In most meta-analysis of odds ratios, the standard errors $se_i$ are based on the log odds ratios $log(OR_i)$. So, do you happen to know how your $se_i$ have been estimated (and what metric they reflect? $OR$ or $log(OR)$)? Given that the $se_i$ are based on $log(OR_i)$, then the pooled standard error (under a fixed effect model) can be easily computed. First, let's compute the weights for each effect size: $w_i = \frac{1}{se_i^2}$. Second, the pooled standard error is $se_{FEM} = \sqrt{\frac{1}{\sum w}}$. Furthermore, let $log(OR_{FEM})$ be the common effect (fixed effect model). Then, the ("pooled") 95% confidence interval is $log(OR_{FEM}) \pm 1.96 \cdot se_{FEM}$.
Update
Since BIBB kindly provided the data, I am able to run the 'full' meta-analysis in R.
library(meta)
or <- c(0.75, 0.85)
se <- c(0.0937, 0.1029)
logor <- log(or)
(or.fem <- metagen(logor, se, sm = "OR"))
> (or.fem <- metagen(logor, se, sm = "OR"))
OR 95%-CI %W(fixed) %W(random)
1 0.75 [0.6242; 0.9012] 54.67 54.67
2 0.85 [0.6948; 1.0399] 45.33 45.33
Number of trials combined: 2
OR 95%-CI z p.value
Fixed effect model 0.7938 [0.693; 0.9092] -3.3335 0.0009
Random effects model 0.7938 [0.693; 0.9092] -3.3335 0.0009
Quantifying heterogeneity:
tau^2 < 0.0001; H = 1; I^2 = 0%
Test of heterogeneity:
Q d.f. p.value
0.81 1 0.3685
Method: Inverse variance method
References
See, e.g., Lipsey/Wilson (2001: 114)
First of all, some terminology. In my opinion, it is arbitrary whether we call it the "Q-test of homogeneity" or the "Q-test for heterogeneity". Under the null hypothesis, we assume homogeneity, so calling it the Q-test of homogeneity would emphasize that we are testing this assumption. But the alternative hypothesis states that the true effects/outcomes are heterogeneous, so we could also say that we using it to test for (whether) heterogeneity (is present). Some may disagree -- but in the end, there is just the Q-test, whatever we call it.
As for using the Q-test when there are only two studies: That is equivalent to testing the null hypothesis that the true effect for the first study is the same as the true effect for the second study. That should be clear if we write down the null hypothesis for the Q-test, namely $$H_0: \theta_i = \theta \mbox{ for all } i = 1, \ldots, k,$$ which for $k = 2$ is identical to $$H_0: \theta_1 = \theta_2,$$ where $\theta_i$ denotes the true effect/outcome for the $i$th study.
We can also demonstrate this with an example. I'll use R with the metafor package for this.
library(metafor)
yi <- c(.14, .75) # observed effect size estimates
vi <- c(.083, .042) # corresponding sampling variances
di <- c(0, 1) # dummy variables to distinguish the two studies
### fixed-effects model
rma(yi, vi, method="FE")
The results are:
Fixed-Effects Model (k = 2)
Test for Heterogeneity:
Q(df = 1) = 2.9768, p-val = 0.0845
Model Results:
estimate se zval pval ci.lb ci.ub
0.5450 0.1670 3.2638 0.0011 0.2177 0.8723 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
So, we find $Q(1) = 2.9768$ with $p = 0.0845$. Now let's fit a (fixed-effects) meta-regression to these data, adding the dummy variable to the model that distinguishes the first from the second study:
### meta-regression model
rma(yi, vi, mods = ~ di, method="FE")
This yields:
Fixed-Effects with Moderators Model (k = 2)
Test for Residual Heterogeneity:
QE(df = 0) = 0.0000, p-val = 1.0000
Test of Moderators (coefficient(s) 2):
QM(df = 1) = 2.9768, p-val = 0.0845
Model Results:
estimate se zval pval ci.lb ci.ub
intrcpt 0.1400 0.2881 0.4859 0.6270 -0.4247 0.7047
di 0.6100 0.3536 1.7253 0.0845 -0.0830 1.3030 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Note that the p-value of the dummy variable is exactly the same as that of the Q-test. If we square the z-value, we in fact get the chi-square value of the Q-test. That is actually the value of the $Q_M$ test, which is the omnibus test of all coefficients (except for the intercept), but since there is only one coefficient in the model (the one for the dummy variable), that is identical to just testing whether the coefficient for the dummy variable is significantly different from zero.
So, "mechanically speaking", that all works out as expected. Is this a valid procedure? I don't see anything wrong with this -- you are just testing whether the effects of study 1 and study 2 are significantly different from each other or not.
Power is of course always an issue we have to keep in mind. And yes, power may be low. So, quite importantly, if you do not find a significant difference, you should be very cautious how you interpret that. In particular, it simply means that you do not have sufficient evidence to reject the null hypothesis, but the true effects could very well still differ from each other.
On the other hand, if you use this approach to carry out lots of tests, it is in fact likely that some significant findings are just Type I errors. So, you also need to be cautious in interpreting any significant findings, especially if you do not correct for multiple testing.
Best Answer
You could try using confidence intervals established using the Q-profile method outlined by Viechtbauer in his 2007 paper "Confidence intervals for the amount of heterogeneity in meta-analysis" available here and used in his R package metafor. I have just tried a toy example with $I^2=0$ and two studies and got confidence intervals.