What is the formula for a normal-theory confidence interval of the mean? 
A psychologist is collecting data on the time it takes to learn a certain task. For $55$ randomly selected adult subjects, the sample mean is $10.5$ minutes and sample standard deviation is $3.25$ minutes. Construct a $99\%$ confidence interval for the mean time required by all adults to learn the task.
a. What formula to use.
b. Plug all values in.

I think I should use a $t$-test formula which would be $(\bar x- \mu_0 )/(s/\sqrt n)$, so plugging in the values would be $(10.5-?)/(3.25/\sqrt{55})$.
What is the $\mu_0$ value? Am I right?
 A: (Since this question is so old, I suspect it is safe to provide an answer.)
I think you're halfway there.  We might often use a $t$-test in a situation like this, but a confidence interval is not quite the same thing as a test.  Since the mean and SD are estimated from the data, you need to take that fact into account.  Thus, we will use the $t$-distribution to form the confidence interval.  The general formula would be:
$$
\bar x \pm t_{(1-\frac{\alpha}{2}\!,\ df)}\ \frac{s}{\sqrt{N}}
$$
The key to using this formula is to find the relevant $t$-value.  First, we need to get the df—it is $N-1=54$.  Then we look up the quantile that corresponds to the $99.5^{\rm th}$ percentile of that particular $t$-distribution in a $t$-table.  I find the value $2.67$.  Hence,
$$
10.5 \pm 2.67\ \frac{3.25}{\sqrt{55}} = 10.5 \pm 1.17 \Rightarrow (9.33,\ 11.67)
$$
A: Let's start saying you can use this tool to check whether you answer is correct or not. 
You should get a confidence interval of $\pm 1.13$.
That said, the correct formula to use is:
$$\bar x \pm z \frac{s}{\sqrt n}$$
Where the $\bar x$ is your average, the $z$ is a constant, $s$ is the standard deviation and $n$ is your sample size. The $z$ varies depending on the confidence interval you are trying to calculate. In your case, because 99%, $z$ is equal to $2.58$.
Plug all the numbers in the equation and you should get 1.130633169350857 which rounded gives you the $\pm 1.13$ I mentioned before. And that's it.
