Interpretation of the Fisher-exact test I'm running some R code, where I would like to check if some data is independent. I could use a Chi-square test, which rejects independence. However I would like to use the Fisher-exact test. Once again p < 0.001 but how should I interpret this odds ratio?
The table consists of male 0 vs female 1 and overweight 1 versus no overweight 0.
> table(nhefs.adjust.nNA$overweight,
+       nhefs.adjust.nNA$sex,
+       dnn = c("overweight","sex"))
          sex
overweight   0   1
         0 407 535
         1 419 295
> fisher.test(table(nhefs.adjust.nNA$overweight,
+                   nhefs.adjust.nNA$sex,
+                   dnn = c("overweight","sex")))

Fisher's Exact Test for Count Data

data:  
p-value = 5.145e-10
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.4376387 0.6554322
sample estimates:
odds ratio 
 0.5358085 

Response to Aaron Zeng's answers
          death
overweight   Y   N
         Y 156 558
         N 165 777

odds ratio 
  1.316281 

Should this then be interpreted as:
The odds of death on persons with overweight is about 32% more then on persons without overweight?
This seems somewhat weird since 156 vs 165 seems to imply less risk?
 A: The fisher's exact test in R by default tests whether the odds ratio associated with the first cell being 1 or not.
That said, you can interpret the odds ratio 0.53 as: the odds of being male for a non-overwieght subject is 0.53 times that for an overweighted subject. Note the p-value is significant and the confidence interval doesn't contain 1. Therefore, you reject the null hypothesis of the odds ratio being 1.
However, you might want to exchange the columns and rows of the 2-by-2 contingency table, so that you could interpret "Sex" as an explanatory variable and "Overweight or not" as the response, which seems more nature. In this case, the estimate of the odds ratio should still be 0.53. But you can more naturally interpret it as: the odds of being non-overweight for a male person is 0.53 times that for a female person.
A: The odds ratio is a measure of how far from independence the 2x2 table is.
In your problem, the odds ratio is P(overweight|female)/(1-P(overweight|female)) (odds for overweight for females) divided by P(overweight|male)/(1-P(overweight|male)) (odds for overweight for males). Being a ratio of two sets of odds, it's called an odds ratio.
If the table is drawn from a population where the factors are independent, those two population odds will be equal, so the population odds ratio will be 1 (or the log odds-ratio will be 0).
The most common estimate of the odds ratio is obtained by replacing the probabilities by the sample proportions (which simplifies to cross-multiplying the counts and dividing the on-diagonal product by the off-diagonal product, or $ad/bc$ in the table $\begin{smallmatrix}a&b\\c&d\end{smallmatrix}$), which gives 0.5356 on your numbers, but that's not the estimator corresponding to a Fisher exact test. That uses the MLE of the odds ratio given the table margins (just as the Fisher test conditions on the table margins). In your case that gives the slightly larger estimate of 0.5358. 
So this particular odds ratio gives a way of estimating how far from independence your table is that more or less corresponds to Fisher exact test (whether the interval for it includes 1 isn't guaranteed to correspond to the result of the test but they generally go together).
