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I have daily visitors data for the last 10 years. I want to do some basic tests like which is the busiest day, which is the busiest month, busiest week etc. I used auto.arima function with argument xreg to find out the coefficients of all the days of the week, week of the month. This is the output I got:

> summary(arima1)
Series: dailysea 
ARIMA(1,1,2)                    

Coefficients:
          ar1      ma1      ma2         Sun        Mon         Tue        Wed         Thu
      -0.1250  -0.4506  -0.3712  -1466.6853  -3623.175  -3895.0555  -3722.146  -3327.4288
s.e.   0.1207   0.1117   0.0891    325.7253    386.738    379.8793    379.883    386.7512
            Fri
      -2146.910
s.e.    325.736

sigma^2 estimated as 7776468:  log likelihood=-6808.5
AIC=13637   AICc=13637.31   BIC=13682.92

Training set error measures:
                   ME     RMSE      MAE  MPE MAPE      MASE         ACF1
Training set 59.63838 2784.809 1952.625 -Inf  Inf 0.8353728 -0.001839015

Can I use these coefficients to conclude that Saturday is the busiest followed by Sunday, Friday etc.? Also I have infinite MAPE which is not making sense to me.

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    $\begingroup$ @Aksakal, why drop AR(1) if the ARIMA(1,1,2) model was selected by auto.arima? That means that the ARIMA(0,1,2) had a higher AIC (or whichever criterion was used) value; recall that auto.arima does search the "neighbouring" models, so it must have evaluated ARIMA(0,1,2) and found it inferior to ARIMA(1,1,2). This makes ARIMA(0,1,2) an inferior choice. Also recall that variable significance is not the right tool for variable selection (Rob J. Hyndman had a post on that somewhere, if I remember correctly). $\endgroup$ Commented Jan 29, 2016 at 19:25
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    $\begingroup$ @RichardHardy You have 3650 data points, and your coefficient comes insignificant. You have a lot of power in your t-test. Cases of which Hyndman is talking about are probably with much smaller sample sizes. $\endgroup$
    – Aksakal
    Commented Jan 29, 2016 at 19:50
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    $\begingroup$ @Aksakal, this is not only about sample size, but I think I get your idea. Still, generally significance testing answers a different question than the one raised when considering variable selection. A variable may be statistically insignificant but still useful for forecasting. $\endgroup$ Commented Jan 29, 2016 at 20:13
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    $\begingroup$ Your model assumes a ton of things ... 1) there are no level shifts in your data 2) there are no deterministic trends i.e. variables of the form 1,2,3,4 ,,,, 3) your parameters are stable over time 4) your error variance is constant over time 5) there are no monthly deterministic effects 6) there are significant holiday effects 7) there are no weekly deterministic effects 8) there are no one-time outliers/inliers 9) there are no special days in the month etc..... etc ...... etc ..... $\endgroup$
    – IrishStat
    Commented Jul 18, 2016 at 0:50
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    $\begingroup$ @Aksakal You are on the right track. The fitting ( no modelling here !) software returned a model that has a near non-invertible ma polynomial (two coefficients add to nearly -1 ) suggesting that the differencing operator might be redundant . Besides that the ar coefficient is not-significant as you mentioned. I would suggest using (0,0,0) and examining the acf of the residuals from the 6 fixed effects in order to assess the need for an arima component. This iterative process of model formulation is what good time series analysis is all about. $\endgroup$
    – IrishStat
    Commented Nov 29, 2018 at 12:18

1 Answer 1

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An answer without testing/p-values, but with roughly estimating confidence intervals: Adding twice the s.e. (Standard error) on your coefficients should give you approximately 95%-confidence intervals for each one. From that perspective, the 95%-confidence interval for Sunday is roughly speaking between -1800 and -1100, which is far away from the Zero influence assumed for Saturdays. Extending the Argument you see, that Sunday is quite far away from all other days, whereas Mon, Tue, Wed, Thur are quite close together.

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