I recently saw a question that gave a 95% confidence interval for one-sample t-interval for means as (5.6, 7) and was asked to find the point estimate and the margin of error.
The solution said that the point estimate was $\frac{5.6+7}{2}=6.3$ and the margin of error was $\frac{7-5.6}{2}$
NOTE: This question included a two-tailed hypothesis.
I wanted to know what equivalent form there is for calculating the margin of error for one-sample 95% C.I. for proportions.
For example, if I am given the interval (0.087, 0.133), I would calculate the point estimate to be $\frac{0.087+0.133}{2}=0.11$
So a 95% confidence interval would be $0.11 \pm 1.96(\sqrt{\frac{(0.11)(0.89)}{n}})$
I could set
$0.11 + 1.96(\sqrt{\frac{(0.11)(0.89)}{n}})=0.133$
and solve for $n$ but I wanted to know if there is a shortcut method like there is for means
Also are there shortcuts for calculating these point estimates and margin of errors for two-sample C.I. for means and proportions?