ARIMA Model - Time-series Plot Gretl Software

Added on -2020-02-05

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Q,3 Use the Excel file “Time_Series.xlsx”. Develop a time series model basedon the data. Provide the process of your model developmentTime series PlotGretl software (WEKA) output fileMODEL-1 ARIMA Model 1: ARMAX, using observations 1:01-22:17 (T = 521)Dependent variable: V1Standard errors based on HessianCoefficientStd. Errorzp-valueconst-0.6477140.175359-3.69360.00022***phi_10.5133560.041564812.3507<0.00001***theta_10.6898320.032964420.9265<0.00001***hour-0.0008259560.000581344-1.42080.15538Mean dependent var-0.865611S.D. dependent var1.008562Mean of innovations-0.000422S.D. of innovations0.579599Log-likelihood-455.8833Akaike criterion921.7665Schwarz criterion943.0453Hannan-Quinn930.1015RealImaginaryModulusFrequencyARRoot 1 1.94800.00001.94800.0000MARoot 1 -1.44960.00001.44960.5000SOURCE: Computed data Output file Gretl software ( WEKA) output file
Normality testDoornik-Hansen test = 53.7725, with p-value 2.10594e-012 Shapiro-Wilk W = 0.954728, with p-value 1.49004e-011 Lilliefors test = 0.0579438, with p-value ~= 0 Jarque-Bera test = 31.2605, with p-value 1.62884e-007Process: Interval estimation point wise Frequency distribution for hour, obs 1-521number of bins = 23, mean = 261, sd = 150.544 interval midpt frequency rel. cum. < 23.636 11.818 23 4.41% 4.41% * 23.636 - 47.273 35.455 24 4.61% 9.02% * 47.273 - 70.909 59.091 23 4.41% 13.44% * 70.909 - 94.545 82.727 24 4.61% 18.04% * 94.545 - 118.18 106.36 24 4.61% 22.65% * 118.18 - 141.82 130.00 23 4.41% 27.06% * 141.82 - 165.45 153.64 24 4.61% 31.67% * 165.45 - 189.09 177.27 24 4.61% 36.28% * 189.09 - 212.73 200.91 23 4.41% 40.69% * 212.73 - 236.36 224.55 24 4.61% 45.30% * 236.36 - 260.00 248.18 23 4.41% 49.71% * 260.00 - 283.64 271.82 24 4.61% 54.32% * 283.64 - 307.27 295.45 24 4.61% 58.93% * 307.27 - 330.91 319.09 23 4.41% 63.34% * 330.91 - 354.55 342.73 24 4.61% 67.95% * 354.55 - 378.18 366.36 24 4.61% 72.55% * 378.18 - 401.82 390.00 23 4.41% 76.97% * 401.82 - 425.45 413.64 24 4.61% 81.57% * 425.45 - 449.09 437.27 24 4.61% 86.18% * 449.09 - 472.73 460.91 23 4.41% 90.60% * 472.73 - 496.36 484.55 24 4.61% 95.20% * 496.36 - 520.00 508.18 23 4.41% 99.62% * >= 520.00 531.82 2 0.38% 100.00% Test for null hypothesis of normal distribution:Chi-square(2) = 53.773 with p-value 0.00000
Source : WEKA gretl output file – Test statistics of normality
Q.1 A) Develop a linear regression model with variable cnt as the dependentvariable and the other variables as independent variables (you mayselect some of them). Provide the process of your model development indetail.Linear Regression model Model SummaryModelRR SquareAdjusted RSquareStd. Error of theEstimate1.628a.395.393141.624a. Predictors: (Constant), windspeed, temp, holiday, weathersit, yr, hr, workingday, mnth, hum, season, dtedayANOVAaModelSum of SquaresdfMean SquareFSig.1Regression52213813.305114746710.300236.657.000bResidual79988564.339398820057.313Total132202377.6443999a. Dependent Variable: cntb. Predictors: (Constant), windspeed, temp, holiday, weathersit, yr, hr, workingday, mnth, hum, season, dtedayCoefficientsaModelUnstandardized CoefficientsStandardizedCoefficientstSig.BStd. ErrorBeta1(Constant)-61260.74540369.191-1.518.129dteday4.534E-006.000.4091.517.129season15.6714.688.0913.343.001yr-52.24494.382-.143-.554.580mnth-8.9557.838-.159-1.143.253hr7.817.350.29422.352.000holiday-24.59814.719-.021-1.671.095workingday3.8804.974.010.780.435weathersit-8.1123.995-.028-2.031.042temp255.92913.361.27219.155.000

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