I have some data that varies over time, and cannot be described by a simple model. In fact we don't care what the model is at all, we just want to know whether it is significantly non-zero at any point in time. Ideally, the result would also tell us at which time points the data is significantly different from 0.

It is possible to do bootstrapping, but I am wondering if there is a more specific way to assess significance (a specific type of ANOVA, or t-test with corrections for multiple comparisons - but how to determine number of comparisons since the data is clearly correlated in time.).

Thanks in advance - I'm sure this is a common problem, but after hours of reading I still can't find a clear answer.


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    $\begingroup$ This is not a common problem because it has no solution. It is on all fours with the probability of being red or any other arbitrary state with no rules given. If I say to you, Tell me the probability that my number is positive, you can't make progress with no assumptions at all. $\endgroup$ – Nick Cox Nov 30 '13 at 1:56
  • $\begingroup$ If a t-test can determine if a sample is significantly different from 0, then there must be some equivalent test for a time series (some form of ANOVA?). (We ended up doing bootstrapping but I just saw that there was a comment on this and am still curious). The null hypothesis is: at every point in time, we are sampling from a distribution with 0 mean. There must be a test to determine whether or not that hypothesis can be rejected. If you have a sample of 100 numbers and 98 of them are positive, then I would easily conclude from a t-test that the population mean is greater than zero. $\endgroup$ – myrtle42 Feb 6 '15 at 0:58
  • $\begingroup$ You pay no attention to dependence structure. Your tacit assumption is that values are mutually independent. That is testable, but there can be no general purpose comparison of means for time series without attention to amount and character of dependence. Bootstrapping doesn't solve this without using more advanced tricks. $\endgroup$ – Nick Cox Feb 6 '15 at 9:09
  • $\begingroup$ The idea that there should be a simple solution because the problem seems simple to state is unfortunately wishful thinking. I'd be more helpful if I could be. Unfortunately, the fact that basic tests require independence assumptions is often glossed over in introductory texts and courses. $\endgroup$ – Nick Cox Feb 6 '15 at 10:38

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