I have a data set with phone calls, that contains the caller id and call duration. I want to test the hypothesis that the call duration during the week is smaller than the call duration during the weekend. I was thinking about using the Welch t-test but as the same caller can do calls on week days and weekends, I believe that I can´t assume iid. The time series is autocorrelated, what complicates the matter. Would ANOVA help in this case?


I would not expect call length to be normally distributed, so I would not jump into a Welch's-t test. Try a non-parametric Mann-Whitney U-test. However, that latter test is less powerful than a paired test which goes to interpretation. That is, if the test is highly significant, then a paired test with the same $N$ would be more significant. Problem is, you have some $n<N$ that are paired, so your test will be less than optimally powerful. You could try a Wilcoxon signed-ranks-W test on the $n$ pairs if $n$ is a substantial fraction $N$, and of those two non-parametric tests one will be more significant than the other.

Which? You might need to run simulations to decide which is best, but if they are both highly significant, it would be overkill. Cannot say whether ARIMA is needed or not, ANOVA probably less needed. Cannot say more without more information.

  • $\begingroup$ Would the autocorrelation of the series invalidate the independent samples that M-W U needs? Imagining that there is one year of data with daily values, one would have 261 week days and 90 weekend days, so it's not a substantial fraction. $\endgroup$ – Ivan Mar 2 '18 at 20:15
  • $\begingroup$ On the other hand, if each call is considered separatedely, the series may be considered independent. So Wicoxon should work. $\endgroup$ – Ivan Mar 2 '18 at 20:25
  • $\begingroup$ Invalidate is probably too harsh a concept to be applicable. Autocorrelation of call length might reduce the power of M-W testing. Again, if M-W testing is highly significant then it doesn't matter. You might start by looking at ARIMA regression. $\endgroup$ – Carl Mar 2 '18 at 20:25

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.