What test to perform if I have percentage data I have the following percentage data from multiple samples, and a corresponding reference percentage. Which test should I perform to test whether the percentage from the samples is different from the reference value.
The data -




Sample1
Sample2
...
Sample9
Reference




0.9
0.85
...
0.91
0.88




Also for all the individual samples I have the actual data from which the proportion has been calculated i.e. I have n_obs(number of observations) and tot_obs (total observations) from which the proportion = n_obs/tot_obs was calculated.
What test should I perform for doing this hypothesis testing.

Use can use a chi-square test for proportions (also called $z$-test). In R you could do it using prop.test. The null hypothesis would be $H_0: p_1 = \ldots = p_9 = \pi_0$ versus $H_1:$ at least one $p_i$ is different from $\pi_0$. In your case, $\pi_0 = 0.88$. Here is a nice explanation of the intricacies of the test.
Here is an example with simulated data with $p_1 = \ldots = p_8 = 0.88$ and $p_9 = 0.75$ with $n_i = 100$ for all samples. So the null hypothesis is clearly false.
pi0 <- 0.88
set.seed(142857)
# Total number of observations
n_tots <- rep(100, 9)
# Number of "successes"
x <- rbinom(9, n_tots, rep(c(pi0, 0.75), times = c(8, 1)))

# Chi-square test
prop.test(x, n_tots, p = rep(pi0, 9))

    9-sample test for given proportions without continuity correction

data:  x out of n_tots, null probabilities rep(pi0, 9)
X-squared = 43.561, df = 9, p-value = 1.7e-06
alternative hypothesis: two.sided
null values:
prop 1 prop 2 prop 3 prop 4 prop 5 prop 6 prop 7 prop 8 prop 9 
  0.88   0.88   0.88   0.88   0.88   0.88   0.88   0.88   0.88 
sample estimates:
prop 1 prop 2 prop 3 prop 4 prop 5 prop 6 prop 7 prop 8 prop 9 
  0.90   0.90   0.90   0.89   0.83   0.90   0.91   0.91   0.68

The $\chi^2$ value is $43.561$ and the corresponding $p$-value $<0.001$, providing a large amount of evidence against the null hypothesis that all proportions are $0.88$.
