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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.

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We need sample sizes to do this. – Michael Hardy Oct 31 '13 at 18:10
Thats what I thought, They never gave me a sample size. Is that something I can find? – shari Oct 31 '13 at 18:11
It can't be found given only the information you've given us. – Michael Hardy Oct 31 '13 at 18:14
In case you don't know about , now you know. – leonbloy May 31 '14 at 2:09

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.

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