Hi Isabella,
First of all, we need to look at our sample size. Is n large? (Some books call "large" 30, some say 50.) If so, we can assume normality. Here, n=580, "large" by all counts, so we can assume normality. Now, moving on to a formula for this confidence interval:
CI=p-hat +/-z*[sqrt(p-hat)(1-p-hat)/n]
Now, let's break this down:
p-hat=probability estimate usually given in problem
z*=z-critical value obtained from z-score table or memory; explained below
n=sample size
For this problem:
p-hat=0.67
z*=1.96
You can get this value from a z-table by looking for 0.975 in the interior, but it is easier to simply memorize since 95% confidence intervals are very common. Just remember z* for 95% confidence is 1.96.
n=580
So:
CI=0.67 +/- 1.96[sqrt(0.67)(1-0.67)/580]
CI=0.67 +/- 0.04
CI=(0.63, 0.71) Answer may vary due to rounding differences.
I hope this helps.
