st: RE: metan command STATA IC 12.1

[Timothy Mak <tshmak@hku.hk>]

<>
The problem is that -metan- is not geared towards doing a meta-analysis of proportions. You can of course do it by supplying estimates and confidence intervals. 

However, the assumption behind these simple meta-analysis models is that the estimates are Normally distributed with standard deviation either supplied by your standard errors, or back-calculated from your confidence intervals. To the extent that your sample is small and your p is near 0 or 1, the Normal assumption cannot be correct. Your estimates, if given in the logit scale, (i.e. if you transform your p and your L95 and U95 by logit(p), logit(L95), logit(U95)) will be more Normally distributed, and it will be more appropriate if you then do your meta-analysis on these transformed estimates and CI. You can derive your overall (pooled) p and CI by back-transforming the overall estimate and CI. (using the -invlogit- function). 

Note that if you assume a fixed-effects model, then a very valid method (the maximum-likelihood estimate) of the pooled p is simply sum(cases)/sum(total) = (9+11+4+3)/(51+45+30+12)= 0.195. CI for this estimate can be derived by  
cii `=(51+45+30+12)' `=(9+11+4+3)'

For a random-effects model, it's a bit more tricky. Note that you can reproduce the ML estimate above by running -glm-, e.g.: 
glm cases, fam(bin total)
di invlogit(_b[_cons])

Therefore, if only there were -xtglm-, you should be able to get a random-effects estimate. But there isn't. 
The easiest way to get round this is to transform your data to the 0/1 format and run -xtlogit-:

gen Study = _n
expand cases, gen(Cases)
gen noncases = total - cases
expand noncases, gen(Noncases)
gen original = Cases == 0 & Noncases == 0
xtset Study
xtlogit Cases if original == 0, re
di invlogit(_b[_cons])

Note that these would all give you different estimates, since you're using different methods. The -glm- and -xtlogit- methods are basically Maximum Likelihood (ML) estimates. -metan- uses DerSimonian-Laird's method of moments. ML is generally more efficient. It also avoids the Normal approximation to the distribution of the estimates of logit(p). Of course, the downside is that -glm- and -xtlogit- do not give you the nice forest plots that -metan- does. 

HTH, 
Tim




-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of shadi Kalantarian
Sent: 11 July 2013 03:55
To: statalist
Subject: st: metan command STATA IC 12.1

I am doing a meta-analysis in which I will need to generate pooled
prevalence rates. For each study, prevalence and its 95% confidence
interval was calculated using exact binomial (cii command, exact level
(95)). I used metan command in two different ways:

1) metan p se, random
2) metan p U95 L95, random

When I use the first command (the one with standard error) the
reported confidence intervals in the forest plot are completely
different from those calculated with cii command. Why does this
happen? and does this mean I should not use metan p se command?

study  total cases SE (standard error) p (prevalence)  L95     U95
one    51     9       0.053382      0.176471  0.084009    0.308726
two    45     11     0.064064      0.244444  0.128823    0.395371
three  30     4       0.062063      0.133333  0.037554    0.307219
four    12     3       0.125              0.25          0.054861    0.571858



. metan p l95 u95 , random

           Study     |     ES    [95% Conf. Interval]     % Weight
---------------------+---------------------------------------------------
1                    |  0.176       0.084     0.309         38.55
2                    |  0.244       0.129     0.395         27.40
3                    |  0.133       0.038     0.307         26.77
4                    |  0.250       0.055     0.572          7.28
---------------------+---------------------------------------------------
D+L pooled ES        |  0.189       0.119     0.259        100.00
---------------------+---------------------------------------------------

  Heterogeneity chi-squared =   1.58 (d.f. = 3) p = 0.664
  I-squared (variation in ES attributable to heterogeneity) =   0.0%
  Estimate of between-study variance Tau-squared =  0.0000

  Test of ES=0 : z=   5.31 p = 0.000


. metan p se , random

           Study     |     ES    [95% Conf. Interval]     % Weight
---------------------+---------------------------------------------------
1                    |  0.176       0.072     0.281         38.22
2                    |  0.244       0.119     0.370         26.54
3                    |  0.133       0.012     0.255         28.27
4                    |  0.250       0.005     0.495          6.97
---------------------+---------------------------------------------------
D+L pooled ES        |  0.187       0.123     0.252        100.00
---------------------+---------------------------------------------------

  Heterogeneity chi-squared =   1.84 (d.f. = 3) p = 0.605
  I-squared (variation in ES attributable to heterogeneity) =   0.0%
  Estimate of between-study variance Tau-squared =  0.0000

  Test of ES=0 : z=   5.68 p = 0.000

Thank you,
Shadi Kalantarian MD MPH
Research Fellow
Massachusetts General Hospital
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