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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 ___?

Respuesta :

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

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

The confidence interval is 
[tex]\hat{p} \pm z^{*}SE_{p}[/tex]
where
[tex]\hat{p}[/tex] = 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%)

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