Correct n to use in calculating confidence interval of a proportion

Question

I know that my 95% confidence interval can be calculated for a proportion using:

$$ 1.96 \times \sqrt{p(\frac{1-p}{n})} $$

where $p$ is the proportion and $n$ is the number of trials.

But if my data is collected in a series of datasets (say, annual data collections), does this change my $n$? For example, if my data is framed:

Year	TRUE	n
2010	14	25
2011	17	25
2012	15	25
2013	11	25
2014	15	25

Do I calculate a total $p=\frac{14+17+15+11+15}{25\times5}=0.576$ and use:

1. $n=5$ because data was collected in five seperate experiments? $$ 1.96 \times \sqrt{0.576(\frac{1-0.576}{5})} $$

2. $n=125$ because data was collected on 125 events? $$ 1.96 \times \sqrt{0.576(\frac{1-0.576}{125})} $$

Furthermore, say there was an extra variable, $x$, for which I wanted to calculate a seperate proportion for ($\frac{\text{TRUE}}{x}$). Say $x$ represents the total number of job openings available and $n$ is the total applications, so $\frac{\text{TRUE}}{n}$ would be the job acceptance rate and $\frac{\text{TRUE}}{x}$ would be the positions filled rate:

Year	TRUE	n	x
2010	14	25	20
2011	17	25	20
2012	15	25	20
2013	11	25	20
2014	15	25	20

Would my $n$ for calculating my confidence interval for $\frac{\text{TRUE}}{x}$ still use the total from the $n$ column (either $5$ or $125$) or would it be the total from the $x$ column (either $5$ or $100$)?

Answer 1

The model for this experiment is $$ X_i \sim \mathrm{Bernoulli}(p) $$ where the $X_i$ are the total outcomes of all of the trials. If you had access to the whole data (ie, the outcomes of all 125 trials), your estimate of $p$ would be $$ \hat{p} = \frac{1}{125} \sum_{i=1}^{215} X_i. $$

Fortunately, you can compute this estimate from the data you do have, which id $$ \frac{1}{125} \Bigl( \sum_{i=1}^{25} X_i + \sum_{i=26}^{50} X_i + \dotsc + \sum_{i=101}^{125} X_i \Bigr), $$ where each of these sums are the total yearly outcomes that you have access to.

Now, by the central limit theorem, $\hat{p}$ has an approximate distribution of $$ \hat{p} \stackrel{\mathrm{d}}{\approx}N\Bigl(p, \frac{p(1-p)}{\sqrt{125}}\Bigl), $$ from which we have the standard error estimate $$ \mathrm{se}(\hat{p}) = \frac{\hat{p}(1-\hat{p})}{\sqrt{125}}, $$ and so the confidence interval $$ \hat{p} \pm 1.96 \cdot \frac{\hat{p}(1-\hat{p})}{\sqrt{125}}. $$

The point here is that $n$ is the total number of independent trials, of which you have $125$. If your data were the outcomes of the $125$ trials, you probably wouldn't be confused, and would confidently use $125$ as your $n$ value.

Well, it turns out that, even though you don't have access to the whole data, you still have all of the information that you need to compute this some $\hat{p}$, so the situation is the same!