I'm doing a meta-analysis of some studies that report results variously as odds ratios, hazards ratios or rate ratios (all with confidence intervals). Is there any way to combine these together/convert between them so that I can do a meta-analysis of all the studies?
Meta-Analysis – How to Conduct Meta-Analysis on Studies Reporting Various Ratios Like Odds, Hazards, and Rate Ratios
meta-analysisodds-ratio
Related Solutions
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)
Hi Erica and welcome to the site. Have a look at this (page 3) document and this paper. The basic formula for the conversion is $$ d=\mathrm{LogOR}\times \frac{\sqrt{3}}{\pi} $$ Applying the delta-method, we get the following expression for the the variance of $d$ (the standard error of $d$ is just the square root of its variance): $$ \mathrm{Var}_{d}=\mathrm{Var}_{\mathrm{LogOR}}\times \frac{3}{\pi^{2}} $$
Where $\mathrm{LogOR}$ denotes the log of the odds ratio and $\mathrm{Var}_{\mathrm{LogOR}}$ denotes the variance of the log odds ratio.
To get the variance of the log odds ratio, you can use the information given by the confidence interval. To get the standard error of the log odds ratio, use the following formula: $$ \mathrm{SE}_{\mathrm{LogOR}}=\frac{\log(\mathrm{CI}_{upper}) - \log(\mathrm{CI}_{lower})}{2\times z_{1-\alpha/2}} $$
Where $\mathrm{CI}_{upper}$ denotes the upper and $\mathrm{CI}_{lower}$ the lower bound of the confidence interval for the odds ratio (as given in the papers) and $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile of the standard normal distribution. For a 95%-CI, $\alpha = 0.05$ and $z_{1-\alpha/2}\approx 1.96$. To get the variance of the log odds ratio, just square the standard error. In your example, the standard error of the log odds ratio is about $0.247$. Hence, the variance of the log odds ratio is $0.247^{2}\approx0.061$. To calculate the confidence interval of $d$, you need the standard error of $d$, which is simply $$ \mathrm{SE}_{d}=\mathrm{SE}_{\mathrm{LogOR}}\times\frac{\sqrt{3}}{\pi} $$
Best Answer
It depends on why they were doing so, and what additional information you might have to work with - like do you have specific cell-counts that you might calculate your own effect measures from. However, some initial thoughts:
cases/person-time, then that is actually a hazard ratio estimate as well. It's just a hazard ratio made under the assumption of both proportional and constant hazards. You can convert those directly to an HR, but again, I'd look at study heterogenity by which measure it reported, as rate ratios and a HR that came out of something like a Cox model are performed under different assumptions.