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 = [tex] \sqrt{ \frac{p(1-p)}{n} } [/tex]
where
n = the sample size.
The margin of error is
[tex]E=z^{*}E_{s}[/tex]
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%.
