The three options that are proposed in riskratio() refer to an asymptotic or large sample approach, an approximation for small sample, a resampling approach (asymptotic bootstrap, i.e. not based on percentile or bias-corrected). The former is described in Rothman's book (as referenced in the online help), chap. 14, pp. 241-244. The latter is relatively trivial so I will skip it. The small sample approach is just an adjustment on the calculation of the estimated relative risk.
If we consider the following table of counts for subjects cross-classififed according to their exposure and disease status,
Exposed Non-exposed Total
Cases a1 a0 m1
Non-case b1 b0 m0
Total n1 n0 N
the MLE of the risk ratio (RR), $\text{RR}=R_1/R_0$, is $\text{RR}=\frac{a_1/n_1}{a_0/n_0}$.
In the large sample approach, a score statistic (for testing $R_1=R_0$, or equivalently, $\text{RR}=1$) is used, $\chi_S=\frac{a_1-\tilde a_1}{V^{1/2}}$, where the numerator reflects the difference between the oberved and expected counts for exposed cases and $V=(m_1n_1m_0n_0)/(n^2(n-1))$ is the variance of $a_1$. Now, that's all for computing the $p$-value because we know that $\chi_S$ follow a chi-square distribution. In fact, the three $p$-values (mid-$p$, Fisher exact test, and $\chi^2$-test) that are returned by riskratio() are computed in the tab2by2.test() function. For more information on mid-$p$, you can refer to
Berry and Armitage (1995). Mid-P
confidence intervals: a brief
review. The Statistician, 44(4),
417-423.
Now, for computing the $100(1-\alpha)$ CIs, this asymptotic approach yields an approximate SD estimate for $\ln(\text{RR})$ of $(\frac{1}{a_1}-\frac{1}{n_1}+\frac{1}{a_0}-\frac{1}{n_0})^{1/2}$, and the Wald limits are found to be $\exp(\ln(\text{RR}))\pm Z_c \text{SD}(\ln(\text{RR}))$, where $Z_c$ is the corresponding quantile for the standard normal distribution.
The small sample approach makes use of an adjusted RR estimator: we just replace the denominator $a_0/n_0$ by $(a_0+1)/(n_0+1)$.
As to how to decide whether we should rely on the large or small sample approach, it is mainly by checking expected cell frequencies; for the $\chi_S$ to be valid, $\tilde a_1$, $m_1-\tilde a_1$, $n_1-\tilde a_1$ and $m_0-n_1+\tilde a_1$ should be $> 5$.
Working through the example of Rothman (p. 243),
sel <- matrix(c(2,9,12,7), 2, 2)
riskratio(sel, rev="row")
which yields
$data
Outcome
Predictor Disease1 Disease2 Total
Exposed2 9 7 16
Exposed1 2 12 14
Total 11 19 30
$measure
risk ratio with 95% C.I.
Predictor estimate lower upper
Exposed2 1.000000 NA NA
Exposed1 1.959184 1.080254 3.553240
$p.value
two-sided
Predictor midp.exact fisher.exact chi.square
Exposed2 NA NA NA
Exposed1 0.02332167 0.02588706 0.01733469
$correction
[1] FALSE
attr(,"method")
[1] "Unconditional MLE & normal approximation (Wald) CI"
By hand, we would get
$\text{RR} = (12/14)/(7/16)=1.96$, $\tilde a_1 = 19\times 14 / 30= 8.87$, $V = (8.87\times 11\times 16)/ \big(30\times (30-1)\big)= 1.79$, $\chi_S = (12-8.87)/\sqrt{1.79}= 2.34$, $\text{SD}(\ln(\text{RR})) = \left( 1/12-1/14+1/7-1/16 \right)^{1/2}=0.304$, $95\% \text{CIs} = \exp\big(\ln(1.96)\pm 1.645\times0.304\big)=[1.2;3.2]\quad \text{(rounded)}$.
The following papers also addresses the construction of the test statistic for the RR or the OR:
- Miettinen and Nurminen (1985). Comparative analysis of two rates. *Statistics in Medicine, 4: 213-226.
- Becker (1989). A comparison of maximum likelihood and Jewell's estimators of the odds ratio and relative risk in single 2 × 2 tables. Statistics in Medicine, 8(8): 987-996.
- Tian, Tang, Ng, and Chan (2008). Confidence intervals for the risk ratio under inverse sampling. Statistics in Medicine, 27(17), 3301-3324.
- Walter and Cook (1991). A comparison of several point estimators of the odds ratio in a single 2 x 2 contingency table. Biometrics, 47(3): 795-811.
Notes
- As far as I know, there's no reference to relative risk in Selvin's book (also referenced in the online help).
- Alan Agresti has also some code for relative risk.
Best Answer
If you do a two-sided level 0.05 test of hypothesis that the relative risk is different from 1 and get a p-value less than 0.05 then this is equivalent to a two-sided 95% confidence interval that does not contain 1. So given the p-value of 0.049 you would expect that 1 would fall outside the interval. What are the possible explanations? 1. The confidence interval is at a level higher than 95% 2. There is an error in the calculation of the p-value or the confidence interval. 3. The confidence interval is not constructed by inverting the Fisher test (e.g. a chi square approximate test was used for the confidence interval).
I think 3 is very likely to be the reason. The chi square interval is easier to calculate than one using Fisher's test. Note that the significance level is very close to 0.05 and the upper limit of the confidence interval is just slightjy above 1.0.