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I have the frequency distribution of muscle fiber size from several control and several problem animals and I'd like to compare both groups, but all the statistical tests I know for frequency distributions are for a single control vs problem.

Can anyone help me with this?

Thank you in advance.

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  • $\begingroup$ Compare in what way? If muscle fiber size is discrete and finite you can use a Chi-square test to compare frequencies of two samples. $\endgroup$ Feb 15 '19 at 9:24
  • $\begingroup$ The problem is that I don't have two samples but instead several control samples (from different control animals) to compare with several problem samples (from different problem animals of the same population). $\endgroup$
    – user237717
    Feb 15 '19 at 10:40
  • $\begingroup$ So is merging all these control samples and abnormal samples possible? Are they similar? $\endgroup$ Feb 15 '19 at 11:08
  • $\begingroup$ It's possible and, if I consider all the controls as a single sample and all the abnormal as another, the chi-square test gives highly significant differences, but I'm not an expert in statistics and I was unsure if that was a correct approach. $\endgroup$
    – user237717
    Feb 15 '19 at 11:34
  • $\begingroup$ We need more details about these samples, how they were obtained, are the animals different, when were they taken, how big they are, .... $\endgroup$ Feb 15 '19 at 11:41
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To summarize from the comments: there are two groups of animals, A and B (control and abnormal). You took 5 different animals from each group and calculated the frequency of 60 to 70 muscle fibers from each of them. You now want to compare the frequencies of the 60 to 70 muscle fibers between the two groups.

Two possible options:

  1. perform 60 to 70 independent t-tests, in each comparing every muscle fiber between A and B; afterwards you should adjust all the p-values
  2. a linear model, this will allow you to use all the data at once, afterwards you will have to build appropriate contrasts to test all your hypothesis (not that hard)
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