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Im new here, also a newbie in statistics. I was told to do forecasting at work by my boss. I used to do a naive forecasting before, and i want that to change. I want to do a real world forecast using the provided statistics method/s.

So here... I don't know if i'm doing it right but here's the measurement of accuracy of my forecast:

MAD: 33.20 MSE(sqrt): 42.38 MAPE: 10.21%

File: "https://drive.google.com/open?id=124VCrJ0AgcY0UTZuu5Mu4qTfnxrt-_1g".

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Is my MSE too high? i think it's too high. I'm a bit satisfied by seeing the trend in the graph. please correct me. Thanks a lot!

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You seem to have an in-sample fit. In-sample accuracy is a very poor guide to out-of-sample accuracy, since you are invariably tempted to overfit. It's much better to use a holdout sample of (say) the last 10 observations, fit your model to the data without this holdout set, forecast into the holdout set and assess accuracy on these holdout observations. (Your in-sample fit looks reasonable, though.)

We can't say whether a given MSE is "high" or "low". It is hard to say whether you have reached the limits of forecastability for a given time series. Ask yourself whether there are any drivers you could use to improve your forecast. And whether your accuracy is high enough for whatever you plan on doing with the forecast, i.e., the subsequent decisions.

You may be interested in this online forecasting textbook, or a book I coauthored, which is less technical.

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  • $\begingroup$ Hi Stephan, Sir what do you mean by "It's much better to use a holdout sample of (say) the last 10 observations" are you referring to Moving Average?(Ave of the last 10 days). Noted on the MSE part, definitely will check your provided references above. My head hurts lol. $\endgroup$ Commented Jan 22, 2019 at 18:29
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    $\begingroup$ No, I mean that if you have 100 historical observations overall, you should fit your model to the first 90, without using the last 10. Then use your model to forecast 10 steps. Now you can compare the forecast to the holdout actuals, which will be far more informative than an in-sample fit. This explains the concept well. $\endgroup$ Commented Jan 22, 2019 at 20:17

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