Stats is not my field, but I had published papers in stats journals. You didn't pay attention to the following text in my message:
It is a relatively small sample with only 24 observations (there is a total of 32 observations there, but 8 people haven't attended loud events, so they don't count). You will be surprised to find out that statisticians would refer to a sample with 30 or more observations as a "large sample:"
"
...at least when large samples are used, such as N ≥ 30."
We are trying to estimate population proportion. Condition 3 on page 106 of
https://projecteuclid.org/download/pdf_1/euclid.ss/1009213286
states that confidence intervals may be used if n*p_hat>5, where p_hat is the sample proportion (in our case the proportion in our sample who got a permanent spike). According to the paper above, condition 3 is a statement that can be found in some popular stats textbooks. In our case n*p_hat = 24*8/24 = 8>5. The paper goes on to critisize this condition. I will come back to their critique momentarily. If that condition 3 is true, then we can use the confidence interval calculators on
http://www.sample-size.net/confidence-interval-proportion/
I set N = 24, x = 8, and CL = 99. I am looking for a 99% confidence interval. The online calculator is providing us with two sets of confidence intervals. One is
Lower bound = 0.119
Upper bound = 0.614
The other one is
Lower bound = P - (Zα*SEM) = 0.085
Upper bound = P + (Zα*SEM) = 0.581
The above are 99% confidence intervals. For an interpretation of the meaning of a confidence interval, see
http://www.mathbootcamps.com/interpreting-confidence-intervals/
To paraphrase, we are 99% confident that the fraction of tinnitus sufferers who will get a permanent spike after attending a loud event is between 8.5% and 58% (I am using the second interval we computed above). This statement takes the sample size into account. As a result of that relatively small sample size, we ended up with a 99% confidence interval that is wider.
[What follows will be of interest to people who took a "baby stats" university course where you learned to use a confidence interval for the population proportion whenever np>5 and nq>5.
Now, if we want to be as conservative as possible, we will look at Figure 4 on
https://projecteuclid.org/download/pdf_1/euclid.ss/1009213286
It gives us "Coverage of the nominal 99% standard interval for fixed
n=20 and variable p". Here, "coverage" means the true confidence of the confidence interval, when you use the formula for a 99% confidence interval. You will see that when the true population proportion p is larger than about 17%, the confidence is not 99%, it is actually somewhere between 89% and 98%. When the true p is between about 10% and 17%, the confidence can get as low as 88%. Note that this is true when the sample size is 20. Our sample size is 24. To be conservative, we can conclude that we are 90% confident that the fraction of tinnitus sufferers who will get a permanent spike after attending a loud event is between 8.5% and 58%.]
You will notice that in my post, I was very conservative - I assumed that my poll implied a probability of 5%. We can be 90%-99% sure that it is actually above 8.5%. Note that 58% is inside of our 99% confidence interval. Taking into account the sample size, we can be 90%-99% sure that given the results of this poll, the probability of getting a permanent spike might be as high as 58%.
We are a subset of T sufferers who are bothered by T enough to register for this forum. This means that whoever is reading the thread with that poll is part of the population the poll is trying to study. In other words, that confidence interval might not apply to a randomly selected T sufferer, but it applies to YOU.