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%.
The answer is B.
172.0 in.²
Answer:74
Step-by-step explanation: first you need to find the ratio so divide 220 by 32 to get 6.875 then didvide 510 by that same number and you will get 74. 18181818 but 74 because you can’t have a fraction of a person
No I do not want to have fun I’m jus typing this cuz I need points
