I’m new to statistics. I'm struggling to find the right hypothesis test.

Data description

My dataset contains 5000 JIRA tickets. Of each ticket it is known within how many days it is closed. When i plot a histogram of the closure days they are right-skewed.

I took 2 random samples:

  1. 125 tickets from year 2018
  2. 125 tickets from year 2019

Sample mean

The mean average time for handling tickets in 2018 = 7.52 days. The mean average time for handling tickets in 2019 = 17.5 days

Data converted

I also transformed the data into a 2x2 table:

  2018  115    10
  2019   88    37

TRUE is that the ticket has been processed within the agreed time of 14 days. FALSE is that the ticket has not been processed within the agreed time of 14 days

How to test my hypothesis?

I want to test a hypothesis if there is a significant difference in the mean or FALSE outcomes between year 2018 and 2019.

I thought two tests would be possible:

  1. Doing a Chi-Square test for the outcomes TRUE and FALSE from year 2018 and 2019.

  2. I also thought it could be an option to compare the average handling time of 2018 and 2019 with a z-test to see if there is a difference.

Is it true that I could use both tests? Or should I look at other tests?

  • $\begingroup$ Is this an assignment in a course? If so, please flag as such. $\endgroup$
    – Nick Cox
    Apr 30, 2020 at 11:46
  • $\begingroup$ @NickCox no, this is for a project I'm working on. $\endgroup$
    – Bluewater
    Apr 30, 2020 at 12:38

1 Answer 1


You seem to be on the right track.

Two- sample Chi-Square on the two distributions is possible (although I guess you are talking about the years 2018/19 rather than 2015/16?)

However, I have two little caveats regarding the z-test. First, you need to make sure (e.g. by repeated bootstrapping) that your sampled means are indeed normally distributed. Most likely this will be the case, given the number of tickets you have.

Secondly, z-test assumes the population variance to be known -or very well approximated by the sample variance, which again depends on the sample size and will most likely be the case. Nevertheless, just to be on the safe side, I would still consider swapping the z- with the t-test, since the latter is unaffected by the population variance problem

  • $\begingroup$ I mean indeed 2018/2019. I changed this in the starting post. So the conclusion can be that i'm safe with the chi-square test or else just take the t-test instead of the z-test? $\endgroup$
    – Bluewater
    Apr 30, 2020 at 12:46
  • $\begingroup$ yes. I would still go for the t-test, since, as a 1D-problem, it is conceptually simpler $\endgroup$ Apr 30, 2020 at 13:56

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