Inconsistency between Chi-sq and CI Estimation using Wald test I am trying to calculate the relative risk for being tested as positive for people aged >25 or <=25, and here is the result.
$data

        Negative Positive Total
  >25        115       11   126
  <=25       117        3   120
  Total      232       14   246

$measure
      risk ratio with 95% C.I.
        estimate      lower    upper
  >25  1.0000000         NA       NA
  <=25 0.2863636 0.08189025 1.001391

$p.value
      two-sided
       midp.exact fisher.exact chi.square
  >25          NA           NA         NA
  <=25 0.03815237   0.05147711 0.03500415

$correction
[1] FALSE

attr(,"method")
[1] "Unconditional MLE & normal approximation (Wald) CI"

One thing that puzzled me a lot is that why I get a statistically significant result from chi.square (I don't use fisher's exact test because NONE OF THE expected value is small than 5), but the risk ratio includes 1, how can I explain? Thanks.
CORRECTION
Sorry for the mistake, I rechecked the data and NONE of the expected cell is smaller than 5
 A: The Wald test for contingency tables is known to be misleading or conservative and the general advice is to prefer the Likelihood Ratio Test:
> library(vcd)
> mat <- matrix(c(115, 117, 11, 3), 2, 2)
> assocstats(mat)$chisq_tests
                      X^2 df   P(> X^2)
Likelihood Ratio 4.730771  1 0.02962760
Pearson          4.444973  1 0.03500415

Wald type tests are of the form $\frac{\hat{\theta} - \theta_0} {SE(\theta)}$ with the assumption that the difference is normal, which is often untenable with smaller samples or low counts in frequency tables.  
An intuitive reason for discounting the Wald test is that you're using the available data to estimate both $\hat\theta$ and the standard error to conduct the test.  Contrast this with the LRT which uses only one estimate from the data.  In a similar vein, I have another answer that tries to explain why the Wald CI can be kooky at times and why the LRT should generally be preferred.
You might also be interested in a paper that explored some of the ways in which the Wald test can be misleading in these types of analyses:


*

*Wald's Test as Applied to Hypotheses in Logit Analysis
A: I think suncoolsu is on the right track, though RR = 1 if and only if OR = 1, so those null hypotheses are in fact one and the same.
Pearson's chi-square test can be derived as a score test. In finite samples, score tests are, in general, better behaved than Wald tests, but less good than likelihood-ratio tests.
If you want your CIs and p-values to look more consistent, you may be best switching to a different (more accurate) way of calculating the CI. I see the R package you're using offers a small-sample adjustment or a bootstrap method as alternatives to the standard Wald method.
A: I am not sure what is the $H_o$ in the case of the hypothesis test. Usually its $H_o: OR=1$. If this is true then, your situation can arise because of the following reasons: 
The asymptotics that you comparing might be very different in the two cases. The test statistic's sampling distribution for relative risk and OR scenario can reach their asymptotic distribution differently, hence the discrepancy.
But if $H_o: RR=1$ then
As you mention yourself, one of the cell count is < 5. This condition should hold for all the cells for the Chi Sq test as well (AFAIK). Due to this the asymptotic Chi Sq distribution may not be completely accurate. 
The p-value may be calculated based on the NORMAL APPROXIMATION where as the CI may be calculated based on the CHI SQ APPROXIMATION. But this is total guess work. Which package are you using?
P-value and CI should agree as they are calculated based on the same $H_o$ (as you expect).
The main reason for the discrepancy may be the Asymptotic Nature of the Wald test. No one can tell when the asymptotics is true, all you can do is hope for the best and assume your sample to be "good enough", so that if you have "large enough" sample asymptotics are true. 
When such a situation arises, it is better to trust Non Parametric test like the Fisher's Exact Test because the parametric tests are based on asymptotic assumptions. FET being non-parametric (you are basically counting all possible tables that can occur as extreme cases conditioned on the present table), is less restrictive, hence more trustworthy.
Your point is valid that FET is not trust worthy as one cell count < 5, but in that case the Chi Sq distribution is not trustworthy as well. 
