Question

A newspaper found that, on average, 7.6% of people stay overnight in the hospital with a 1.5% margin of error. Construct a 95% confidence interval.

Answer 1

Given:
p = 7.6% = 0.076, the percentage of people who stay overnight at the hospital.
E = 1.5% = 0.015, margin of error
95% confidence interval.

The standard error is 
Es = $\sqrt{ \frac{p(1-p)}{n} }$
where 
n =  the sample size.

The margin of error is
$E=z^{*}E_{s}$
where
z* = 1.96 at the 95% confidence level.

Because the margin of error is given, there is no need to calculate it.
The 95% confidence interval is
p +/- E = 0.076 +/- 0.015 = (0.061, 0.091) = (6.1%, 9.1%)

Answer: 
The 95% confidence interval is between 6.1% and 9.1%.