Answer:
CI 90 % = [ 0,165 ; 0,235 ]
Lower limit 0,165
upper limit 0,235
Step-by-step explanation:
Sample information:
sample size n = 225
number of dissatisfied individuals x = 45
p = 45/225
p = 0,2 and q = 1 - p q = 1 - 0,2 q = 0,8
p*n = 0,2*225 = 45 and q*n = 0,8*225 = 180
p*n and q*n big enough to use the approximation of binomial distribution to normal distribution
90 % of Confidence Interval then a significance level is α = 10%
α = 0,1 in z table we get z(c) for that significance level
z(c) = 1,28
CI 90 % = [ p ± z(c)*√(p*q)/n ]
z(c) * √(p*q)/n = 1,28 * √ ( 0,2*0,8)/225
z(c) * √(p*q)/n = 1,28 * 0,027
z(c) * √(p*q)/n = 0,035
CI 90 % = [ 0,2 ± 0,035 ]
CI 90 % = [ 0,165 ; 0,235 ]
Lower limit 0,165
upper limit 0,235
