Undergrad Need clarification on the margin of error in two different cases

[jldibble]

TL;DR

Sample Proportion vs. SD from Sampling Distribution of Sample Proportions

I need this in simple terms. Here's what I think I know so far (assuming 95% confidence level):

MOE from a sample proportion is 2√(p(1-p)/n) and I think this is assuming the sample proportion is close to the population proportion.

But then there is a sampling distribution of sample proportions which gives a standard deviation. The MOE in this case is just 2σ

Let's say I take a single sample proportion and want to compare it to the average from the sampling distribution. A lot of questions will do this and then ask if the value of the sample proportion is consistent with the data from the sampling distribution. It seems like these questions ignore the MOE for the sample proportion and just worry whether or not it falls within the MOE for the sampling distribution.

Couldn't there be some overlap between the two MOE? Am I missing something or not understanding this properly? I can't find examples when to worry about one and not the other.

Thanks

 

[Stephen Tashi]

You're more likely to get an answer if you use standard terminology from mathematical statistics. I think you are using terms peculiar to opinion surveys. (e.g.https://en.wikipedia.org/wiki/Margin_of_error )

But then there is a sampling distribution of sample proportions which gives a standard deviation. The MOE in this case is just 2σ

Presumaby you wish $\sigma$ to denote the standard deviation of some random variable. Perhaps the random variable is the mean of a sum of N identically distributed bernoulli random variables.

If the probability of "success" on each realization of the bernoulli random variable is $p$ then the variance of a single realization is $p(1-p)$ and the variance of the mean of a sample of $n$ independent realizations is $\frac{n p(1-p)}{n^2} = \frac {p(1-p)}{n}$ If you intend $\sigma$ to mean the standard deviation of the mean of N realization, then $\sigma = \sqrt{\frac{p(1-p)}{n}}$.

You are using the terms "sample proportion" and "average of the sampling distribution" as if they designate two different random variables. I don't understand how you define them.

For example, if I toss a coin 10 times and get 3 heads, the "sample proportion" could refer to 3/10, but 3/10 is also the average of 10 things, 3 of which are 1 and 7 of which are zero. So 3/10 can be called "the average of the sample". Do you intend "average of the sampling distribution" to mean that?