Question

If p is a given sample proposition n is the sample size, and a is the number of standard deviations at a confidence level, what is the standard error of the proportion?

Answer 1

The standard error of a proportion p in a sample of size n is given by:
`s = \sqrt{(p(1-p))/(n)}`

Explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

In this question:

The standard error of a proportion p in a sample of size n is given by:
`s = \sqrt{(p(1-p))/(n)}`