A question about survey sample size

Question

I am looking at the formula on this page, it says the formula for estimating the sample size of a survey is:

$$Sample\;size = \frac{\frac{z^2 \;\times \;p(1-p)}{e^2}}{1 + (\frac{z^2 \; \times \; p(1-p)}{e^2N})}$$

where
- $e$ is the margin of error
- $N$ is the population size.

I have the following questions:

1. What does $p$ represents?
2. Why is the variance of a Bernoulli distribution (i.e. $p(1-p)$) used here?
3. Let's say
   • N= 10,000
   • e=3%
   • confidence interval = 95%
   • based on the numbers above, the sample size is 965

How would you interpret the result? I am 95% confident that by surveying 965 people, it will be enough to present the entire population? (what about the margin of error?)

Answer 1

This sample size calculator seems to assume you're trying to determine the proportion of units in your population has a particular characteristic.

1. $p$ represents the proportion in the population that has the characteristic you're studying. The calculator probably assumes $p=0.5$ since this gives the most conservative (largest) sample size.

2. The variance of a Bernoulli random variable is used because each unit in the population is either has the characteristic you're studying or doesn't. If you're not just calculating proportions - for example if you're trying to estimate average height, then you replace the $p(1-p)$ with an estimate of the variance of the characteristic you're collecting (e.g. do some initial research to estimate the variance of heights in your population).

3. That result is saying that if you repeated the survey many times, then 95% of the time, the proportion of units in your survey that have that particular characteristic will be within about 3% of the true proportion in the population.