0
$\begingroup$

I have two time-series datasets that I'm trying to compare between years (see attached examples). I'm particularly interested so see whether overall levels (means?) per month were lower in one year compared to the other, ignoring the day-to-day variability.

I have no experience with time-series data yet and don't know how to approach that and which statistical test to use.

Thank you very much in advance.

Sample dataset 1

Sample dataset 2

$\endgroup$
  • $\begingroup$ This question may throw some light: stats.stackexchange.com/questions/144745/… $\endgroup$ – Arun Jose Nov 29 '16 at 4:36
  • $\begingroup$ Thanks. I just had time to quickly scan it, but it seems he's more interested in seasonality and general trends than comparing two years with the same/similar trends. At this point I really only need to know if levels from one year were lower than the other while following a trend, and as you can see from the graphs they both follow the same trend. Thank you. $\endgroup$ – Anke Nov 29 '16 at 18:35
  • $\begingroup$ Chi-square test comparing 12 means of one year to the other could be a start? $\endgroup$ – Arun Jose Nov 29 '16 at 18:54
  • $\begingroup$ I'd already thought of that but wasn't sure whether that was appropriate, especially since I do have those increases/decreases in some months. Also, doesn't chi-square test frequency distributions? And I'm not sure the assumptions are met, namely random samples, since I do have time-series data and as such autocorrelation. $\endgroup$ – Anke Nov 29 '16 at 19:15
  • $\begingroup$ You are right that it won't satisfy all assumptiosn as far as test of independence is concerned. However, if you treat each monthly reading as proportions of a yearly event, then the chi square test will at the least detect a variation in proportions. This will be evidence of firstly a geometric difference between the two years. You can support this with other tests and not rely on just this test. (This is an opinion, so I'm not posting this as an answer. There could be problems with this approach that I haven't considered yet.) $\endgroup$ – Arun Jose Dec 1 '16 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.