Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 stats.stackexchange.com , now you know. –  leonbloy May 31 '14 at 2:09

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.