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?

  • $\begingroup$ Out of 100 people 117 have >= 1 incident? I'm not sure how to make sense out of it. (It sounds like "12 out of 10 dentists recommend Colgate toothpaste.") If you just want see which groups seem to have more incident occurrences, why not just 100 / 500 for M and 20 / 200 for F? $\endgroup$ Jun 23 '20 at 20:51
  • $\begingroup$ Good question, and I understand why this was confusing. I was attempting to incorporate the size of the populations into the analysis rather than just the proportions. $\endgroup$
    – seth_plunk
    Jun 24 '20 at 21:16

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.

  • 1
    $\begingroup$ This is excellent thanks! I am running the real number now in R. $\endgroup$
    – seth_plunk
    Jun 24 '20 at 22:08

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