How can I find the Margin of Error E the professor hasn't gone over this section but I want to understand it so that I won't confused in class. I want to learn so that I can do the other problems.
So this is the problem, If anyone could teach me so I could take note I'd deeply appreciate it. Thank you for your time.
Question: 
Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.
n = 550, x equals 330, 90 % confidence
I know how to obtain the alpha/2 to find the confidence.
I just went over the Binomial Distribution and Limit Theorem, which understand now this is next. If anyone would care to spent their time explaining I'd be forever grateful, thank you again for your time.
 A: Confidence intervals are characterized by two things: sample means and standard errors. The sample mean gives you the center of the interval, and the standard error helps define the radius. We shall write $\bar{x} \pm z_{\alpha/2} \sigma_{\bar{x}} $ for the confidence interval. Remember in this context we're talking about proportions, so we need the material from the previous question you asked, of which I will use freely.
Recall that the standard error is $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, where $n$ is the sample size and $\sigma$ is the standard deviation of the associated distribution. As we're using a binomial distribution,
\begin{align*}
\sigma_{\bar{x}} &= \frac{\sqrt{npq}}{\sqrt{n}} \\
&= \sqrt{pq}
\end{align*}
Now that makes things a great deal easier. The second part of the radius relies on this $z_{\alpha/2}$, the z-score associated with a particular alpha-level. What is this alpha-level? Take your confidence level and subtract from one, done. For a 90% confidence level, the alpha-level is 0.1, so we must search in a normal distribution table for a z-score whose left tail area is 0.05. I'll leave it to you to see that this is approximately 1.645 (one of the magic numbers to remember).
The radius is then $1.645 \cdot \sqrt{\frac{330}{550}\frac{220}{550}} $, which, according to some texts I've seen, but this may vary on the source, is the margin of error.
