Testing for significance of models parameters (Non-Seasonal Components) ARIMA (2,0,1):

1/ State the null and alternate hypotheses.

H0: Øi = 0

H1: Øi ≠ 0 , i = 1,2

H0: θ1 = 0

H1: θ1 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø1) = 0.000 (there is a small chance that the null hypothesis is true)

P-value (Ø2) = 0.004 (there is a small chance that the null hypothesis is true)

P-value (θ1) = 0.077 (there is a chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that the first and second parameters of AR(2) are contributing significantly into explaining the variation of the response variable (the monthly total (in thousands) of persons unemployed in Canada from January 1956 to December 1975). However, the first parameter of MA(1) is not contributing significantly into explaining the variation of the response variable.

Checking the stationarity conditions:

│<1

< 1

< 1


│-0.635│<1

< 1

< 1

So the non-seasonal model is stationary.


Checking invertibility conditions:

│<1


│0.461│<1 so the non-seasonal model is invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 34


The normality assumption seems to be satisfied. However, there are two observation at the top and the bottom of the graph above may decrease the P.V of the goodness of fit test.


Constant Variance:

ARIMA-MODEL BY"IBM spss statistics 24" 35


The constant variance of the residual term seems to be satisfied since there is no megaphone shape is observed in the graph.


Independence of residuals:



ARIMA-MODEL BY"IBM spss statistics 24" 36


The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model is so close to White-Noise Series








Model Description

Model Type

Model ID

Unemployment

Model_1

ARIMA(2,0,2)(1,0,1)



Model Statistics

Model

Number of Predictors

Model Fit statistics

Ljung-Box Q(18)

Number of Outliers

Stationary R-squared

RMSE

MAPE

MAE

Statistics

DF

Sig.

Unemployment-Model_1

0

.966

26.975

5.841

20.066

23.993

12

.020

0



ARIMA Model Parameters

Estimate

SE

t

Sig.

Unemployment-Model_1

Unemployment

No Transformation

Constant

433.010

290.024

1.493

.137

AR

Lag 1

1.227

.284

4.314

.000

Lag 2

-.278

.273

-1.017

.310

MA

Lag 1

.118

.280

.422

.674

Lag 2

-.208

.085

-2.463

.015

AR, Seasonal

Lag 1

.972

.015

65.785

.000

MA, Seasonal

Lag 1

.490

.078

6.307

.000



Forecast

Model

Mar 1919

Apr 1919

May 1919

Jun 1919

Jul 1919

Aug 1919

Sep 1919

Oct 1919

Nov 1919

Dec 1919

Unemployment-Model_1

Forecast

827.88

793.11

727.08

691.14

659.01

619.32

587.40

581.99

621.33

682.89

UCL

875.29

863.90

820.66

802.56

784.25

755.45

732.29

733.99

779.20

845.65

LCL

780.48

722.32

633.50

579.72

533.77

483.18

442.52

429.98

463.45

520.14

For each model, forecasts start after the last non-missing in the range of the requested estimation period, and end at the last period for which non-missing values of all the predictors are available or at the end date of the requested forecast period, whichever is earlier.



ARIMA-MODEL BY"IBM spss statistics 24" 37

Testing for significance of model parameters (Seasonal Components) ARIMA (1,0,1)12:

1/ State the null and alternate hypotheses.

H0: Ø12 = 0

H1: Ø12 ≠ 0

H0: θ12 = 0

H1: θ12 ≠ 0

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø12) = 0 (there is a small chance that the null hypothesis is true)

P-value (θ12) = 0 (there is a small chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that both parameters of ARIMA(1,0,1)12 are contributing significantly into explaining the seasonal variation of the response variable (the monthly total (in thousands) of persons unemployed in Canada from January 1956 to December 1975)

Checking the stationarity conditions:


│<1


│0.972│<1 so the seasonal model is stationary.


Checking invertibility conditions:


│<1


│0.490│<1 so the seasonal model is invertible.


Testing for significance of models parameters (Non-Seasonal Components) ARIMA (2,0,1):

1/ State the null and alternate hypotheses.

H0: Øi = 0

H1: Øi ≠ 0 , i = 1,2

H0: θj = 0

H1: θj ≠ 0 , j = 1,2

2/ State the decision rule, report the p-value.

level of significance (α) = 0.05. Since it is two-tailed test, reject H0 if P.V < 0.05/2 otherwise fail to reject H0

P-value (Ø1) = 0.000 (there is a small chance that the null hypothesis is true)

P-value (Ø2) = 0.310 (there is a big chance that the null hypothesis is true)

P-value (θ1) = 0.674 (there is a big chance that the null hypothesis is true)

P-value (θ2) = 0.015 (there is a small chance that the null hypothesis is true)

3/ What is your decision regarding the null hypothesis? Interpret the result.

Since P.V < 0.05/2 at 0.05 level of significance, we reject H0.

We conclude that the first parameter of AR(2) is contributing significantly into explaining the variation of the response variable (the monthly total (in thousands) of persons unemployed in Canada from January 1956 to December 1975) while the second parameter is not significant. However, the first parameter of MA(2) is not contributing significantly into explaining the variation of the response variable, while the second parameter is significant.

Checking the stationarity conditions:

│<1

< 1

< 1


│-0.278│<1

< 1

< 1

So the non-seasonal model is stationary.


Checking invertibility conditions:

│<1

< 1

< 1


│-0.208│<1

< 1

< 1

So the non-seasonal model is invertible.


Checking model assumptions:

Normality of residuals:


ARIMA-MODEL BY"IBM spss statistics 24" 38


The normality assumption seems to be satisfied. However, there are two observation at the top and the bottom of the graph above may decrease the P.V of the goodness of fit test.


Constant Variance:


ARIMA-MODEL BY"IBM spss statistics 24" 39


The constant variance of the residual term seems to be satisfied since there is no megaphone shape is observed in the graph.


Independence of residuals:



ARIMA-MODEL BY"IBM spss statistics 24" 40

The graph above depicts a potentiality of having negative auto-correlation.


Reaching White-Noise Process:


Looking at the Residual ACF and Residuals PACF, the model is so close to White-Noise Series


  1. Based on the results obtained from parts (a and c) or (b and d), fill in the tables below to evaluate the potential models. Which model do you prefer for estimation purposes? Which model do you prefer for forecasting purposes?

Model/ Measure

ARIMA (2,0,1)(1,0,1)12

AEIMA (2,0,2)(1,0,1)12

Normality

Moderately violated

Moderately violated

Constant Variance

Satisfied

Satisfied

Independence

Moderately violated

Moderately violated

White noise

Moderately violated

Moderately violated

Model Stationarity

Satisfied

Satisfied

Model Invertibility

Satisfied

Satisfied

Both models perform equally well and suggest further refinements.

Model/ Measure

ARIMA (2,0,1)(1,0,1)12

AEIMA (2,0,2)(1,0,1)12

RMSE

27.290

26.975

MAE

20.340

20.066

MAPE

5.950

5.841

Number of Parameters (Complexity)

Number of Significant Parameters

The highlighted model is the best model for estimation purposes since the two models are really close to each other in terms of accuracy; however the first model is less complex. .

Model/ Measure

ARIMA (2,0,1)(1,0,1)12

AEIMA (2,0,2)(1,0,1)12

RMSE

13.01525067

10.59290376

MAE

11.599

8.981

MAPE

0.017343794

0.013070704

The highlighted model is the best model for forecasting purposes.


  1. Use the entire data set and based on your best model for forecasting, forecast the observations: 241 to 252. Provide both a point estimate and a confidence interval for each forecasted observation.

Forecast

Model

Jan 1920

Feb 1920

Mar 1920

Apr 1920

May 1920

Jun 1920

Jul 1920

Aug 1920

Sep 1920

Oct 1920

Nov 1920

Dec 1920

Unemployment-Model_1

Forecast

852.54

847.45

825.01

780.70

708.01

686.01

646.41

612.44

579.33

572.89

623.17

681.14

UCL

899.24

917.02

917.17

890.31

831.13

819.83

788.89

762.04

734.87

733.39

787.87

849.41

LCL

805.84

777.89

732.86

671.08

584.89

552.19

503.93

462.83

423.80

412.38

458.46

512.87