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

Answer:

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

Step-by-step 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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

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