I have a situation like this:

          Sample1   Sample2   Sample3    .....    
Gene1      0.38        0        0.13     .....    
Gene2      0.11       0.12       0       .....     
Gene3      0.55        0        0.9      .....    
.....     .....      .....     .....     .....   

I need a statistical test able to compare Gene by Gene and Sample by Sample the numerical values reported. In other words I would like to perform a statistical test able to tell me how statistically significant are the measurements: 0.38 (Gene1) and 0.11 (Gene2) for Sample1, as well as 0.55 (Gene3) and 0.38 (Gene1) for Sample1, and 0.55 (Gene3) and 0.11 (Gene2) for Sample1 and so on. 0 means "not_reported" so I suppose I will ignore comparisons with 0.

Can anyone help me please? I'm not an expert statistician.


1 Answer 1


You cannot get the significance between 1 sample on 1 pair of genes. You can get the significance of 1 pair of genes across all the samples, or one pair of samples across all genes (although this is probably not useful). Also, don't use 0 to mean "missing" - that is an invitation to disaster.

  • $\begingroup$ Hi Peter, thanks a lot for your suggestion..So in other words, the best way is to compare the profile of each gene across all samples with the profile of the other genes across all samples.Right? $\endgroup$
    – Fuv8
    Aug 28, 2013 at 11:58
  • $\begingroup$ Genetic analysis is a whole complex sub-specialty and I am not an expert. But what you initially asked to do is not possible $\endgroup$
    – Peter Flom
    Aug 28, 2013 at 12:05
  • $\begingroup$ Hi Peter, I may have misread you but it seems like you're saying that it's impossible to do a hypothesis test with a single observation. That's certainly not true. A simple example is testing a difference in means between two independent normally distributed observations with known variance (perhaps from knowledge of the measurement apparatus). $\endgroup$
    – Macro
    Aug 28, 2013 at 14:33
  • $\begingroup$ @Macro Certainly, if you know the variance and the distribution, you're right. But that's not the case here, as far as I can tell. (I think it's quite rare to know the variance - except where it is known by force). $\endgroup$
    – Peter Flom
    Aug 28, 2013 at 19:34

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