Correct method for this should be by using all numbers in one test as follows:
> prop.test(c(817, 842), c(1551, 1446))
2-sample test for equality of proportions with continuity correction
data: c(817, 842) out of c(1551, 1446)
X-squared = 9.1169, df = 1, p-value = 0.002533
alternative hypothesis: two.sided
95 percent confidence interval:
-0.09175405 -0.01932406
sample estimates:
prop 1 prop 2
0.5267569 0.5822960
Distribution is generally not an issue in test of proportions. The P value is 0.002533
If you want to test one set of values with a given proportion, you should use binomial test:
> binom.test(817, 1551, p=0.582)
Exact binomial test
data: 817 and 1551
number of successes = 817, number of trials = 1551, p-value = 0.00001188
alternative hypothesis: true probability of success is not equal to 0.582
95 percent confidence interval:
0.5015559 0.5518565
sample estimates:
probability of success
0.5267569
>
> binom.test(842, 1446, p=0.527)
Exact binomial test
data: 842 and 1446
number of successes = 842, number of trials = 1446, p-value = 0.00002477
alternative hypothesis: true probability of success is not equal to 0.527
95 percent confidence interval:
0.5563847 0.6078723
sample estimates:
probability of success
0.582296
The P values are similar to what you got with prop.test. These also gives confidence intervals of both proportions. There are no means here, only proportions and they are significantly different.
Read more about these tests in R using commands ?prop.test and ?binom.test