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I have N tissue samples, labelled as either A or B. 2 measurements are taken from each sample.

The data is not normally distributed.

I'm interested in 2 things: firstly, whether there is significant difference between the means of the 2 classes A and B (treating the repeated measures as independent data instances). This seems simple enough.

However I'm also interested to know how well the 2 measurements taken from each sample tend to agree. This sounds like intra-class coefficient to me, however each "class" in this case only contains the 2 repeated measures.

Would the stndard calculation of ICC suffice here? Should I repeat the test independently for the groups A and B or can I lump them together?

Thanks so much, I hope this isn't too basic. let me know if I can provide any other information.

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  • $\begingroup$ When you say the data are not normal, is this known a-priori e.g. the data cannot be negative? If not, how non-normal are the data? The following histograms: A data (n = N), B data (n = N) and the N means of A and B for each case; are helpful? And what is the value of N? $\endgroup$ – Heteroskedastic Jim May 16 at 16:37
  • $\begingroup$ There isn't an obvious reason why the data shouldn't be normal, but the histograms don't appear to be. I was just hoping for a method that didn't require normality. For simplicity, let's say N=50, |A|=|B|=25. Then there are a total of 100 data points, 2 measurements for each element of A,B. I'm trying to keep things simple/abstract - I just want to know if ICC can be applied to a large number (50) of small (2) groups of data. $\endgroup$ – Will May 16 at 17:58
  • $\begingroup$ What is wrong with a simple correlation coefficient in place of the ICC? $\endgroup$ – Heteroskedastic Jim May 16 at 18:11
  • $\begingroup$ Ok I guess it was a dumb question and the ICC does apply. I guess I'd only seen ICC used when the data are organized into a smaller number of large groups, but I guess there's no reason why it shouldn't extend to my case $\endgroup$ – Will May 16 at 18:18
  • $\begingroup$ thanks for your help $\endgroup$ – Will May 16 at 18:20
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As a measure of agreement, a simple correlation coefficient should suffice.

For whether the data have different means, you should not ignore that they are paired. So the simple strategy replicating the paired t-test would be to subtract measurement 1 from measure 2 and test whether the difference is statistically different from zero. So a reasonable strategy to adjust for non-normality would be:

  • Calculate the difference score
  • Perform a one-sample t-test on the difference score using bootstrapping
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  • $\begingroup$ Thanks! when you say a simple correlation coefficient, you do mean ICC though correct? $\endgroup$ – Will May 16 at 18:26
  • $\begingroup$ I mean the Pearson correlation coefficient. $\endgroup$ – Heteroskedastic Jim May 16 at 18:28

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