confidence intervals: A sample of 126 employees showed....

I need help with this problem:

A sample of 126 employees at Ford Motor showed 14% with advanced degrees. Construct a 99% confidence interval to estimate the proprtion of advanced degrees in the employees of high-tech companies.

What is the equation for this? I am completely lost on this one.

\(\displaystyle \L\\E=z_{c}\sqrt{\frac{\hat{p}\hat{q}}{n}}\)

answer

Thanks for the help-I got the answer as: 0.14+/-0.08 (I can't underline the plus sign like needed). Is this right?

Yep