Question 1175556
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Hi
 n = 200, 28 used debit card
 sample p = 28/200 = .14
ME = {{{z*sqrt((p(1-p))/n)}}}
95% confidence interval
ME = {{{1.96*sqrt((.14(.86))/200)}}} = ± .0481
CI = .14 ± .0481

b) Yes, np = 28 > 5

c) n = {{{(z/ME)^2 (p(1-p)))}}}
 90% confidence interval
  n ={{{(1.645/.03)^2 (.14(.86)))}}} = 362.006
Sample Size 362 needed.
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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