Re: Is constructing a contingency table from odds ratio's (including 95%CI) possible

On Jul 30, 9:20 am, nyika <NDKr...@xxxxxxxxx> wrote:
On 29 jul, 21:20, nyika <NDKr...@xxxxxxxxx> wrote:



On 27 jul, 21:18, Bruce Weaver <bwea...@xxxxxxxxxxxx> wrote:

nyika wrote:
For a meta analyses i was able to subtract the following information
from an article:
1) the Odds ratio on poor outcome, with 95% confidence intervals
2) the number of patients with treatment and the number of patients
without treatment. (so, also the total number of patients).
I was wondering if anyone knows if it is possible to reconstruct a
contingency table with this information. (is there any software that
can do this? )

regards
ND Kruyt

On the off-chance that you want the contingency tables to plug into your
meta-analysis program (and you don't need them for anything other than
that), you can do the meta-analysis without the cell counts. All you
need for each study is a measure of effect and its variance. In your
case, the measure of effect would be ln(OR), the natural log of the odds
ratio. And its variance can be worked out from the 95% CI.

Let Y = ln(OR)
SE(Y) = [ln(95% CI upper limit) - Y] / 1.96
Var(Y) = SE(Y)^2

The Fleiss article cited below gives details on how to perform the
meta-analysis. I have a Windows program I can send if you wish--just
drop me a line via e-mail. (Unfortunately, I don't have the article as
a PDF, or I'd send it too.)

Fleiss, JL. The statistical basis of meta-analysis. Statistical
Methods in Medical Research 1993;2:121-145.

--
Bruce Weaver
bwea...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx/wv/bwhomedir
"When all else fails, RTFM."

Thank s to all of you,
I'm not a mathematician myself, so i will need some time to work it
out. Perhaps i could give the exact figures from the article?
I will look them up this week.

thanks again
regards
ND Kruyt- Tekst uit oorspronkelijk bericht niet weergeven -

- Tekst uit oorspronkelijk bericht weergeven -

Dear all,
I tried to work out the equation, but my lack of mathematical
backround was too much.
The numbers i have are the following: (from a contingency table):

A=?
B=?
A+B=36
C=?
D=?
C+D= 67

Odds ratio: 2.4; 95% CI: 1.0 - 5.4. (these figures are probably
rounded of).
If it is easy for you and if you would have some spare time i would
appreciate it if you could show me how to work this out. Regards
NDKruyt

It is easy to develop a table-generating formula:
for any A, you can calculate B=n1-A, then find D so that the OR works
(n1=36, n2=67, or=2.4):
D=or*n2*B/(A+or*B), then C=n2-D. Of course, you will have to round C
and D. I rounded them to the nearest integer, but perhaps due to the
funny nature of discreteness, rounding the "wrong" way could also be
interesting to look at.
Now for each of the 37 possible tables calculate the OR and the 95% CI
(setting up a formula in Excel, for example).

I found two realistic candidates:
A=18, B=18, C=20, D=47, OR=2.350, 95% CI: 1.018 - 5.427
and
A=21, B=15, C=25, D=42, OR=2.352, 95% CI: 1.029 - 5.378
None of the other options would give 5.4 for the upper limit when
rounded. This calculation assumes that they used the asymptotic normal
formula for the confidence interval without any corrections.

However, as noted by a previous poster, you might not need the actual
values for the meta-analysis at all.

Aniko

.

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