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I have two groups I want to compare, however, I only have one data point in the first group compared to 450 data points in the second group.

Here's the story. I have a patient that has a mutation in one of his miRNA, and I'm trying to find if it affects the expression of the miRNA's target genes. So, I took one of the target genes and I compared its FPKM value (normalised expression) (group 1) to the genes that are expressed in patients that don't have this mutation (group 2).

I thought about doing a T-test, but obviously can't perform a T-test with what I have. Do I need to generate "artificial" data in my first group in accordance with the elements from my second group? Is there a package in R or a tool that could help me generate this data? I tried searching for solutions but I only found options for re-sampling already existing data or creating new random data, which I do not want to do.

Thank you for your help!

Aaron left Stack Overflow
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snowy_squirrel
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    Well, you can't really do a t-test on generated data either. Perhaps you could back up and give the whole story; that's more likely to result in reasonable answers. (And whether or not you could perform a t-test on what you have may not be as obvious as you think, either.) – Aaron left Stack Overflow Aug 19 '13 at 18:30
  • @Aaron: I added more details. I hope it offers a more complete story. – snowy_squirrel Aug 19 '13 at 18:40
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    It helps, but the people who might answer probably won't find it without a new title. Also, it would help to put the edit first. (I'm going to edit it for you; if I get anything wrong, please feel free to edit back.) – Aaron left Stack Overflow Aug 19 '13 at 20:48
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    "*but obviously can't perform a T-test with what I have*" ... not so fast; that's [not so obvious](http://stats.stackexchange.com/questions/44475/is-there-a-statistical-test-to-compare-two-samples-of-size-1-and-3/44476#44476)! Your problem has already been solved, in an even worse case than yours. No need to simulate, you can do a t-test if you're prepared to assume that under the null, the variance for your patient is the same as the variance in the other group (which is just saying 'under the null, this patient is like everyone else' - if the mutation has no effect, that'd be reasonable) – Glen_b Aug 20 '13 at 00:14
  • @Aaron: Your edit is so much better than mine, thank you! – snowy_squirrel Aug 20 '13 at 14:58
  • @Glen_b : Thanks! The link to the other question is very useful – snowy_squirrel Aug 20 '13 at 14:59
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    (Too slow answering => comment): the problem can be sees as either one instance of the mutation group - or as an "outlier"/"novelty" problem: comparing a group to single case(s) which you do not assume to belong to a well-defined second group. With that point of view, have a look at distance metrics like [Mahalanobis-Distance](http://en.wikipedia.org/wiki/Mahalanobis_distance). I think you may be able to learn from the projection involved in calculating that distance. A more general approach would be [one-class classification](http://en.wikipedia.org/wiki/One-class_classification). – cbeleites unhappy with SX Aug 20 '13 at 21:19
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    (@whuber: I would have linked the other question as related but not exactly as a duplicate - after all it was possible to replicate experiments there. Obtaining more patients with this mutation may be really impossible. So maybe that other approaches are needed here.) – cbeleites unhappy with SX Aug 20 '13 at 21:28

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This reminds me of situations where a reference interval is often used. You basically just take the 5th and 95th percentile of your data (for a 90% reference interval), and look to see if your patient of interest is inside that interval.

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