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I am measuring two unpaired variables $x$ and $y$ in two different conditions ($x$ and $y$ are magnitudes of some special magnetic signals). In the first condition, my hypothesis is that $\bar{x} > \bar{y}$ and in the second condition that $\bar{x} < \bar{y}$. Now that I have $N$ samples from both variables, how can I test whether my hypotheses are true? I am not sure if I can safely assume that $x$ and $y$ are independent from each other. Neither do I know from what kind of distributions they are sampled from. The sample size I have is small. I have read several introductions to statistics for the past few days, but never saw a worked out example for this kind of situations. All help appreciated.

Edit: Like Michael Mayer wrote, there is a binary grouping variable "condition". Sorry for a bit unclear question.

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user31197
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    I have a high degree of belief that exactly one of your hypotheses is true :-). Your data might be able to cast doubt on one of them. But what is "small"? To some people that means $N\le 2$; to others, it might mean $N\le 10^6$. – whuber Oct 07 '13 at 21:34
  • My sample size is $N = 10$. – user31197 Oct 07 '13 at 21:39
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    Hypotheses are about *populations*, not samples. You can check whether $\bar x < \bar y$ at a glance - no need for p-values or anything. – Glen_b Oct 07 '13 at 23:20
  • Given your small sample and lack of knowledge about the distributions, I'd suggest a permutation test. – Peter Flom Oct 07 '13 at 23:53
  • @whuber "N le 2"? :-) – Peter Flom Oct 07 '13 at 23:54
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    @Peter See http://stats.stackexchange.com/a/1836/919 for why $N=1$ can work. $N=2$ is usually needed in order to estimate the variability. In applications where observations are sufficiently expensive, $N \gt 2$ may be considered extremely *large*. (I work in a field where (a) private parties pay for observations which are (b) required by government regulations that (c) are viewed as a burden and, in the worst situations, (d) an observation (actually a monitoring station) can cost \$100K or more. If you want to tell my clients they need a larger $N$, you had better have a *great* reason!) – whuber Oct 08 '13 at 03:10
  • @whuber: Maybe there is an additional binary grouping variable "condition" involved which is expected to change the order of the true means(?). – Michael M Oct 08 '13 at 10:27
  • @Michael Thank you! Your interpretation sheds new light on the question. – whuber Oct 08 '13 at 14:46
  • Okay. To provide help, we would need to know much more about $x$ and $y$. What do they measure? Are they paired or unpaired? – Michael M Oct 09 '13 at 08:01

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A simple approach would be the following:

1) Take all observations sampled at random under condition A and obtain the relevant one sided p-value from Wilcoxon's rank sum test.

2) Do the same for the observations sampled under condition B.

3) If the smaller of the two p-values is below the level $\alpha/2$ and the other p-value is below $\alpha$, then your claim holds at the $\alpha$ level. (This would be the Bonferroni-Holm correction for multiple testing.)

Since the sample sizes are extremely low, you will get a "significant" result only if the signal is very strong.

Michael M
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