I assume that the $2 \times 2$ frequency table with marginal totals is as follows:
In Sub Out Total
------------------------------
Succ 200 300 500
Fail 800 3700 4500
------------------------------
Total 1000 4000 5000
It is better to compare those in the subgroup with those outside the subgroup
than to compare those in the subgroup with everyone in the total group. (The
latter comparison counts those in the subgroup twice and so muddies the comparison. Also see links in Comments by @Glen_b.)
In R, the function prop.test tests the null hypothesis that those in and out of the subgroup
have the same Success rate (against the alternative that Success rates differ).
The null hypothesis is rejected at any reasonable level of significance, with
a very tiny P-value.
prop.test(cbind(c(200,800),c(300,3700)))
2-sample test for equality of proportions with continuity correction
data: cbind(c(200, 800), c(300, 3700))
X-squared = 137.5, df = 1, p-value < 2.2e-16
alternative hypothesis: two.sided
95 percent confidence interval:
0.1767413 0.2677032
sample estimates:
prop 1 prop 2
0.4000000 0.1777778
The function binom.test provides 95% confidence intervals. [The tests themselves are not directly relevant; I have not installed the procedure binconf on my computer.] The CIs for those in and out of the subgroup
are $(0.176, 0.226)$ and $(0.067, 0.084),$ respectively. These intervals
are quite far from overlapping, so it seems reasonable to conclude that
the success rates differ.
binom.test(200, 1000)
Exact binomial test
data: 200 and 1000
number of successes = 200, number of trials = 1000, p-value < 2.2e-16
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.1756206 0.2261594
sample estimates:
probability of success
0.2
.
binom.test(300, 4000)
Exact binomial test
data: 300 and 4000
number of successes = 300, number of trials = 4000, p-value < 2.2e-16
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.06702528 0.08360192
sample estimates:
probability of success
0.075
Because each of the intervals above has a 5% chance of not covering the
their true Success probabilities, comparing them is roughly equivalent to
a test at the 10% level.
To conclude a difference in Success probabilities at the 5% level of significance
it would be better to use the Bonferroni inequality and compare two CIs at
97.5% level.
The code binom.test(200, 1000, conf.lev = .975) returns the interval
$(0.172, 0.230),$ rounded to three places, and binom.test(300, 4000, conf.lev = .975) returns
the interval $(0.066, 0.085).$ Again, no overlap.
Notes: (a) I believe that the R procedure binconf gives Wald CIs, now deprecated,
while the procedure binom.test gives Agresti-Coull CIs (also called 'plus-4' intervals). Agresti-Coull CIs have been shown to have more stable coverage probabilities than Wald intervals, especially for small $n;$ for the current
data the difference is negligible. Perhaps see this Reference and its references.
(b) Procedures prop.test (shown above), chisq.test, and fisher.test (all with with similar syntax) could all be used to test whether the Success probability is the same in and out of the subgroup. Fisher's Exact Test uses the hypergeometric distribution. The other two use normal and chi-squared approximations and
typically give very similar P-values.
