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.

Because some of my proportions are next to 0 or 1, the Wald-type test can give biased results. To overcome this issue, I simulated two beta distributions with specified parameters and applied a bootstraped difference test to obtain the p-value.
Beta.t.test <- function(x1, n1, med2, lb, ub, M, Alpha){
  #Function to estimate Beta parameters
  estBetaParams <- function(mu, var) {
    alpha <- ((1 - mu) / var - 1 / mu) * mu ^ 2
    beta <- alpha * (1 / mu - 1)
    return(params = list(alpha = alpha, beta = beta))
  }

  #Data from group 1: absolute frequency
  med1 <- x1/n1 ; var1 <- med1*(1-med1)/n1

  #Data from group 2: meta analysis result with 95%CI limits
  var2 <- ((ub - lb)/(2*qnorm(1-Alpha/2)))^2

  #Estimating beta parameters for both groups
  Beta1 <- estBetaParams(med1,var1)
  Beta2 <- estBetaParams(med2,var2)

  #Ploting
  X <- seq(0,1,0.001)
  B1 <- dbeta(X, Beta1$alpha, Beta1$beta)
  B2 <- dbeta(X, Beta2$alpha, Beta2$beta)
  plot(X,B1,type = "l"); lines(X,B2,col = 2)

  #Bootstrap test
  set.seed(1)
  RB1 <- rbeta(M, Beta1$alpha, Beta1$beta)
  RB2 <- rbeta(M, Beta2$alpha, Beta2$beta)
  if (med1 > med2){
    DifB <- RB1 - RB2
  } else {DifB <- RB2 - RB1}
  #Percentile 95% CI for the difference
  quantile(DifB, probs = c(0.025, 0.975))

  p.value <- 2*sum(DifB <= 0)/M ; p.value
  return(p.value)
}

When applied to the data far from 0 or 1, we get a very close p.value as reported by Scortchi
> test(12, 76, 0.20, 0.12, 0.29 , 0.05)
test.stat   p.value 
0.6988176 0.4846661 
> Beta.t.test(12,76,0.20,0.12,0.29,99999,0.05)
[1] 0.4823248

But with proportions near 0 or 1, the results can be a little distinct
> test(2, 50, 0.10, 0.07, 0.13 , 0.05)
test.stat   p.value 
1.8952016 0.0580657 
> Beta.t.test(2, 50, 0.10, 0.07, 0.13, 99999, 0.05)
[1] 0.09390094

