Question 1174171
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Hi,
    x̄  = 415/500 = .83   n = 500
 Using a 95% confidence level, compute a confidence interval estimate
 for the true proportion of adult residents of this city who have cell phones. 
 
   ME = {{{z*sqrt((p(1-p))/n)}}}

   ME = {{{1.96*sqrt(((.83)(.17))/500))}}} = .0006

    .83 - .0006 < {{{mu}}} < .83 + .0006
       
Wish You the Best in your Studies.

CI	z = value
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576



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