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.

You could use $x = 24$ successes in $n = 76$ trials to test $H_0: p = .53$ against $H_a: p < 0.53,$ where $0.53$ is the lower end of the CI from
meta analysis. In R, the relevant exact binomial test
is as shown below. The null hypothesis is strongly
rejected with P-value nearly $0.$
binom.test(24, 75, .53, alt = "less")

        Exact binomial test

data:  24 and 75
number of successes = 24, number of trials = 75, 
 p-value = 0.0001944
alternative hypothesis: 
 true probability of success is less than 0.53
95 percent confidence interval:
  0.0000000 0.4195974
sample estimates:
probability of success 
                  0.32 

There may be more elegant solutions, but your data
with $\hat p = 0.32$ are so far from matching the results of the meta
analysis that the decision is hardly in doubt. [Turns out that there is a "more elegant solution"; see @Scortchi's Answer.]
Even a 99.9% one sided CI for $p$ indicates your
point estimate $\hat p = 0.32$ is not consistent with $p > 0.5035,$ below $0.53.$
binom.test(24,75, .53, alt="less", 
           conf.lev=.999)$conf.int
[1] 0.0000000 0.5034918
attr(,"conf.level")
[1] 0.999

