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Just a quick introduction, I'm rubbish at stats. I asked a question before which was very well answered but I have since modified the experiment in a way that confuses me as to which test I should use.

So in the previous question we finished concluding that an independent t-test would be suffice as I am comparing the effect of irradiation on each leaf so, one half would be the control or the "before", and the irradiated half would be the "after" treatment. Well, I got my results, but I didn't map out which irradiated half went with what control as it was very difficult to keep track so I ended up with a jumbled set of controls to a jumbled set of irradiated.

So I made the following changes to my experiment:

1) Instead of having 5 circles per half of leaf, I instead opted for just 1. This meant it was quicker to get results as things like overlapping circles reducing light intake for the leaf discs on bottom didn't happen. So I would initially cut all my leaves in half. take one sample from one half, and whilst waiting for photosynthesis to take place I would irradiate the other halve(s), which brings me on to my second change..

2) I did 3 syringes at the same time instead of just 1 quadrupling my speed. (That's what I meant when I said I did four at the same time), so 4 control, whilst waiting, irradiate other halves of leaves for a minute and switch once photosynthesis finished, then switch again to another batch of 4 leaves and repeat. I ended up with 16 results either side

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  • $\begingroup$ If you have jumbled control and treatment labels, nothing can rescue it. On your last sentence - standard deviation of what? $\endgroup$ – Glen_b Apr 15 '13 at 0:22
  • $\begingroup$ Oh damn :L Ok, well I'll explain what I did in further detail. So I have to work out "If applicable" things like mean and Standard Deviation. So can I do that with my results? All my values are above 0 (since I'm using a stopwatch to measure). Again, so that means I can't do a statistical test? Surely if I've got n number of results for non-irradiated with an equal n number of control, there can be another test I can do?? $\endgroup$ – Adam Apr 15 '13 at 0:32
  • $\begingroup$ I didn't really follow your explanation in further detail, sorry, perhaps you can explain in more detail in an edit to your question. Where you say - "surely if I've got n number of results for non-irradiated with an equal n number of control" - you can compare them only if you know which is which. It's possible I misunderstood something, I suppose. Do you have some of each group where you do know which treatment group they're from? $\endgroup$ – Glen_b Apr 15 '13 at 0:40
  • $\begingroup$ Ohh yes yes, I 100% know which is control and which is irradiated. It's just, for my test I would have had to compare a before and after for EACH INDIVIDUAl leaf (paired t-test). But I didn't keep track of which control/irradited was for which leaf (as I did 4 at a time). So what's a broad statistical test I could use to compare the difference between the control and irradiated (Again I can explain if it's not clear. It is difficult to describe, sorry). $\endgroup$ – Adam Apr 15 '13 at 0:52
  • $\begingroup$ Oh, you mean you just lost the pairing? Then you do a two-sample test instead of a paired. You'll lose some power but all is not lost. I can probably give an answer now. Still not clear about computing standard deviation from a Mann-Whitney. Why not compute them from the samples? $\endgroup$ – Glen_b Apr 15 '13 at 0:58
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It sounds like you've lost the information that would let you identify the inherent pairing in your data. This is unfortunate and will likely cost you power, but you can still do a test.

With the loss of pairing, you're in a two-sample situation.

You seem to suggest you're thinking of a Mann-Whitney test. That may well be suitable depending on the specific hypothesis you're interested in testing and the specific additional assumptions you may be willing to make.

If you're willing to assume identical shapes apart from possible location shift, you can use a Mann-Whitney to test means, for example. If you want a more general location test, you don't necessarily need to assume identical shape. It sounds like your data are potentially fairly discrete - are you likely to have lots of ties?

Another possibility is - especially with such small samples - to consider a permutation test of whatever is quantity is of interest.

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  • $\begingroup$ The link I gave you was a little wrong. So I initially was going to do the technical replicates thing, but I realised that would take a long long time. So instead, I made the following changes: $\endgroup$ – Adam Apr 15 '13 at 1:14
  • $\begingroup$ Perhaps you should edit the information into your question. $\endgroup$ – Glen_b Apr 15 '13 at 1:16
  • $\begingroup$ I think I've lost you >.< $\endgroup$ – Adam Apr 15 '13 at 1:40
  • $\begingroup$ Thanks for the extra info. What would be the null and alternative hypothesis you'd be most interested in testing - in plain words rather than formal notation. (Are you able to put up the data, by the way?) There are several further questions in my partial answer. $\endgroup$ – Glen_b Apr 15 '13 at 2:46
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    $\begingroup$ Yes, I agree, if by "equal shape" to mean identical distribution apart from ... shift Then of course, the test is just the test of that shift (and hence of any quantile, including that one corresponding to mean). $\endgroup$ – ttnphns Apr 15 '13 at 10:52

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