How to combine population proportions and confidence intervals? Say I've looked at three random samples, one each from three populations of students. I found that a few students in each sample didn't turn in a fieldtrip permission slip, leaving me with the following point estimates:
| Population name | Population size | Sample size | Missing Slips (Sample) | Estimated Missing Slips (Population) |
| --------------- | --------------- | ----------- | ---------------- |---------------- |
| 6th graders     | 1,200           | 200         | 2|12 |
| 7th graders     | 800             | 100         | 5|40 |
| 8th graders     | 1,000           | 200         | 4|20 |

Can I say that an estimated 72 of 3,000 (2.4%) 6th, 7th, and 8th graders didn't turn in field trip permission slips? If I can, how do I go about calculating a confidence interval for my estimate?

You can think of this as a Bernoulli trial process in which returning the slip is "success" and not returning the slip is "failure". Your estimate of the failure rate is:
$ \hat P_{failure} = n_{failure}/sample\,size = 2.4\%$
You can take this observed value to calculate a confidence interval for the probability of failure in the population using the following formula (95% confidence interval, z = 1.96):
$ \hat P_{failure} \pm z \sqrt{\hat P_{failure}(1-(\hat P_{failure}))/sample \, size} = 2.4\% \pm 1.3\% $
