How to compare proportions by having only the mean proportion and 95%CI (reported by a meta-analysis) I have a proportion of 31.6% (24/76) and I need to test whether this proportion is significantly different from a proportion reported by a meta-analysis study equal to 59% (95%CI: 53%; 65%). The meta-analysis found that proportion and the 95%CI from linear mixed models, pooling data from several other studies.
I suppose I cannot use the classical methodology for testing two proportions using normal approximation and estimating the standard deviation as p(1-p). I believe that the 95%CI reported by the meta-analysis should be used somehow.
Any suggestions on how I can perform this comparison?
Thank you very much.

An estimate of the standard error of your estimate $\hat\pi_1$ from a sample of $n$ is given by
$$
\hat\sigma_{\hat\pi_1}=\sqrt{\frac{\hat\pi_1(1-\hat\pi_1)}{n}}
$$
Assuming that in the meta-analysis the upper & lower $1-\alpha$ confidence bounds $\pi^\mathrm{u}_2$ & $\pi^\mathrm{l}_2$ are calculated in reasonably conventional fashion, then an estimate of the standard error of their estimate $\hat\pi_2$ is given by
$$\hat\sigma_{\hat\pi_2} = \frac{\pi^\mathrm{u}_2 - \pi^\mathrm{l}_2}{2\Phi^{-1}\left(1-\frac{\alpha}{2}\right)}$$
where $\Phi$ is the standard Gaussian distribution function. A Wald-type test of the null hypothesis that the difference between the two proportions is zero seems the most straightforward approach: asymptotically, under the null
$\frac{\hat\pi_2-\hat\pi_1}{\sqrt{\hat\sigma_{\hat\pi_1}^2+\hat\sigma_{\hat\pi_2}^2}}\sim\mathcal{N}\left(0,1\right)$
test <- function(x1, n1, p2, lb, ub, alpha){
  p1 <- x1/n1
  v1 <- p1*(1-p1)/n1
  v2 <- ((ub - lb)/(2*qnorm(1-alpha/2)))^2
  se <- sqrt(v1 +v2)
  z <- (p2-p1)/se
  pv <- 2*min(pnorm(z), pnorm(-z))
  return(c(test.stat=z ,p.value=pv))
}

test(24, 76, 0.59, 0.53, 0.56 , 0.05)
#   test.stat      p.value 
#5.090600e+00 3.569321e-07 

test(12, 76, 0.20, 0.12, 0.29 , 0.05)
#test.stat   p.value 
#0.6988176 0.4846661 

