Proportionality of discipline incidents between student groups My goal is to see if one student group has a disproportionate number of discipline incidents relative to the another group/s, to what degree, and with what margin of error. I will be speaking mostly in lay terms, and my intended audience is a lay audience.
I am going to simplify the problem as such:
I have two groups (not samples, complete populations) of students, M and F, for a given school.
There are 500 M students and 200 F students.
Of the M students, there are 100 with >= 1 discipline incident.
Of the F students, there are 20 with >= 1 discipline incident.
So:
500/700 (~71.4%) of students are M students and
200/700 (~28.6%) of students are M students
And:
100/120 (83.333%) of total students having >=1 incident are M students, while
20/120 (16.666%) of total students having >=1 incident are F students
Are the following calculations then valid?
Ratio of M students with >=1 incident per 100 students =  83.333% / 71.4% * 100 = ~117
Ratio of F students with >=1 incident per 100 students =  16.666% / 28.6% * 100  = ~58
If this is valid, with what margin of error, given the total students in each group and the number of students with incidents in each group, can I say that this accurately represents the proportionality of incidents between these groups?
 A: For M students, you have 100 out of 500 of with disciplinary incidents. So the estimated proportion of M students with incidents is $\hat p_M = 1/5$
Similarly, for F students you have 20 out of 200 with incidents for
an estimated proportion $\hat p_F = 1/10.$
So the two estimated proportions are $0.2 = 20\%$ for M and
and $0.1 = 10\%$ for F. The sample proportions are different, but
the difference is not necessarily statistically significant.
You can't just look at percentages without taking sample sizes
into account.

For example if you had 1 out of 5 M students with
incidents and 1 out of 10 F students with incidents, that might
be an interesting story, but it would be an anecdote rather than
serious evidence. (Suppose a box has 5 blue marbles and 10 red ones;
if I take out two at random, it would hardly be surprising to get
one marble of each color.)

A proper statistical test will let you test the null hypothesis whether the rates $p_M$ and $p_F$ are the same against the alternative hypothesis that they are not
equal. In R statistical software the procedure for conducting this test
is called prop.test, illustrated below:
prop.test(c(100,20), c(500,200), cor=F)

         2-sample test for equality of proportions 

data:  c(100, 20) out of c(500, 200)
X-squared = 10.057, df = 1, p-value = 0.001517
alternative hypothesis: two.sided
95 percent confidence interval:
 0.04561321 0.15438679
sample estimates:
prop 1 prop 2 
   0.2    0.1 

The P-value 0.001517 means that of the proportions for M and F
were equal, then the probability of a difference greater
than $|p_M = P_F| = 0.1$ or greater would happen with probability
$0.001517$ (that's less than 1 chance in 600),
Thus, if the null hypothesis is true, a very rare event
has happened. This description of the data causes us not
to believe that the null hypothesis is true. So we say that
the difference in observed rates is statistically significant.
The printout from the test also gives a 95% confidence interval
for the population difference $|p_M-p_F|$ That confidence interval
is $(0.046, 0.154).$ or $0.1 \pm 0.054.$ The number $0.054$ is
called the margin of error of the estimated difference $0.1.$
Note: The argument cor=F in prop.test is to turn off 'continuity correction', which is not necessary with the number of students you have.
