BookmarkSubscribeRSS Feed
WilliamRoberts
Calcite | Level 5

This is seamingly a trivial question, but several have looked at my question and are also perplexed.  I created a simple moving average prediction based on 100 data points in JMP.  I would like someone's assistantce in replicating, in Excel, the forecast provided by JMP using the coefficients provided by JMP.  Writing out the equation using actual numbers would be perfect for my needs. I have looked for sources with more than generalized formulas for hours now. The coefficients are as follows:

TermLagEstimateStd Errort RatioProb>|t|Constant Estimate
MA11-0.998490.433763-2.30.02350.179165
Intercept00.1791650.2483440.720.4724

The first 10 data points, forecast and residuals are provided below:

ObservationActual YPredicted YResidual Y
11.29170.17921.1125
23.50040.73542.7649
32.15952.02250.1370
42.24760.28191.9657
52.27911.75170.5274
62.35010.61871.7315
71.24911.6633-0.4141
80.9596-0.18321.1428
91.20701.19500.0120
100.16680.1899-0.0231

Could someone provide the math using the numbers in the tables that yields the Predicted Y for Observation 5 and 9 as an example?  Thank you!

4 REPLIES 4
Reeza
Super User

You should post this in the statistics portion of the forum Smiley Happy

user24feb
Barite | Level 11

I'm not a JMP-User and can offer only a part of the answer (at best). If you put your 1st 10 observations into proc arima:

Data A;
  Input Observation Actual;
  Datalines;
1 1.2917
2 3.5004
3 2.1595
4 2.2476
5 2.2791
6 2.3501
7 1.2491
8 0.9596
9 1.2070
10 0.1668
;
Run;

Proc ARIMA Data=A;
  Identify Var=Actual;
  Estimate q=1;
  Forecast Lead=0 Out=Result_ARIMA (Keep=Actual Forecast Residual);
Run;

The Result is: My=1.67681, MA1=-0.25377 and

Actual  Forecast       Residual
1.2917 1.6768075816 -0.385107582
3.5004 1.5790802232 1.9213197768
2.1595 2.164373954 -0.004873954
2.2476 1.6755707359 0.5720292641
2.2791 1.8219693795 0.4571306205
2.3501 1.7928119662 0.5572880338
1.2491 1.8182285508 -0.569128551
0.9596 1.5323818872 -0.572781887
1.207 1.5314547933 -0.324454793
0.1668 1.594471862 -1.427671862

The results calculate as I think they should: 1.579=1,6768-(-0.25377)*-0.3851, etc.

If you double-check your JMP-Input?

WilliamRoberts
Calcite | Level 5

Thank you.  I'm getting different predicted values from JMP, which likely is the source of my confusion, no doubt caused by me.

Reeza
Super User

Can you post the full data so someone can replicate the results/model?

sas-innovate-2024.png

Join us for SAS Innovate April 16-19 at the Aria in Las Vegas. Bring the team and save big with our group pricing for a limited time only.

Pre-conference courses and tutorials are filling up fast and are always a sellout. Register today to reserve your seat.

 

Register now!

What is ANOVA?

ANOVA, or Analysis Of Variance, is used to compare the averages or means of two or more populations to better understand how they differ. Watch this tutorial for more.

Find more tutorials on the SAS Users YouTube channel.

Discussion stats
  • 4 replies
  • 2917 views
  • 0 likes
  • 3 in conversation