Question

the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5 %. assume that a sample size of 40 people was surveyed from the population an infinite number of times. 95% of the sample mean occurs between ___ and ___?

Answer 1

Given:
Population proportion, $\mu _{p}$ = 57% = 0.57
Population standard deviation, σ  = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%

The standard error is
$SE_{p} = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.57*0.43}{40} } =0.0783$

The confidence interval is 
$\hat{p} \pm z^{*}SE_{p}$
where
$\hat{p}$ = sample proportion
z* = 1.96 at the 95% confidence lvvel

The sample proportion lies in the interval
(0.57-1.96*0.0783, 0.57+1.96*0.0783) = (0.4165, 0.7235)

Answer: Between 0.417 and 72.4), or between (41% and 72%)