Not that strong in statistics, so I need a little help getting started with some data I've gathered.

In my experiment subjects had to perform tasks while I took time for them to complete it. Now, I want to do some t-testing based on various parameters, but.. My data ranges from 0 to 60 seconds. The experiment stopped at 60 seconds, so about 7-8 % of the data is '60'. Same amount around 0 seconds. Rest is highly skewed towards right with an overall mean time at 16,89 seconds.

So my question is: Can I do a t-test on this, when the data is so skewed? If not: What test can I do to check if groups/parameters has an effect?

I also have two different groups with an assumption that they're different (p < 0.05), but same problem again. The data is not normalized.

Could a Mann-Whitney U test be suitable for this?

Sample size: ~380

I've attached a picture of the histogram. X-values is time, Y-values frequency.

enter image description here

I've tried to do a log10 transformation. Before that I removed the 60's, since those are not actual times of completion. When I do the log10 then the first group (sample size ~380) is somewhat normalized:

http:// imgur. com/XDEv74V

But the group for which I try to do the t-test is not (sample size 52):

http:// imgur. com/pfrAV8H

(cant upload 2+ pictures due to reputation < 10)

The two-tailed t-test of difference in variance gives a p-value much lower than 0.05. But can I count on this being correct on a log10 with only one of the samples being "normalized" ?

I've done the wilcox.test (Mann-Whitney-Wilcoxon test) in R and i get the following results:

Rank: W = 13840, p-value = 0.001673

Signed: V = 895.5, p-value = 0.5861

What's the difference between rank and signed?

  • $\begingroup$ It will be useful if information about objecives and study design is indicated including sample size s and parameters. $\endgroup$
    – user10619
    Apr 25, 2016 at 11:10
  • $\begingroup$ You said that 7 to 8% of your data are 60 but your histogram has no one at 60 $\endgroup$
    – Peter Flom
    Apr 25, 2016 at 12:07
  • $\begingroup$ I included the wrong histogram, now the correct one is uploaded. I excluded the 60's because that's when I stopped the subject due to the time limit of 60. It's displayed as 'Mere' here $\endgroup$
    – Msjohansen
    Apr 25, 2016 at 12:13
  • $\begingroup$ This question might be related to : stats.stackexchange.com/questions/110801/… or stats.stackexchange.com/questions/69898/…. $\endgroup$
    – A Gore
    Apr 25, 2016 at 12:50
  • $\begingroup$ Your data are right censored at 60. You can't simply ignore that issue by treating them as being the same as an observed time of 60, nor can you omit those observations. $\endgroup$
    – Glen_b
    Apr 25, 2016 at 16:42

1 Answer 1


This is basically a failure time model. Subjects who did not complete the task by 60 seconds are right-censored (you know it took them longer than 60 seconds to finish but you do not know how long it would have taken them).

This can be modeled using log-rank tests, accelerated failure time models, proportional hazards, etc. Standard graphical presentation is the Kaplan-Meier curve (survival curve).

You could also do a conditional analysis. Analyze the effect of the predictors on the chance of success in 60 seconds. Then analyze the effect of the predictors on those who did succeed 60 seconds. I would recommend one of the models above, however.

  • $\begingroup$ So, if I were to use the proportional hazard model (Cox's) I would do it like this?: Time: the time datapoints in my data (how long it took them to complete the task - from 0 to 60) Event indicator: The censoring (0 for < 60 and 1 for 60) Variable: Group (is it group 1 (sample 380) or group 2 (sample 52)) Then run the Cox with those parameters and see if the group variable gives a p < 0.05? Is it "that easy" ? $\endgroup$
    – Msjohansen
    Apr 26, 2016 at 11:20
  • $\begingroup$ That sounds right to me. $\endgroup$
    – StatNoodle
    May 3, 2016 at 22:02

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