Margin of error and $98\%$ confidence interval question for stats people :) The question is (this is homework, for an online class, no teacher so at times confusing) 
Carl conducted an experiment to determine if there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in sample was $91.1$ with a population standard deviation of $.52$ and mean body temperature for women in sample was $97.6$ with population standard deviation of $.45$.
-Assuming population of body temperatures for men and women were normally distributed, calculate the $98\%$ confidence interval and the margin of error for both.
*I have a bit of experience with confidence interval, but only have $90\%, 95\%,$ and $99\%$ and the course gave me a "confidence interval calculator" and has only that. Also, I have never before heard of margin of error, when I looked it up I didn't understand it. Could someone please explain to me in a way that I would easily be able to understand?
(I asked the same question yesterday, but no one replied. I hope someone can respond today, I wasn't sure I could refresh the old one"
Thank you. 
 A: It is not realistic to say that the population mean is known when one is finding a confidence interval for the population mean, but those are being handed to you, and when one has a large sample size one can often procede as if there were no uncertainty in the estimate of the standard deviation.  You have probably see this:
$$
\bar x \pm 1.96\frac{\sigma}{\sqrt{n}}
$$
That gives you a $95\%$ confidence interval for the population mean if $n$ is fairly large (say $\ge100$), where $\bar x$ is the sample mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.  For small values of $n$ one takes into account the uncertainty in the estimate of $\sigma$ by using Student's $t$-distribution with $n-1$ degrees of freedom, and then you'd use a bigger number than $1.96$.
The question is what to use instead of $1.96$ when you need a $98\%$ confidence interval instead of a $95\%$ confidence interval.
Often textbooks have a table that includes a row that looks something like this:
$$
\begin{array}{c|cccc}
& 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & \cdots \\
\hline
& \vdots & & \vdots & \vdots & \vdots & \vdots & \\
2.3 & 0.9893 & 0.9896 & 0.9898 & 0.9901 & 0.9904 & 0.9906 & \cdots \\
& \vdots & & \vdots & \vdots & \vdots & \vdots &
\end{array}
$$
For a $98\%$ confidence interval, you'd find $0.99$ in the body of the table and thus get $2.33$ instead of $1.96$, so the endpoints of the confidence interval are:
$$
\bar x \pm 2.33\frac{\sigma}{\sqrt{n}}
$$
The reason you look for $99\%$ is that if the table appears as above, it's giving you the point where $99\%$ of the probability is to the left of that point.  That means $1\%$ is to the right of $2.33$ and $1\%$ is to the left of $-2.33$, so $98\%$ is between $\pm2.33$.
So plug in your sample mean in place of $\bar x$ and your given standard deviation in place of $\sigma$.  If you find out the sample size, put it in place of $n$ and then do the arithmetic.
Many software packages give more accurate answers, but there's no practical importance in cases like what you're looking at.  I'm getting $2.326348$ from a program that has no warranty but has always been right as far as I've been able to check it.
