According to the American Medical Association, Physicians in the United States w

Can anyone provide some insight on how to do this?

This is a one-sample problem in which we do not know the population standard deviation and we are given sample data; thus, we must use a t-interval, the formula for which is:

\(\overline{X}\pm t^*(\frac{sd}{\sqrt{n}})\)

In other words, we are going t* sample standard deviations from the mean in both directions, with n-1 degrees of freedom. I'm assuming this is elementary stats, so I'll grab t* from a table.
\(
\overline{X}=59
n=25
sd=3
t^*=2.064\)

which gives us

\(25\pm1.2384
\)

23.76-26.24

For part b, we merely set the standard error equal to .05 and solve for n. With this in mind, I'm wondering if the topic creator didn't intend to write margin of error instead.

\(\frac{s}{\sqrt{n}}=.05\)

n=3,600

Awesome!